Complete Power Electronics Course in Electrical Engineering

1. Power Electronics Promo: Hi, I'm the welcome everyone to our course for power electronics. This course is for electrical engineering students. For the electrical engineers, for anyone who would like to learn about with power electronics. For complete beginners. Let's start. What are we going to learn in this course? Complete AAC supports, of course, for electrical engineering. This course is designed for anyone who would like to understand the AAC shoppers. It's an AAC shoppers are basically used the two convert the variable EC into another variable. Easy, or to be more specific, we change the magnitude or the value of the AC voltage. The AC shoppers again increase the AC voltage or decrease the institutional part, AC voltage depending on the circuit itself. In this course, we will learn about the AC shoppers which are connected to on our load. Another issue, part circuits with a load or pure inductive load, parallel load and our LLC, it is lewd. We will learn about AC shoppers connected to AC capacitance. We will learn about an important control which is called the integral cycle control of the AC shopper. We will learn how to use the astute shoppers in transformer tap changing. We will also learn when the equations of connection of an AC shopper was an AC motor with a back EMF or electromotive force. Also, we will learn about the applications and the disadvantage of using the AAC shoppers. Now of course, all of these types, which we have discussed, all of these we'll be solved with, with a step-by-step examples. So don't worry, we will go to each of these circuits with equations, with every single was there. What did we get this equations? And we will learn step-by-step examples. Another course so which we are going to learn in this bundle is a complete DCI shoppers course. So the DC shoppers are, what are their functions they are used to convert is at DC voltage or a constant DC voltage into Amazon, constant DC voltage is, or to increase the voltage or decreases voltage. We can say it is a step-up and step-down DC converter. We will learn about the definition of a DC shoppers and applications of DC shoppers. We will learn about that meaning of the duty cycle. We will learn about the different types of circuits in EC shoppers on, in DC shoppers use the two step up increase or step-down degrees is a DC voltage. We will analyze a step down this issue of power which are on the RLE load. We will learn also about the stem up DC shopper with different connected loads. We will learn about words or puck regulate or bolster regulator and the buck-boost regulate or circuits. We will learn the equations needed to get the minimum values of inductance and the capacitance required in these three circuits, which is a buck converter, the converter and the buck boost converter. This part we will learn about DC shoppers we sold with examples. Don't worry, we will analyze each circuit with a step-by-step equations. Next course, we will learn about the inverters. Inverters are used to convert that DC into AC is a fixed voltage into a variable voltage. We will learn that definition, importance and applications of inverters. We will learn about the single phase half bridge inverter connected to an onload and RLE load. We will learn about with ZAP performance parameters of an inverter. We will learn about a single phase bridge inverter. And in case of the R and the RLE loads, we will learn about the three-fifths inverters and how to obtain the line and phase voltages. We will learn about a single pulse width modulation, multiple pulse-width modulation, sinusoidal pulse width modulation. All of this will be step-by-step lessons. Again, don't worry about anything. We will understand each of these lessons with a step-by-step examples. Is this course. For, this course is designed for electrical engineering students. Electrical power students. Complete beginners, hold onto know anything about board electronics and would like to understand what is the meaning of power electronics and what is the function of power electronics and what are the different circuits used? We will also, it is used for electrical engineers holed like to refresh as a knowledge Empower Electronics. I would like to thank you for watching this promo video and I hope to see you in our course. 2. Introduction to Power Electronics: Hi, and welcome everyone to our course for power electronics. In this course, we are going to learn everything you need to know about power electronics from scratch without any previous knowledge. So in the first lesson we are going to have an introduction. Ai powered power electronics. So first, what is the meaning of power electronics? So power electronics simply representing a branch of electrical engineering or electrical engineering that deals with power conversion from one form to another using several components, such as inductors, capacitors, and semiconductor devices such as diodes. So restores MOSFET or IGBT as GTO and et cetera. So the power may be from milliwatts when in a very small loads or a very small applications to megawatts, which is a power system, power rating. This energy conversion can take place in any form. It can be conversion from e.g. in AC to AC or what we call AC shoppers. It can be also DC to DC or the DC shoppers. It can be also DC to AC, which is inverters, and AC to DC, which is a rectifiers. So simply it's a power electronics are used in electrical power system mainly for this application, for conversion from one form to another. As you know that in electrical systems, we have two types of waveforms. We have the DC waveform or the direct current. And then we have the AC form, which is the alternating current. If you already took my OnCourse for electric circuits, you already know the difference between these two, DC and AC. Now e.g. when we are using AC to AC conversion, or the AC shoppers, they are used to convert the AC from one value to another. So e.g. if you would like to change the voltage of the input signal, we an AC waveform. We use an AAC shopper that will convert it from one value to another. And other types of ISI shoppers. Our used it to change even the frequency of the waveform. Second pi, which is DC to DC, if we have a DC voltage is a certain value and I would like to step up this voltage or increases voltage or decrease this voltage using a step down DC shopper. Then we will use this conversion system or the DC short bars. Then we have conversion from DC to AC, which is called the inverters, which converts DC to AC. As an example, inverters are used in solar energy system. They convert the power coming from the panels, which are DC power into alternating current for our AC loads. As our types, which is AC to DC conversion from AC to DC. We use this conversion in e.g. any charging of our batteries. We have an AC supply and we would like to challenge our batteries. So we need to convert AC power from our electrical source into DC power for a charging batteries. Now, what's the difference between the power electronics? The electronics? The power electronics representing an area of electrical engineering that involved with Zach control, air conditioning, and conversion of electrical power using the semiconductor switches, such as diets or IGBT or MOSFET. And so he restores that we will talk about in this course. On the other hand, electronics is a vast area of electrical engineering that includes all applications of semiconductors. So it includes everything about semiconductors. However, here in power electronics, we are concerned with only a certain branch which is controlling, conditioning and conversion of this wave forms from one format to another, AC to DC, e.g. or DC to AC or changing AC value or changing is a DC value. Now here, there's a summary of the power electronics field in the electrical power engineering. As you can see, the fault conversions that we talked about in the previous slide. And as you can see here, these are the symbols. You can see this shape. You can see here this waveform representing our sinusoidal waveform or sine wave. You can see a sine wave which are representing the alternating current or AC. Here. As you can see, two dashes here, or the equal sign, this equal sign or resenting that DC. So when you see the sample, it means that we are converting from the AC waveform is your alternating current waveform into DC, or constant voltage waveform. Now another here we can see DC to DC. You can see equal and equal, which means conversion from DC to DC. You can see here this one which is using this usual pores that we will talk about in the course. Now another way of convergence that DC to another value of DC STRs forest a conversion from DC to AC. Then we will use another conversion which takes a C and converts it back to DC. So this is another way of converting the DC to DC. Two ways by using, let's say back-to-back converters in order to change the value from one to another or directly using DC sharper. Here, as you can see, we have inverters, inverters which you take AC power and convert, it, takes DC power and converts this into AC. It, it changes the constant or the direct current voltage into alternating current voltage. And as you can see here, AC to AC, AC form to another easy forms. This is just a general idea of the different conversions that we have in the power electronic systems or the power electronics field in general. Now let's talk about some of the applications of the power electronics to understand why this course is important for every electrical engineering, electrical engineering student. So e.g. let's talk about an application which is automobile or the electrical vehicles, electric vehicles. So you can see here in the electric vehicles which are working with electricity, you can see inside it we have batteries. Batteries, which are used to store electrical power so that it will be used later for running the electric motor of the car. So the batteries here are usually lithium batteries. Lithium batteries which are designed or form it in major companies like Tesla, e.g. so as you can see here, we have inside this cow, we have our charger. This is Georgia is used to connect this. This scour will be connected to a power supply, e.g. in the current day, technology, which is a Tesla super charger. Teslas over chosen, which is used it to his fast charging of our calf. So this charge or here, it's going to add this power supply is connected to a charge, or here. Now what does this charger do? This a charger is used to regulate the charging of the battery itself. So here we need some kind of conversion. This conversion is done using the power electronic equipment. So e.g. if this one is an AC supply connected to our home socket, e.g. to our home socket. Then we will have an AC power. We need to convert this AC power into DC power for our battery, we need what? We need a rectifier. Another time. If we are talking about a DC source, then in this case, we will have a charger that will regulate the charging of the battery. It will take DC and convert it into another DC. Now, you have to understand that this type of batteries has its own cycle of charging. That's why we need a charger to regulate the charging of this battery. Another one here, which you can find also support electronic here for this electric motor. This power electronic is used to regulate the torque and the speed of the engine of this cow. You can see this is just one application that involves many convergence of the DC to AC or AC to DC and so on. Now, another applications for power electronics or renewable energy. You'll find power electronics in every renewable source of electrical power, such as solar energy. When the energy, wave energy and etc. E.g. the solar panels will generate a DC power. And for general applications, e.g. in our home, we need an AC power. Hence, our power electronic converter is used to provide this AC power. Now, let's look at this schematic. If you have joy into my own course for solar energy, you already know about this. Now, if you don't know, I will explain right now. So first, here you can see this is a schematic for something which is called the stand alone PV system or an off-grid PV system. You can see we have our solar panel that are used to convert sunlight or some irradiance or sun irradiation into DC power. The output power coming from the panels is in the form of DC power. Panels are used to convert sunlight into DC electric power. Now, as you can see, we have in our system or the stand-alone system, or the off-grid PV system. We have group of batteries. Batteries are used to store electrical power during the day. And it will help us to provide electrical power at night when the sun is not available. So here we have between these two, we have our charge controller. This is charged to control ends up bv system is used to regulate the charging of the batteries. So it takes DC and convert it into another value of DC. So the charge controller has two functions to, let's say e.g. I. Maximum power point tracking charge controller. This type of controls are used to control the voltage of the panel itself to produce or generate the maximum power. And at the same time it's used to control the voltage across the battery is the current and the voltage according to the charging cycle of these, the type of the battery itself. Now you can see also we have an inverter is this inverse, or its function is to convert the DC power coming from the battery into AC power for our home. Because as you know that all of our appliances inside our home or all of our devices are working on AC power. That's why we need another power electronic converter, which is an invert. So you can see in sample application like this, like BB panels, we have controller here and we have inverter, or we have DC to DC converter. And we have also an inverter. Don't worry, we will talk about these converters inside the course itself. How another application is called the high voltage DC system. Now this type of systems is used to interconnect between two countries of different frequencies. Let's say e.g. if we have a country with a 50 hortus working on a 50 hz AC power of 50 horses. And another country working on AC power 60 hz. And I would like to e.g. transfer electrical power from this country to 50 health is what I would like to transmit electrical power from the 50 hz 2 s charts. So we need to have a conversion between these two. We can not just connected them directly because they have different frequencies. So we need power electronic equipments in order to connect between these two countries. Now we have something which is called high voltage in DC, which means that we have a DC voltage. Transmission system is a DC voltage. However, it has a high voltage. It means it has a very high value of voltage, can be e.g. 1,000, 1,000 kilo volt, e.g. very large voltage on selves and kilovolt on thousand 500 kilovolt, very large high voltage. Now as you can see, e.g. we have this system here. This one here. This is one country and this one is our mother country. Okay? Now, this country, e.g. working on, let's say e.g. 50 hz. This one working on that 60 hz. Okay? Now, between them there is a high voltage DC transmission system to interconnect between these two countries. So first, we take the AC power and convert it to DC. So we have AC to DC conversion using power electronic equipment. Then we have the high voltage, which is working on DC, which carries DC power, and goes to another substation, which takes us at DC and converted to AC with a suitable frequency of the system. So it takes is a 50 hz is the first one, takes 50 hortus and converts it to DC. Then after transmitting electrical power, we will take DC and convert it back to AC. That's why it's called the high voltage DC transmission system. And you can see it in one of these two types of converters. You can see how important or power electronics for electrical engineering. Now and also applications which are at home as a live demo, air conditioning, UPS or uninterruptible power supply. It can use that as a battery chargers, as you can have seen in the PV panels, such as incest charge controller. Here is, this is an image of for UPS system or the uninterruptible power supply. If you look inside the uninterruptible power supply, what does, it's a function. It is used to take the power coming from the AC source or the power grid, e.g. and convert it from ac into dc, which is acting as a battery children to charge batteries which act as a back up for our system. Now, when the power grid is not available, what will happen is that this battery will start providing electrical power to our building. We will have another converter, which is DC to AC converters that will take DC power coming from the battery. And to give AC power required for our load. The UBS is a really important component that is used inside the residential buildings or not residential buildings. Commercial buildings to be more specific. So it acts as a backup for important or critical critical loads, such as emergency lighting for computer system and so on. Now, also, public tongues are used in industries. Almost all of the Motors employed in industries are controlled the pi power electronic drives. So we have something which is called the variable frequency drive, which is used to control the speed e.g. of an induction motor. So as you can see this image here, we have a three-phase system, a three-phase supply that will be passed to a rectifier, which is used using a power electronic circuit. Then we have a DC pass to smooth the rebels of the voltage. Then we have an inverter circuit using power electronics also to reduce the required frequency and voltage for our mode. Also, the product lungs are used in power supplies in aircraft, satellite, in space, shuttles, missiles, and other defensive equipment. And this image here shows a summary for all of the applications of power electronics. So if you don't know anything about where your power electronics is, then this course is for you. We are going to understand that. So now in this lesson, we, we understood the importance of power electronics. In the next lesson, we will start talking Apollo. It appears it power electronic circuit components and different switches that we use inside the Power Electronics. And from here we will start talking about words out rectifiers, inverters, AC and DC shoppers. 3. A Power Electronic Circuit: Hi and welcome everyone. In this lesson, we will talk about the power electronic circuit, an example for a power electronic circuit. So what you can see in this figure is a power electronic circuit representing one type of GC shoppers. Dc Chappelle's show pores which are used to convert DC value from one value into another. So as you can see here, we have our supply, which is a DC source. And do we have here our loot, which will be also DC. So as you can see, we have a supply, a DC supply. And using this, what you see here is this part, which is DC shopper or a power electronic circuit, 1/1 of the types of power electronic circuits. Using this, we can convert a DC value, let's say e.g. 100 volt to any value, let's say 20 volt as an example. Or it can be also added step up. So get e.g. from 100 volt due to 100 volt, depending on the type of the circuit here. Okay, so what do we can see here is that our circuit or the power electronic circuit, consisting of two or three main components. The first one which is our input, second one which is our output or our loot. And between them we have our power electronic circuit that is consisting of several elements. You will find that a power electronic circuit is consisting of switches such as diodes, so restores or IGBT or MOSFET, BGT, and et cetera. So what you can see here is that this one here, this one is called the dial. This diet allows the current to flow in one direction and the block so that current from flowing in a reverse direction. Then we have something which is called the switch here. This is switch which is used to open the circuit, which does not allow any current to flow here. Or it can be closed and allows the cannon to be float. Okay? However, you will find that here. This is which is not an ordinary switch. It as a swatch which is made of, or it can be as high restore or an IGBT or MOSFET or BGT is. These are different types of switches, e.g. the aldehyde is called the uncontrolled switch. We can not control. It is a soluble story is called a semi control. The switch, which we can control its spotting or its conduction. However, IGBT, mosfet and validity are called fully controlled switches. Now, each of these switches we are going to talk about in this section of the course. And then we are going to see how we are going to use these different switches in forming the different circuits such as rectifiers, AC shoppers, DC show pores and inverters. So here we are just learning about the components of n or an electric socket or a power electronic circuit to a PMO specific. We have diets. We have a switch. Now why do we use as switches like so restful or IGBT or MOSFET or obesity. Because they can be switches, switch it very fast. E.g. when we're talking about the AC system, we are working with 50 hz or 60 hz, which means we need to turn on and off, switch it 50 times in 1 s, or seconds two times in 1 s. So this cannot happen mechanically, however, it can happen electronically. That's why we need a special type of switches to be turned it on and off. Another thing which you will find in the power electronic circuits, as you e.g. in this circuit here you can see we have a storage elements such as capacitors and inductors. You can see we have an inductor, which you can see here. In other circuits. We have capacitors. So we have a storage elements that can be used in power electronic circuits. And you can see is that we have also power dissipation elements such as resistor. So sometimes we can add a resistor, e.g. you can see here our load can be a pure resistive load or insides their composition of the upon electronic circuit. We can add a resistor to limit the current or to prevent short circuit condition. So all of these are components of the power electronic circuit. Now, here we said before that these switches here, which you can see here, this one and this one are different types of electronic switches. They are divided into three main categories, which are uncontrolled switches, such as diets. We cannot control them. They operate depending on the voltage and conditions in our electric circuit. Then we have also the semi controlled switches such as heroes stores. We can control their own condition. Then we have a fully controlled device or fully controlled switches, such as the BJT or MOSFET or IGBT and GTO, we can control Wednesday or on Wednesday or off. Now we're starting from next lessons, we will start talking about these converters, the tides. So restores BGT MOSFET or IGBT, and G20. 4. PN Junction: Hi and welcome everyone to this lesson. This lesson we will talk about a very important definition which will help us understand some of the power electronic devices such as diodes. So here we will talk about the p n junction. So first, we have to understand that power electronics, or electronics in general are made of a semiconductor material. Now, silicon, silicon, of course, is that currently the most used semiconductor for power devices? Of course, you already know about this is that we use silicon in, as a semiconductor material for the power device or power electronics. Now, semiconductors are commonly made of silicon because the material is easy to find and it has an optimal electronic structure. The outer orbital of this silicon atom has four electrons. So if you look at the construction of the silicon atom, you will find in the outer orbital, which is the last orbital here. You can see 1234. So this four electrons, they form ionic bonds with other atoms. The we will see how can we form that be p-type and n-type materials. Now the problem is that this type of semiconductor materials is near and near insulator, so only a very small amount of electricity can pass through it. Now why is this? Because if you, if you learn it and our course for electric circuits, we said that the electric current in general, electric current, electric current is forming due to the flow of electrons. And when I say electrons, three electrons, now you can see that the silicon does not have free electrons. Now, that's why we need to make some adjustment to this structure. So what I mean by this, we need to put the silicon material through a doping process in order to increase the amount of electricity allowed to flow through them. Now by you doing is that doping process, we will be able to produce more holes or more electrons inside them, which will allow the flow of electric current. Doping refers to a chemical reaction that introduces impurities to a silicon crystal. We add, we don't add silicon only as a semiconductor material. We add an additional and additional material or another element which will help in reducing free electrons or free holds. Now, this impurities allows a crystal atoms to form ionic bonds with them. This process will lead to two types of impurities, the n-type material and the B type material. That type itself, such as N or B, the bend is on the materials used to create the chemical reaction. Now let's understand what does it mean and what does B type material mean? So the inner thigh, and you can think of this like in this en, abbreviation for negative, b, abbreviation for positive. So N has excess number of negative electrons. That's why we call it negative type or n type. Now, be time has an excess number of holes. Holes are both positive Z act as opposed to be charges. So B type material, which is a positive material or has excess number of positive holes in p-type, has excess number of negative electrons. Now, how can we form an inner thigh material with excess number of electrons and how can we form a B type with EXOS number of holes. Now, first, for the inner thigh material, we add phosphorous or arsenic in order to create an n type semiconductor material. So phosphorous or arsenic are having five electrons in their outer orbital, having five electrons. And the silicon crystal has four electrons in their outer orbital. So when z are forming ionic bond with each other, for electrons are forming a bond with other four electrons. However, there will be one electron which is does not have anything to bond with. So we wouldn't have a free electron that will increase the flow of electric current through the silicon. On the other hand, for the p type, we have gallium or boron added as a catalyst or as an impurity inside the silicon. Now, these materials have only three electrons in their outer orbitals. As x, y, z are forming bonds with the silicon atoms, you will find that we wouldn't have holds. We will have an additional hole or law positively charged inside this material. Now in order to understand this, let's look at this figure. We have here as the first type, which is n type material, and the second one which is B type material. So the inner thigh material here, you can see n pi, which we just talked about here. Phosphorous or arsenic are used to create an inner thigh muscles. You can see here we have a phosphorus atom. Now, this phosphorus has how many electron in the outer orbital? It has five electrons. Do we have silicon, which has 12341234, and so on. So what will happen is that this phosphorus, one electron to form a bond with other electrons of the silicon. It will form an ionic bond here and here, and here. Now it has only four. You will find that there will be only free electron. You can see here a free electron here that is not bonded to any other at the phosphorus forms ionic bond 0 is for silicon atoms. One phosphorus with four silicon atoms. And then we will have one electron which is completely free. Now free electrons contribute to the flow of electric current. So if you've done this process a lot, we have thousands or millions of atoms. So we will have millions of free electrons which will allow, which will allow the flow of electric current. Now the same procedure for the B type, we have an element with only 123123. Now, you wouldn't find that here. The B type semiconductors can be gallium or boron, which have only three electrons. Here we use Donald's or materials which is called Indian indium. Similar to pour on or gallium. All of them have three electrons in their outer orbital. So you will find that here we have 12344 silicon atoms, and we have 123. So you will find that here. We don't have any electrons, so we have a whole year, a hole. This hole is a positive charge. So anytime gives us negative electrons, negative, three negative electrons. And the B type, we have a boast of holes. Okay? So here we will have posted the electrons and the positive hole. We have an inner thigh material and the P-type material. So by taking these two together, we have a beat or even bacterial and we have n type material. You can see B type contents testing of lots of holds. All of these are holds. Both the volts and the genotype. All of these dark dots are electrons. Now we have a B material and any material when we combine them together, we will form something which we call the b n junction. There be an injunction is really important because it B type in the type it can be be an injunction or it can be be NAB or n, maybe n or whatever. B type materials and n-type materials are used in the construction of power electronic devices such as diodes, so restores and so on. That's why this lesson is really important. Because it helps you understand what does a B type material mean and what does anytime material mean that we use inside the power electronic device? So what will happen is that wins a p-type and n-type layers are joined together when we make them approach each other. And interesting phenomenon, well, okay, there is a B type semiconductor has excess holes and the postal charges, as we said in the previous slide here. And n-type semiconductor has excess electrons and of negative charge. So you can see both the poles and the negative electrons bolster which orders and negative source. So what will happen is that these are going to each other, right at the point of contact as the electrons will go and default. This holds. Right, go and fill this holds. So what will happen is that as a point of contact as a whole in the B type will attract the electrons in the n type material. Hence, the electron will diffuse and occupies the holes in the B type. So all of these electrons will go here, here, and here. Not all of the electrons, but adds a point of contact. So it will be a region here which will be formed. This region, which is a very thin region. In reality, it is very, very thin region. Convert that to be an injunction. Very small region. Now with this small region, well contain only all stuff ions and negative ions. Now why is this happens? Because here we have another type and B type. Now with this region, let's say it's a spot, loses electrons because electrons are free. Electrons here go to the other side. So when this atoms lose electrons, they, they will form positive ions. Both the winds, the winds, this excess electron go and fill this holds, this part. We'll get more negative electrons. That's why it will become negatively charged it or negative ions. Now, due to the presence of both ions and negative ions, there will be an electric field formed between them and prevent any electron from going into the other side. And they prevent also any holes from moving to the other side. So it acts as a block for electrons and holes. So here you will find that after their electrons diffuse and go to the other side, you will find that the small region near a small region of the time starting to lose electrons and they behave like intrinsic semiconductor material. In the B type, a small region will also be filled by holds and behave and intrinsic semiconductor. We have here two reasons. One which contains a positive ions and the one which contains negative ions. So let's endorsements on you can see here this, you can see all of the electron gun to the other side forming, be in a junction forming a post of ions and negative ions and the between them there will be a magnetic and electric field is that prevent any transfer of holes or electrons from one side to another. So I hope what we just explained here is clear for you. You can see anytime material and the B3 electrons is negative means electron goes to the other side. And the field as a whole, you can see when we are approaching them together, the electrons will go and deferrals are holds. Now when the electrons move from the left aside from the right side to the left-hand side. We will form of boast of ions. And on the other side we will have negative ions. So we will have in the end electric field that prevent anymore transfer of electrons or holes. Okay? So this is what we call the b integer function. Now someone will ask me why did we just explained this be injunction in this lesson? Why did we talk about there be any junction specifically? You will find that in the next lesson when we talk about diets, you will find something which is really interesting. Diodes are results being the junction. So what did I just explain that P injunction is simply diodes. So when I'm going to talk about bytes, I will represent them in the form of a b n junction. So this may in a junction will help you understand how does dilute allows the current to flow in one direction. And the book is a current from another direction. Okay? So I hope that lesson about there being a junction is clear for you. 5. Diodes in Power Electronics: Hi and welcome everyone to this lesson. In this lesson we will do as a first power electronic switch, which is diet. So what is exactly a diode light is something which you, or the power electronic switch that you see in this figure. So the light is a semiconductor device that acts as a one-way switch for cat. So what I mean by this, it means that it allows as I can to flow in one direction. And the blocks is a current in the opposite direction. So as you can see, it allows the current to flow easily in one direction, but restricts their current from flowing in the opposite direction. The lines themselves have polarity, which are determined by an anode which is positive lead, and the cathode which is a negative late. What I mean by this, it means that this, this canceled or this diode has two terminals. One which is a bowl stiff. And the warnings and negative, there's a bolster is called the anode and the negative is called Zach caseloads. And you can see it's all rely gives us. You can see this one is a simple light which we use when we are drawing our electric circuits. As you can see here, the diode, you can see anode and cathode. And as you can see, you can see this, this triangle is pointing to this side, right? So if you look at here as if it tells you that the current is flowing like this in this direction. So it means that it's flowing from anode going like this in this direction. So when you look at any died and see is this sample here, it means that it flows from the positive to negative or flowing in this direction. Okay? It allows the current to flow in this direction. If the current comes in this opposite direction, it will look at, okay, so this is a real valued and reality and this is a symbol that we use in electric circuits. Allow current to flow when suppose the voltage is applied to the anode. And what I mean by this, so if you look like this, the x is positive and the negative, let's say I connected it to a battery like this with a positive terminal and the negative. Now in order for this dye to work as a voltage across it, mostly people stiff. The difference in voltage between here and there must be a positive value. Or it means that I'm going to connect this outpost of terminals a battery here, and negative terminal of the battery here. So what will happen in this case is a current will flow like this, right? Going out from supposed evokes the battery through the diode and it will get back to the negative of the battery. Okay? So in this case, the light itself acts as a short circuit, nearly as holistic as if it does not exist. Now, also, diodes are known as rectifiers. Why? Because they can exchange the alternating current or AAC into a pulsating direct current, DC. Okay, so what I mean by this, Let's look at this circuit. This circuit is called the half wave rectifier have, which we are going to discuss in the rectifiers section, have wave rectifier. So what you can see here is that we have an input supply is AC voltage. You can see an AC voltage on an alternating current. You can see it has a port which is positive, then negative, then positive, then negative, and so on. Now what will happen by using this diode here? This is our input, okay? This part is our input. And the output is through a resistor, as you can see here. Now, what does this guy do during the positive cycle of the sine wave? It will allow current to flow. So the boast of cycle will appear on the output, as you can see here, during the negative part of the cycle, it will block the current. So you will see that on the hours there will be zero current. Now, how does this even work? So during the post of cycle, remember here we have our diet positive, negative. Or as you can see, it allows the current to flow in this direction only. Now let's look at the supply during as opposed to psych, during the positive cycle. It will be like this plus minus. So it will allow the current to flow like this, right? Or to be more specific. Suppose stuff is connected to the bolster wolves alive. And negative is connected to the negative. So that line will be in the mode which is called the forward mode. It will be operating as a short circuit. The circuit would be like this, it will be like this. And this diode will become a short circuit like this and the resistor like this. So during the positive half, the output voltage across the resistor will be equal to the input voltage. So you can see the input similar as our during the post of psych. Now what will happen during the negative cycle? During the negative cycle it will be like this negative, false. This one is a bowl stiffening, so it will allow the current to flow like this, right? Normally is a guarantee during the negative cycle will go like this. However, this is not the direction of the light. So the light will block this current and it will act as an open circuit like this, like this. So during the negative cycle it will block that Karen. So it will be an open circuit. So the voltage across the resistor will be equal to zero during training dataset. So as you can see, zero, z naught will become positive agains in zero. Now, what you can see here is that it converted using a diode. We converted the alternating current, which you see here. Positive, negative, positive, negative. We convert it into a DC. But to be more specific, pulsating DC, you can see a pulse, pulse like this. So we have only poles the value. That's why it's called a rectified output waveform. It is in DC because it is unidirectional and it has only boast of direction. That's why we call it DC. So by using our diet, we converted this, posted this positive-negative into only Apple stuff. Most of wave form. Aldehydes are rated according to their time, which we are going to discuss, voltage, current capacity and another, another, another rating or another port, which is an exact reverse recovery time. Reverse recovery time. This is another factor which will, we are going to divide our lives based on it. In addition to, of course, is the operating frequency. Now the question is, how does avoid work? Now, we said before in the previous lesson, we talked about powers that be in a junk Tran. And I have said in the end of csa lessons that the diode itself is add a new junction. So basically light is just the Abbey in the junction, as we discussed before, with a B type material and inner thigh material between them zeros at the Appalachian region. And we said n type is filled with negative electrons, as you can see here, lots of negative electrons. Here we Kitoi material field always boast of holes. Now is a B type here, which is connected to the ANA, shows up all stiff terminal of the diode itself, which is the anode is the one which is connected to B type material. You can see both positive anode connected to the B type material and the castle which is negative, is connected to the anytime material. Okay? So what happens exactly when the voltage across it is greater than a certain value, which is called as a forward voltage. This is a small value of voltage. Now, when the voltage is greater than this very small value, the height is becoming a forward biased. And what I mean by forward biased, the electrons will have enough energy to move from the n-region to the b region to fill the holes. Let's look at here. You can see here forward biasing voltage. So what I mean by this, you can see here as opposed if terminal, this is a battery, e.g. a battery or stiff terminal and negative ten. What you can see is that we have B n junction. This is a bureau region, and this one is the n region. Now you will find that we connected z are all stiff with the B region or the anode of that valued. And here negative connected to the negative three and negative region or the inner thigh material. So if you look at the circuit here, you will find that here we have the supply. She is a battery e.g. positive-negative. Both have connected to the anode of the diode itself. And negative, which you can see here, is connected to the negative of the diode itself. So here in this mode, the voltage across the diode will be greater than VF. The voltage will be greater than VF. When the supply is greater than V F or the forward voltage, which is a voltage Eric wired to make this by the start walking or becoming a short circuit. So what will happen is that when the voltage applied here is greater than the forward voltage, what will happen is that here we have B region, right, which is filled with holes, those holes, and the n original which is filled with negative electrons. Now what will happen is that when this voltage is greater than Vf, you can see we have a bolster V of, right, and we have post of all, so that there will be a repulsion force between these two, between the all step and the poles, the votes. So there's a whole, so we'll start going to the left. And the negative will be repulsion force with a negative electrons in the n the time. So they will start to going to the right. So this motion from the left, from here to here, or from here to here for the motion of the holes or the motion of the electrons lead to generation of electric current. So when does this happen? When the voltage here becomes enough in providing enough energy for the electrons to move from the n region, JOB region, or for the whole the tumor from B region to n region. In this case it to the light will act as a short circuit. Now, this motion causes of this movement causes some flow of electric current. Now, what will happen if that voltage is less than the forward voltage is applied in this case will be something which is called the reverse biased. And the depletion region will start increasing. The current will be allowed to flow. So if you look like here, you will see that here this is a case of the reverse bias. So what you can see here is that a positive connected to negative and negative connected tools or bolster. What will happen is that this holds or this whole supposed to volts will try to go, will be attracted to the negative of the battery. And this negative electrons here will go to the left. Two as opposed to the much soil are going like this. So what will happen is that this region in the middle will start becoming bigger. So becoming a bigger means it will block current. So you can see we have a wider depletion region. All of the electrons go to the left. All of the hole that goes to the right shows up looking area will be bigger here on the, unlike the forest as the second case, you will find here is that the depletion region becomes very small. So it acts as a short circuit. Now here, once you can see impulsive connected to the negative of the diet and negative connected to the pole stiff. So here, current will be zero. This one will act as an open circuit because it blocks the current. Now, let's understand this with this video. As you can see here, this case, which is a forward bias. Now let's see exactly. Diseases are forward biased. As you can see, n no type material, B type material, anode and cathode forward bias positive connected to the anode, negative connected to the castle. Now what will happen is that this pole stuff. We'll do a repulsion force will be with holds, those negative. We'll do a repulsion force with the electrons. So the electrons will go to the left and the holes will go to the right. So as you can see here, let's look at it. You can see going to the left and gone, sorry. So this motion itself leads to generation of electric current. Okay? Okay. Because what will happen exactly is that when this holds, go to the right, this electron go to the left. This electrons flow inside the circuit itself. So it leads to generation of electric cars. Now in Zooniverse supplies, you will find what here we have posted and do we have negative our how we have here holds. Okay? So the most volts are attracted to the negative ends are negative electrons are attracted to the positive part. Okay, because it is a reverse biased. So these electrons will go to the right, and this also goes to the left and the depletion region becomes bigger. As you can see, this is a forward biased. Now look at the reverse, but you can see all of them holds going to the left. Electrons are going to the right. So you can see there is no transition of an electron does not move from here to here, or volts does not move from here to here. Z are going away from the depletion region. So the area itself is a blocking area becomes bigger. Okay? So as you can see here, again, let's look at all for the pies. Electrons holes are moving in the reverse supplies. You can see here that depletion region will start becoming ubiquitous. As you can see here, the reverse biased. Okay? So this will lead us to diets, modes of operation and the IV characteristics. Now, what we said right now is that we have two modes, the full row, two pints, which allows a flow of electric current, and the reverse bias, which blocks that electric current. Now we have, of course, the anode and cathode. And here what will happen exactly if we connect it to as anode and cathode positive voltage soon as opposed to voltage across it, let's say e.g. 0.7. Okay, is this the changes from one material to Amazon? Because this voltage which is across the diode is 0.7 v, then it will become forward biased or it will allow the current to flow through it. Okay. Now, if that reverse happen, if we connect it to the bowl stuff to Zach caseloads and connected to the negative and the current will not be able to go in the reverse direction because it is a reverse supply, so it will be an open circuit. Now, this will lead us to the IV characteristics which help us understand more. Now, you can see here, this is our bite host of negatives or voltage across it and the direction of the current flowing. Now, what you can see here is that this is a forward direction and this is a reverse direction. How did I know? Because the voltage here, which is our bolster voltage, means that the voltage between here and here is a positive value. Now what you can see that here is a forward the voltage which we just talked about. It was a voltage applied here. Less than vf, Let's say like this. Plus minus. If it is less than the forward voltage, it will be zero current. Now before we forward a little bit, that current will increase a little bit. Starting from V forward. If the voltage reaches V forward, then the current will increase exponentially becomes a very, very large. Now, this is theoretically, of course, the current will depend on the circuit itself, okay? But it will become a short circuit. Now, what is the value of V forward itself is made of silicone, then it will be approximately 0.7 volt. If there is a germanium, it will be 0.3 volt. Okay, So it depends on the material itself. Now in the reverse direction, what will happen in the reverse direction? And the reverse direction if we apply the voltage like this. Why is this false? Negative? In the reverse direction? The light itself will start, will block there. So the voltage here is in the reverse direction, negative, which means it is both negative like this. As a voltage applied is in this direction, opposite to the voltage of the diode itself. Now, if the voltage here is a negative value, what will happen is that you can see this x is representing Zachary and right, you can see in the reverse direction it blocks Zach current. It prevents the current from going through it. However, you will find something which is really interesting. You will see that the current here in the reverse direction, we have a very, very small value of current, very, very small value. Now what does this guarantee is called leakage current, because our devices are not ideal, then there is a very small leakage current that will go through in very, very small, not big like in the forward the mood, but very, very small, flowing in that evolves direction. Now until a certain point, if this diet is subjected to the voltage, which is the breakdown voltage, which will destroy this byte completely. Here, in this case, 50 volt. If we apply a 50 v, then this diode will be completely destroyed. And in this case it will become a short circuit, but it will be damaged it completely. So you can see here, when we reach the voltage or the breakdown voltage in the reverse direction, it will be completely destroyed. Okay, so we have now understand using this characteristics, we have the forward mode and then we have the reverse mode. Forward means allowing the current to flow when the voltage is greater than a certain value. And as a reversible looks as a current, excepted that there is a very, very small leakage current until the breakdown voltage which destroys the die. So I hope this lesson helps you understand that composition of the diet and the two modes of operation ends the IV characteristics of this device. 6. Types of Power Diodes: Hey everyone. In this lesson, we will talk about the types of diets. We talked in the previous lesson about without died, the composition of the Allied, the VI characteristics. And now we will talk about the types that we use in illegal power electronics or in the electrical system or electrical power applications. We have serine main types. Number one, general purpose diets, the fast recovery diodes and Schottky diodes. So we have general purpose tides, faster recovery dies and Schottky diodes. Now let's first, before we compare between them, there is a very important definition that we need to understand, which is called the reverse recovery time. What does this exactly mean? This is representing is the time taken by the light to switch. It's a condition that is from the forward biased or the unconditioned, the reverse biased or the off condition condition. So the light or the rectifier has stored charge, thus must be discharging before the guide blocks the reverse current. Now this, this is charged, it takes a finite amount of time, certain amount of time called the reverse recovery time. Now, let's understand what does this even mean? So let's say we have our diet, which is walking in forward mode. So we have the current of the light in the forward mode. The current flowing like this normally Sousa bite. Now let's say that at a certain moment, let's say at this time here, this axis is representing the time. So as a test server ten moment at a certain moment. And instead of having plus minus, the supply is reverse it. So it'd become plus, minus likes us. So it should be reverse biased diode. The current will try to go in this direction, so the light should block it, right? However, what happens exactly? This is our desired responses. So since at this instant, I would like it to look like this, right? To become zero and continue like this. So it'd be this forward and then adds the reverse or the reverse bias, it becomes a zero. Right blocks is account. However, this does not happen exactly. What happens is that when we try to switch from on to off state, you will find that the diet itself will allow current to flow in the opposite direction. So the current will flow like this rose or I'd normally throws all right. For a certain amount of time called the T S or the storage time. Then after it, as this current will start, decaying, goes time until we reach zero, until we reach the z-score, that transition time. Okay? So again, we are in the forward, the more derived. So when we switch it off, it will block the current and it becomes zero. However, this does not happen. By itself, allows the current to flow through it. The reverse current flows through it for a certain amount of time called the storage time. The storage time representing water, representing the time taken to this charge as a store, they charge inside it. Okay. So as you can see that it has stored, the charges must be discharging before the light blocks is a reverse current. So this time is called the storage of time. In addition to a transition to start the transit, to transmit or transit from the reverse the current value into a zero. So this will take a certain amount of time to go from maximum value in the reverse direction to zero, Okay, called the transition time. Okay? Now our diets itself, now this, this time itself is different from one type to another. The reverse recovery time. So I hope the idea of reverse recovery time is clear. So it's simply presenting some time. If we switch it from the old mode to the mood, there will be some time before the diode will start. Blocking is a current. Now this will lead us to the three types of lights. General purpose behind this one. As this possesses TRR quite high, so that recovery time, or the reverse recovery time is quite high compared to what? Compared this to the schottky died and convert it to that faster recovery. So this one has a very slow response between point 1 μs and wife microphones. So you can see very small time in general. So the TRR or the time reverse recovery time is between 0.1 micro to macro. Very, very small, right? However, compared to fast recovery diode and Schottky diode, it is a very slow response. Now this type of bias is used in low frequency applications. So it's used as a voltage regulator at DC to DC converter, AC to DC converter inverters, oscillators that will operate up to 1 khz. So if we are operating up to 1,000 tortoise or 1 khz, we can use this diode easily. Now, the rating of this slide are available up to 6,000 volt and 4,500 and bears. Now, this will lead us to the second type, which is called the fast recovery night, which you can see in this figure. Of course, all of the lights have several shapes. You can see this one is a faster recovery light, and this one also is a fast recovery diode. Now what you can see in this figure here, you can see forward direction by exist. Then we would like to turn off. So it will start going into the reverse direction. Then again, it will take some time. Then it will go like this. And blocks are current, so all of this is a current or the reverse recovery time, right? However, the blue one, blue one can represent as a general purpose diets, and the green one represents a faster recovery lies. You can see a big difference. You can see here from this point. It goes down and blocks, however, is a blown goals like this and take some time. And the blocks, you can see this gap in time, this difference in time. So the faster recovery has characteristics which is having low fat, low recovery time or low reverse recovery time. Now, this as a construction of this slide is similar to the Gen Z are both be n junction. So the general purpose is or being junction. The fast recovery diode is also at the injunction hours that the difference is that with a semi conductor material, we add little bit of gold. Gold is added to the construction of the semiconductor material or the construction of the diocese is a difference between it and the general purpose enlight from the construction perspective. Now, reverse, reverse recovery time here is low, very low reverse recovery time. It can be from tens of nanoseconds to 100 ns compared to the other one, which is up to 5 μs. Now this time is used in high frequency applications, and it has a rating up to 1,400 amperes and 6,000 volt. It is used in radio signal detectors, analog and digital communication circuits, industrial and commercial applications. This will lead us to the third one which is called Schottky byte. So Schottky diode construction is different from the other two diets. What's the difference between it in this light? Instead of having a b and junction, we have a metal semiconductor junction. So we have an n Pi material. And we don't have a Beta of material, but instead we have a metal region. So we have a metal semiconductor junction is formed where generally in the metal itself or the metal itself is used, is aluminium and is preferred for metal. And silicon is a semiconductor material to have that fast switching speed. So this is the function of it. We have a p-n junction. Instead of z be any junction. We have a metal semiconductor or an N metal junction. Now, the response time here is extremely low, less than two Winnie nanoseconds, less than 20 ns. And the junction form, it is short and you known as the MOS junction, why it's called the MS junction because it is a metal semiconductor junction. It is a metal semiconductor junction. And the one important neurology here, or one important inflammation here is that this Schottky diode is also known as single. To be more specific, unipolar junction. Okay? Uni-polar, uni, uni polar junction. Now why unipolar junction? Because as you can see here, we have n only type only. Now with this genotype is a source of electrons. So it has unipolar because it has one source of electron, which is the n type material. That's what's called uni polar. Compare the two there be any junction, she has two poles or two source of electric current, the holes and electrons. Now this one has a rating up to 100 v and 300 amperes. You can see it has a very small rating. Combine the two, the first two types and the very low voltage drop. That's why we use it as a blocking diodes in the PV system or a stand-alone photovoltaic system to prevent the batteries from DC charging and rectifiers and power supplies. Now what does this mean if you look at the BB system as a baby, sustained construction is that we have solar panels that is used to provide electrical power during the day to batteries right? Through a charge controller to charge the batteries. However, at night, this panel does not generate any electrical power, right? So what will happen is that this battery will start to providing electric power the reverse direction back to the panels. How to prevent this phenomena from, phenomenon from happened? We add the blocking died here, which is a Schottky. By this blocking glide prevents the current from flowing from the battery to the solar panels. Okay? So in this lesson, we talked about with the different types of diodes, such as general purpose slides. The Schottky diode is a faster recovery died and we talked about the reverse recovery time. Hope this lesson was helpful and informative for you. 7. Thyristors in Power Electronics: Hi, and welcome everyone. In this lesson or in this video, we will talk about XOS, how you restores in power electronics. So in the previous lesson, we talked about diodes. In this lesson we will talk about what those are erased, those in power electronics. So what is answer a restaurant? So what is the difference between us, I restore and dilute. So that's our restaurant, is a four layered semiconductor rectifier in which is a flow of current between two electrodes is triggered by a signal at assert electrode. So restorer is a current latching the why's it operates by using a gate current poles and remain conducting after removing the gate pulse. It is known also as an S SCR or a silicon controlled rectifier. So what you can see here is the shape of the Cyrus, Cyrus door here. This is our diode. If you remember. This is the shape of the die. And the difference between that we will have an additional terminal called the gate terminal. So you can see we have the anode which is positive, and we have the castle which is a negative. And do we have now a so determinant, which is the gate terminal. So this work is how does it work? The difference between eating or diet is that if the voltage across it is greater than voltage is greater than a certain value, the thyristor will not work. However, this time the cyrus to work is by providing a gate current. So by providing here at current, it will start operating. And even if we remove this current, it will still be in operation. That's why we call it a current latching device because it operates I providing a current or a gate, current pulse through this gate terminal, which will make it operate as a short circuit in the ideal case. And it's also as a silicon controlled rectifier because we can control, it is a semi controlled rectifier because we can control this, this operation without any gate signals, as Aristotle would not operate. When we give it a signal, it will start or break. Now, if you look at here, it's called also afford layer semiconductor because inside it, this composition as simply a full layer, you can see 123.4. So it's consisting of B type material than n type material, then B type material than n type material. The anode is connected to the forest B type material set, which contains a post of holes. So it's called the anode or the positive terminal. Then the n type here. The last one is connected to a castle or scalds a castle because it is a negative terminal, which has lots of electrons or negative electrons ends. The gate is connected to this B layer. The B layer here, or the material here. So this is a composition of the restaurant. And if you remember, the light was only p-n junction, right? It was only be any junction. But this one is a two as a full layer which is b n, b n. Okay? Okay, now the important question here is that how it has this iris or walk, okay, so as you can see, this composition is more complex, Zan Zach composition of the diet. So what I would like to tell you is that the composition in general for us as Power Electronic Engineers is not really important. What I mean by not important, not important. Understanding how do the walk we explain how as I rest of work or how iodide work. Because it will help us understand the idea behind this device. However, when we are dealing with solvers, stores, or diets, we are dealing with XOS to assemble and we are dealing with voltage across them. And that gate balls with respect to a restaurant, as you will see in the EC shoppers and in that rectifiers. This explanation is additional explanation to help you understand what happens behind what happened behind the scenes. So how does silos to work for us? If you remember, this is a p-n junction. P-n junction which are representing diet. And we said that the B type material consisting of large number of holes in the toy material consisting of larger number of free electrons. And between them, when we start connecting them together, we will have the depletion region between them, which is a region does not contain any free electrons and does not contain any empty holes. So this region does not have any electrons or holes. Free electrons or empty holes. That's why we call it that the ablation region. Now as you can see here in this figure for the thyristor, we have b n, b n. Now, between each of these two, there will be a depletion region. Between p and n. We will have a depletion region here. Okay, VeriSign, the ablation region and between P and then we will have also another depletion region. And between b and then we will have also another depletion region. Okay, So we have how many ablation region 123, this three or during, when we just keeping this device, as you can see here, without doing anything. Okay? So if you look carefully here, you can see here that figure, as you can see P and P n material. And between them there is a depletion region. Let's understand how does this virus to walk. Okay, so the first thing here is that let's talk about that if they reverse mode, the reverse bias mode. The reverse bias is simply as the reverse bias of the light. So the anode here and the canceled, we connect this to our supply here would be like this is the positive terminal and the negative term. The anode is positive and the cathode is also negative here. So positive and the negative you can see we can make the dollar battery with the reverse the signs. What will happen in this case? Let's look at here. We have negative and then we have all stuff like this. So let's look at here. Now, how can we analyze something like this? Okay, between each two layers? You can think of this as we have negative here and then we have all severe. You can think of this layer first as negative pause between each two depletion region, negative. Four steps similar to the supply negative posts. Between these two, the Appalachian region, we have negative poles it between these two negative poster. To help us understand what happens. Now at first, let's look at the first one. You can see here B type material connected to the non-negative. So what will happen to this holds this hole. So let's try to go to the left. Will go to the left like this. What happens to the electron? Electrons will go to the right like this, because you would like to go to the positive terminal of the battery and the holes would like to go to the negative terminal of the battery. So what will happen to this depletion region? This depletion region will become bigger. Okay? So here we will have this depletion region will become bigger than before. Okay? So since this one is bigger, it means it will block current, block that current. Now let's look at the second one. We have N type and B type. We have here negative posted. What I mean by negative negative, which is negative of the supply of the battery and the cost of the battery. Now let's look at here we have negative electrons and the negative of the battery. So what will happen to this electrons? They would like to go out away from the negative, right? And we would like to go to the post office. They will move to the right. Like this. They will move to the right. And the water powered the holes. Holes will have a repulsion false for so with the post of the battery. And they would like to go to the negative terminal. So they will go like this. So again, czar, boast of holes will go away from the post of the battery. So you will go to the left. The negative of the battery causes repulsion force with the electrons here. So that electron is, or will it go to the right as they would like to go to the positive terminal of the battery. So what will happen is that this one goes to the left. Then this one goes to the right. So what will happen to the depletion region? Depletion region will become smaller. So this depletion region will allow current to flow. In this case. Now what about the depletion region? Now, as you can see, negative of the battery and the boast of that battery, right? You can see we have electrons here and we are connected to the positive of the battery. What will happen to the electrons? Electrons will be attracted to the post of the battery. So they will go to the right. What about this holes? These holes are connected to the negative of the battery. So they will go to the left because they would like to go to the battery, to the negative terminal of the battery. What's you can see here is that N, or the electrons go to the right holds, go to the left because they would like to go to the negative of the battery and the electrons would like to go to the post of the battery. So what happened to the depletion region? Depletion region will become bigger. So here it will become bigger. Okay? So now we have, we already have how many depletion region 123 in the reverse mode when post above the battery connected to the negative canceled. And the negative of the battery is connected to the pole step of the anode or the ball step anode or the poles towards the sorrow restore. You will find that there are do the ablation reasons that will become bigger and the block, the current. However, the woman in the middle will allow the current to flow. So in the end, we have to blocking depletion region. And the one allowing the current, which means over all this soil Restore will block the account, okay? Because we have at least one blocking layer. Okay? So Walter is also set right now is the explanation of the reverse mode. Now, let's understand the four words Demode. We have again, anode, positive, canceled negative, or stiff connected to cancer, connected to here positive, negative. So suppose of the battery connected to the anode, negative of the battery connected to the cathode. Now, let's see it here. We have our battery like this and like this. Okay? So positive, negative, both have connected to that boast of all is the anode and negative connect to the negative terminal. Now, let's analyze our depletion region. So we have both positive on the left and negative ones. Alright, so at every depletion region we have here positive, negative, positive, negative, positive, negative. The potential difference between any two is positive. So we have here positive connected to here and negative connected here is this helps us understand what happens inside the B and B and junction, or that's how you restore itself. Now let's look at the first one. So we have ads. First one here, not this one here. This one here. P type sows are bored stiff connected to post the right and negative connected tools. And what happened to this layer? You will find that this holds what we'll have a repulsion force, repulsion force with the positive of that battery. So they would like to go to the right because you would like to go to the negative of the battery, is a boast of holes are attracted to the negative of the battery and the Z are also having the repulsion force was opposed to, would like to go away from the positive terminal. And then sometimes they are attracted to negative. So if a girl to them, right? For the end the thigh material, we have here negative connected to the negative terminal of the battery. So we have negative electrons connected to negative of the batch. So they will have a repulsion force. And at the same time, the electrons are attracted to the positive terminal of the battery would like to go to the left. So we have this one going to the right, this one going to the left. So what happened in this case, the depletion region will become very, very small. So they will allow current to flow. Now let's look at this one, n type. And the B type, n-type, and p-type. You can see that n type here is connected to as opposed to the battery. So that negative electrons here will be attracted to the positive. So Z will it go to the left? Okay. What about B type? B type material? What about them? For the B type, they will be attracted to the negative, so z will go to the right. So n here we'll go to the left, go to the right. So this depletion region will become bigger. So this one will become bigger. My exists blocking Karen, so they will block the account. Now let's look at the last one here. So we have B type and anytime negative electrons and the poles, the volts. Now, this negative was this one here, which is this negative. Negative electrons have a repulsion force with this negative of the battery. So they would like to go to the left. What about the pollster of all stiff having repulsion force with suppose we would like to go to the right. Okay, so what happened in this case, in Z m, the whole finds that this depletion region will become very, very sun, which means it can allow current to flow. So what we learned in z and in the end we can see is that we have do the ablation region, this one and this one allow can, however, the one in the middle here, but prevent the current from flowing, right? So over all, the current will be, also be blogged. Okay? So the current will be clogged. So the question is, how can I solve this problem? Okay, so how can I solve this problem? So we have two depletion region that allow current. But the problem is that in this depletion region. So how can I solve with this? You can solve this by converting, converting this be material into an immaterial. What? Yes, by converting this into n material, you will have a big n. Zeros is a big B, which means similar to the construction of the diet. So how can I convert this b into n material? This happen? You can remember that beam material assembly filled with holes falls the volts. So in order to convert this into an n type material, I will inject electrons or inject electric current by injecting negative electrons, lots of negative electrons. This B type material will be converted into N material or n-type material. In this case, all of this will be considered as one big unit. So they will be n material. That's why in normal operation or in the forward the moon. When we are connecting this to the whole system of the battery and this one to the negative. Nothing will happen because this one is a big depletion region and the blocks is a current. So when I start giving gate current the vowels. So when I start injecting the current here or injecting negative electrons, what will happen exactly is that this B will be converted into n. In this case we have n and n and n. So this depletion region will be completely that it will be removed. So in this case, the current will go like this as a bolster will go like this. And negative electrons will go like this. So that Aristotle will behave as abide in the forward movement. So I hope the idea is clear. So in the reverse biased, we have two depletion regions which are very big. And the blocking is a current. In the forward mode, we have one depletion region which is blocking is a current. By giving or injecting current gate pulse, we will be able to convert this be material into n material that will allow the virus to behave as a buyer. In this case, we have the forward mode. Now of course, of course, this isn't the forward direction. Now when we remove the gate current, this B type will still be n, right? Because we injected lots and lots of negative electron. Now when does this one return this back to become B? When we apply a reverse voltage, when we apply a reverse voltage, this n new material, we'll return back to becoming Gobi, which means it will block current again. Or to be more specific, before even the reverse voltage. When the current becomes lower than when the current flowing here. Flowing here, is lower than a certain value called the holding current. This n new material, the new and the material will return back to become B material. So I hope in this lesson you understand now how does Osiris to work and how does, by, how, how does this virus to work? By injecting assault and current balls through that gate. We converted the p and n material, which will allow the current to flow normally through it. Okay? So this is the difference between our diet and also restore from the construction perspective. 8. Thyristor Mode of Operation: Hello and welcome everyone. In this lesson, we will talk about what a restore modes of operation. So in the previous lesson, we talked about with towels I restore walks. In this lesson, we need to know that different mode of operation, similar to what we talked about with lights. Now you will find that the characteristics of the restaurant are, as you can see in this figure. We can see this psi a restaurant has three modes of operation. Number one, the forward blocking mode. So it can block the current in the forward direction because as you can see, voltage is applied in the direction that it goes to conduct. Because here, as you can see, this is Osiris right? Pointing downwards. It means that it allows a current to flow like this. So the anode should be connected to a bolster, the castle to the negative k plus, minus. So the current should flow like this, right? However, if we don't apply any gate signal, any gate Karen two balls, this thyristor will not operate. So as you can see here, you can see that here this is a voltage and this one is a current. This is the voltage across the diode. And this one is Zach can. So let's draw it in another form to understand. So let's say we have here like this plus minus. And do we have here our restaurant like this? And we have here a resistive load. A resistive load. This is a DC voltage, and the voltage across it will be positive like this. So we have a DC supply, let's say e.g. any value 10 v 100 volt, whatever. Now when we apply a DC voltage you is that with any good the bus, we didn't apply any pulse here. This one will act as an open circuit like this. Plus, minus and open circuit. The current flowing here will be equal to zero. No current flows through the circuit. So our circuit will be like this. We will have a soul, so like this. And we have open circuit representing our lives, our psi restore plus, minus and the resistance, right? So you will find that the voltage across society Restaurant itself by applying KVL here. By applying KVL, you will find that the voltage across this iris torque is equal to the supply voltage and no current flows. So here you can see that this is a voltage across that restaurant. Here. Voltage across this error. So this is a count of lung social scientist. So in the forward, the mode you can see that as we increase the voltage, as we increase the voltage across, That's why restore. We will have increase in current, but very, very small, increase in current, very, very small. Like this, very, very small. Okay. Why? Because we have a leakage current. It is blocking, but there is a small leakage current flowing through it. Okay? Now, the wrist, they're very important part here is that at a certain voltage, V, which is the forward break over voltage, as the forward break over voltage. What happened in this case, when the voltage in the forward or move becomes a very high at a certain value. This semester will break down what will happen in this case, this thyristor will be converted from open circuit and become short circuit like this. So the voltage here, voltage will drop from V and goes down to a very small value. So the voltage across it will be lower, Okay, why? Because it became a short circuit and all of the supply voltage will go to the resistor. Now, not only this, the current towards zero, right? Now, the current year will become very large because Osiris to break down, it became a short circuit. Now, if we translate what we just said right now, you'll want it like this. The voltage, as the voltage across it increase, the current will start increasing with a very, very small value, leakage current until a certain point. Here, all of this, the gate current is equal to zero. All of this gate current equals zero at a certain point when the voltage reaches a break over voltage. That's why roster will break down and become a short circuit. So suddenly you will find that the current will start increasing and the bone goes very large. Okay, so as you can see here, very small, very high voltage at which the virus to break down and the current will increase very large soundly. So this is the characteristics of the restaurant at the break over voltage. Now, of course, we don't operate at this value. We don't operate as this value. However, we operate at a lower voltages, lower values. This voltages will be corresponding to several gate currents. So what I mean by this, so the higher the gate current I give to the side restore a lower voltage required for the rest of to become in forward mode or in the forward conduction mode. So you can see that here, IG three, arg2, arg1. So here we have voltages which is equivalent to certain current. Okay? Now, if I apply a large gate current, it will go from here to the characteristics, then it will go like this. And it will conduct. If I apply higher voltage but lower current, it will also go from here to here and conduct. If we apply a higher voltage but lower count, it will do the same. What I mean by this. So the higher the gate current, the lower voltage required to make that Cyrus store in the on condition. You can see here at, this is a smallest voltage, right? This is a largest voltage. The smallest of all chill is corresponding to a large gate current. The largest volts corresponding to a small gate current. So as we increase the gate current, we will need lower voltage to be in the conduction mode. Okay? So in the forward conducting mode, so Cyrus or has been triggered into conduction and will remain conducting until the forward Karen two drops below a certain threshold, a threshold value, no one has a holding account. So you will see that here at this instant here, when the, when the voltage at a certain voltage ends, I get current. It will start to go into the conduction mode and the current will increase through it until a certain point and it will go up. Now when the current, what is the value here? What is the minimum value are required to make it goes into some conduction mode. Minimum current flowing through the diode as Roza I restored. Minimum current is called latching account. As we will see in the next lesson. It is the minimum current required to make. The rest will go into the conduction when applying the gate signal. Now, the holding current after we remove the gate signal, no guarantee here. In order to keep the current flowing through it, this current must have a minimum value called the holding current. This is the minimum value that will keep the rest are in the conduction. Now what about the reverse blocking mode similar to the diet in the reverse block mode, we have a very small leakage reverse canon. And until the reverse big, big voltage or the breakdown voltage that it will go to something which is called reverse avalanche region, at which it will break down and allow current to go through it in the reverse mode. Okay? So this representing the characteristics of the psi a restaurant. 9. Natural and Forced Commutation: So here it will lead us to the definition of the latching current and holding current, which is also set the latching current. This is the minimum forward the current that flows through those Iris tool to keep it in the forward direction or the conduction mode at the time of triggering. To be more specific nodes I get current. The gate current should be greater than a certain value. Okay? I get should be greater than a certain vy naught Zara latching cannot but a certain value depending on the design of the store itself. However, the most important thing is that when we apply this value of gate current, is the current flowing through it must be must be greater than the latching current to keep it in the forward direction, in the conduction or at the time of triggering. Also, as you can see, if the forward, the current is less than the selection currents or service will not turn all the current flowing through it at the time of triggering. If it is less than the latching current, it will not operate. The candidate must be greater than latching cat to make this thyristor operate. Now, the latching current is in the order of ten to 15 milli and bear very, very small amount of current. Now, what about holding current? Holding current is the minimum forward the current which is the current flowing through those I've restored as, as follows suit Cyrus, Cyrus store to keep it in the forward conduction mode, it is a minimum currents that must have laws, rules, or thyristor to keep it in this mode. So latching, what does the difference between latching and holding? Latching current is at the instant of triggering. When we give a current, the poles, this soy restore. So they have a current flowing, so it equal to the latching current at least which is 10-15 million bear. If this Kant does not flow through it, it will not latch and it will not operate. Now after removing the gate signal, after removing this gate signal. In order to keep those iris or insect conduction mode, we should have a minimum value of current called the holding current must be flowing through. When the forward current reduces below the holding current. Then what will happen is as Osiris store will turn off. Now, this will lead us to two important definitions which we are going to use in the analysis of our circuit, which is a natural commutation and forced commutation. So the natural commutation or commutation is simply that if we consider we have an AC supply like this one. And then we have a pure resistive load, as you can see here. And there between them we have as high a restaurant. Okay? Now what happens exactly if we consider an AC supply is they can untold fruits rules as zero crossing line while going from positive to negative beak. Okay, so what does this even mean? So if you look at here, this is the AC source, which you see in this figure, this wave form. Now we have a voltage during this positive cycle, this cycle here, you can see here during the cycle, the voltage across the resistor is like this, plus minus, or the source itself is plus minus as if it is in this part at DC voltage source or a bolster DC voltage as if it looks like this one will give us plus minus supply. Now this one would give a current in this direction, right? Direction, which is in the same direction of the restaurant. Now let's say that the triggering itself, triggering itself occurs at time equal to zero, is just an assumption. So when the supplies or blight this side rest or would become a short circuit. Why? Because it has the signal and the bolster, the voltage across it is a boast a value. So it will act as a short circuit like this. Okay? So the output voltage will be like this. Okay? So all of this, the current will flow through the resistor like this. At this point here. At this point, at that instant of zero crossing line, while going from positive to negative big as we go like this. At this point, at this point, the voltage is also will be zero, which means that the current itself will be equal to zero. So add this instant ones, the current is zero. Current is zero. It means that the current flowing through the thyristor I, which is equal to zero, is less than the holding current. So this thyristor will be turned off and they become an open circuit. So that is what we call the natural commutation. Now, this Cyrus or turn it on by giving appalls, by giving a pulse here. And we have as opposed of cycle giving up balls at this instant or here, or here, as you will see in the rectifiers. But at any instant you will find that this solver store will start operating normally. At the zero crossing the year. It will be turn it off naturally without doing anything, without forcing it. That's what we call a natural commutation because it is turning off or turn it off naturally due to these aspects of the circuit. You can see that the reverse voltage will appear across the device simultaneously, which will turn off the thyristor immediately. So we have two reasons. Why does this virus so turns off. The first reason is that at zero and instant, the voltage across it will be zero. As a voltage of the sublime is equal to zero. So the current is equal to zero, which means it's lower than the holding current. So the thyristor is turned off. Another reason here is that when we are going to exist in the negative cycle, what will happen is as the voltage across it will become negative or reverse voltage is applied. So it will be also turned off due to this reason. Now this process is called Zen natural commutation. As a service story is turned off naturally without using any external components or a circuit or supplier for communication purpose. So it is naturally due to the negative cycle, it turn it off. Now what about the force commutation, which is different from the natural commutation. So it's also a restaurant here, can be turned off by reverse biasing. That SSE r, which is a silicon controlled rectifiers that we talked about before, which is Osiris store itself. So what I mean by this in the previous slide, we reverse biased it by using, by using the natural, natural cycle of the AC source. Here we have DC supply, what I mean by DC supply, DC voltage. This voltage can be e.g. 100 volt constant value. So Cyrus or cannot be turn it off in DC circuits, we need to do something in order to force it to be turn it off. So by using active or passive elements, those aerosol can be reduced it to a value below is a value of the holding current in order to turn it off. Since the zoster virus or external forcibly it is expressed as for communication process, the force commutation can be observed using DC supply. Hence, it is also called the dc commutation. So that dc commutation or the force commutation, both of them are similar to insurance as this process happens. So when we are using DC supply. So we use an external circuit for the force that commutation process is called the commutation circuit. And the elements used in the circuit are called the commutating elements, such as L and seen. Now let's understand what happens exactly. So as you can see in this circuit here, we have our supply here, not here exactly. Here. We have a DC source plus, minus, and we have our thyristor here. And the gate signal, which we will apply at time equal to zero. We will apply it at the beginning of this circuit for simplicity. And you see here we have an L, which is the inductor, and we have a capacitor, and then we have our resistor. What does this mean? This one representing our load, this one representing our loop. Now, as you can see here, that commutation circuit here, which is circuits that we added, we added 0 and we added. See, if you get back to the previous one here, you can see we have an AC supply store and the resistor three elements. Now as you can see here in Zack commutation here, we have not only three, we have DC source, the DC source here. We have the resistor. That's why restore. And we add the two elements, which is L and C, which are the commutating elements of the suck. Okay, so let's understand what happens here. Let's talk e.g. for this circuit, this, these two circuits will be operating similar to each other. Here we have the lute in parallel with the resistor. As you can see in almost all the way here is a capacitor is in series with a resistor. Ok? So that's why here we have our load pattern with capacitor here, load were in series with the capacitor. Now, let's understand e.g. the first one here. So as you can see here, all Eugene representing the gate current, the balls. So we applied a pulse here, a pulse to make this one operate. And at the same time, you can see that the voltage here, which is always positive plus minus, you can see voltage across the SCR is always post them. So since this voltage is positive, as you can see here, as you can see here. And we apply a gate current topology. This one will become a short circuit like this. Okay, So it is, turn it on, turn it on. Now we have an inductor here and capacitor, this are two storage elements, so the current will flow like this through that capacitor and the solute-solute, right? So since it's a current will flow through the capacitor. What happened to the capacitor is the capacitor itself will start charging, charging. So if this one is started at zero volt, it will start increasing, increasing. So as you can see here, v c, which is a wave form of Zach capacitor. You can see the voltage starts increasing like this, keeps going up until a certain point. Okay, so the voltage across the capacitor here starts building up due to the flow of current. So as you can see here is the current flowing through the thyristor is a start flowing like this increase in glycolysis, okay? Okay. Now what happens exactly now, since we said that this capacitor voltage starts to build up. Now what happened is that this source, which you see here, e.g. let's say it is 100 volt, okay? Now the capacitor will start from zero. And since the current going through it, it will start a charging until it reach 100 volt. So when it reaches 100 volt, we have a source here with 100 volt, which opposes that DC source, which is 100 v to this one, pushing a current in this direction, and this capacitor pushing a current in the opposite direction. So these two sources will cancel each other. So the current flowing through that Aristotle will be equal to zero. So as you can see, the counsels of zeros are at the instant when the voltage of the capacitor is equal to the DC source, it will become zero, the current will be zero. So in this instance, Cyrus store will be the net of it will be turn it off. Okay? So here, what will happen after this? After this, you will see that this one is completely charged and this one is off. It is an open circuit. This one is like this open circuit. This load here will only take electrical power from the capacitor like this. And of course, the inductor or the storage element. So the voltage across the capacitor will start going down again. Now why does it go down here? Because the voltage across the capacitor start decaying. Why does decay? Because it is giving all of its energy to the resistive load. The voltage will start decaying like this until the voltage becomes zero. What will happen next? We will give another good bolts. So we will give us one another gate pulse. So it will become a short circuit. And again, it charges the capacitor. As you can see, the current will start flowing like this, do to the bolts which we gave. What you can see that here we call it four sit commutation for set because we added Ireland c in order to turn off this circuit. Okay, So as you can see, start operating, then off, operating them off, and so on. Okay. This is what we call the force commutation. So I hope this lesson was clear for you in understanding about the natural and forced commutation and the basic definitions and the modes of operation of Osiris. 10. Types of Thyristors: Hi and welcome everyone. In this lesson, we will talk about what types of psi restores. The first type is called the phase controlled thyristor, which we are going also to use in the rectifier is part of the course. Now, the phase controlled thyristor, which you see in this figure here. This one is used in electrical power system. So when we are talking about the back-to-back converters that are used in wind energy system, e.g. we use this psi restores as our rectifiers. And as in voters. This type is used in the electrical systems. So it's typically used as a phase control. The current evolve. What I mean by this phase controlled can't evolve. If he's controlling the carbon-12, it means that this psi restore here, if you remember, it's shaped like this. Like this, which is, this is our gate. So we provide current, we control it using a current by applying gate current at a certain phase. Okay? So that's why it's called the FAS control the current. Okay, so what does this mean to, so if you have a waveform like this, this one is the source. Now, you can see that e.g. if I would like this thyristor to operate at this instant here. Okay? So this instant is a time omega t, and this is our voltage. Let's say this is our supply and our cellular. So to conduct at a certain instant, like say here, exactly. Since I would like it to conduct at this instant here, this instant is is can be translated into as certain fees or the orange angle. So let's say e.g. this one is at 20 degrees or 20 degrees. So let's say e.g. if I would like to control my own restaurant to operate at 20 degrees. So this will detect as a phase. And we apply a current at this phase at sea, e.g. a. Equal to 20 degrees. As an example. You will understand this more when we reach the rectifiers section. Now, this one is used for AC to DC conversion and vice versa. Now it is overrating at low frequencies, most often around the AC line frequency, which can change from one country to another. It can be 50 hz or 60 hz. Now, the first control is the most common form. Common form of this I restore power control. Now this one is used for high power application. You can find this in the electrical power system. The second one, which is called the inverse or grid Cyrus tone, which you've seen in this figure. This one is used in the four set commutation applications in which we will have only DC voltage or DC to AC, e.g. it is used in DC to DC shoppers. And it is used in DC to AC inverter. The inverter, great. That's what's called the inverter, DC to AC inverter grade. So restaurants are turn it off by forcing the current to be zero using an external commutation circuit similar to the L and C that we used in the previous lesson. Now, this requires additional commutating components, thus resulting in additional losses in the environment. Now something which is I would like to mention here is that in the inverter which convert this DC to AC, usually will find that the, that the component or the power electronic device used. Usually it's something like MOSFET or IGBT. Mosfet or IGBT is this true, are commonly used in this inverters or the inverters aerosols or not, not commonly like them. Because as you can see here, in order to turn them off, you have to add additional commutating components. That is a problem of the inverted great Thai restaurants or list or in general, it is a semi control the device, because you can control when it is on, but you cannot control when it is off. You can control it off using force. It can mutation circuit or an external commutation circuit. And you can not control when it is off in a natural communication circuits such as in the AC system. However, the MOSFET or IGBT are, are known as them. Fully controlled switches. We can control when they are on and when they are off. That is the best single power. Okay? Now, this type of cirrhosis is quite fast. Combine the two, the phase controlled, which is slow, a little bit slow. That's why it's operating at the AC frequency, or 50 or 60 hz. Then we have the light activated Cyrus tone, which you can see in this figure, similar to a restaurant, but it is activated using the photons of light. So the light activated silicon controlled rectifier or photo SCR. It is similar to an ordinary CSER, which is a normal, ordinary Thai restaurants that we talked about in the previous slides or in the previous videos. But the difference is that it can be triggered, and instead of giving a current, it can be triggered by photons of light like this. Okay? So it is a semiconductor device, turns on which expose it to light, as you can see here. So the light-activated swan restore the type of services that is triggered by photons present in the light rays. This photons will lead to activation of the current and lead to operationalize. Those are restaurants or own mode. So instead of applying a gate current, we apply our photons of light. Now, it is used in optical light controls. A relay is motor control and computer applications. Then we have the try a C psi restaurant. Now, this one which you've seen in this figure has something which is really interesting. It operates in both modes, in the forward and the reverse. So you can see that here we have the gate balls. This gate current OR gate pulse can be positive or it can be negative as you would like. Okay. If I would like it to Britain's afford the mode, I will apply positive current. If I would like it to Britons and reverse mode, I will give negative current. And as you can see it is here. You can see two directions. Bully direction or y direction. You can see here can flow like this or applying this can allow current in both directions. Now try EC, which is a droid, throw it for alternating current. It is a three terminal electronic component that conducts current in either direction when triggered. That's right. You see are a subset of cirrhosis and related to the silicon controlled rectifiers. This one is different from the normal thyroid stores or SSE owls, that allows a current flow in one direction or unidirectional. But this type is allowing the current in both direction, in the forward direction, in the reverse direction. Now, most of the Troy is all can be triggered by applying either a positive or a negative voltage to the gate. So we can apply boast of walls or negative balls to the gate. Now once a trigger that Troy AC, similar to the normal side restaurants, will continue to conduct even if the gate current is ceases until the main Cannon drops below the holding current. So in the normal Cyrus tools, they turn off when the current goes down below the holding current, or the voltage is reversed. It here in the Troy you see it turns off when the current goes below the holding current. But the biggest advantage of this type that allows the current in both directions. Bi-directionality of czar Troy, he sees ME exam convenient switches for alternating current because they operate in both directions. It's called that dual polarity. So the rest of us are back-to-back stores. They are used as an AC regulators of all lighting and fan control. Okay, So let's take some blame them. Characteristics also try easy. So if you look at here, you will find the characteristics here different from the single one, single, single one or the unidirectional to be more specific. In the SSC all or the interface controls or restaurant or the inverter grid Cyrus tool. You'll find that we have this part, right? Which we talked about before, which is the beak or forward peak, peak break over voltage. The forward break over voltage. And we said before that this is a characteristics until we reach this voltage, it will break down and the current will be very high. And we said before, depending on the gate current, we can have lower voltage. So at lower voltage, we can add high, at high value of gate current and low voltage. We can conduct this restaurant here. Lower current, but higher voltage here, higher voltage and lower current. In the end, we will have conduction mode. Now, this isn't a forward direction in the reverse, so we had it like this. We had a line leakage account like this until that avalanche region at which we will break down. Now with this stuff will not happen here. This is a pi direction and it is used for both directions from here or here. That's why it has the same characteristics. But in the reverse direction. We have he'll also the break over voltage. But here we have negative break over voltage. So it is the same characteristics but in the opposite direction. So I hope that and listen, how would you learn about the different types of psi restores? 11. Bipolar Junction Transistor (BJT): Hi, and welcome everyone. In this lesson, we will talk about with the bipolar junction transistor or BJT. The BGT is simply as semiconductor device that can be used for switching or amplification of signals or electric signals. So as you can see in this figure here, we have this one which is BG team. This is our device bipolar junction transistor. This one which we are going to use in some of our circuits, such as inverters, in order to generate AC voltage from DC voltage, as we will see inside the course. So this is a BJT in real life, it has three terminals. Number one, it has a metal, it has base and collector. So it is a three-terminal device, as you can see here. B, which is means base. C, refers to collector, E refers to the emitter. Now, as you can see here in this shape here you can see this arrow means as the direction of the current will be coming out from the image like this. You can see guarantee will go out here. And since it goes out like here, so the current will be also entering from here and here, like this. So we have collector current, base current. And the outer one, which is the emitter, is a metal, is equal to 0 eBay's plus our collective submission of these two currents in the BJT here is a control signal that will control is a switching on and switching off of the BJT is simply the base current. So pi controlling is the base current. We will be able to control if this transistor is on or off. So it is used as a digital switch. And it has two constructions. It can be NPN or B&B. What I mean by this, we can have Abby and be like this one here, which means a P-type material connected to an end, the pipe material connected to a B type material. Then we have n b n, which is n connected to, be connected to N. So it is NPN. And you will see that here. If it is p and b, then b is connected to the collector. The last one connected to the metal, and the middle one is connected to the base. As you see here. Similar to here, N will be connected to collector and the end connected to emitter, and the middle one to the base. This is really important because it will affect our analysis. Okay? So as you can see, your n p n, which means collector base emitter to collector base in metro. Similar to here, p and b collector base in Metro. Ok. Now you can see that here is this figure here similar to this one, because these two are in that configuration of NBN. This configuration which will I am going to use inside the course, okay? I'm going to use all of the BJT in this form. Okay? So what's the difference between NPN and PNP enzyme construction itself? Another thing is that NBN becomes in their own mood or operates as a short circuit between c and d. So if this switch is walking, it will be like this short circuit. We can remove this and add a short circuit between what, between c and d. Like this. If this switch is off, it will be open circuit like this between C and D. So in the end, I connect four. Here I'm talking about the application of switching inside our electric circuit, not amplification. Amplification has a different connection. For switching. It will be connected to, this one will be connected to the circuit. And this one would be connected to the rest of the circuit. And we add the control signal from here. Similar to a restaurant or the dye so that it will be like this. Or let's say so rest of, okay, So we have that connected to the circuit, the first terminal, similar to that collector, and second term now connected to the rest of the circuit. And here we have the gate current or the signal similar to. Here's the base current will be the signal. Okay? So what's the difference between again, NPN and PNP? In order for this one to become an in the on-state or a short circuit. The voltage between collector and emitter. Vce must be positive. Voltage between collector and emitter must be a positive voltage between base and emitter. Vbe must be also post the value. Vce must be positive, and VBE must be both. Vba is the voltage here like this, and VCE voltage like this. Okay? So these two voltages must be equal in order to become a short circuit or in the on-state of the switching mode. For the PMP, it is a reverse voltage. Here. Vce should be negative and the VBE should be also negative. So you can see all of the condition is reverse it. So in NPN, we have certain conditions. These conditions are reverse it in B&B. Okay? Now what about the currency? So we select, we said that in the NPN, which is similar to this, this is our direction of current going downwards like this. So I collected also downward, I exist and I base entering. Now I emitter is going inside, so it will be entering like this. So I could lecture will be leaving also same direction of the metal. You can see collector going down same as metal and the metal going on similar to that collector. For the base, it will be going out. Reverse the direction of this state. Okay? So the control signal here is the base current. The base current which we apply here will affect the state of the bipolar junction transistor is in the on-state, is it in the other state? Is it in the switching mode or amplification mode? The problem of this type of transistors or a switching devices, is that it requires a very large value of base current. So we need large amount of base can order to make this power electronic switch start working. Now, you will find that in the next slides, Zanzibar, I collect them is equal to beta or eBay's. So the higher the IBS, as we increase IBS, more collector current will be produced, which means more emitter current, right? So, so that's why here, the problem is that we need very large amount of base current. Why? Because high amount of current will lead to high amounts of emitter and the collector currents. Now it's advanced, it has low conduction losses and we will understand the meaning of conduction losses in the losses, power losses lesson. It's a problem which is a complex and expensive surrogate combines the tools or other types. Now let's add our senses are working principle of BGT. So here, if you look at this circuit here, because that's what I'm talking about when we are going to discuss the inverter. One of the circuits is a single phase full-bridge involved. So what does this even do? It takes a DC supply, VDC supply, and convert it into AC supply. Okay? Now, the input will be something like this, a constant value of voltage, and the output will be a square wave like this. Okay? Like this. And it repeats itself. So you can see in the single phase full bridge inverter, we used how many, how many BGT? 123.4. And as you can see here in HB z, t, We have the three terminals, 123-12-3123 and so on. You can see the first terminal here, which is the collector. And the metal has an arrow sign, which is like this. And here we have our base. Here also collector, emitter and base. Remember when you are using it in the switching action, when the transistor is turned on, it will become a short-circuit legs is between c and d, like this. Like this. If it is, turn it off, it will be an open circuit between C and D. Okay? Very simple. And as you can see, all of the BJT is that I use here is npn. You can see the direction of the arrow is going out of the image, going out from the sample itself, which means this one is an NPN, which I'm going to discuss how it will work, which is similar to B and B and B but with a reverse conditions. We have here NBN. As you can see, n b n, which is on the base I see, and a metal which is this symbol that we use in that circuit. Voltage between C and D called the VCE. Voltage between B and D is called the VBE. Now, let's take this one and understand how can we make this one work. Okay, so first here, since we have NBN, whatever it is, we need a bolts the voltage. Remember since this one is NPN, the VCE must be posted. So what I mean by this, this one should be connected to the positive of the supply. And the negative is connected to this one. So we have both positive and negative. Okay? Now what I would like you to concentrate, concentrate with me. We have n p, n. So between N and B, we have a depletion region right here. Between P and then we have a depletion region. Now we have negative electrons here. Y exists, lots of negative electrons. Here we have both the holes exist. I exist, and negative electrons like this. Okay? Now we have suppose that what was the supply and negative will supply all stiff connected here. What will happen to this poster boast of would like to attract all of the electrons. So the electrons here will go upward, the lines, this will go like this. And because I like to go to the ball stuff. What about this one? This one would like to go like this. Okay. Would like to go like this. What about would be B would like to go to the negative of the supply source would like to go downward like this, okay? Okay. What does the problem here? You will find that here, as opposed to Paul's going downward, the negative electrons going upward. So the depletion region here will become very small as if it does not exist. Okay? So this one will allow the current to flow because electrons go like this and the holes like this similar to the aldehyde. So this one will not exist. The depletion region will become very, very small and the supply will cause this electron to pass this system. Okay, it will go like this. So where's the problem? The problem is this, in this depletion region in here, we have electrons going like this, and we have holes going like this. So what will happen? This depletion region will become bigger. So it means it will block the flow of current. So you will find that here this depletion region will prevent any current, this electrons from going into here and go like this and performing a complete cycle like this. So how can we solve with this problem? In this problem, we add an additional supply like this between VBE, which is a post of supply. Like this one here. This is baby will produce a candidate. Or let's make it more clear. Is that what does n b n mean? N. N applied material means that we have large number of electrons, right? Or to be more specific, the majority, majority. Majority is negative electrons. Electrons. And the minority. Minority is positive holds. So we still have moles in the holes here inside the B type material or n-type material. So Enter type contains negative electrons and has both the volts, but the majority of the charges here are negative electrons. Hence, a very small percentage or minority is both the volts same as for this beam. For this B, what will happen is that here it has large number of post-divorce or the majority is both the voltage and the minority. Negative electrons. So we still have some small electrons. These electrons will be attracted by the positive y-axis. And the holes. There will be a repulsion force in-between holes and bolts stuff. So it will also go downward like this. Okay? So we have two sources. This source VCE that causes a whole to go downwards, and this source which causes the holes to go also download. Now for the electrons, we have an additional source, is a positive connected to P and negative connected to N, similar to our dye it, okay? So due to this one, this supply will absorb and negative electrons will Bush them here like this, in this region. It will push them like here. So here we will have lots and lots of negative electrons. So in the end, we will have a region filled with negative electrons. So this depletion region does not exist anymore. So to make it more clear, due to the presence of this supply, it will produce electric current, or i base. The more AI-based we have, the more negative electron in this region, which will be driven by the external force or the external supply here. It will be driven, be absorbed by this supply. So they will push like this, going like this. So in the end, this current will flow through it like this. Okay? So the higher obeys, the higher is the base current. More electrons present here, which pushes which we are as they will be poached using Vc. Okay? So that's why Zach collector current increases with the increase of base. So how does the BJT as a switch? If we look at this graph, this graph shows us the regions of operation. So this is the characteristics in-between voltage, VCE, all a base and collector. Now let's look at here. First. We have saturation region, we have active region and the breakdown region, and we have also got off region. Now let's start the simplest case, which is AI-based equal to zero. Now when I base is equal to zero, very, very small amount of current will flow. So you can see here as, as VC increase. As VC increase, the collector current will start increasing, as you can see here, at 00 base current. So it is a, there is a current, but it's a very, very small current until the breakdown region. As VCE increase, as the current will start increasing until the breakdown point. So here in this region is called the cut off region because the current is very small. Or we have leakage current. So it's called the cutoff region. So when we have this very, very small current, we say that our switch is in the off state. It is not operating. Okay, now what about the breakdown voltage when VCE becomes a very, very large, what will happen in this case, when the voltage becomes really large, it will lead to break down. So you can see at this point, this device or this power electronic switch will break down and become a short circuit. So you can see that the current will start increasing suddenly, as you can see in this regard. Okay? This is same for different characteristics. As you can see, the breakdown point current will increase exponentially very large. This is what we call the cut off region. Now, as you can see here, that we have obeys, if you remember, obeys or I collector is equal to Beta, which is called the current gain, multiplied by 0 eBay's. Okay? This is a relation between collector current and IPAs. So as we increase the current, more negative electron will present here. Okay, which will be attracted and lead to more collector current. Now someone will ask a question or a question here. Here we say that the negative electrons will go in this direction due to the poster boards, the supply. Okay? But we see that the collector here going downward, it is opposite to this current. Now we have to understand that that direction of current. We have two directions of current, zeros, that conventional current and the real current. So in reality, the current due to the flow of electrons move towards the positive. So if we have a battery like this, and then we have a resistor. In reality, the current is the flow of electric alpha electrons, flow of electrons through the wire. So in reality, the electrons inside the conductor itself will move toward the Zappos stuff. So the current direction should be from negative to positive. That is what happens in reality. However, scientists agreed, or all of the scientists agreed that they will select as a conventional current direction going out from the positive. So they say that the current to go without forms of positive to negative. This is what they agreed on, but it is not the real current. So the current goes from negative to positive. That's why here when we have positive and negative, we said, does the electron goals in the upward direction or the collector current increases? However, since we are talking about with Zach, conventional current, which is IC, it will be going downward as the signs agree from Boston to negative. Okay? So anyway, as obeys increase in electric current will increase. So as you can see here, 123456. So AI-based here start increasing. So as I increase, you can see that the collector current increase. So AI-based to greater than 13, greater than two, and so on. So you can see that the value of IC here, as IB increases, also I see increases and so on. Okay? Now, this will lead us to two final reasons. Number one is the saturation region and the active region. Now, as you can see here in the active region, which you'll see here, that the current is almost constant, almost IC is almost constant. So as we increase VCE, the current will increase a little bit. Not very much. Okay? So this one is called the active region. This region, this region here is used for what? You use the folds, the amplification process. Amplification process, which we are not going to use in this course. We are dealing with the switching, the process or the BJT as a switch. We have another region, which is called the saturation region. This region which we are going to work in representing the region in which is obesity will work as a switch. That's why it's a voltage drop here. Maximum voltage drop due to this device is 0.7 v. Okay? So in reality, we will make here exists as a switch. We can remove it when it is in the on-state. We can remove it and add plus -0.7 v likes us. So we assume that there is, we can add our resistant like this. Okay, So we can replace the device with a resistor with a voltage drop of 0.7 volt as a replacement for the device because it has a voltage drop. So as you can see here, as VC during, in this region or the saturation region as we see increase, you will find that the current IC will increase. You can see here, as we see increase, I see starts increasing until a point at which will become constant. This point is the beginning of the active region. So when we are dealing with the switch as a Winsor device as a switch, we are dealing with this region. Now why is this? Because here, if you use any value here, you will find that VCE is very large or the voltage drop is very large, and the power losses is very large. In this region. If we use this as a switch in our power electronic circuit. However, we use this one, this point exactly at which we reach a maximum value of IC and the minimum value of Vc. So the maximum value of IC, VCE minimum. Okay? So we try to operate at this point. So to summarize what we just said is that the IV characteristics of obesity describes the relation between current flowing through the transistor. The voltage applied to its terminal. So current, IC and voltage applied through this two terminal C and D. Or VCE. This characteristic and represented graphically as a blot of collective current versus collector emitter voltage, or a metronome voltage with different curves for different values of base current. So as you can see, I base two greater than one by three greater than two, and so on. So ions are eBay's increase. The collector also increase the saturation region, which is the first region here, is the collector current will reach a maximum value and remains relatively constant with increasing collector emitter voltage. So as you can see here, after reaching a maximum value, like here, or here, or here, depending on the value of the base current, you will find it becomes a relatively constant, almost constant. So as VC increase, it increases by a very, very small value. So it's called the, in this region, the transistor is said to be saturated and the actors like a close the switch and this here, the cutoff region, the base current is zero. Here, as you can see here, I be zero. And the collector current is very, very small, very small leakage current. So it acts as an open switch. So when we are dealing with it as a switch, we are using the saturation region as a closed switch state. And we use them cut off region for the off state. Now, the exact shape of the IV characteristics can vary depending on factors such as temperature, operating conditions, manufacturing variations, and so on. Also will find that the second or the region which is the active region is the active region. You will find that there, all you see is controlled by IEP. As we said before, that I see dependent on IEP. As you can see here in this region. V1, V2, V3. So as I increase, hi collector increase, and you can see that it increases linearly. Linearly with increasing the collector emitter voltage. So you can see it is increasing linearly as VC increased, but this is a very, very small value, okay, in all that much. Now we have a factor which we called Zan current again. When we said that collector is equal to beta or eBay's, right? So this beta, which we talked about is called the current. Again, this is representing how much a transistor, well amplifying an input signal in terms of current. Here I'm talking about the amplification process. It defines the ratio between collector current. The base current, IC divided by IB gives us beta or them amplification ratio. So e.g. if Zack of that transition has occurred, indicator of hundred or be take 100 means that for each 1 million and Bayer current flowing into the base. So if we have IPAs of 1 million pair and b to equal 101 million pair. So if you multiply these two together, you will get 100 million pair. You can see it's 100 million bed will flow out of the collector terminal here. Now what are the applications of BGT or the bipolar junction transistor is used as an amplifier to amplify weak signals, such as in microphones and sensors. It is used as a switch to control the flow of current in an electric circuit. It can be used to switch on and off different electrical components or electronic components, such as LEDs or motors. It can be used in oscillators to generate signals such as sine wave or a square waves. It can be used in the signal generation or generation. It can be used also as a voltage or renewal later in order to keep a constant output voltage despite the variation in the input voltage or the load current. It is used as an applications to as power supplies or battery charger. So you can see BJT or it's a bipolar junction transistor, is used in many applications. I hope this lesson was clear for you on understanding what is the meaning of BGT, how it works, and the main applications or applications of BJT. 12. Gate Turn-Off Thyristor (GTO): Hi, and welcome everyone. In this lesson, we will talk about another power electronic device, which is a gate turn-off thyristor, or abbreviated as G, G, T. Or we told in the previous lesson a power BJT. Something which I didn't say is that BGT is one of the fully controlled switches. Because when we apply enough base current, you will be able to turn it on. So it is a fully controlled switch. Compared to something like psi restore, which is a semi controlled switch you can control when it is on, but you don't control when it is off unless you're using a natural commutation or forced commutation is uncontrolled switch because it's the characteristics of it depends on the circuit itself. Now, the third category, which is a fully controlled switch, one over 1/8 is one of them is BGT. The other types. We are going to tell you about them right now. So as a fully controlled switch can be used, can be turned it on, your zinc can be turned on and off using a very small control signals. So the G20, which I'm going to discuss in this lesson, and the bipolar junction transistor, which we talked about before in a previous lesson, we have another one, which is called the metal oxide silicon field effect transistor, or abbreviated as MOSFET. We have also the insulated gate bipolar transistor or IGBT. So these are the family of fully controlled switches. In this lesson, we will talk about the G20 or the gate turn-off psi restaurant, which you can see in this figure. It is similar to as I restore from what it looks like. It is the same shape as my restaurant, but it is considered as a special type of Cyrus store. This one is a high power semiconductor device. It is invented by General Electric Company as Arg2 OLS, as opposed to normal Cyrus tours are fully controlled switches, which can be turned off and on and off by applying current into the gate to the gate terminal or lead. This gives it a unique capability within this I restore family of devices. The device is turned on by providing into as opposed to if current pulse between the gate and the castle nuns. And it can be turned off by applying a current to the gate in the reverse direction, which means applying negative current. Does that require to turn it on? So as you can see, it is the shape. You can see the shape of it. It is, although it has a terminal OR gate, terminal similar to what? Similar tools or psi restaurant. But how can we differentiate between us I restaurant and the gate turn-off Cyrus tone by using this tool signs is a line like this, inclined line like this means this one is get an offsite restaurant, or it has this bi-directional arrows. These two representing G20 Cyrus stores OR gate turn-off side restaurants. They call the getter of Cyrus stores because we can turn it off. We turn this Cyrus tore off by applying a negative current, a large negative current. This is a biggest disadvantage of this device is that it requires very large revolts, the current or very large and negative current to turn it off. Okay? So it's similar to a sigh resto, but it can be turned off by applying a negative current. So let's talk about the construction of the G20 and the principle of operation. So if you look at construction here we have three terms. As we have the anode. The anode, we have Zach canceled, and we have the gate, which are the three terminals that you have seen here and not canceled AND gate. Now, this street elements inside the construction itself, we have different components. Here. We have p plus n minus b and a plus b plus, what does this even mean? N and p, which we already know. B means b type material, p-type material, and then negative type material which is doped. The boys are silicon is doped by B39 material and silicon here is doped n-type material. But the difference is that both means heavily doped. Heavily doped. Negative means lightly doped. Okay? So here we have heavy doping here. Here, we have light doping here. And B is moderate, which doesn't have positive or negative some moderate doping. Now why does this happen inside the G20? Doping levels of different regions are carefully chosen. To balance is a trade off between various performance characteristics. So what I mean by this, I mean that by using the heavy doping of the gate region, what I mean by heavy doping, p plus such as this one. This one improves that turn off time, improves the turn off time. Wildlife doping increases the emitter efficiency from the cathode region. The cathode is attached to two heavily doped n plus layer, n plus layer to have a higher emitter efficiency. What are the cost of a reduced breakdown voltage? So what we need to understand is that all of this dopamine levels are used to balance between the various performance characteristics. They are related to the design of the G20 itself. But what is important for us right now is to understand how does this even work. So again, we have the anode, which is a bold step, and Castle, which is an act. Now the first thing is that we are going to apply a boast the voltage again between anode and cathode by exist. So we have positive here and negative here. Now, let's look at what will happen between this spot between cathode and anode. Okay? We don't care about the Get Right now. We are carrying a power between anode and cathode, this region. So we have positive here, positive of the battery. And then we have here that are highly doped region of P-type material. The hole here, what does the holes would do? The whole thing here. Loves this holds. And here, negative electrons, here holds. And negative electrons. Okay? So first the whole year we'll have a repulsion force with the positive z will go like this. And water power negative electrons, negative electrons. So we'll have our repulsion force with the negative of the battery and attraction force in-between. It ends up almost all symmetric. So they will go downwards like this. So in this region, this region or resign and depletion region here, will allow current to flow because the whole Zavala exist and the electrons go like this. So this one will do nothing for us. Well, let's look at this region. We have B type, so we have here holds between it and the negative terminal of the battery. So they are going upwards like this. And the negative electron will have a repulsion force with the negative of the battery. So they will go downwards like this. So the depletion region between n plus and B will not exist. Depletion region will not exist, allows the current to flow similar to this depletion region. Now, the only problem we have is this depletion region. This region here, p. And the n. You will find that n goes down, the electron goes down and the whole year are going up, as we just explained. So the depletion region here, here will become bigger. Which means there is no electrons or holes here. So the current will not flow through the circuit. So we need to fix this problem. How do we fix it? You can see we have p. And then we added here another battery or oppose the voltage between gate and castles like this plus, minus. So what will happen is that this bowl stuff. We'll have this hole here. This holds here. And here of course, we'll have a repulsion force, so this positive, so this holds. We'll go quickly here, like this. Go quickly like this, like this. And what about the electrons here? The electrons here will have a repulsion force with a negative socially will go down like this. Go down like this. Now, this electrons which are here, we'll go downwind. To the negative electrons here, which are also going down due to the post of the battery. So what you can see here is that in the end, the current will flow like this from here. Like this. Current negative electrons will go through all of this by exist. So what happens when we add the supply here? When we add them know supply here, this box stuff causes the holes to being moved. Those are depletion region. So we have here lots of holes. And we have here negative and the negative electrons. So they will go down. Now when this electrons become down here, they will be attracted by as opposed to the supply. So they will continue to go down like this. So in the end, we will have a current flowing here. Okay? Now when older too, if we remove this supply, because this is one of the disadvantages of the G20. If we turn it off this supply, this process would continue. This process will continue and they will be flowing lines. The current will be moving like this, and the holes, of course, would be moving like this in the opposite direction. Now in order to stop this process, we need to add a very high negative current like this. So we need to apply high negative current which will return to its original state. Okay? Now, another important note for us is that the G20 itself, this GTO can be represented by two transistors, like us, children's us to live. This is just a small note for you. Now, the G20 characteristics, Let's look at the reverse bias mode. So if we connect this here, if we connected here, these are both positive and we connected here is the negative. Ends are reverse mode. It has the same characteristics of Zara. And as I saw your story, you can see here very small leakage current in the reverse and then it will break down at a reverse breakdown voltage. Now adds, adds a forward biased mode. You will find that here what will happen is that at a certain applied voltage, voltage between anode and the castle, you can see anode and the castle. If this is a boast of voltage, at each value, there is a corresponding gate current required. So the higher voltage applied between the anode and the castle, the lower getCount required. So as you can see, e.g. here we have high voltage applied. You can see gate current is the minimum value. If we apply three, which is the highest to get current, then we need lower voltage applied. What will happen exactly is that they will go to this characteristics similar tools or as Iris tool. And it will go like this. The voltage will start decreasing and the current will go up. So it will become in the on state. Now when the office state, when we apply high negative gate current, what will happen in this case? You will find that let's say e.g. we are operating at this point. It will start going like this. The voltage will decrease and the current will go down like this. Okay? So if we look at here, you can see the same characteristics here. Ands reverse mode. Here is a forward mode for the blocking because we don't have enough gate current. You can see very small leakage current as we increase the gate voltage, we will have very small leakage account until the breakdown at which the current will go like this, and they become very, very large, similar to a restaurant. Now we're, let's say at this point we apply it enough gate current. So it will go up like this and it will stop. The voltage across the G20 will start decaying, and the current will start to becoming a very, very large suddenly. Okay? Now we are in the on-state. If we apply negative gate signal, then the voltage will start increasing. Voltage will increase across the GTO, and then suddenly it will start, the current will start decaying, returning back to the office mode. So what you can see here is that when it is turn it on, it takes some time at which is a voltage will start decaying through it. So the voltage across is Osiris who started decaying. And suddenly current will increase. Then when we turn it off, the voltage will start building up and then suddenly goes down to zeros or volt is our current goes down to z. So what are the applications of G20? G20 or high power semiconductor devices that are used in a variety of applications. This includes the high-performance drive systems, such as the field oriented control, the schemes that is used in rolling mills, robotics, and Machine, machine tools. The field oriented control scheme. You will understand what does this mean. When you have more advanced power system course. In a more advanced control, a power system, the field oriented control, the rate to something which is called the ID or our direct, the excess current and IQ quadrature axis current, which is found in the electrical machines such as possessing comas machines and induction machines. Another application, it is used in the attraction applications because of it's lighter weight. It is used as in boy in the inverters to convert DC power to AC power. It is used also in the DC shoppers or DC drives, which is used to provide variable voltage to control the speed of the DC motors. It is also used in the induction heating induction heating systems. So in this lesson, we talked about with the G20 or the gate turn-off Cyrus. And we understand now that this thyristor can be turned on using a high gate count. However, an old using a gate current, however, in order to turn it off, we need to apply your high reverse or a negative current. Maybe three to four times. That. Turn on Charon. 13. Metal-Oxide-Semiconductor Field-Effect Transistor (MOSFET): Hi, and welcome everyone. In this lesson, we will to cooperate with another type of power electronic devices, which is a metal oxide semiconductor field effect transistor, or abbreviated as MOSFET. So what is the MOSFET? The MOSFET is a type of field effect transistors that has a metal oxide semiconductor structure. So what you can see in this figure is our MOSFET. And what I mean by field effect, you will understand what does this mean. When we go to the principle of operation. This type of devices has three terminals, gate, drain and source. So we have three terminals, gate, drain and source. The voltage is applied to the gate to control the amount of electricity that can flow between the drain and source. So here is a control signal is voltage in the BGT and in the thyristor. The control signal was the gate current and the base current. So in these two devices, we control them using gate currents or not get karen get current and base current. Here in the MOSFET device that we will discuss in the next lesson, which is our IGBT. In the most words, a control signal is the gate voltage. The gate voltage. This device is widely used for switching and amplification of electronic signals, similar as the previous devices. Switching action and amplification of course. So we are going to use this in our course for switching action. There are two kinds of mosfets. Any channel and the p-channel mosfet depending on the polarity of that channel. So if you look at here, this are two types of the MOSFET. As you can see, your consisting of n type material, n-type material and the p type. Then we have P-type and N-type material. And here we have the three terminals, drain, gate and source, drain gate and source. And you can see we have the metal, metal contact between the metal gate and here we have gate oxide. So you can see it is a metal oxide semiconductor field-effect transistor. Metal oxide, because you will see that the gate inside, below the metal itself. We have here. Although electric material, which is the gate oxides, this one preventers the current from flowing from the gate through the device itself, as you will see in the principle of operation. Now, what's the difference between n-channel and the p-channel? Now when this one is start walking, you will find that we have n type material and honest or n-type material. Now, due to the principle of operation, you will find that we will have a channel like this for mid between genotype and phenotype. Is this a channel will be n type material. We say is that this type is called n channel because it's forming a channel, n-channel between genotype and phenotype. Here we will call this one is a B type because it will form API type channel between this one and this one, as we'll see in the principle of operation. Now if you look at the construction or the sample of the metal oxide semiconductor field effect transistor or MOSFET. You will find that the n channel looks like this. We have drain, gate and source, drain gate and source. Now when the n channel, you will see that for the source, you will find that the direction of the arrow inside it, here, inside the p-channel direction of that arrow is going outside. So what does, what is the difference between these two? We will understand right now. So this one has a high losses due to the presence of these are on-state resistance. Now let's look at the n-channel and p-channel and understand the difference between them. So an n channel MOSFET is a type of MOSFET in which is a channel through which is a current will flow between source and drain is made of an n type semiconductor material. So if you look at here, at this, this one is an n channel. Because in order to make the current flow between drain and source, we will need a channel between them. This channel will be like this, which will allow the current to flow line exists between them, between drain and source. So we will have a current flowing like this. In the p-channel. It will be like this. Okay, So in an N-channel MOSFET, what do we do? We apply a bolster voltage to the gate terminal relative to the source terminal, which will allow the current to flow from the drain. Joe source in a p-channel type of MOSFET in which is a channel is made of a p-type semiconductor material. In the p-channel mosfet and negative voltage is applied to the gate terminal relative to the source terminal. So the n channel is opposite to the p channel and the channel we apply both the voltage of the gate in the p-channel. We apply negative voltage to the gate, which will allow the current to flow from source to drain. So here, as you can see these two figures here. We have the drain, we have source, we have source and drain here. You can see they are opposite to each other, okay, so you can see the current going out from the drain like this. Here. The current coming from the drain to the semiconductor device itself. So you can see current going out from that semiconductor device or the drain current. And here drain current entering their device. Okay? So this one here, this one here is called n channel kn China. Why? Because you can see that the current will flow from drain going downward through the source like this, going like this Darwin Award. You can see any channel. Both the voltage apply. So the current will flow from drain to source from here to here. So it flowing downward. Now, this one is at p-channel. Why? Because the current, as you can see, will flow from source to drain like this. Remember that these two are reversed it okay? So the current will flow from source to choose a dream like this. You can see here B channel, source current will flow from source to drain. So they are opposite to each other. Now one thing which is really important in order to, if you would like to know what is the direction of current, it is really, really simple. When you look at here, if you look at this one, e.g. you can see that the current, this arrow points upwards, right? Points upward. This means that the current will flow in the opposite direction downward. If you look at this one, you can see the current entering. So it is actually going out. So the direction of current is always e to this R. Now, what is the benefit of this arrow? This arrow representing the direction of what? Direction of electrons. You will find that the direction of the conventional current and solid are going downwards. That drain current, we call it the drain current. Here we can see drain current going into the device. Here, drain current coming out from the device. Now, you can see that the electrons going out towards, going outward from the drain, going outward from the, from the gate, from the gate, from there get exactly. But we will see right now what I mean by this, okay? To avoid more confusion. So an n channel MOSFET, the current will flow from drain to source when I post a voltage is applied. And p-channel Canada flow from source to drain when a negative voltage applied to the gate. Now here are the same two samples. As you can see, p-channel. Is this one here similar to this one. You can see that the current will flow from source to drain like this. Opposite to this are going like this. So this is our IED. Look at this one currently exists or is the arrow pointing to the gate? So it means that the current will go down Hold opposite to it. Okay? This is very simple way in order to understand where's the current flowing. Okay? So let's talk about principle of operation, e.g. here you can see that we have a DC supply and we have how many MOSFET here? 123456. You can see that the arrow here is pointing to the gate. So as it currently exists, right from here to here, however, it is opposite to it, so it's going downward like this. So this is a drain current. Okay? So it means that for this one to operate, we need a boast of voltage between gate and source, and also boasts the voltage between the drain and source. So we have two voltages here. First, this one should be forward biased or bolster voltage between them potentials, the potential differences between drain and source, and the boast of voltage between gate and the source, similar to the PCT when we have VBE and VCE, most of them were positive. So this is the same idea. Now what's the function of this one? This configuration which you see here is a conversion from DC to three phase AC. So it converts DC to AC or uninvolved. Okay, let's understand how does the most fit walk. So if you look at the construction of a MOSFET here, this one is called an n channel. And channel. Similar to what? Similar to this one. This one is also an n channel, the arrow pointing to the gate. So it means that this one is an n channel. Now let's understand how does this work. So we said that we have two conditions. Number one, we need voltage between, so we have here source, as we have here, drain and source AND gate, right? So we need three conditions. Number one, vds should be positive. Vgs should be also positive. So let's start with the first condition. Vds should be posted, which is depending on the supply, depending on the supply. Whereas D and S, This is our drain and this is our source. So we will apply a positive voltage like this. That's a voltage like this between them. Here. Positive. Here, negative, likes this. False negative, or positive voltage between drain and source, right? Which is the first condition. Now let's look at here what will happen to the system here. Now you can see we have n, p and p. And so we have a junction here between P and then we have a junction here between n and p, right? So we have two junction here and solids, a MOSFET. We have the whole stuff connected to negative electrons or n type material. The negative, if we go that one here, we will find that the voltage here, like this negative post of both deaf coming from here and the negative coming from cells. As you can see that negative connected to a P-type material and the post if connected to an n-type material. So this function will be what will be reverse biased. So this one will be reversed biased. Or we will have a bigger depletion region. Okay? So no electron flow through it. Okay? No electron. Therefore we have a lot depletion region without any electrons or holes. Why? Because if you look at here we have negative electrons, right? Negative electrons which will be attracted like this, Jews or bolster wolves or supply. We have here both the holes which will go to the other side. So we will have here a region without any carriers or any charges such as holes or free electrons. So here we will not have any current will flow between them. Okay? For this junction we have p and the n. So n will have a repulsion force, will have negative electrons here. This electron, so we'll have a repulsion force with the negative of the supply. So they will go like this. And the holes here will be attracted to the negative of supply. So they will go like this. So this junction will be forward biased. Okay? Now one thing which is really important, you have to know that the body itself of this device is grounded or has a zero volt. Now, same for this as a source itself is also zero volt because it has the same charge as the body itself. Okay? Now what is an extra step? So the problem here is that we have, we have what? We have this junction which is reverse biased, so no current will flow, right? So this will lead us to the second condition, which is the gate signal or not that good signal, the gate voltage that will start making this as this, most of it as a short circuit. So we need the condition that will make this MOSFET as a short circuit. So how can we do this? We need to apply voltage between gate and the source, which is a poster. So we have gate and source, okay? So we will apply oppose the voltage like this plus minus, like this. Negative. Okay, now you will find something which is interesting here. If we look at the gate, at the construction of the most fit here, you will find that the gate, this is a metal, right? It is not in contact with the B3 material. You will find that we cannot attract or not attract. No current will flow through the gate. Why? Because we have this gate oxide. So what's the function of this gate oxides? This gate oxide act as though electric material, dielectric material. Okay? So here you have body with a zero voltage and the gate with a post to voltage with a boast the voltage between them. There is what? Zero is an dielectric material. So as if we have like this a material, a conducting material, and another conducting material between them, poles two of the supply and the negative of the supply by exist. Between them. They'll die electric material, which is the gate oxide, like this dielectric material. So no current will flow through it. Through the gate oxide, okay? No current will flow like this. However, you will find that here we have the bowl stuff. Here we have the negative. So we will have what? So we will have electric fields. Electric fields. So this is the gate and the body, or the zero voltage we have between them. Electric field lines is going from positive to negative leg, just electric field like this. To make it more clear. Positive, negative electric field likes us. So if you look at here, you can see that electric field. So we have electric field, likes as electric field like this, from positive to negative or zero voltage. Okay? Now another thing is that this electric field will cause the electrons to be attracted upward. So remember that the electrons move opposite to the direction of electric field. Direction of electric field from positive to negative. So all of the electrons, remember, remember something which is really important. A beam material means that the majority is both still holds. However, we have some minority of electrons. This minority of electrons will be attracted by this electric field. They will go here like this and the go upward. So due to the presence of the electric field, all of the negative electrons inside the p material will be accumulated like this here. Negative, negative, negative, like this, close to the Perrier. They will not pass. Why? Because we have dielectric material or an insulator. So we will have here, as you can see, negative, negative, negative, negative. What will happen is that we will have a channel like this. Lots of electrons which will form a bridge between this and the post and this post. So we have electrons here, and electrons here adds this region. Okay? So what will happen is that all of the electrons will move like this, throws up or stiff and go like this. So now our MOSFET is conducting. So why do we call this one and n channel? Because as you can see, that channel format here is from negative electrons or n channel n as if it is an n type material. That's why we call this one, is an n channel material. Now when we remove the supply, everything will go back to normal. Okay, when we remove this, the gate voltage, make it zero. Then this MOSFET will beat on and off. This channel will disappear and it will not conduct. Okay? So what will happen? Again? All of these negative electrons will move like this. Those are both positive. Or it will be a repulsion force in-between it ends of Boston. Okay? Now as you can see here that I'm here for the, for the N channel type. You can see here that the current going from where to where the direction of electrons are moving from source to drain, right? Like this bidirectional fourth, electrons. Election coming out from source, going into the drain. However, we are talking about in electric circuits about with the conventional current. So we say that the current to go from drain to source. So in reality you can see that here, the source, the electron goes from the source like this. Choose a tray, right? That's why is this R0? So was that the electrons go from source to the drain, going to the device, like this, from the source. However, since we are talking about the conventional current, which has a direction opposite to that Over the 2s electrons. So we will say that the current will go down when in reality, electrons move through the drain. However, in the conventional current which we use, it will move from drain to source like this. Okay? So I hope this explanation is clear for the MOSFET. Now, let's talk about the characteristics of the MOSFET. So if we plot the characteristics of the MOSFET, you will see that it is almost similar to BJT, similar to the beach deposit difference is that instead of having the control signal as IEPs, remember that the cut-off region, cut-off region or the office State of State when I paste was equal to zero inside the PG team. However, here in the mosfet, the state when VGS, when VGS is lower than a certain value called the threshold. So at least we need a certain value of V G S to form an n channel. The minimum value of voltage for men and a channel or a bridge in order to be in the on-state. So if the voltage is lower than a certain value, it will be in zone of state as VGS increase, similar to as obeys increase in the Apigee team. So as VGS increases, more current will flow. More current will flow between drain and source or ID to be more specific, I dream. So as VGS increases or the increase and also as VDS increase, ID also increased. You can see as VDS increase, you can see all these stores increasing. Now, here we have two reasons. Number one, that linear region and saturation region. You can see that this is different from what from BGT. In BJT, we had the first region as saturation region. And the second reason was the active region. Here is the most weight. The first region is called the linear region or the ohmic region. Second one is called the saturation region. Now why do we have different names? We will understand this in the next slide. Okay? So here in order to be in this region, you will find that VDS as lower than VGS minus V threshold. In order to reach this saturation region, VDS must be greater than VGS minus V threshold. Now why do we walk in this region? Why do we work in this region as a switch? As a switch similar to the PCT, PGD, we work it in this first region as a switch. Now why is this? Because if you look at here, you will find that the relation between voltage and the current is a linear, linear line like this. So we have voltage and current. We have a linear region. Linear relationship exists between voltage and current. This is similar to what? This is similar to the resistor. So you can see that this relation or similar like this, similar like this as v like this. So in this region, they say is that this MOSFET acts as what acts as a resistor. That's why we walk in this region. Because it is a MOSFET itself behaves as a resistor. And this saturation region will not help much. It will lead to power losses. That's why we use this reason we use linear region in the office or in the on-state as a resistor. And in the office state, we use this voltage. We use voltage or lower than a certain value. So in general, any MOSFET is seem to exhibit three operating regions is the cutoff region, which is MOSFET in the off state. The ohmic or the linear region. The saturation region, which is the final one. The cutoff region is a region which is our MOSFET will be off as there will be no current flowing through it. In this region, as are most of it will be an open, as an open switch. So by controlling the voltage between gate and the source or vgs, we can make the switch on and off. Okay, so this one is used when we are required to be function as an electronic switch. The Army Corps, the linear region is where the current IDS increase with an increase in voltage. In this region, it will behave as a resistor. And it has a low voltage drop across its terminals, which will minimize our loss as compared to the, to the saturation region. In the saturation region. So most people will have a constant current IDS, or current flowing from drain to source constant, despite increasing the VDS or the voltage across it. When the mosfets are made to operate in this region, they can be used as an amplifier. Again, the second region, or the constant current region. This is the region at which we work as an amplifier. Similar to what? Similar to BG team. Here this is the same region at which we will work as a switch similar to the Apigee team. Now, let's understand the difference between saturation in most width and BGT. If you look at the two figures here for the BJT and a MOSFET. Because I know that I will get this question a lot. Here in the MOSFET, we have the first two region called the saturation region. Second one called the active region. In this region we work as a switch, right in the PCT, because the relation between current and voltage, similar to as a resistor, a linear relationship. Here we work as amplifiers right here for a MOSFET, the forest region called the linear region. At what tool would work as a switch. And second reason for saturation, we will work as amplifiers. Now, you will see that there is a difference in, in names between these two. This region called the saturation region. However, the first division is called saturation region. So why do we have different names? Saturation in the BJT is and the mosfets refers to the front phenomena. In a PCT is a term. Saturation refers to the region where both the base emitter and base collector junction are forward biased in the PCT in that BGT here. Here we will have the voltage base in metal. Vbe and VBC are forward biased and the collector will be limited by the base current. So as base current increase, the value of z I collected will increase. Saturation here means what means in the regions that we will have a and metal and base or a base collector are forward biased. Now, another thing you will have to understand that in the amplification region or in this region, you will find that VBE is forward biased and VBC is reverse biased. So one is forward and one is reverse biased. This region, what we call active region or amplification region in the mosfet, however, the term saturation refers to the region where the drain current saturates. You can see drain current becomes constant. It stops increasing as VDS increases for some. So here when we call saturation, we talk, we are talking about Id becomes constant. Here. When we say saturation, we mean that we are talking about that base emitter and the base collector junction are forward biased. This difference in terminology can be a source of confusion because it has becomes a norm in, but it has become the norm in electric circuits. Yes, it makes some confusion really. Because if we look at here, this one should be called saturation because I collector is almost constant. However, we can't do anything about this. This one is called saturation. Okay, anyway is all single bonds this lesson is, and the applications of MOSFET. Mosfet are used in switching, which is in digital and electronic circuits as we will use in this course. It is used also in amplification by using the saturation region. It is used as a voltage of regulators in order to keep a constant output voltage despite input voltage variation or load current variation, which is similar to the applications that we discussed before. You can see that all of that power electronic switches almost have a similar applications to each awesome. It is used also in power conversion circuits such as DC to DC converters, inverters, and motor controls. Controllers as they can efficiently switch high currents and voltages at high frequency. They can be used in the microprocessors, memory chips, and other integrated circuits. So in this lesson we talked about Zara more sweat, the application of the applications of the MOSFET, the principle of operation, construction, pipes, and the characteristics. 14. Insulated Gate Bipolar Transistor (IGBT): Hi, and welcome everyone. In this lesson, we will talk about with another power electronic device, which is insulated gate, bipolar transistor or IGBT. So as you can see in this figure, we have the IGBT, which is simply as three terminal power semiconductor device, which is used as an electronic switch. Similar as what we discussed before is the most fit. That BGT is G20. All of the other device have three terminals, except of course is aldehyde. One is a three terminal power semiconductor. It is used to combine high efficiency with fast switching. It's consisting of four alternating layers of P and P and that are controlled by a metal oxide semiconductor gate structure. And it has our characteristics of obesity. But it's controlled as a MOSFET or IGBT or the insulated gate bipolar transistor, is used to combine the properties or the characteristics of Beijing team and the MOSFET. That's why. Here what you can see here is our IGBT. Are IGBT. This one is, can be represented as a PNP transistor with an N-channel MOSFET. So it is as if we combine both of them together and we form it one device. So what you can see here at this terminal is the gate with, you can see the shape of the capacitor here. Similar to the most fit, similar to the mosfet, no current will pass. So we will have here a dielectric material or a metal oxide as insulator. We have here gate. Now similar to the PGD, we have collective a method and then instead of the base, we have the gate of the MOSFET. So as if you take some of the PCT and some of the MOSFET characteristics. So it is a voltage controlled device, controlled by voltage Vg. Vg, similar to the voltage between gate and source in the MOSFET. As before. This one requires a very small voltage zones again, to keep conduction, unlike BJT, which requires large base current in order to keep the saturation region. Now, this one combines the characteristics, as we said, between MOSFET and a BJT. That's why it has low losses, SPECT and easy to turn on and off as a MOSFET. Now let's look at the working principle. We said that we have this to the PNP transistor or that BGT and the MOSFET. So if we look at the construction of this device here, you will find that the way in order to make this one a short circuit between collector and emitter, same as a PGD. We need that short circuit like this. When it is on the, when it is off, it will be an open circuit between collector and emitter. So in order to have this VCE voltage, Vc mostly people stiff. And in order to turn it on, we need to apply a voltage of VG. And of course the collector current is equal to the emitter current because we don't have any gate current. Now if you look at the construction here, you will find that we have gate and the emitter here divided into two regions, the metal. And we have here the collector. Now what you can see here is that we have P and P and P NP, this part p and b here. And this BNB. Similar to what? Similar tools or transverse total P NP. And if you look at here we have the internal. If you look at this part, n plus n plus NP, this sport representing a MOSFET and this one representing a MOSFET. That's why we say it is. Its composition is a combination between an n channel MOSFET and the PNP transistor. Now we will find that how many junction we have, we have a P plus, we have n minus between them. There will be junction on per one for depletion region between n minus np junction here and here. Between P and then a plus another junction. Now, you will find here we have insulator is all U2 or silicon dioxide. This is a metal oxide used as insulate. So you can see that the gate itself is insulated from the material itself or the inner construction. That's why no current will pass through the gate because we have this insulator or a dielectric material that insulates between the gate and construction itself. Okay? However, for affords the metal, you can see the metal is a metal connected directly to the junction so current can flow through it. Okay, so let's look at the working principle first. We need a voltage, VCE, VCE positive voltage. So the current flows from collector to emitter, as you can see, similar to the direction of this are, as you can see here, from a metal from collector to emitter. So we need a post of voltage. So we will get like this. We have a host of, of the battery, negative of the battery and connected it here. We have plus minus. This is a voltage of V C E. So let's look at what will happen. So we connect, it's up a whole step and negative here. So both stiff connected to that be junction and the negative, as you can see, connected to this negative. So between this junction we have all stuff. Well, push the holes. So the holes will go like this. Holds, will go upward. Repulsion force between it ends up bolster for the electrons, negative electrons here. What will happen? They will go downward like this because they will be attracted to the positive of the battery. Attracted to. Suppose that all symmetric, okay? So that junction number one here is forward biased. This depletion region is very, very small, which means they will allow current to flow. Water powered junction Number two, the negative electrons here will be attracted to post if Suzanne negative interactions will go downward here, similar to this. Okay, go downward. Water Power, BI, BI, or the holes here. The whole thing here, will it go up or go upwards? Now why is this? Because they will be attracted to the negative of the battery. So they will go up four. So we have electrons going down. The holes are going up, so this function will become bigger. Okay? So we have a big gap junction here, which will, which will not have any type of carriers. So no electrons here or more holes will exist here. So this will prevent the current from flowing. So it is a reverse biased. For this tool, we have n with a negative, so this negative electrons will go downward. And do we have this holds it going upward due to the effect of the battery. So this junction here and here will allow current to flow. So we have one junction which is reverse biased. So how can we solve with us? We can solve this by simply applying a voltage between VCE, voltage between gate and the emitter or bolster voltage between gate and metrics. So we will have like this, we have a battery. Like this. Law exists. And this negative will also be connected here. So it's a boast of V G, V G E. Okay? Or stiff connected to the gate, negative connected to the emitter. Now, let's see what will happen. Now as you can see this post of what they will do. They will do something which is interesting here. This first is this negative here, Let's look at this one. Negative first. As a negative, what it will do, it will push these electrons because they will have a repulsion force whose electron? So the electrons will go away like this, will go away like this. Now, due to the presence of this positive here, positive here at this region. And we have a metal with a negative, with a negative in metro negative. What will happen is that these electrons will be attracted to this region. So we will have electrons in this region here. Like this negative electrons. So this negative electrons will form a bridge between here like this, a bridge like this, similar to the n channel that we talked about in the MOSFET. So what will happen is that we have VCE difference in voltage. This difference in voltage will lead to, or to be more specific, let's look at it in another way. You can think of this as negative here, or the emitters connected to zero is the lowest voltage zero. And then we have here a poster volts. So you can see we have here both the voltage and the metal, which is in this region here. Negative. So there is a magnetic field here that will attract all of these electrons at this region. This electrons here, all of the electrons that will be attracted from the right and from the left, they will form a channel here. So when we have this electrons as a channel, what will happen exactly is that the electrons will go from here by exist, electrons will move like this, goes down. All the twos are positive of the battery. So we have electrons going downward. And at the same time we will have this holds that will go into the other direction lie exists through the negative of the battery. So we have motion of electrons from this region going like this. And the electron is coming from here. Choose a postal symmetric. So due to the presence of this voltage, this will lead to breakdown of this junction. So you can see we have n Nepal step and holds. However, this junction is reverse biased. So by applying a voltage between these two, we will be able to break down this junction. Okay, so this is how an IGBT works. You can see that all of these devices all almost have the principle of same principle of operation or the same idea behind them. The same idea is to convert a reverse biased region into a forward bias by applying a voltage or by applying a current or any other message. This will lead us to the characteristics of the IGBT, which you can see in this figure. Again, similar to the previous divisors. Here we have, instead of the drain current, we have Zach collector current, and we have voltage VC, which is voltage here. So as we see increase, the current will increase. However, Vg must be equal to or greater than a threshold value. Vg must be greater than a threshold value to start giving. What you can see here is that the control signal is VCE, as we've just seen, which is the voltage between gate and a metal. It can be done by applying a small VGG voltage. So what you can see here is that we have two regions, linear region and the active region similar as before. Linear region is the region at which we will work as a switch. Here, because it behaves as a resistance or a resistor. Here, the same idea exactly. Now in order to turn it off, VCE must be lower than a certain value. So it will be in the cut-off region or the folly of region. Now of course, as we increase VCE, as we increase VCE, the current will increase until the saturation. Saturation of the current. Also, as we increase VCE, that curve will go upward, current will increase. So if you look carefully, you can see that this characteristics is similar to the characteristics of BJT. But the difference is that instead of IPs, we have voltage VG. Okay? Now what all the applications of the IGBT? Igbts are used in a variety of application due to their high voltage and high current characteristics. They can be used in a switched mode power supplies, which is used in medical equipment and computers, used in the UBS systems, can be used in the AC and DC motor drives, which can control zero speed. It can be used in Zara shoppers, DC shoppers, and AC shoppers. And in Wolters that will convert DC to AC. It can be used in the solar inverters, used also in the induction cookers, induction rice cookers, and micro waves. So these are some of the applications of the IGBT switch. So you can see that all of these switches in z and z are controlled by applying a current such as BJT or restore. And others are controlled by applying a voltage such as IGBT or MOSFET. Okay? 15. Types of Switches: Hey everyone. In this lesson we will talk about the different types of switches. And what I mean by this. The difference between an ideal switch and practical switch. Now similar to transformers, e.g. we have an ideal transformer which does not have any kind of losses. And we have the practical transformer in which we add more leakage current or leakage reactance and resistance and so on. So in the ideal switch, as the ideal switch which is not exist in reality, it can block infinite voltage in the reverse bias mode. It doesn't even break down. Reserve has no inverse or reverse breakdown voltage. It can block infinite voltage in the reverse mode and in the, in the forward the mode, so it can not be damaged it at all. However, this is not in reality. In reality when we have talking about a practical switch, it has a certain finite voltage. Second enzyme, they'll switch. We don't have any leakage current in the reverse mode or in the forward blocking mode. However, in the practical, which is what will happen in reality is that we will have a leakage current, very small leakage current inside it. Here for the interests which it can provide infinite current and during conduction, it can give infinite current, unlimited current. However, this does not happen in reality. In reality is a practical switch, has a certain current rating, e.g. it's a rating can be, let's say 5,000 amperes wasn't and bear whatever. This, however is the idea. It means if it is infinite for the Either, it occurs as it provides zero voltage drop. However, this doesn't happen in reality is a practical switch have a forward drop. When it operates. Here, the zero transition between on and off in state zero time, which it does not happen in reality. Here in the practical switch, we have a finite time between on and off state, similar to if you remember, the reverse recovery time in psi restores. So of course, there is no zeros, there is no zero time for transition. This talks about ideal switch or ideal case. However, this is a practical case, which is a practical which happens in reality. If you combine between them, sometimes when we do that circuit analysis, sometimes we deal with switches as ideal switches. So when I say ideal switch, I say that this e.g. the light will be converted into a short circuit without any kind of voltage at all. However, is the royal or the practical switch. It has a voltage drop across it called the V F, let's say 0.74 v for silicon and 0.3 for germanium. Here. This is a characteristic or characteristics of the light which we talked about before. So this is the practical cases, a real case. This is the ideal case. So you can see that here, when current is equal to zero or when the voltage is greater than zero by a very small volume, it can give infinite current. However, this does not happen. In reality. There is a small leakage current, then after certain value it will start increasing. Also in the ideal case we have infinite current. It can give maximum current what I mean by like 1 million and Bayer, which does not happen in reality. However, in a real practical, it has a limited maximum current. And also in the reverse current, zero reverse current or zero leakage current. However, in the reverse mode, there is a small leakage currents. And you can see there is a breakdown here. We don't have any forward the breakdown or any reverse breakdown. Okay? So in reality, this what happens in reality. However, what's the difference between these two? The ideal helps in simplifying or making simplification for the circuit analysis itself. So when we deal with idealists, which I know is that it acts as a closed switch orders assault, set it without any volts drop, without anything. However, in reality it is in that short circuit case or in the close the case, it has voltage drop with a forward resistance. That's why is this forward the resistance causes voltage drop and the cause is power losses. Okay? 16. Comparison between Controlled Switches: Hey everyone, In this lesson we will summarize what we is offset or a comparison between controlled switches that we talked about. So we talked about these switches, I restore, IGBT, mosfet, GTO, and BGT. So let's look at this. Does this have a latching case or not? So e.g. latching for Cyrus or so when you give a certain amount of current, it will be operated on. So it is a latch. It, it means that I don t need to give continuous gate current exhausted by giving a certain amount of current, it will start operating and the key operating in the same mode. However, in these devices, must have a continuous gate OR gate signal or base currents such as BGT. So PGP requires all he pays all the time. So if I base drops to zero, BJT will be off so it cannot be latch it. Gto, the same idea, MOSFET, the same idea when we apply our VCE, or a voltage between gate and VGS, VGS. When we apply a voltage VGS, Allston value greater than uncertain threshold, the value, it will be in their own mode when we make this voltage equal to zero, this one will be turn it off. For the IGBT. Same idea. We have the voltage Vg e between gate and in metal. So if this voltage drops to zero, this one will be turned it off. The control signals Cyrus or as I said before, controlled by a gate current, BGT control the white base current. Gto also get current, most of it. And LGBT voltages for the switching frequency Cyrus has a small switching frequency, operation frequency 50 or so seconds. The artist BJT is same idea. Low frequency or LGBT can work with high frequencies, high switching frequencies up to 100 khz, this MOSFET in the megahertz range and G20 in 5 khz range. The power rating of the several stories high. Gto is also high. Igbt, medium, BGT, medium. Most of it is a low power rating. Then h of z is representing the maximum voltage rating for each of these devices. You can see upto five kilo volt, BJT, one kilovolt, G20, five kilovolt NCO, and so on. Maximum current for kilo ampere 1.2 and so on. Now one thing which we have to understand is that Cyrus DOE, e.g. five kilovolt. If our system e.g. operating, let's say at, at 20 kilovolt. Then we will connect several Cyrus towards the end zeros to increase the voltage rating. Now what are the factors that we consider when we select the power electronic switch? Number one, we look for the voltage rating, such as a forward voltage drop and the peak inverse voltage. What is the maximum voltage that can block in the reverse bias mode? And what is the voltage drop in the forward the moon? And it's a voltage rating itself. How much it can with stand in the forward mode. And also we have the current ratings such as the average current, the root mean square current, maximum current, and leakage current. These robots are called the voltage current power capacities of Zahn power electronic switch. So when we are looking for a switch, we say, how much voltage it can withstand, how much current rating or the current rating, how much current can flow through it, and the power capacity, or how much power it can operate in. Now, the four is the maximum allowable temperature for the device. Because since we have current flowing through support electronic switch and this power electronics, which has its own internal resistance, it means that we will have power dissipation, or heat energy. Another factor is the variation of current with time and variation of Bolchoz time. The variation of current d by d t. Each power electronic device has its own rating of D over DT. Variation of current with time, and voltage times d v by d t. How much is the voltage can change with respect to time through this or across the terminals of this power electronic switch. 17. Power Switch Losses: Hey everyone. In this lesson we will talk about with the power switch losses for power electronics, which we have z's types of losses that can occur. Number one, conduction losses. This losses will occur during the period or conduction period. So if you remember that when we represent the practical switch, which says that it has a certain resistance, which we call the on resistance when the current flows through this resistance. So we will have losses in the online mode. This is what we call the conduction losses. Second velocity is the switching losses, losses virtual care between transition between on and off state. Another type of losses, leakage losses in the forward blocking mode and reverse blocking mode depending on the device. Of course, e.g. you can see this switch here, this figure here representing zoster virus store, the forward blocking mode. We have a forward leakage, you can, and in aversive blocking mode, we have a reverse leakage current is this current which should, when it flows through the circuit, it will cause power losses, which we call leakage fluxes or blocking losses. Then we have that driving losses or gains or losses. This is due to the driving signal. What I mean by this, let's say e.g. if we talk about it. Otherwise, let's say e.g. BJT, Let's say e.g. BJT, we have Bayes, collector and emitter by exist. So if you remember that the driving signal, this is representing our gate, get losses inside something like a MOSFET or IGBT. They have gait. So the driving losses representing those occurs due to the application of the signal. Here is application of the control signals. So the base current, as we remember, the base current must be high in order to allow high current flowing, high current flowing from collector to emitter. Now since there's IPAs is high, this is the one which have high power losses. That's why this losses which is due to the driving signals such as oil-based or the current is throws us I restore, or voltage due to voltage across the terminals of the MOSFET or IGBT. All of these can be neglected. Very small losses except as a BJT, as it has a high base current causing power losses. Another orange is the terminal or lead losses as a terminal or lead losses, or lead, lead losses, terminal or lead losses. And those losses due to the terminal and resistant. You can see this terminal here, this term here. These terminals have its own resistance. When the current flows for them, they will have a certain power losses. Now, the rest of e.g. have this, all of these losses, exceptions are driving losses as the gate current is very, very low. Now why it is important to consider these tools that occurs inside our power electronic device. Because this will have to ensure that this system will operate efficiently under the prescribed, prescribed ambient conditions. So probably also designed a heat removal mechanism as a heat sink, e.g. order to dissipate the switching losses. So the first statement, what does this mean? Remember that our power electronic device or any electrical equipment will be installed in an ambient condition at a certain temperature. Let's say e.g. 50 Celsius degrees, which is different from a location like 30 Celsius degree. So the higher the temperature. In addition to the losses, this may damage power electronic device. So we have to add heat removal mechanism such as a heat sink in order to dissipate this switching losses, these losses, we have to dissipate them, remove them from the power electronic device to ensure that this power device will not be damaged it. So what you can see in this figure inside the power electronic circuit, you can see all of this representing what a heat sink, which will take heat energy from the power electronic device. This a bit out to there. So this is a function of the heat sink. And the why should we consider this losses that occur inside the power electronic switch? 18. Introduction to Rectifiers: Welcome everyone to this section in our course for power electronics. In this section, we will start talking about the rectifiers or the EC two DC converters in power engineering. So let's start by understanding the definition of rectifiers and why do we need them? So recti fars are electrical component or electric circuit that is used to convert the alternating current or the IC current AC voltage to a direct current or a DC voltage. A rectifier is analogous to a one way valve that allows the electrical current to flow in one direction only. So the process of converting this AC current or AC voltage into DC current or DC voltage is known as rectification. So let's understand what I mean by this sentence. As you can see in this figure, we have an AC source, an AC voltage source, right? So what does AC mean? AC or alternating has two properties. So the first property is that you will find that we have AC source, we have positive and negative or we have switching of directions. Sometimes we have a positive part and another portion of the cycle, we have the negative part. Then again, positive and negative. This the first property regarding AC voltage number two. You'll find that the magnitude is changing throughout all of the time. So as you can see, from zero starts to increase till reaching a peak value then starts to decrease. So here, as you can see, we have positive, negative switching of directions, and at the same time, we have a variation in magnitude. Okay? Okay, so what about DC which we would like to achieve? So we would like to convert from this into something like this, DC DC. So DC has two properties. Number one, it is uniditonal, uni trictonal. Okay, unidirectional. What I mean by this, it has one direction either to become only positive or only negative. That is the most important feature regarding the DC source. Number two, you will find that most likely DC source is having a constant magnitude. It is not changing with time. Unlike AC source, which is changing with time and its magnitude is changing throughout the whole cycle. So what we would like to achieve using rectifiers is that we would like to convert from this AC form into something which is constant like this, which is a DC source. Now, we will try as we can. We have different rectifier circuits. That will try as possible to become close to the unidirectional or DC source. For example, you can see here this circuit, which is a full wave rectifier, which we are going to discuss. As you can see here, it's consisting of instead of having positive negative, we have all positive. Positive, positive, positive. Still, the magnitude of the voltage is changing with time. However, now we have the first feature, which is the most important one, which is a unictonal. Now, as we advance to more complex circuits or we add filter like capacitors, we will be able to reach very, very close to constant DC. So what are the rectifier circuits? So rectifiers are divided into two main categories. Number one, the single phase circuits, and number two, the three phase circuits. So what's the difference between these two? First, the single phase? It means we are talking about a very simple source consisting of two terminals, the line and neutral. Line and neutral. When we are talking about a three phase system, we are talking about three phase in electrical power system, A, P, C, or R, S T, or the three phase system, and we can have the neutral or we cannot have the neutral depending on the connection, star or Delta. So here when we talk about single phase, we are converting this one source, the alternating source into DC. For the three phase, we have the three wave forms that we would like to convert all of them into just one DC form. Under these two main categories, we have several families. You can see here, uncontrolled rectifiers, half controlled rectifiers, fully controlled rectifiers, similar to the three phase, the three categories, and many, many other circuits. I will try to explain most of these circuits inside the section of the rectifiers. Rectifiers are used to convert AC to DC. Why do we need rectifiers? For example, if you have an AC source, an AC voltage source, you cannot use something like this to charge a battery. For example, at lv volt battery. With positive and negative terminal, you can't charge with using AC, this battery. Why? During the positive cycle, this voltage will start increasing or charging the battery. During the negative cycle, we are discharging the battery. So in order to if we convert this into something like this, half wave like this or like this, full wave like this we will we will be able to charge this battery despite the variation, is still the most important feature, which is unidirectional. Okay? So let's start in the x lesson poi explaining first the single phase, uncontrolled half wave rectifier. 19. Single Phase Half-Wave Uncontrolled Rectifiers – R Load: Welcome everyone to this lesson. We will start in this lesson by talking about the single phase half wave uncontrolled rectifiers with an R load. We will explain what does this even mean? So this is a circuit that we are talking about in this lesson, we are going to explain. And in the next lessons, we have several examples and different definitions regarding rectifiers. So first, single phase, half wave uncontrolled rectifiers are loud. What does this mean? Number one, single phase. What does this mean? It means that our source that we would like to convert into DC is a single phase source with one phase and neutral. Number two, what does uncontrolled mean? Uncontrolled, it means that we are going to use uncontrolled power electronic switches that we are talked about in the first section for the power electronics course. So uncontrolled here, we are referring to diodes which are uncontrolled and we cannot control them. Number three, what does half wave mean? Half wave means that the AC voltage to DC voltage is converted in only half cycle of the main voltage. What I mean by this, if you look like something like this, you have the AC supply, and if you use a half wave uncontrolled circuit, you will have something like this. During the positive half, we will be able to pass this positive half and we will block the negative part. We will have positive, zero, positive, zero, and so on. As you can see here, we only since it is half wave, it means that only half of the sine wave is converted into DC. So we allowed the first half and we blocked the other half. So we used only or converted only half half of these waves. Number four, what does R load mean? A load means that our circuit or our single phase is connected to an R load, or purely resistive loud. So let's translate all of these sentences in just one diagram. Number one, single phase, a single phase source, AC. Number two, uncontrolled. We have only one diet, very simple circuit. Number three, R loud. As you can see here, a purely resistive loud. So by using the circuit, you will be able to convert AC like this into rectification or rectified out voltage. This voltage is a voltage across the resistance, which is positive only and negative blocked. Now, let's understand how does this circuit even work? So what happens exactly? This circuit is similar to this one. So during the positive half cycle of the input voltage. Now, remember this is our dit, right? And what we learned before is that the current flows in the same direction of this did simple. So the direction or the dite itself is looking to the right side like this to this side. So it means that the current this dite allows the current to flow from here to like this in this direction, right, similar to the direction of this arrow. Or the symbol of the dite itself. So you can see that the voltage of the dite is positive negative voltage across the die as we learn it in the first section. Now what happens during the postive cycle? Okay. So during the post of cycle, what happens exactly? You can see that during the boost of cycle, here we will have positive, negative, right. So the voltage here will be positive. So this positive, which is voltage here positive, and this negative is connected like this to here, like this, right. So the positive is connected to the positive of the diode and negative is connected to the negative, right during this positive cycle, so what do you think what will happen? This dit, since it has a post voltage across it, let's just assume for now that this diet is ideal. For simplicity, ideal, it means that V forward is equal to zero. So it just require any post voltage, it will start operating. Okay? So V forward is equal to zero, so it means starting from this point, it will start induction. So this dide will become a short circuit like this during the positive cycle because it has a positive voltage drop or postive voltage across it. Okay. Now, when this one becomes a short circuit, what do you think is the voltage on the output? So if you apply KVL in this loop, you'll find that the V supply is equal to the voltage across the resistor. So we have our source like this, this sine wave with a maximum value of VM. As you can see here, V output during the positive cycle here. The output is equal to supply. So as you can see, during postive cycle, it is similar to it, right from zero to Pi. Starting from Pi or 180, the sine wave is reversed. So I think now you understand this part. Due to the postive conduction, this one has the same voltage, similar to the supply. What about the negative cycle? During the negative cycle, you will have or the polarity of the supply will be reversed. The polarity will be reversed. Negative, it means that this one will become positive and this one will become negative. Here, the supply is reversed. During the positive cycle. The positive is connected here. To the dit, and negative is connected here. As you can see the voltage here is now applied in the reverse direction. What will happen to the die? The dite will be working in the loing mode in the locking mode as we learned before. Why? Because it is reverse paced. Since it is reverse pised, it will become an open circuit. Our circuit during the negative cycle will be like this. Like, open circuit like this, and resistance so since the die is an open circuit, what is the current flowing through the circuit? The current will be equal to zero. So the voltage across the resistor will be also equal to zero. So in this part, the die will block, so as you can see here equal to zero. Okay? So it allows only during the positive and block during the negative. Very easy. Now, what about the current? So as you can see here in this circuit, during conduction, during induction when D one is in the forward mode, you will see it will become a short circuit like this. It will be a short circuit. So the current of the supply Is, which is a supply current will be equal to similar to the current going to the loot. Now, during the blocking mode, IS will be equal to zero, which is similar to the loot current which is equal to zero. What is the benefit of this? I would like to just clarify or want you to remember that the current of the supply is equal to the loud current in this circuit. Remember this. So what is the supply current? If you would like to find the current going through the circuit, it will be simply equal to during the conduction, it will be V output voltage across the resistor, divided by the resistor, which is supply divided by the resistor. Supply, divided by the resistor. As you can see current starting from zero goes to maximum and goes down. So you can see it has the same waveform, similar to the previous, similar to the supply, and similar to the voltage across the loot. So this value, which is the maximum value of the current max will be equal to V, which is the maximum voltage of the supply divided by the resistance. This resistance can be the Road, or it can be Road plus the resistance across the diet if the diet has a resistance, if it is non ideal or if we consider it as a non ideal diet. Okay, so this is the explanation of what I just said regarding the positive and negative cycle. Now, what about the voltage across the voltage across the dite? During the positive cycle, it is short circuit, right? So when it is a short circuit, the voltage across it will be ideally, ideally, equal to zero. Why? Because it is ideally a short circuit. However, in reality in reality, there will be a V forward, like, for example, 0.7 volt as an example. So in this case, we will make this in this way form, we will just make it like this, a constant value of 0.7. Okay? If it is non ideal, here I am sobbing, I'm assuming that it is ideal, so it will be zero. Now during the negative cycle, this voltage, all of the voltage of the supply will go to the blocking dit. So the voltage across the blocking dit during the blocking mode will be equal to the supply. All of this voltage is going into this open circuit. That's why you can see this negative part. Voltage of the supply is equal to the voltage across the diet. Okay? If you don't understand it, it is very, very easy. If you just apply. Let's delete all of this to help you understand this part. You can see that let's say we have the supply here and we are talking about the diet during the ploging mood here diet. Okay, so it will be open circuit, right? The voltage of the dite is plus minus. And we have here the resistance, right. So during the blocking mode, if you apply like this, plus minus, this is the general signs, okay? So if you apply here at QVL, you will find that voltage drop across the resistor is equal to zero, and V dit V di minus VS will be equal to zero. By applying this KVL. If you don't know about KVL, you have to go, of course, to the electric circuits course. So VD will be equal to VS. Voltage across the diet will be equal to the volte of the supply during the blocking mode. So the voltage of the supply is a sport negative right, negative sign. That's why diet, it has the same value of negative. Okay? I think it is pretty clear right now. Now let's understand some important values that we need to get. Number one, we need to get the average value of the outward voltage. Average of the outt voltage. Now, why is this because the average value, the average value of voltage or current of any way form representing the equivalent DC value. For example, if you look at this wave, this is a pure sine wave, pure sine wave, and pure AC wave form. If you try to get the average of this wave, you see that the positive is equal to negative. That's why DC is equal to zero. Because DC representing average. Now, this is the out wave from this one for the voltage and this one for the current. If you get the average value of voltage and the average value of current, you are simply getting the equivalent DC value of voltage and DC value of current, which I'm concerned with. So we have our form, VS supply, Vmax, sine omegaty. Remember, this one is V max, sometimes or maximum value. Sometimes in the problem, it is given as VRMS. So in this case, you will take VRMS and multiply it by root to get the maximum value, as you already know from circuits. Now, what is the out value or the average value? It will be. Let's just delete all of this for the voltage here. It will be one verte. The average of the whole period 0-2 Pi. That's why one verte is equal to 1/2 pi, the whole period. After it, it will repeat itself like this, zero, like this and zero. So it repeats itself every two Pi. Okay. So one of two Pi, integration from zero to T, which is a whole period 0-2 Pi, right? However, if you look at this waveform, you will find that we have integration from zero to Pi with a certain value. However, from Pi to two Pi is equal to zero. So we don't need to integrate this part. It is already clear that it is equal to zero. For this part of integration from zero to Pi, zero to Pi for the waveform, which is the input sine wave V max sine omega t with respect dOmega T. Now, if you integrate this waveform, you'll get Vmax over Y. This is the integration with a simple steps. This one will give us VMX over Pi after doing the mathematical steps. What do we learn from this is that the average value or the DC value of the output voltage is equal to the maximum value of the supply divided by Pi. Okay. Now, this is what load for R load. This is pretty important as we have RL, we have different loads that we are going to see. So as you can see, V average, will be Vmax over Pi. What about current average? It is very simple. You can see it is similar to this waveform, however, divided by R. So as you can see, I average will be average over R. Okay? Okay. Now, what about the root mean square. Another value which is very, very important is the root mean square value or the effective value. So first, root, as you can see, root and mean means average one T, integration from zero to T of the waveform, D omegaty. However we have here mean square, square, it means square of the waveform, V max, sine omegaty all square. So what I'm going here is that I'm applying this would mean square to this waveform, which is integration from zero to Pi, similar to the average. But the difference is that we have here square and we have the square root. If you do this mathematical formula, you will get Vmax over two. So we have now V max over two representing root mean square value of the output voltage of the output voltage and V average, which is equal to V average, which is equal to V max over B. Now what about root mean square current, it will be voltage over the resistance. So it will be VRMs over R, which is V max over two R. V max over two R. V max over two. The sport is a Vrms. Now the final thing that we would like to get is the average of power absorbed by the resistor. Average of power absorbed by the resistor. Power is equal to voltage multiplied by current, right? So we can get the amount of power absorbed by the resistance here by using the RMS square multiplied by R or VRMS square divided by R. 20. Understanding the Difference between AC and DC Powers: Welcome everyone to this lesson in our course for rectifiers. And this lesson, I would like to explain something which is very important, which is the difference between AC power and DC power, okay? To avoid any kind of confusion, okay? So I'm going to explain this on the external tablet and everything will be shown here on the screen as you see. So let's start. First, I would like to understand what does AC power mean when I'm getting the effective power, the power consumed. Let's say we are talking about, let's just write here as if we have a resistor. Okay? This resistor, I would like to know how much power consumed. Okay, consumed inside this resistor. So first, I will start with AC. In AC system, we have first, let's just draw like this, very bad drawing. Okay. Just let all of this again here like this. Mm hmm. And we have first the AC voltage, like this. This one is AC voltage. Number two, let's go down here and draw the current waveform. As we know that in the resistor, the current waveform is similar to the voltage except it is divided by the resistance. So it will be also a sine wave. This one is AC current. Now what is power? What is the definition of power power or the instantaneous power? Instantaneous power is equal to voltage as a function of time, multiplied by current as a function of time. Okay, so let's go down here and multiply these two waveforms together. To get the instantaneous power power as a function of time. If you multiply these two with forms together, you will find that all of the negative part negative multiplied by negative, give us positive signal and positive with another positive will give us positive. If you multiply these two together, you will get like this, which is a product of voltage and current, right. Okay. So when I say when I say the power consumed P consumed, which is IRMS square, blood by the resistance or V RMS squared divided by resistance. What kind of power I'm getting, I'm getting the average of power. As you can see here, this wave form is changing with time, right? So when I'm getting the average, I'm getting the average. So B consumed here, it means that I am getting average. Average of the instantaneous power. Now, what about DC? If we go here and do the same for DC. We have a voltage with respect to time. For DC, it will be constant value, VDC, VDC, which is a constant value. Similar to it, the current will be V over R, voltage divided by the resistance, it will be also a constant value I DC. If I would like the power consumed power consumed in DC system in DC, it will be voltage, multiplied by current, which is a DC voltage, multiplied by DC current or V, DC square divided by R, or I DC square multiplied by R. All of this will give us the same value. Now, what does VDC mean here? VDC is the DC value, which is here constant. Okay, constant value. Now remember that in DC system, the average value V average is similar to VRMS similar to the V constant, which is a constantive value. Similar to here, if you go here like this, here, I average is equal to I RMS equal to V equal to I constant. So the constant value of current its average is similar to RMS, similar to this constant value. For example, if this value is 2:00 A.M. Pairs, it means that the average value two pairs, RMS, two ampers and so on. Similar to the voltage here, okay? So in DC in DC system, Average is similar to RMS. However, if you go here to AC system, as you can see here in AC voltage, you will find that VRMS is equal to a certain value. However, V average is equal to zero. Okay. Now, the RMS here is used to give us the effective value. So here in AC system, when we say AC power consumed, AC power consumed, we are talking about the effective value of power consumed inside the resistor, using the RMS or to be more specific RMS or AC Power here, the power consumed, representing average of the instantaneous power. Average of the instantaneous power in AC is obtained using the root mini square. In DC system, the average is similar to RMS, similar to constant value. All of this VDC. If you go here, here, VDC, IDC, which is constant value, multiplied by constant or V RMS, multiplied by RMS, V average, multiblo by I average. All of them will be similar to each other. Now, when we are talking about rectifiers, the power the AC input power, which is coming from the supply, which is represented by I RMS square or RMS square multiplied by here, the total resistance or lot plus plus the resistance of the dite. This is the AC input power that is going to the rectifier circuit, including the lute. However, part of this AC power will be converted into DC power, which will be calculated using V average I average of the lute. Okay. So I hope you now understand the difference. So in DC system in DC Power, you'll find that AC is similar to DC Power. I average multi blood by V average is similar to VRMS RMS. But in rectifier system, since the power, all of the AC power here, which is RMS square, not blood by this is not completely converted into DC. That's why we have something which we call efficiency of rectification, which we will learn in the next lessons. But for now, when we talk about the real power consumed, how much real power consumed, through the resistor, we are talking about in general, all of the power consumed. When we say how much power consumed, we are talking about all of the. If you get back here, in this one here, you can see average power consumed. Why average? Because in any wave average, for example, in the DC system, the average of this wave average power is the power consumed inside the resistor. In AC, we use RMS square multiplied by R. Now, how much power is consumed by this resistor? This resistor has both power DC and AC. Okay? So AC, which is the RMS, has two kinds of power. One, which is the AC power and the other one, which is DC power. So DC is just a part of the whole power received by the loud. So when I say how much power consumed by the resistor, in general, in general, all of the power, the real power, the effective power, we use root mean square. However, if I'm talking about just DC power, DC power, then I will use V average square divided by R or I average square, not blood, but I remember there is a big difference between these two. So I hope the point is clear for you. 21. Example 1: Let's start with the first example, example number one on rectifiers or to be more specific on the half wave rectifier. This is a very straightforward and very simple example regarding rectifiers. So if you look at this circuit, we have this circuit here. Which is shown in figure. This circuit is consisting of an EC source, sinusoidal EC source of 120 volt RMS. Remember, here, this is a supply voltage as RMS, not Vmax, however, RMS. This is very important. Number two, the frequency of the supply is 60 hertz. Number three, the load resistor is five ms. We need to find number one, the average loud current I average, I average, which is similar to IDC at the loud site. I average or IDC number two. We need to find the power absorbed by the loud. Here I'm talking about the average power absorbed by the loud, the total power, which means we are going to use the RMS relation. Here, I'm not talking about DC power. I'm talking about the whole power which is RMS. Number C, we need also to find the power factor of this socket. Let's start. First, you have here as a source of voltage, VRMS 120 volt. In order to convert this into Vmax, as we need it to get the current, VMAX is equal to VRMS multiplied by root two, 120, multiplied by root two. If you don't know about RMS or the relation between them, you have to go to the course of electric sockets. So V max, 120 root two, which is 169.7. Now, how can we get average root carreter? So I average, if you remember I average is simply equal to IMAX divided by Pi, right? Similar to us, similar to V average. Equal to V max over buy, we average across the load, which we learned in the previous lesson. We average Vmax over Pi, so this is Imax over y, and Imax itself is equal to Vmax divided by the total resistance in the circuit. It would be like this. It will be the average current or the outt current will be V outt over R. V Out here since we are talking about average, it will be V max over pi, as you can see here, V max over Pi. Multi divided by R, as you can see here, V max over Pi R. This will be root 220 divided by R, which is a resistor five and Pi gives us 10.8 am pairs. That is the first requirement. I average equal to V average over R, and V average itself is Vmax over Pi. Second one, which is power absorbed, as we said before, we will use the RMS values. So power will be equal to VRS squared divided by R or IRS squared multiplied by R, and RMS of the output as we learn it in the previous lesson or the previous previous lesson, RMS equal to VMX divided by two. From the relations that we explained when we talked about the half wave rectifier. So RMS, we max over two. It will give us 84.9 volt. So RMS squared divided by R gives us the whole power consumed by the lute. As you can see here. The last requirement, which is a power factor of the circuit. Here when we are talking about power factor, power factor exists in AC part. So whereas AC, we are talking about the supply. So when we say power factor of the circuit, I'm saying power factor of the supply. So let's just delete all of this. So what is power factor? If you remember, power factor is simply equal to active power consumed, divided by the apparent power. So what is the amount of power consumed inside this circuit? So the power consumed is similar equal to RMS square over R, which is 1,441, power factor B input over S input, and the input power, which is going to the rectifier part, rectification plus the loud, P input, similar to B output. Why? Because the resistance of the diet is equal to zero in the circuit or we assume ideal diet. So we assume that this diet does not consume any kind of power. So all of the active power supplied by the supply going to overload. So PN is equal to P output active power, which is 1441. Now, what is S input, the input, apparent power. And as you know, from electric circuits, input power is equal to VRMS Multiploid by RMS. Remember, here, VRMS is the input RMS input, which is the RMS of the supply, which is this value. This one here representing the RMS of the output across the lot. Here, RMS here input voltage, and multi blood by the input current, and as you can see here in the circuit that the input current is similar to the output current. So IRMs here will be this value for current, which is VRMS 84.9 divided by the resistance, which is five arms. Now, where did you get this? Here's the current going through the loot as an RMS will be VRMS the voltage across the loud, divided by resistance. So the voltage RMS across the load is equal to 84.9 divided by the resistance gives us RMS. This current or this root mean square current is similar to the input current coming from the supply. That's why RMS output is similar to RMS input. So let's just collect all of this. As you can see here, V RMS of the current. Is equal to V max over two. This part is the VRMS of the output divided by the resistance, gives us 17 apairs. Now power factor is the power active power provided, which is 1441, as we said, and VRMS as I said, input voltage and the current output is similar to the input current, which is 17 apairs. The power factor of this circuit is 0.707. Okay? 22. Ripple Factor of Half-Wave Rectifier: Welcome everyone to this lesson. And this lesson we took about a crucial or important definition, which is called the rebel factor of half wave rectifier. This lesson and all of the next lesson regarding half wave rectifiers, representing some definitions that we will use in every circuit, in every rectifier circuit. All of these definitions are used to express the efficiency and how well the rectifier will convert the AC into DC. So starting with the ripple factor, what does ripple mean? Ripple assembly unwanted AC component remaining when converting AC voltage into DC waveform. So as you can see here, we have this original waveform, which is AC. And this was form, which is considered as DC. However, not completely DC. If you remember DC, what I'm talking about, I would like it to be like this constant value. However, this variation going from zero to up and down zero up and down all of this variation is what we call ripple. Okay? Similar to like this, you'll find that, um, if we have a signal like this, I would signal like this. Okay? This is also considered the DC. However, with ripples, this variation called ripple going up down, up, down like this. This is called ripple. However, a pure DC will be like this constant value. Now, what we see here it has also another This is called also the pet pol setting, pul setting. DC waveform, DC waveform, as you can see in form of pulse like this. Okay? So this is what we ripple here. This is what we call the AC component. That is presented inside the Out waveform. So even if we try our best to remove all of the EC components, there is still some left on the outside, which pulse saves DC waveform. The one which causes like this is the AC component inside the oututwavefm. Now this undesirable AC component is called ripple. Now, to quantify how well the half wave rectifier can convert E into DC, we use something which is called ripple factor. One of the definitions that will help us understand how efficient is this rectifier. The ripple factor is very, very simple. It is a ratio between RMS value of the AC voltage and the DC voltage of the rectifier. So here, Gamma will be RMS value of the AC component divided by value of DC. Now, value of DC component, here we are talking about the output. Output, DC component. What is the output DC component? It is simply average VDC is simply V average, so the average of the output component is considered as the DC part. Now, this first one is very, very important, very important RMS value of a component. So it is VR RMS. What does this mean? If you look at the signal here, this signal is basically basically consisting of VDC plus VAC. So it consisting of two components, part of it, which is a DC, which is average value, and other one which is the AC which is causing this pulsation or causing the ripples. So as you can see, it is RMS. So the first one here is RMS of the AC component. Okay, the AC port of the signal is VRRMS. Okay? However, VRMS which we obtained before in the previous lessons when we talked about consumed power or output power or anything, VRMS in general is the whole RMS of the signal. So again, we have DC and AC. DC only is the average apart. AC here is the one which is causing ripples. Okay? So RMS of this only this component is the first one here. However, RMS of all of these will give us the RMS of the whole signal. Okay? I don't think you understand anything. But let's see this with equations so you can understand. So in order to get this AC part, which is causing the ripples or pulsations, it will be like this the V output signal as a function of time. This signal is consisting of two components, part of it, which is AC or the one which is causing the ripples or pulsations, and the DC, which is a constant value, which is the average value. Now let's say I would like to get VRMST component only. The root mean square of the AC component only. It will be like this VRMSOT integration from z root to T for this part only, VAC, which is the one which is causing pulsations, square all under the square root, as you can see here. Okay, so we talk only RMS, root mean square of just the AC component. Now, how can we get this value assembly from this equation V Albut minus VDC, like this. So if we square both sides in order to eliminate this one like this, square, RMS square will be one over T, integration 0-3 to four the output signal minus the average. This difference will give us only the AC component presented inside the output waveform. Okay? So here o minus VDC all squared, it will be VO squared minus two, VDC, plus VDC squared. Now, let's divide this one t integration from zero to T for Vo square, the sport, and one art integration from zero to T of negative two out VDC, negative one art, the sport is this one. Okay. The last one, which is the integration, zero to T, one ttegration from zero to T for the DC component. The DC component is a constant value. So the integration of a constant value from zero to T for this one, let's just write it. For this part, it will be one vert, integration from zero to T V DC square DT DT. Okay, like this. So integration from zero to t, it will give us one of our tes this part, multiplied boy. This VDC is a constant value. V average is a constant value. So it will be VDC square as it is. Multiplied from zero to t, it means multiplied by T. T will go with T, and we'll have VDC squared, likes plus VDC square. So we have these two components remaining the swan and the swan. So again, what we did is that we divided this integration into three parts. First part, second part, and third one. Now, one t integration from 02 of the whole signal square. Do you remember what this is representing? This one, one t, this part. This representing VRMS square, the root mean square of the whole signal, one T integration from 02 of the square of the voltage. Under the square root, it will give us VRMS of the whole output. However, since we don't have the square root, it will be only squared. It will be squared of this voltage. Minus two VDC of T, integration from zero to t o DT. Now, integration of this part, one of T, integration from zero to T, V. What does this represent representing V average? One of T, integration from zero to T for V DTT representing V average, right? It will be average multiplied by two VDC over T and V average is similar to VDC. So we can say average as VDC, right? So VDC multiplied by two VDC over T, it will give us what two VDC square, two VDC square, right? One T integration is the average that we obtain. So there is no one t. So we have RMS square minus two VDC squared plus VDC. If you subtract this from this, you will have negative VDC square. So here we are RMS. It will be like this. How did we get this? Very simple. Gamma is equal to. VR RMS, which is root mean square of this part. So VRMS is equal to root. If we remove this square, it will be root of Vrms, square minus VDC square divided by VDC. Right? If I would like to get VDC under the square root, I will put it under the square root and they make VDC as square. So if we divide this to VMs square over VDC square minus VDC over VDC, it will give us minus one. So as you can see, this part will be minus one, and this part will be VMs square over VDC or VMs over VDC or square, as you can see here. So this formula is the one which is used to get the ripple factor. Okay? Now, if you apply this, this is a general rule. This one is a general rule that will be applied to any rectifier circuit. Now, if we apply this rule to the half wave rectifier, we have VRMS equal to Vmax over two, and VDC is equal to Vmax over B. Okay. So like this, max over two, we max over P. If you substitute in this equation, you will get finally one point to 21. So as you can see here, this is ripple factor 1.21. This the higher this number, the more ripples that you have in the outut. So we will try to minimize this value as possible. So in general, how can we minimize ripples inside rectifiers using capacitors and inductors as a filter. So to be more specific capacitors, capastors will be connected in perl like this. When we connect a capacitor in peril with lute, you will find that this waveform will become like this. It will be let's say the first one, it will go like this, go like this and then go down like this. So it will be something like this, go from this point and start discharging and then going up, then go down. This in the end depends on the value of the capacitor used. The higher the value, the lower the ible that you will have at the out waveform. Okay? 23. Efficiency of Half-Wave Rectifier: Welcome everyone to this loson and this loson we took a poet, another factor, which is efficiency or another definition, which is the efficiency of how wave rectifier. So when I'm talking about efficiency, we are talking about the ratio between output and input. So efficiency in general of anything in device or in equipment is output power with respect to input power. Now, since we are talking about efficiency of half wave rectifier, of a rectifier, I would like to know how much of the input input AC power is converted into DC power. This is my own goal of the efficiency. So as you can see, it is the ratio of the DC power available at the loud compared to the applied input AC power. This is known as efficiency, and mathematically, as I just explained, as you can see here, Efficiency is DC output power. How much power as a DC converted with respect to the input AC power supplied from the supply to this circuit, including diet and the lute. B, DC over BEC. Now, first, let's just assume that our diet here is a non ideal diet. So when this diet is in conduction mode or in the forward mode, it can be represented by a resistance like this. With this resistance is called RF, and we have here our loads. RF representing the forward resistance or the resistance when the dite is in the conduction mode, as we explained before in the first section, is that when the dite in the conduction mode or any power electronic device, in the conduction mode, it leads to a voltage drop or it has an on resistance. So if you would like to get the current at any instant, the current flowing through this circuit, it will be simply equal to the supply divided by RF plus RL, QVL, V over the total resistance. And as a function of time is Vmax sine Omega T. So if we get Vmax over RL plus RF, it will be Imax, sine omega T. Okay? So our maximum current max of Rf plus RL being the maximum current. Now, the first thing that I would like to get is AC input power. Input active power, input active power. The power EC input active will be, of course, RMS square multiplied by the total resistance in the circuit. So we are giving power from the EC supply. To both of the diet and the loot. So it will be the root mean square since we are talking about AC power. So it'll be root mean square, squared multiplied by the total resistance in the circuit. We have diet and we have resistance. In the previous example of the first example, example one, we assume that the did was equal to zero. The resistance of the diet is equal to zero. So all of the power coming of the supply is going to only the loud. That's why the power input, when we calculated the power factor, it was equal to RMS square multiplied by the resistance of the lot. Now, since we have here, power input coming from the supply will be divided into power consumed inside diode and power consumed inside load, it will be RMS square multiplied by the total resistance. Or VRMs square divided by the total resistance. Now, the second thing that we would like to get is BDC or the DCO power. Remember, I'm not talking about the total power or the average consumed power by the resistor. I am just concerned with DC power. When I say DC, I'm going to use the DC laws, which is V average square multiplied by L or divided by L or I average square multiplied by L, as we told you before. You can see here BDCs IDC which is the average cn square, divided by is our resistance, VDC square over R. Now, someone will ask me, is this actually actually all the power consumed inside the resistance, I will say no, this is not all the power consumed. We are just concerned here in the efficiency with how much power of the AC input converted into DC. That's why I'm using DC as an average value. If I would like the whole power, all of the power consumed, it will be using the root mean square law. Okay? Okay, so let's divide these two together. So the efficiency equal DC power divided by AC, which is power DC will be VDC square over RL and divided by this will be RL plus RA VRMS square. If you get this one here and take this one here, you will get on the right side one plus RA over RL and VDC over VRMS all square as you can see here. VDC is V max over Pi and VRMs is V max over two. This is depending on the circuit we are dealing with. Now, if we get the square of this, we will get 0.4 053 multiplied by one plus RF of RL. Now, of course, in reality, the resistance of the diet or the volts drop even on the diet is very, very small. So we can neglect R F with respect to RL. RL is very large. It can be, for example, 100 Os and this one can be 10. So one divided by hundred is a very, very small value compared to one. So we can consider this one as zero or simply we neglect R F. So the total efficiency or the maximum efficiency of the half wave rectifier will be 0.4 053, which is 40.53%. What does this even mean? So the AC active power coming from the supply, AC power input, coming to or going to the A going to the resistance, only only 40% will be converted into DC. So there is a power losses of about 59.47%. This is lost in rectifier circuit. So as you can see, very large amount of power lost in the circuit. Now why is this? Because if you remember that the half wave rectifier only only take advantage of half of the wave. And the other cycle is completely blocked, so we only take 50% of the power. That's why the efficiency is very low, 40.53% and why there's not 50% exactly not 50 because there is a difference between average and RMS. The larger the gap between them, the lower the efficiency of the half wave rectifier. 24. Form Factor, PIV, and Peak Factor of Half-Wave Rectifier: Welcome everyone. In this lowson we will talk about some other factors or definitions regarding rectifiers, form factor, big inverse voltage, and big factor of half wave rectifier. Starting with the form factor, what does form factor mean? It is a ratio between the root mean square voltage and the average root voltage. So it will be like this Vrms, divided by VDC, these definitions are constant for every type of rectifiers or every circuit. The only difference is that the values of VRMS and VDC will change from one circuit to another. So the VRMS in this circuit of the half wave rectifier is V max over two and average is V max over pi. So it will be 1.57. What does the form factor actually represent? It representing the gap between RMS and average value. The larger the gap between them, the more worse or the worse the rectifier is. What I mean by this, actually, if you remember that in DC, we have average equal to RMS, VMS, similar to v average. Right. Average equal to VRMS. So in this case, the form factor in a pure DC circuit is equal to one, right? So the larger the form factor, the larger or the further the form factor away from one, it means that the circuit is much worse than before. The lower this value or closer to one, it means the rectifier is much better. The average is close to VRMS. Now, remember the bel factor. Remember the bel factor Ribble factor, if you don't remember was root and we obtained it in a previous lesson. S of VDC, all square minus one. Now, VRMS of VDC simply the form factor that we just explained. So we can say there is a relation between ripple factor and form factor. It will be equal to root form factor square minus one. Another one which is important when selecting the diet, which is a peak inverse voltage. Now, what I mean by this is the maximum voltage that the diet can withstand during the reverse pise conditions. So if you look at this diet here, now when does it work in the blocking mode? It works in the blocking mode during the negative cycle, right? So my question is to you, what is the maximum value of voltage? That the diet must withstand in the blocking mode. This is the maximum value. This value, this is the maximum value that the light must withstand in the reverse pius mode. This maximum value in the inverse direction or in the reverse pis conditions is called the big inverse voltage. Now the problem is that let's say, for example, this dite can withstand ten volt in the reverse mode. If this value reaches 15 volt, what will happen is that the dite will be destroyed because 15 volt is greater than the big inverse voltage of the diet itself. Now, the inverse rating of the diet is a very important or a primary importance is a design of the rectification system because this is a value that the diet have to block the maximum stress applied to this dite. As you can see here, the whole input voltae appears across the die during the reverse Pius mode. The same I just explained. Okay, so the maximum voltage appears across the die is equal to big inverse or the big value of the supply voltage, which is this value. Okay? Okay. Now, what is the peak factor of a half way rectifier? This is another definition. So the BC factor is simply a ratio between big value of the output voltage with respect to RMS value of the output voltage. So it will be like this, V max. So if you look at here in this circuit, which is a halfway rectifre, what is a big value, big value is V max. With respect to the root mean square of the signal. Now, in this circuit, which is a halfway rectifre, it will be equal to two. Now why is this Because simply V max, which is the maximum value of the output signal, which is equal to V max of the supply, right? VM now what about RMS RMS is we max over two in the half wave rectifier. Dividing these two, this will go with this one divided by half gives us two, which is this value. That's what we call the big factor ratio between a big value of the output voltage with respect to the RMS. Now let's talk about applications of half wave rectifier. Of course, this is the simplest and very simple circuit that is used as a rectifier. Usually, we don't use this half wave rectifier, but I will just explain what's the application that can be used despite the large power losses and the efficiency of 40% that I have talked about before. So number one is used for the detection of amplitude modulated radio signals as used in welding purposes. It gives a polarized voltage or a voltage with one polarity. For example, as you can see here, positive only one polarity. Since the trick f a signal, it can be used in many signal demodulation processes. All of this this one and this one are related to communication field, electronics and communication fields. We don't care about this from the electrical power engineering perspective. What are the advantages? As you can see, it is a very simple circuit, just one die to convert an EC or an EC signal into a rectified or a DC outbot. Number two, it is a very cheap signal, cheaper upfront cost as it requires very less equipment. However, there is a lot of power losses due to the presence of efficiency of 40%. The disadvantages include number one. I only take advantage. It only takes advantage of just half of the signal. The other half is completely wasted, which leads to power losses, 40% efficiency. Number two, the out voltage is very low. If you take the average value here for the volta, it will be very low compared to something like full wave rectifiers. Now, also the output current, as we said before, our out volt is not purely DC as it contains ripples. Remember the ripple factor that we talked about, and for this circuit, has a high ripple factor. As you can see, large ripples compared to even full wave rectifier. 25. Example 2: Hey, everyone in this los on, we will have the second example on the three phase rectifiers. In this example, we have a step down Delta star transformer, Delta star transformer with a pre phase tons ratio of five. This transformer is fit from a three phase 1,100 volt, 50 Hertz source. The second ray of this transformer through a rectifier feeds a load of our equal ten. We need number one to find the average value of output voltage, average and RMS values of diet current bow delivered to the luid in case the rectifier is a three phase, three pulse type, a three phase, six type. Number one, we have our supply Delta star transformer, right? Now, this transformer we started from Delta. Input voltage, 1,100 volt. This representing the line to line voltage, 1,100 volt, line to line voltage, and RMS now the ton of ratio between this transformer is what is equal to five. Okay? Now, we have the secondary is connected to a resistive load of ten oms. Okay? So let's type this. Okay, number one. We need first to convert 1,100 to the secondary. Now, from transformer, we know that V two over V one equal to N two over N one. Now, since our transformer is a step down transformer with a ratio of five, it means that this voltage, which is the line to line voltage, and at the same time, since it is a delta connection, it will be phase voltage. If you remember from a connection of transformer, Delta connection has a line to line voltage equal to phase voltage. Now, this ratio for phase voltages, so V two will be equal to voltage phase of secondary will be equal to V one phase of primary, multiply it by turns ratio. V one phase is equal to phase voltage of the primary 1,100 volt, and we multiply it by the turns ratio. Now, is our transformer a step up or step down. It is a step down transformer, it will be 1/5, as it will lead to reduction in our volt. So what you can see here is that secondary RMS phase phase voltage, phase and phase, phase voltage, secondary RMS will be V phase primary divided by turns ratio gives us 220 volt. Okay. Number two, we need average value of output voltage and the average and all of these. Number one, in order to get average value, we will use the equations for average. So if you remember average using the general formula for pulses, Remember that we average B over two pi integration from negative B over Pi to B over By V max phase, cosine omegaty. Here, this is a mistake. I have to change this after finishing this lesson should be cosine omigaty. As we learned in the previous lesson for the general formula for multiphase, we use a cosine function instead of a sine function. To do this integration. As you can see here, I already added goes on here correctly. Now, how many phase we have a three phase, three pulse, number of pulses will be three, three, and three xs. V max phase. This is our phase, root mini square value. In order to convert this into a peak value or maximum value, we will simply multiply this by root two xs. To become a maximum phase voltage. By doing this integration, you will get this value for V average or VDC. Now, what about the average average and R mess of diet current? Number one, in order to get diet current, we need first to get average lud current, I average output or I output average. Which will be equal to V average divided by the loud resistance of a ten Ms like this I average, we average over R, gives us 25.73. Now, what's the next step? If we remember, since we have three pulses, I average, I average four dived will be simply I, I would average divided by what P or number of pulses. So it will be like this I output average over P, which will be 25/3 gives us 8.57. What about the RMS exactly the same as this function, I RMS of dt will be It RMS divided by root B. To get I outbut RMS, we will do the integration for function like this I put RMs, root P over two P, same integration like this one. As you can see, we put it as cosine, but divided by R to get the current equation. By doing this integration and substituting, you will get 2,060.15. Now divide this by root P number of pulses, which is three, like this, 15.1, you'll get the diet RMS current. Now the last requirement in this problem is getting power delivered to the lot. In this case of the three phase three plus type, we need power delivered, power delivered here, since it does not say average power delivered, it says power delivered in general. Since it says in general power delivered, then we are going to use the root mean square rules. Like this power reaching the loot will be I mass squared multiplied by resistance. This value squared multiplied by ten OMs. It will give us this value. Now, if we say average power delivered, it means it will be power of DC, which means I average, multiplied by V average or I average squared multiplied by R. Now, let's apply the same equations for three phase six type. So number one, we need V secondary phase. Remember that we said V secondary phase in the previous, one, 220 volt. Now remember, remember this phase triples, which was like this. Let's write it like this. Like this. Now, this is very important. All of this the total 220 volt phase voltage, RMS for secondary. Remember that in the three phase M six type or the midpoint type, what did we do? We took half of the urns. We divided the urns by half, like this. Write A one and A two. What does this mean for me? This means that the output voltage will be also divided by two. Now, why is this Because remember that voltage voltage is directly proportional to the number of turns. As number of turns decreased to half, voltage voltage for the phase will also decrease to half. That's why, in this case, the phase voltage will be 220/2, like this. Because our tons for each phase is now reduced to half. Okay. Now, what the second requirement average value of out voltage. Again, the same equation, but we are going to use six pulses. You can see Vmax cosine Omega T here Vmax 110 root two. Pulses here, 66 type, which has six pulses. So it will be six, six, six, like this. Us 148.5 55 volt. Average and mass of die current, we need average lot current. It will be this value divided by ten ms to get average current, we will have to divide this by three B six, not three, B six because it will be I average, divided by number of pulses, which is six pulses, it will be 2.47 58. For RMS, we first need the RMS current, which will be root, voltage divided by R like this. Same equations for the previous slide, except that we replaced the pulses with six instead of three. This will give us 14.868. Now, if I would like the RMS of diet, it will be IRMS divided by root P, which is root six, like this. The same equations that we talked about in previous slide and in previous lesson of the multiphase. You can use these rules for multi phase, or you can also use the rules that we discussed for each type as before. Okay? The lost requirement is getting power delivered. It will be out current square multiplied by resistance like this. In this example, we learned how to apply the multiphase rules in order to get values of average, root RMS current and power delivered in a three phase repulse and three phase M six type. 26. Single Phase Half-Wave Uncontrolled Rectifiers – RL Load: Welcome everyone to the sloson in this sloson we took a poet. The single phase, have wave uncontrolled rectifiers with an RL lute. So in the previous lessons, we talked about the same circuit here, except that we had only a resistance without any inductance. However, in reality in reality, we have several loads, such as industrial louds that typically contain inductance as well as resistance. So we can represent the industrial load as RL loud. So what are we going to do? We would like to analyze the, the current waveform in the presence of our loud. We would like to get the average current, the root mean square current, the voltage, and so on. First, let's understand what happens when we have an additional element, which is the inductance in our suct. So let's start step by step to understand how the circuits work. First, we will understand this by logic, and then we will start doing the mathematical equations. First, as you can see here, we have supply. Which is from zero, going to peak value V max, then going down to Pi at which is the voltage will be zero, then going like this and like this still repeating itself every two Pi. Now, what about the current? So when the voltage here in the postive region here, Okay. When the voltage is in the postive region of the supply, this dide will start conduction or it will start conducting or providing electrical current. It will become ideally a short circuit like this. So in this case, we will have a current flowing through the circuit, right? So as you can see here, the output current, it will start flowing through the circuit. Start increasing with the increase in the supply. However, you will find that there is a phase shift phase shift between current and voltage. If you look carefully here, you will see that this is a peak voltage, and this is a peak current. You will see there is a phase shift between them or a lagging of the current from the voltage. Now, this lagging or this phase shift is due to the presence of the inductance. So as you know, from electric circuits, that inductance cause lag between voltage and current or causes the current to be lagging from the supply voltage. So as you can see here, this is a phase shifted between them. So when the voltage becomes zero, there will still be value of current and current will go to zero at a different angle. At instead of Pi, it will become at bit. Now, let's understand why does this even happen? So during the post of cycle or during the first part here, the post of cycle, the diet becomes a short circuit, as you can see here, and the current will flow and start charging the voltage or start charging the inductance, store the energy, start accumulating inside the inductance. Okay? So we will have a value of VL will start increasing due to the presence of the voltage which supplies current through it. So the voltage of VL will start increasing. Now, when the voltage drops down to zero, when voltage drops down to zero and start switching to the negative cycle, start switching to the negative cycle. What happens here is that if you remember that the inductance revents prevents or does not allow the instantaneous change of current or large change in current. So if you remember that the voltage across the inductance is equal to L, D, over dt. Okay? So here we have a value of current like here. Let's say, for example, two a pair, okay? And when the volte goes to zero, it should be automatically goes. The current should be automatically goes to zero, right? So if this happen, di over dt will be like this, it will be the new current which is zero minus 2:00 A.M. Pairs. Divided by the time at which this change happened from 2:00 A.M. Pairs to zero. So it will be very instantaneous, let's say in one millisecond 0.1, for example, you'll find that the value of the voltage here one, two, three, so it will become like this 2000 apre. Or 2000. This value is 2000. Multiplied by the inductance, it will give us a very large amount of voltage. Diardt 2000 or let's say, very, very small current like this, increasing this in a very, very small time, I would like to say zero, but it is not zero. It's a very small time. Anyway, this spot will be very, very large leading to very, very large voltage dused across the inductance. That's why the inductance does not allow instantaneous change of current. So what will happen is that this inductance will cause the current to keep flowing in the circuit. So how does this even happen? Simply, let's say this voltage, this supply is ten sine omegaty ten sine omega t. Now, let's say at this point at which the supply equal to zero, this inductance has a voltage, let's say eight volt, eight volt plus minus due to the charging of it from the supply itself. So we have eight volt. Now, when the supply becomes in the negative part in the negative part, Okay, so the current will not go to zero. Why? Because this inductance will switch its popularity in order to prevent the instantaneous change of current. So instead of having plus minus eight volt, it will be switched like that becomes minus plus eight volt. So it will be switched. Okay? And this supply, let's say, at this point, gives us two volt, okay? To volt like this, here is a positive and here is the negative. This supply to volt and this supply ad volt. Now, you will find that the voltage across the dite will still be positive. That's why in this part, it will still be conducting despite the presence of negative supply because the voltage of the inductance will start giving current through the circuit like this, flowing through the dite. That's why you will find that we will still have a current, but it will start decaying to zero. When this inductance completely ends all of its energy or completely de energized. Okay, I hope that the idea is clear. If you'd like it in simplicity, simply during the postive cycle like this, till Pi, we have a current. Okay. And then at starting from the negative part here, this inductance will switch it to polarity to prevent the instantaneous change of current and allow the flow of current as if it is a supply, allow the current to flow like this, through the dt and keep it in conduction. Okay. That's why here during from zero to an angle called extinction angle called Peter from here to here, it will still conduct. Okay? Now, what you will find is that since the dit is conducting from zero to Peter, right, from zero to pet. So what will happen to the voltage across the out out? Remember that Out, when this diet is conducting, V outbut is equal to V supply. As you can see here from zero to Beta. All of this from here to here, this is a waveform, going like this and then going to the negative part till better. This is our supply. That's why you will find that outt from zero to Beta like this to beta. At Beta when current goes to zero, this one will become an open circuit, so output will return to zero from here to here to zero. Okay, and cycle repeats itself. Okay, so this part due to the conduction from here to here, due to the conduction of current or the presence of current through the circuit from zero to extinction angle pit. Then goes to zero when this dite stops conducting. Okay? Now, what about the voltage across the dite? So from here, to here, what happened exactly is that the current exists, right? This diet is conducting, so it will become a short circuit, ideally, ideally, it will become a short circuit. So that's why from zero to pea it is a short circuit because it is a period of conduction. Starting from Beta two Pi, the reason at which we don't have any current, this one will become an open circuit, so the voltage across it will be the voltage of the supply from KVL. That's why this part from Beta Pi is similar to this part from Beta two Pi. So what you can see here is very, very easy. If you look at V outt and voltage across diet from KVL here, supply is equal to V diet plus VO so output here, which is a sport. If you add to it the voltage across the dt, you will get the original waveform, which is supply. Okay? So I hope you now understand how does the circuit works? So let's start by learning how can we get the mathematical equations regarding the circuit. So first, we will apply the Kersev voltage law equation, which describes the current in the circuit for the forward biased ideal dit. Here we are dealing with ideal dites. If you are not dealing with ideal dite then it will be. And instead of zero, it will be a constant value equal to V forward. And this Vout will be V O minus V forward or shifted by a certain value of V forward, okay? So let's start first during this conduction mode when this one is conducting like this, ideally a short circuit, then V max sine omega t will be equal to the voltage across the resistance and induxans which will be IR plus L di overdt or I plus LDI overdty. This is from Kersev's voltage low now, first, let's say I would like to get VO average or the average DC component of the Out voltage means we need the average of the outed voltage. So let's look at the out wave form, which is this one from zero to Peter. So if I would like its average, it will be one over T, one over t, integration from zero to T, V max, sine Omega T. D omega T. Now, the period here, what is the period at which this repeats itself 0-2 Pi T here will be two Pi. Now, integration 0-2 Pi 0-2 Pi. As you can see, from beta to two Pi, we don't have any voltage. It's equal to zero. So the integration will be just from zero to Peter, right? This is the region at which we will have the output. So it will be from zero to Beta Instead of the whole period, it will be just this part. And by integration, you will get the average, as you can see here, 1/2 Pi, 1/2 Pi. As you can see, Vmax, as you can see, integration from zero to pita the region at which is out exhaust, sine Omega T D Omega T. So it will give us one minus VMX over two pi, one minus cosine Pita. Now, if I would like to get the average current, it is very, very easy to take the average voltage and divod it by the resistance. Like this, I average will be VMX over two Pi or the same equation here, but just added the resistance. Average over R. Now, someone will ask me where the inductance go or where the inductance in this equation. Now, we will find that V output, the output voltage here is consisting of voltage across the resistance plus the voltage across the inductance. So this will form. This will give us this wave form. Now when we talk about average, Okay, you will find that if you draw the voltage across the inductance, you will find there is a positive part and another negative part, positive and negative. We will see this when we go to the controlled rectifiers, okay? They are similar to each other, except with a small difference. Okay? That's why some concepts, I will discuss them in the controlled rectifiers instead of uncontrolled rectifiers, okay? Anyway, when you draw the voltage across the inductance in this circuit, you will find that the positive part, which is during charging during the charging mode will be equal to the negative part during the discharging mode when this inductance starts providing current through the circuit. So when you get the V average, it will be equal to zero. That's why the average of this waveform is only the average of the resistance, average of the resistance because the average across the inductance is equal to zero. That's why average of V output is similar to average of the resistance. That's why this representing V output average or average of the resistance. That's why when I divide it here, when I get I average, I only divide it by the resistance since the average of inductance is equal to zero. However, of course, this is not the case when we try to get the root mean square current or root mean square voltage. We have to add the inductance in our case. So our current what we would like to get here we obtain the average. If I would like the root mean squared, then I have to get the VRMS, VRM, and voted by R or get the current equation and do the integration. Okay? However, what I would like, what is the point which is missing here, which I'm looking for is the value of Peter. Okay, which is the extinction angle. I don't know this value. So in order to get this value, I need the equation of current. By getting the equation of current, we will get beta, which will help us to get cosine beta, which will help us get average, I average, and the rest of all of the values I would like to get. So in order to get beta, I would like to express our current in the form of summiion of two parts. Number one, a forcet response and natural response. As we learned in the electric circuits course, when we said that the current can be suppressed as I forced response and natural response. Forced response due to the presence of a supply and natural response when the supply does not exist and the circuit discharge naturally. So here we have the forcet response, number one, which is representing the part at which the circuit is charged by the supply. So when the supply is connected to when the le is conducting and supplying electrical power and charging the inductance. This is the first part which is called the forces response because it's forced by the supply. Then we have the second part, which is called the natural response when the inductance started discharging through the circuit. So first, the forced response. First, when the facet response exhaust, we would like the current. So what is the current flowing through the circuit? When we have forced response, we max sine Omega t. So we would like to get the current. So the current will be voltage. Divided by the total imbedance since we are talking about an AC circuit. We have an AC supply and this one is conducting, right. So what we will have now, we will have an AC circuit voltage, AC supply, and resistance and inductance. So it will be V over that, right? V, which is a voltage, V max sine Omegaty right, which is a root or square plus l square. Now, the voltage here, don't forget about something which is very, very important. V over that, which is V max over the magnitude representing the magnitude of the current or the maximum value of current. However, don't forget about the angle. The angle which is ta or negative sea which is a delay angle of the current. Since we have an inductance here, so the current is delayed from the supply voltage. That's why when I write the current, it will be voltage divided by Z, or, Vbx sine omega ty. However, since we have a phase shift between current and voltage, we need to add instead of sine Omegaty we have to say sine omegaty minus Sta as you can see here. So this is due to the phase shift between current and voltage is due to the presence of the inductance. So our values here that will be root or square, real part, plus imaginary part, which is XL square, which is omgal square and Theta, which is phase shift equal to ten minus one omgal over R or the imaginary part over the real part. So that is the first part which is the forced response of our circuit. Second one, which is a natural response, natural response. Natural response of the circuit is when we don't have any supply, this supply as if it does not exist. So we have only inductance that will start discharging through the circuit to through the diet like just start discharging. Our current here start decaying like this part here. If you get back here, Remember this we have charging part going into charging, keep chargingtll a big value. Then starting from Pi it will start decaying. Decaying to PT. This decaying can be represented by a decaying exponential, like this. The force response R and L equal to zero as if the supply does not exist. This is a natural response that we learned in the electric circuits course. It will be like this, I natural will be certain constant E to the P negative T over t, decaying exponential. This will be like this. At the gangs ponential going to zero, where A is a certain constant and tau is L over R in the RL circuit. Okay? Now, if we combine these two together, we will have the final equation, which is V max over Z sine Omega t minus eta plus A E to the power negative t over tau, as you can see here. A. Okay. Now, second thing that we would like to say is that how can we get first problem here is that we need to get the constant A. In order to get the constant A, we need the initial conditions. So as you know that from this circuit that at a time equal to zero, Omega E equal to zero, we know that the value of current is equal to zero, right at the beginning of each cycle or at the beginning of the circuit, I omegat equal to zero, and I equal to zero. So we are going to do this time equal to zero, that current will be equal to zero. So by substituting current equal to zero, and Omegaty equal to zero, and we have the rest of this problem, so we can get the rest of this equation, so we can get the value of the constant, which is A will be equal to negative VM over that, sine negative theta, which is like this. So we are going to take the sport and substitute it here to get the final equation of current, which will be V max over that sine omega t minus Theta plus sine Theta e to Z power negative T over t. Okay? Okay, so this is a function. So this is the equation of the current. Now, what I would like to get is the value of Pita or the extinction angle, extinction angle. Now, as you can see that the extinction angle at Omega equal to Pita, the current will be also equal to zero, right? So as you can see here, the value of Omega T. So we have two values of omega T that will lead to zero current. The first one, which is omegat equal to zero, current will be equal to zero, and at Omega T equal to Beta or angle Beta, the current will be also zero. So what I'm going to do is that I'm going to take this angle and substitute here, omega T equal Peter The angle equal to beta and substitute in this equation and the current will be equal to zero to get the value of beta. So it will be like this. So Omega t equal to Beta. Omega equal to Beta, and time here. What I'm going to do in order to substitute with omegat since it is an angle, we will just multiply by Omega and divide by Omega. So Omega t divided by Omega tau, it is similar to the previous one. So Omega ty will be better like this, and Omega ta will be as it is. All of this at Omega ty equal Beta here, current will be equal to zero. Our equation can be simplified like this, sine beta minus eta plus sine theta, e zero negative Beta over Omega tau equal to zero. So this equation which you can see here can be solved numerically using a calculator or any other method or by iteration, I will try to show you how can you do this. But for now, if you would like to if you solve this equation, you will get the value of Peta required. Okay. Now, finally, if we would like to summarize what we have said, the current has two waveforms, starting from zero to Peter, which is initial equation, this one, as you can see here, from zero to Peter and starting from Peter to two Pi, it will be equal to zero. And that, as we said before, and Statominus one omega L over R and t over Okay, so we have the equation of current and we can obtain beta by substituting with the value we have inside the circuit. Now, what about the average current? Average current can be obtained by two methods, number one, average of the voltage. V max about two p one minus cost take this one and divide by R, as you just said before or by integration, one of our T integration from zero to Peta for the current current D omegaty. Now, this equation is the one which we just explained. This one, Okay? So if you use the equation or use the equation of the voltage and wide prior, it will give you the same solution. Okay? Now what about RMS and RMS? You have to use this one or RMS as you would like. So RMS will be root, root mean, mean, which is the average, one of our T integration from zero to T, zero to two pi from here to here. Here it is equal to zero, so it will be from zero to Pita like this from zero to Pita for the square of the current take the previous equation and square it. You will get finally the value of the RMS and RMS similar to it. Integration from zero to Pita Root 1a2p integration from zero to pita V max on Omega ty squared d Omegaty. Now this is a value of VRMS and this is a value of RMS. We can get RMS by using this equation or by using VRMS value, VRM value, and divide it by. So remember, here we are getting VRMS VRMS the root means square value, so we are talking about AC, so we'll use the whole Z R plus XL square, which is R squared plus OmegL square. If we talk about V average, it will be only the resistance. Okay. So from this equation, we can get RMS, VRMS over z or by integrating the current function as you can see here. So this was a pretty long lesson, as you can see here. But I hope you now understand the RL single phase un half wave uncontrolled rectifiers with an RL load, and you now understand how does the circuit work? 27. Example 3: Welcome, everyone to example number three, in which we are going to learn how can we solve or get the average current, the power factor, and any other values for an RL loot. So in this example, we have R L half wave rectifier, with the resistance value 100 OMs and inductance 0.1. Omega Omega is 373 77 radians/second, and V max is 100 volt. Maximum voltage of the supply is 100 volt. So what we would like to get determine number one, suppression for the current, we would like to get the current as a function of time. Number two, we need the average current. Then we need the RMS current, and we need absorbed power pi Rolute and finally, the power factor. Okay. So let's start with the expression of the current. So first, as you remember, these values are important as we use them in writing our current equation number one equation that or Z is equal to R square root of R square plus omegL squared Omega 377 and inductance 0.1, resistance 100 Os. By substituting, you will get the imbedance of the circuit is 106.9. Number two, we need the phase shift between current and voltage. So the phase shift will be ten minus one Omega R. So it will give us 20.7 degrees in radians, 0.361 radiance. And we need omega T, which is Omega 377 multiplied by time. Or to be more specific, here we are not looking for omegat Omega tau, Omega tau. This is taunt T, and tau is L over R, L divided by R gives us multiplied by omega, which is 377 will give us 0.377 radiance. Now by substituting the equation, I as a function of Omega T will be like this. Now, where did we get this equation? If you get back here like this here. This is I as a function of time, V max over Z, V max was given in the problem. Sine Omega T minus Theta, Theta, which we just obtained and omigata which I just calculated. So by substituting this equation, you will get finally this one. This final equation. Suppression for the current from zero to Pita. Now, if you would like to get the value of Beta, it will be using this equation, which is a condition when the current is equal to zero, when the current is equal to zero, at omegaty equal to Peta, and omegaty equal to Pita, you will get this equation similar to the one which we discussed in the previous lasson. So by using this and by using the calculator, as I'm going to show you in the next lesson, how can use a calculator in order to get any value, you will get Pita to be 3.5 radiants or 201 degrees. Okay? Okay, now, this is a value of Pita, which means it will occur here at 201 degrees after the pi which is 180 degrees, right? Okay. So this is expression of current. Now, if I would like to get the average current, as you know, average current is average over R. And as you remember from the previous slide we talked about like this, we said that average is V max over two Pi one minus cosine peta. Now, by using this, we can get 0.308, average over R. Okay. Similar to it, you can simply say I average is equal to 1/2 Pi, one over integration from zero to Pita. Beta here, of course, in radians. We use all of the values we use in radians. So from zero to Pita, which is 3.5 and integrating this one in the calculator, you will get 0.308 similar to this one. This one is correct, similar to this one. So we get the average current. For the root mean square current, you will use the same rule, root mean, which means average one over T integration, and square means the square of the function. By using the calculator also directly, you will get 0.474 or by getting RMS and divide it by R, divide it by Z by Z, not resistance, only you will get the root mean square current. Then we need the power absorbed by the Rute, how much power absorbed by the Rolute. And as we know that, inductance does not absorb electrical power. What I mean by this not absorbing does not consume electrical power. Only the resistance that consume electrical power, so we will say IRMS square multiplied by R, the consumed power by the resistance. For the inductance, the average power across it will be equal to zero because in half of the cycle, it absorbs power, and on the other half, it will provide electrical power. In the end, the average power is equal to zero. For the resistance, RMS squared multiplied by the resistance. So it will be RMS squared multiplied by R, as you can see, 22.4 what? Now, what about the power factor? Power factor is simply equal to. As we remember power factor of what? Power factor of the supply. So it will be power input. Divided by input. And the input power of the supply is equal to the consumed power through the resistance, right? Since the dite is treated here as an ideal device, without any kind of power losses. So all of the power input from the supply to this circuit is consumed inside resistance. So input power is equal to the Bo power which is 22 w, which we just obtained. Now for the input power for the supply, it will be VRMS multiplied by RMS, VRMS of the supply, which will be 100 divided by root two, since this one is VMAX. Okay. And I RMS or the root means square current will be similar to the output RMS since the current from the supply, similar to the current going to the loot in this circuit. Okay? This is very important, as when we go to the next one, which is with a free wheeling guide, you will find that the current of the supply is not equal to the output current. So here, during this conduction mode, the current of the supply equal to the current going through the loot, and when the current is equal to zero, the current through the load zero and current of the supply zero. Okay. So the RMS is the one which we just obtained here, 0.474 and pairs. So as you can see here, power output, 22.4 and current 0.47 RMS and supply is 100, which is the peak or the maximum. Voltage divided by root two to convert it into root mean square current. So the power factor of this circuit will be 0.67. So we learned in this lesson, how can we obtain the output power, the root menis square, the average, the peta angle, and all of this using using a very simple example. 28. How to Solve Numerical Equations Using Calculator?: Welcome everyone to this lesson. And this lesson, I'm going to show you how can you use a calculator in order to solve any equation in order to obtain the value of Beta in this example or any other example. Okay? So we have this equation that I would like to solve. So I'm going to use this calculator or any other calculator in order to with the property of solving equations, okay? So I'm going to show you how can you do this? First, we have sine. We would like to write this equation inside the calculator. So first, we have sine, so I will select sine. Then we have Pt. Pita is our variable, which we would like to solve. So I'm going to choose Alpha, which is alphabet and choose any sample for Peta, which is X or Y or Z, whatever inside the calculator. So we have X and Y. So I'm going to choose X like this. Alphabet and X. -0.361, then close this bracket and plus sine 0.361, and exponential, Alpha Alphabet. And here you can see the exponential, like this. Then to the power negative Peta or Peta 0.377. So to the power, negative Peta, which is X Alpha, X, and dvded by 0.377. Okay? So now we wrote the left side, we would like to say equal to zero. So I'm going to go like this and then choosea alphabet. This is very important. Not this equal, not this one, but this one. Say alphabet and equal like this. Then equal to zero. Okay. So now we wrote our equation. Now, what is X X is simply the angle, but we would like to get it in radian. Remember that the angle here inside, sine beta -0.36. This is in radian. So this one is in radian. Also, the angle outside is in radian, not in degrees. That's why you have to get the value of Beta in radian first and convert it into degrees, okay? So first, before solving, you will see that the calculator itself is in the degrees mode. So we have to convert this degrees into radians. So how can I do this simply by going to shift, clicking on shift and mode like this. Then you will see that three means calculator will be in degrees and four, it will be operated in radiance. I will click on four because I would like it in radians like this. So you can see now the calculator is in radiance mode. So when we solve this equation, it will give us values in radian. Okay? Now, what is the next step? Next step is that we will click on Shift and solve. You can see here solve like this. Then solve four X, solving for X. What is the value of X you would like to solve four? You can say equal directly, and it will give you the value like this. Or you can simply heft, solve and then type a very close value to X. For example, if you just click on equal, sometimes it will give you zero. X equal to zero. Because we have two solutions, two solutions for this problem. We have current will be equal to zero at the beginning, and current will be equal to zero at BT. So you would like to add Peta to the calculator in order to very close value to Peten in order to get the correct solution, which is 3.5. So when it says solve for X, I will put a value very close to Peta, which is, of course, Pi, as you know that Pi or Peta is greater than Pi. So Peta greater than Pi and very close to it. So I will type Pi, which is 3.14 right. And then this is an estimation or approximated value close to the correct solution. So equal two. So it will give you X will be equal to 3.5, which is, as you can see, 0.5 radiants. And you can see L minus R, which is a left side, minus right side will give you 8.39 multiplied by ten to zPower negative 15. So what does this represent? Is representing the error. When you substitute with 3.5, the error will be very, very small, which means this is a correct solution. So 3.5 ingredients. If I would like to convert it into degrees, I will multiply by 118 and divide by 3.14. Okay, not like this. 3.5 multiplied by 118. Okay. Then divided by 3.14. Okay? So it is 200.6 or 201 degrees. Okay? So what I did in order to convert from radiance to degrees is that I multiplied by 180 degrees and divide by Pi. Okay? So this is how can you use a calculator in order to solve any equation like this or any other equation. 29. R-L Load with Freewheeling Diode: Hey, everyone. In this lesson, we will took a pet the half wave rectifier with an Rude. However, this time, we added something which we call the free wheeling dit or the commutating dit. So what happens here exactly in the previous circuit, which is supply with one diet, and we have Rute. So what is the problem exactly with having an Rut? The problem is that if you compare compare this circuit, here, this is the outw forms of circuit and compare it with the first case, which is like this, we have a diet and a pure resistive lute. We would like to compare between this and this one with respect to V output. So VO, as you remember, in the B resistive loud was like this from zero going like this, till Pi, then becoming zero to Pi, again, and like this and zero, and so on and repeats itself. Okay? Now, what is the difference here? If you look at this is the first one, which is a pure resistive ute. And this one with the negative part with a negative portion here is the Rlute right. Now, if you obtain V average, of the pure resistive lute and we average of the Ruote we average of this wave, the red one which representing the R Loot and obtain the average voltage for this wave which representing the RL loot, you will find that we average for the pure resistive loud or VDC is higher in the pure resistive lot than the RL load. Now, why is this due to the presence of this negative part? This negative part made average become lower because we added a negative part to the wave. So if I would like to remove this average, what I'm going to do is that I'm going to add another diet in parallel with the lute, which is called the free wheeling diet or the commutating diet. The circuit will be like this. We supply D one, d2r and L. Okay? Now, a free wheeling dite which we call D two, can be connected across an Road as shown in figure. So that both of these dites cannot be forward pised at the same time. What I mean by this, you will understand this when we go to the figure. So die D one will be on when the source is positive and D two will be on when the source is negative. Now when we apply a post the voltage, D one will be on and D two will be off and the equivalent circuit will be like this. So let's start with the first one. So we have our waveform, supply and with respect to time or Omegati. So the supply voltage is like this, right? We have positive and negative. During the positive part, what will happen exactly is that this diet will become a short circuit. So the current will move from the supply through this load, like this. So our circuit, this one will be turned off. It will not be operating. All of the current will go from the supply to the R load and charge the inductance. So the circuit will be like the short circuit and this one will be open circuit. And L will be charged it, right? Now, when we start going to the negative half here, the sport starting from here, remember that D one or the inductance will start giving electrical current, right? So the inductance has two options. Number one, to go through the diet easily like this, D two and go like this. It will rotate like this from inductance to the dite like this, or it can choose to pace the supply like this and go through D one. Which one is easier for the current? The easier path is to go through D two. Now, why is this? Because this one during the negative part, it will be like this, positive and negative. It will reverse it polarity and would like to give current in the opposite direction. The inductance, if the current goes through this path, it will be much difficult than going through this pass. That's why the current will go through D two like this and D one will be turned off. The circuit will be like this. D two will become a short circuit, parallel to RL. So our circuit is divided into two parts. The first one during the postive cycle, D one will conduct and D two will be off. So be short circuit like this, going to the load. During the negative half or when inductance has stored energy, it will go through D two like this instead of facing the supply. So it will go through D two and like this. I will become a short circuit, parallel to RL. D two will conduct and D one is turn it off. Okay, similar to what we just said. Now, if we look at the waveform. So here during the postive cycle, from zero to Pi, V output is equal to V supply, right, out equal to V supply. So from zero to Pi, O equal to V supply. What about from Pi? To Peter, the extinction angle, the extinction angle, if you remember in the previous example, we had the current going from zero like this and then going down to Peter at which we have current equal to zero. Then it will become zero like this and then start increasing again like this and then goes down to Peter plus two Pi, right here from Pi to Peter, the inductance will give stored energy through D two, right. So this one will be a short circuit. So the circuit will be like this. So the voltage across the abut will be equal to what? Equal to zero, since it is parallel to a short circuit. During this part from Pi to Pita it will be zero. After this supply or this stored energy finishes, D two and D one will be postned off. So the voltage will be also equal to zero. That's why in general, we will have during the circuit from zero to Pi, we will have the out and from Pi Pi will be zero. Okay? Now, number two, during here, if you look at this current, this is the current waveform. We will understand now the difference between this and the previous one. Okay. So anyway, you will find that from zero to Pi when D one is conducting, so the current of D one, this one representing current of D one and this representing D two. So for D one, it will start from zero to Pi during its conduction from zero to Pi. And for D two, it will start from Pi to Peta or here in this example from Pi to two Pi, from Pi to two Pi. The summon at any instant, this is very, very important. We have I output. So I O equal to ID one from KCL at this point, ID one plus ID two will be equal to I output plus ID two. So I Out equal to the submission of the two currents. Now, this is very important. Why? Because as you can see here that in this when we have a free wheeling diet, IObut is not equal to I supply, right? I supply, I supply is equal to ID one. So in this circuit, they are not equal to each other. If you add these two wave form, you will get I out. Now, the very important question here is that why does the current looks like this? If you get back, let me just show you this is very important too. For the current I out in the first circuit, it starts from zero to Pi, then start discharging to Peter. Okay. So this extinction angle extinction angle depends on what depends on the values of R and L, value of inductance and resistance. And also the higher the value of inductance, the larger the angle. So p increasing inductance, pi increasing inductance, L. Instead of having Pita like here, we can make it pita here exactly. So for example, the current will start discharging slowly till a two poi then start charging again, then discharging a two pot. And even if we if we increase the inductance too much, add a large inductance. Instead of having from here like this, it will start from here and start a discharging and does not reach zero at two pi. Starting from two pi, the supply will start charging the inductance again. I will start charging like this and then there's a charging, charging, discharging, and so on. This is what we call a continuous current waveform. Continuous current because you can see here that the current does not go to zero. You can see value then goes down, DC charge. Then you still have a value charge, DC charge, and so on. It's a continuous wave form. However, in this case, in the first case here, we had discontinuous start exhausting. Then here, discontinuous becoming zero, then charge again. Then zero. So this is a discontinuous waveform. It is not continuous. Okay? So the continuous or discontinuous waveform depends on what. It depends on the values of inductance itself and resistance, especially the inductance. The higher the inductance, the slower the current will start discharging through the circuit. That's why in this example here, you can see, the current waveform can be as before. It can be like this from zero to Pita then start decaying to Pita, for example, starting to pi, then goes down to Pita. Then charge again, then this charge, it goes to zero, similar to the previous slide. However, here, the current in this waveform is assumed to be continuous. It means that this inductance is very large and led to this continuous waveform, you can see that the current does not go to zero. Okay, because it is a continuous waveform. So it is not specific or not related to the free wheeling dit. However, the continuity of continuity of the current depends on the inductance and resistance. Okay? So we can draw it like this or draw it continuously like this, depending on what, depending on these two values, okay? 30. Example 4: Now let's have an example number four, which will help us to understand the free wheeling diet. Very easy example here. So we would like to determine the average load voltage and the current for the circuit in which we have the resistance to OMs inductance 25 milli Henry. Here's the space between them should be here. V max is 100 volt and frequency six sets. You will find that it is very, very easy. We would like what? Average load voltage. If you look at the waveform, out from zero to V max. And it goes down to Pi, the n zero, V max and so on. So if you look at this waveform, you will find that this waveform is similar to the R lute, right? R lute, the pure resistive lute. So we average, we can apply the laws of average since we would like averaged voltage and current, not the RMS. So we average will be like this VMX over Pi, right? We average V max over same low of the pure resistive low, pure resistive loud because it has the same waveform, right? So we max over poi, which is 100 over poi. Now, what if I would like to get the average current? Average or I average average current through the ****, as we remember that the average voltage of the inductance is equal to zero. Average voltage. Here I'm talking about average, not root mean square. So since the average voltage across the inductance is equal to zero, so the resistance itself will be or the current through the load will be out average. The average voltage divided by the resistance like this. Why? Because we are talking about average. You will find that this is similar to even the same law for RL loud without any free wheeling glide. Okay? These two laws will change when will it change if I'm looking for RMS and RMS. Okay? For VRS, it is same as it is similar to the PR resistive loud. You can see that the waveform didn't change at all. It will be VMAX. Overt, similar to the pure resistive lute. However, RMS will be VRMS divided by Z or, the total impedance of the circuit. Or if you'd like RMS, you will have to use the equations which we talked about before in the previous lesson about Root, a large equation which had Beta or Peta as a large equation, V max over Z, sine Omegate minus eta, or eta, and so on, okay? So this was a very straightforward example regarding the RL load with a free wheeling light. 31. Single Phase Half-Wave Controlled Rectifiers – R Load: Good morning, everyone. In this video, we took a pet, another circuit that is used to convert AC into DC, which is called the single phase of wave controlled rectifiers with an R load. The difference between the circuit and the previous one is that this one is controlled. The previous one is uncontrolled rectifier. So starting with the first circuit, which is loud or a pure resistive loud. So what's the difference between this circuit and the previous one? No difference except one thing, which is the presence of the switch. And this switch is not died, but this time is a si restor. So we have the same circuit or supply, then we have the switch which was died in the previous lessons, and we have aud which is a pure resistive loud. Now we have our si restor with gate control signal. We are going to give it current in order to turn it on. Now before we continue this lesson, I would like to mention something which is important. Remember these samples, this one and this one. This is something which is not related to the course at all, but it is related to the pronunciation of these samples. These Greek samples. You'll find that this one like this and this one like this. In the British accent or in the British pronunciation, this one is pronounced as Sita and this one is pronounced as Peter. Okay? This Greek letter, previously, it's called in the Greek alphabet. It's called Vita. When we pronounce it in the British accent, it will be beta. So sometimes you will see Sita and Beta. In the American accent, these two samples are pronounced as Seda and Beta Seda and Beta. If you hear Sita or Seda, beta or Beta, you now understand these are the two symbols, because they have different pronunciation in American and British accents. Okay? That's what I would like to say, okay? So anyway, let's get back to our circuit here. So we have here our supply, the same AC source, and we have this sample. So what's the difference? The difference is that this is a way to control the half wave rectifier circuit. Is that this circuit or this switch will not convert AC into DC unless we give it a signal. So we use a CIstor in a set of a diet, and there must be two conditions that must be met or fulfilled before the SIRstor can conduct or provide electrical power. Number one, the iistor must be forward biased, similar to the dite, the voltage across it must be greater than zero, similar to the dite. And the second condition, which is that the current or the gate current must be supplied, or the current must be applied to the gate of the restor so that this iistor will be switched on. So during the positive half cycle here, the first condition will be fulfilled. The voltage across the soy restor during the positive half will be positive. So the Systor is a forward pased. However, without gate signal, without giving any gate signal or gate current, this soistor will be still turn it off. The second condition which is providing current using the gate control circuit, it will give us a gate current that will turn on this systor. So the restor will not begin to conduct a source becomes post because we need the second condition, which is the current applied to the gate of the soy restor. So the conduction is delayed until a current is applied to the soy restor. Once the is conducting, we can remove the gate current signal. So we can just give only this gate current is just a pulse, giving it just a pulse. By giving the pulse, it will be switched on and become a short circuit, and it will remain on even after removing this gate current, until the current goes to zero. So let's see what will happen here. So the gate current. So the gait current blight here is represented by a certain angle or it is provided at a certain angle. So as you can see here, this gate control signal is provided at Alpha. So Alpha here, this is representing the angle at which we are going to provide the gate current. So Alpha is called the delay angle. So as you can see here, this is representing the input supply, right? Okay? So this is the output. So when the voltage is positive and we don't have any gate signal, the output will be right. It will be zero like this. Okay. Why? Because in this period, we didn't give any kind of signal or any gate pulse to the y restor. So this istor is off until we give the angle Alpha or the pulse at angle Alpha. So add the instant Alpha omegat equal to Alpha, which is a certain angle that I choose by giving this pulse, giving pulse to the systor at Alpha, this istor will start switching on and become a short circuit. So the VOut will be equal to V supply. So Alpha giving when we give pulse, it will be switched on and the Vout will start becoming equal to V supply. This is a V supply signal, and this one, which is a circle line, this one is representing the output signal. So Vout will be equal to V input after appllying the Alpha or the gate signal. Before it, the Vout is equal to zero. Then when we reach Pi, when we reach after Pi, the voltage will become negative. So the first condition which is rest must be forward Pi or the voltage across a greater than zero. When we go to the negative part, the voltage will be negative, so this cistor will be off again. So Cystor will become an open circuit, similar to the tight, and out will be equal to zero. So as you can see, starting from Pi, it will be equal to zero like this. And even if we have a positive signal, we will still be zero until the firing angle Alpha is called the delay angle or firing angle Alpha plus two Pi. So the Alpha is provided every two Pi. The first one at Alpha and second one is at Alpha plus two Pi. This is the Abut signal. Now, if you compare this to the diet, the diet you can think of it of the diet as Alpha equal to zero. If the sister angle is Alpha equal to zero, what does this mean? It means that it will start conducting, starting from here. Out will be like this. Like this, similar to what, similar to the diet. So we say that if the firing angle is equal to zero or delay angle is equal to zero, then the diet or the crestor will become a dite, okay? Okay. Now, another thing if you apply Alpha equal to Pi, it means that VO will be always equal to zero. So if you make the firing angle at this instant, which is Pi, after giving the firing angle, the voltage will be negative, so nothing will happen. So the VO if Alpha is equal to Pi, VO will be always zero. Add Pi because after it, the voltage becomes negative, so it is a reverse pies. So VO this switch will be always turned off. When Alpha equal to zero, it will become diet. The difference as you can see here between the diet and ostore is that we just delayed delayed. That's why it's called the delay angle delayed, the turning on of the switch. Instead of starting from zero, we have more control on the circuit. We start from here or here or whatever. What is the benefit of this? This will help us control also the average. We can control the average output voltage, the average current, the power, and so on by controlling the firing angle. Now if you look at the iistor here, as you can see the circuit, supply from KVL, equal to voltage across the crestor plus voltage across the resistance. So when the cistor is off, then VO is equal to zero when the crestor is off and supply all of the voltage of the supply will go across the cistor. That's why during this period, you will see that V across the cistor it will become equal to the supply. And when the Vo equal to supply, stor is equal to zero. Okay? And here, zero, then equal to V supply and so on. So if you look at these two signals for VO and voltage across the Istor, their submission at any instant will be equal to V supply, okay? Now, in order to understand the conduction, as you can see, this square or this rectangle representing a pulse. Okay? So here we are providing at Alpha equal to zero, we give a pulse for a very short time, and then it goes down to zero. So we give a current in the form of a pulse like a small pulse. Starting at zero, you can see it start conduction all of the post of cycle then zero, then conduction then zero. This is similar to what when Alpha equal to zero, similar to the dit. Here, another Alpha at Cat here. We give a pulse like this and you can see goes from zero and zero like this. So these signals the difference the angle between them, the time between them or the angle between them is two Pi. So this one, for example, this one is Alpha, this one is Alpha plus two Pi, this one is Alpha plus four Pi and so on. Here at another angle Sta three. So as you can see, as the firing angle Alpha increase as the firing angle alpha increases, the V output will start decreasing. As you can see here, firing angle equal to zero, lots of voltage. Cita two when Alpha increased, you can see voltage here is smaller than here. And Cita three, you can see very small part. So Pi controlling Alpha, we can control the output voltage. Okay, so let's understand the equations of this circuit. So as you can see, if the get signal of light at omigat equal Alpha, the angle Alpha, we call it the firing angle or the delay angle. Now, if you would like to get the average voltage or average current or RMS values and power, this is very easy. All you have to do is similar to what we did in the uncontrolled rectifiers, which is integration. So VDC or the voltage average voltage will be equal to one over t the whole period, which is two pi, 1/2 pi integration from zero, to two Pi, right? This is the integration of the average. However, if you look at here, you can see that. Our voltage, our voltage does not exist except from Alpha to Pi. This part is zero. This part is equal to zero, from here to here. This one is zero. This one is two Pi, right? So this is a complete cycle from here to here, from here to this part. So you can see that the output Out exists only in this part. Okay. That's why the integration is from Alpha to Pi. So you can see integration, integration from Alpha to Pi. For what for the signal, Vmax, sine Omegaty d omigaty. So if you integrate this, you will get Vmax over two Pi, one plus cosine Alpha. So as you can see, Pi controlling Alpha, when Alpha becomes zero, this part will become cosine zero, which is one. One plus one will give us two. Now, let's say Alpha equal to Pi. It will be one plus cosine Pi and cosine Pi is negative one. So it will give us zero. The output will be equal to the average voltage will be equal to zero when Alpha is equal to i similar to what we said if Alpha is here, it will not turn on at any instant. It will be turned off all the time. Okay? Now, this is for the average voltage and since we have only a resistance, the current will be the same. It will be voltage divided by resistance. I average we average over R, as you can see. Now, what about RMS? You will have to do the root 1/2 pi. You can see RMS, root mean one of two Pi integration 0-2 Pi for the VO square, which is V max sine omegaty all square. Now, as you can see here, this is integration. Integration will be also from Alpha to Pi. Here from here to here. Why? Because everything else is equal to zero. As you can see here, from Alpha to Pi. For the function squared, it will give us, as you can see this equation. Okay? Now, what about IRMS? The root mean square current. It will be the VRMS divided by R, like this. So this is the equation here. It will be like this. So RMS will be VRMS over R. So it will give us this equation. Finally the power power consumed inside the resistance. Here I'm talking about power in general, the EC power, not just the average power. I'm talking about the real consumed power, all of the power consumed. So it will be the root meani square current. It will be RMS square multiplied by the resistance, or VRMS square divided by R, as we learned before. So it will be VRM square over R or IRMs squared multiplied by R. Now, another thing here which you can see, let's first lead all of this. You have to understand that when I was in college like you or when I was a student learning all of this, during exams, what I have done is that I do the integrations. Okay? I don't memorize any of these functions. Okay? I don't memorize all of this. I just go inside the exam and start doing all of the integrations because it's much easier, okay? So there were some people who memorized these equations, but I don't advise this. My recommendation is that you in any kind of circuit because there are lots of circuits regarding the rectifiers. In any circuit, you have just to do the integrations in exam 1/2 Pi integration, max from the output signal as you have seen and you learn it from here. In both of the average voltage, RMS voltage, current, and so on. Okay? 32. Example 5: Let's have the first example example number five on the single phase, have wave rectifier, control rectifier. I put all of the formulas that I have explained or drived in the previous lesson. So as you can see here, we have in this example, we have a single phase half wat fire with a pure resist blue R and the lay angle is Alpha equal Pi over two. Find that average I average VRS IS. What I'm going to do is that I'm just going to substitute in these equations. So average, you can see equation average is V max over two pi one plus cosine alpha. So I'm going to take Alpha equal Pi over two and substitute it here. Like this, V max over two pi one plus cosine Alpha, which is Pi over two gives us this value. For the I average, it will be this value divided by R. You can see same value divided by R. Then we have VRMS. It will be this equation here, so we will substitute with Alpha equal to Pi over two and sign to Pi over two. To Alpha Alpha Poy over two, and here we have Alpha Poy over two gives us this value, and IRMs will be this value divided by R like this. This is a very straightforward and easy example. 33. Example 6: Now let's have another one. Example number six. Example number six, it says, find the delay angle. So we need here to find Alpha. That will lead to an average voltage of 40 volta, so we average average or VDC is equal to 40 volt in this circuit. And we have the resistance equal 100 Ms from 120 volt RMS, 60 hurt is AC source. So this supply is 120 VRMS. Also determine the power absorbed by the resistance and perfex we need the active power, and we need the power factor, which is power factor. Okay. First, what are we going to do is that we are going to use the V average. So we already know that average in this circuit which is integration from 1/2 Pi, integration from Alpha two Pi, VMAX and Omega t will be like this. And remember, VMAX is equal to VRMS hundred 20, multiplied by root two. VMAX, VRMS hundred 20 root two. All of this will be equal to equal to V average, which is 40 volt. Okay? So by solving this equation here, as you can see 120 root two, here, cosine minus one, the inverse angle for all of this. So it will be 61.2 degrees. And remember, when we are working with these circuits, we have to use the radiance, okay? No degrees radiance. So 61 degrees is 1.07 radiance, okay? So now we have our Alpha. What's the next thing we would like to get the power absorbit. So power absorbit will be VRMS square over R or RMS square multiplied by R. So we're going to use the VRMS or VRMs from the one which we learned before. It's VMX of two root, one of our Pi Pi minus Alpha and two Alpha plus divided p two. Alpha here is 1.07 radians and here Pi minus Alpha, which is 1.07 radiance. And here, this is very important. Don't use any angles here in degrees. You have to use radiance, okay? So if you put 61 here, it is completely wrong. You have to use the radiance, okay, radiance. Okay? So by substituting with the angle, you will get the value of the voltage. Let's just delete all of this. Okay, VRMS, you can see root 220 V max and substitute 1.07. As you can see radians, you get 75 volts. This is the VRMS. So VRMS is square over R. This value squared divided by the resistance, 100 Ms, you will get 57.1 watts. Now we need the PR factor. Pow factor, as you remember, is equal to active power or the real power consumed, divided by the apparent power. So the active power consumed is 57.1. Okay, it's 57.1 and S, which is apparent power of the supply is VRM, IRMS. Okay. So VRS of the supply is 120. Okay? Now, what about the current current RMS, arent? So current RMS is equal to I I here RMS will be VRMS which is a voltage across the loot, divided by the resistance. Voltage across the loot divided by resistance. So it will be VRMS Not the supply. Not the supply. The voltage across the luid divided by its resistance, gives us the current flowing through it. So it will be RMS, divided by R, which is 0.756 a pair. You will use this current, multiply it by the RMS of the supply, you will get a parent power. So you can see 120 VRMs of the supply multiplied by the current going out of the supply, gives us volta and pair of the supply itself. Now, we will take 57, which is a consumed power divided by the apparent power of the supply. So it will be 0.60 29. Okay. So this was another example on the circuit with control directifre halfway control directfre with an R loot. 34. Single Phase Half-Wave Controlled Rectifiers – RL Load: Welcome everyone to this lesson. And this loon, we too with a single phase of controlled rectifiers. However, this time with an RL luid. Let's start. Here we have our circuit, but instead of having the uncontrolled rectifier, we had a diet. Here in the controlled rectifiers, we will use a presto. So you can see the circuit is like this. Now let's look at the waveform of this circuit. Okay. So what you can see here first, this will be very, very significant in understanding how does this circuit work. So as you can see first, we have our supply, VS which is a supply voltage, as you can see a sine wave, as you can see here with a value of VMX maximum value of the sine wave. Okay? The first thing that we have is the gate signal, the gate current or the gate pulse is provided as you can see at a time or at an angle called Alpha, the delay angle similar to the previous circuit. So you can see, we just give a value of current, just a pulse of current, it can be like this, or it can be like this. For a very, very short time, this pulse will lead to making this rest start conducting. As you can see Alpha, before Alpha, the Vout is equal to zero, starting from the firing angle Alpha from here, VT one will become a short circuit and VO Out will be equal to V supply. So at Alpha, O equal to V supply, and it will continue like this till pi right. However, similar to the circuit of the uncontrolled rectifiers. What will happen is that this VL does not allow instantaneous change of current. So it will start reversing its gularty and giving current to the circuit and keeping this iristor in the conduction mode till the extinction angle bet this one. You can see it will still conduct a short circuit. So V output will be equal to V supply, so it will keep continue until BT Okay, until bit. So VO, this is V supply. So VO, as you can see, will be equal to V supply due to the energy coming out from this inductor and keeping this systor in the conduction mode. Until extinction angle beta, after which this i restor will be turned off and voltage will be zero until the nexstfiring angle at two Pi plus R. Alpha and two Pi plus lf. We will give another pulse here every two Pi. Okay, so that is the first thing. What you can see here, what is the difference between this circuit and the uncontrolled fire. Here we have a si restore and instead of a diet, number two, you will find that the conduction mode starts from Alpha and instead from zero. From zero, it was like this. Like this in what circuit in the uncontrolled rectifier. Here in the control, it will start zero and start like this. Okay? That is the only difference between them. Now let's look at the current. So the current, as you can see, from Alpha, to Pi current going through the circuit, you can see from Alpha to Poitll this point. You can see current increasing O out start increasing. Till Poi and after Poi this inductor will give its own energy to keep the sistor in conduction mode. So it will start the lute current will start decaying as energy of the inductor is starting to feeding or start decreasing till zero. You can see current charging, then start ds charging. So from Alpha to PT. Okay. Similar to the uncontrolled circuit, we started from zero to PIT. Here we started from Alpha to bit. Okay? Now, what about the cistor voltage? Now, remember supply is equal to V of the crestor voltage across the restor plus V out from KVL. We supply this plus this one. As you can see here, this is our VO and this is our V supply. Supply is equal to VO plus voltage across the ist. So if you add the signal with this one, you will get supply. As you can see, during the duration when Vistor is an open circuit, all of the voltage of the supply will be across the solstor. And when it is in the induction mode, it will be a short circuit, so the voltage is zero, and during this period at which voltage we will be equal to zero, it will be equal to the supply. Okay? Now, what about the voltage across the resistor? Voltage across the resistor will be equal to the current flowing through it. Multiplod by the resistance, right? So if you look at here VR, which is the resistance voltage from Alpha to Beta, from Alpha to beta, similar to this one, same waveform, but multiploid by the resistance. Here again, same wave form, similar to this one, like this. Of course, it will continue like this, go to zero, and this one will go to zero, similar to this, this part, similar to this one. From Alpha to Peter, as you can see, IL, which is a current, this current or the ut current multiplied by the resistance which is a constantive value. So it has the same form. Now, as you can see, there is a V average average for resistance. What about VL voltage of the inductance? Remember, this is very important and it will clarify what we have said before in the uncontrolled rectifier. I told you before that V average across the inductor is equal to zero. Average across this loot will be out average will be will be current, multi blood by the resistant. I average multi blood by the resistance. So let's explain why is this first VL, is equal to L di over dt, right? So L is just a constant value. What is important for us is di over dt, right? So here, as you can see, let's start. We have our current like this. So di over dt is the derivative of this function. So what does that derivative mean? Drivative means the slope of the line slope of the line. Right? Derivative means slope. So as you can see, during this period t here, during all of this period, you can see here, the slope here, the slope here, and here like this and here, all of this is positive slope, positive slope. That's why from here from Alpha, till Pi, till Pi. You will see during this period, the VL will be a postive value because the ardity is a postive value. As you can see here from Alpha till Pi, you can see all of this a postive voltage from here to here. Why? Because we have a positive slope or postive variation of current with respect to time. Now, starting from this point here, you can see that the slope of the current start going in the negative direction. You can see here. If you get the slope of this line, here, slope of this will be zero. That's why this point is zero voltage. Now, let's continue at here like this. Negative slope, this one like this, negative slope, and this one like this, negative slope. That's why from here, all of this gives us negative value of di over dt because it is a negative slope. That's why due to the negative value, we will have negative voltage, as you can see here. All of this is negative, because di over dt is a negative value. Now, as you can see, if you look carefully here, we have a positive part and a negative part. If you get the average average of the signal, During one complete period two Pi. What you can see here is that the positive part is equal to the negative part. That's why V average across the inductance is equal to zero. That's why when we get V average, we get V average, which is a voltage across the resistor. This case is similar to the uncontrolled richtfre. I hope this clarify what we have just said before in the previous lessons. Now what we would like to get is the same process, the same process that we have done in the uncontrolled rectifiers, we are going to follow the same steps, which is the current is the sum of the faucet and natural responses of the current. So as we remember that I faucet plus I natural, I faucet is the one which is coming from the supply. Which is V max over the total imbedance of the circuit sin omegaty minus the phase shift. Phase shift due to the presence of inductance, plus the natural response when this inductor start giving its own current. Now, before I continue this, as you can see here, this is the equation that we have explained before when we talked about uncontrolled rectifier, but as you can see here, here, during charging during this first period, you can see voltages positive, to the polarity, positive, negative. Now when it goes to the other side, it means that the polarity is reversed. That's why the current goes like this from the energy stored of the inductor. This is just explanation of why the polarity is reversed because the current is decaying, so the Di validity negative, so the VL polarity is reversed. Let's continue. What is the only thing that I'm looking for? I'm looking for is a constant I, I need an initial condition. This is our lot current, we form this Okay, from here to Pi, at this point, to Pi. So the current exist from Alpha to beta, right? So I need an equation. This equation is a general equation. So we need to add our boundaries, which is beginning and end the beginning, which is Alpha, and which is Pita. So at Alpha, the current starts from zero, right? Current is equal to zero, and at Vita, current is also equal to zero. So these two conditions will help us get the constant A, and it will help us get the average current and so on. So we will start from the first condition, which is initial condition here, Omega y equal to Alpha. I Alpha will be equal to zero. The current will be equal to zero at Alpha angle, right? So we will substitute this in this equation. So zero will equal to V max over sine Alpha minus t plus A E to the power. Omega t which is Alpha divided by Omega T which is L over R, similar to the previous circuit of the uncontrolled rectifiers. So we will get A will be equal to the constant will be like this, and we will substitute in this equation here to get the final formula. So you can see I as a function of Omega T with respect to the angle from Alpha to Beta from here to here. It will be this equation, which is this one with a constant A. And otherwise, here or here it will be zero, right? Now, what is beta beta? It is the angle at which the current returns to zero. So what I need is beta. Why do you need beta? Because when we get I average, I need to get the value of integration from Alpha to Beta, so I need beta. So as we remember, the second boundary condition, which is Beta equal to zero or Beta is equal to zero. Okay, Beta equal to zero. So you can see at zero equal to V max over that from this equation here by substituting here, omega t equal Beta minus Alpha minus pita. Here, this should be minus t as it is. We replace each omegaty with PT. It is p Alpha minus Pita, as you can see, divided by Omega tau. So by solving this equation, which we have all of these values except PT. We will solve this by using the calculator as we have done in the uncontrolled rit fires. As you can see, it can be solved numerically for Beta. Okay. Then we need to another thing which we have to say is that Pita minus Alpha. This difference in angle. This difference from angle, Pita minus Alpha. This difference, this period is called the conduction angle, conduction angle. So the conduction angle conduction angle gamma, representing the period at which is a current will conduct, from Alpha to PT. Now what if I need average? V average will be 1/2 pi integration 0-2 Pi, right? Now, what about load voltage? Load voltage exist from Alpha to Beta. That's why the integration will be from Alpha to Beta. We maxline Omega t, it will give us this equation. For I average, it will be the same 1/2 buy integration 0-2 Pi, and the current exists only from Alpha to Beta, it will be Alpha to Beta for the current equation, which is this one, this one, from Alpha to Beta. Okay, so it will give us the value we need or you can say that I average is equal to V average over R, right? V average, which is this value divided by R. Now, we neglected VL. Why? Because the average of VL is equal to zero, as we said before, positive equal to negative, average zero. What about RMS? VRMS will be root, similar to this, but square of the signal, right from Alpha to pit. So it will be like this for RMS, we can use the same formula, root 1/2 Pi integration from Alphabetap square of the waveform, or you can say RMS equal to VRMS over the imbedance. Why? Because RMS here we are dealing with the AC signal, VR and VL or the resistance and inductance. So it will be VRMS over Z, which is this equation, the sport, all of this, representing this one. Okay, divided by root or square plus Omi L square, which is Z or the imbedance. Okay? So I hope you now understand the control direct fire circuit with an R load. I think it is pretty straightforward and easy. That's why I added both of the half wave controlled and uncontrolled in one section because as you can see, they are very, very close to each other. They are almost similar to each other, except that instead of starting from zero in the did, we start from the firing angle alpha, okay? 35. Example 7: So let's start by having the first example or the example number one in the RL load or the control dictfre with an RL load. So for the circuit, as you can see here, we have supply 120 VRMS at 60 Hertz, resistance 20 ohms, inductance 0.04 henry, and the firing angle or delay angle is 45 degrees. So we need to find, number one, expression for the current, the root mean square current, the power absorbed by the loud and the power factor. So first, starting with the expression of the current, for the expression of the current, as we know that we need all of this, if we get back here, this is the equation of the current, this one, V max over there so we need imbedance which is root R L square plus Omega L square. We need CtA which is ten minus one Omega L over R. We need Alpha, which is 45 degrees, but we will convert it into radians which is pi over four. We need TA, which is L over R. By getting all of this, you will be able to write the current equation. So we have VMX, which is VRMS Multi blood by root two. Which is root R square plus Omegel squared, CT which is temus one omega over R, Omegata which is tau L over R and Omega which is two Pi multiplied by the frequency. Frequency ct hortus, you will get this and Alpha, which is 45 degrees, which is Pi over four, which is 0.785 in radians. And you said that all of the angles we use in radians. So I as a function of time will give us this formula from Alpha to Pita. Alpha, which is Pi over four, Pta we still didn't get it. Ok? So in order to get Peta, simply, we are going to substitute with Beta as a function of Omega t and put all of the values of Pt and Peta we don't know yet. We max over that, Sita and every value we know except Pt. This equation, which is the general equation for the current that we discussed before. By solving this equation, you will find that Beta will be equal to 3.85, as I remember, okay? So 785, okay? 785, like the radians. To get the root mean square current, I root mean square, simply, it will be the square root of one of T integration from Alpha, which is 0.785 till beta, which is 3.785 or 3.79. For the equation of the current, which is one all square, it will give us 3.20 6:00 A.M. Pairs. Okay, so now we have the RMS current. Now we need the power absorbed by the fluid. It will be root mean square current squared multiplied by the resistance. So it will be IRMS squared, multi blod by resistance, which is 213, what? Then we need the power factor. As you know, power factor is simply equal to the power absorbed by the load, divided by the apparent power. So power absorbed by the load is equal to 213, right? 213. Supply, it will be VRMS of the supply, which is 120 supply, multiblo by the current RMS going out of the supply. So IRMS going out of the supply will be 3.26, right. So it will be like this power factor, BOS, it will be 213 real power divided by apparent VRMS which is a supply. And multi blood by RMS, which is the current of the supply in the root mean square. The bar factor will be 0.54. So this was another example regarding the RL loud or the circuit of the controlled rectifier Hwy rectifier with an RL. 36. Half-Wave Controlled Rectifiers – RL Load with FWD: Welcome everyone to this lesson. In this lesson, we will too the half wave controlled rectifiers with an RL load. In this case, we have a free wheeling dite or a commutating dite. So if you understood all of the previous lessons on how to get the current wave form, you will be able to understand this, and this will be a piece of cake for you. So let's start. Here we have the same circuit, the same circuit as the half wave uncontrolled with a free wheeling dite. But instead of having diet, we have here a sol restor. We have our supply side restor, free wheeling diet, and our loud, which is RL lute. Now, in this example or in this explanation, we will assume that the current is continuous. The current does not go to zero. It is always continuous. So we will understand what does this mean when we look at the waveform. So again, similar as we said before, we have our supply, which is this waveform, sine wave, right. So during the postive half cycle, this soy restor will be forward biased, right, forward pised. However, we need to provide a gate current or a Git pulsy to the istor so that it will switch on at angle Alpha, which is a delay angle. So as you can see here, before Alpha for V output, before Alpha, output is zero. Here before Alpha because this restor is turned off. Starting from Alpha, this restor will be switched on and be a short circuit like this. The current will flow like this from supply, go through the stor, through the loud and get back like this. This free wheeling dt will be turned off or reverse pies, reverse pis. In this case, output will be equal to supply. As you can see here, out similar to from here. This is V out like this, equal to supply till angle Pi. Now, what will happen in this case, due to the presence of energy stored inside the inductor, energy stored, this energy stored will go through the free wheeling like this. The current will go through this circuit, and this erstor will be turned off, right. So when this free wheeling dt becomes a short circuit, short circuit, parallel to the RL load. It means that V output is equal to zero, right, because this load is parallel to a short circuit. That's why starting from Pi till two Pi plus Alpha, the next firing angle to Pi plus Alpha. From here to here, all of this will be zero. Why? Because from here to here, this is a reverse psedsistor and this stor does not have any firing angle from here to here. So all of this region is zero voltage, okay? And cycle repeat itself. So you can see that this waveform is similar to the uncontrolled rectifier with a free wheeling tide except that instead of starting from zero like this to Pi, we started from the firing ang. Okay, so it is a controlled circuit. Okay. So now we understand that waveform, the outward we form supply. Now let's look at the other waveforms. First, iistal voltage across the istal. This sister will be turned on from Alpha to Pi, right, in the forward mode. So during this period, the voltage across it, it will be a short circuit. So the voltage across it, it will be zero from Alpha to Pi. You can see here from Alpha to Pi, all of this is equal to zero. And the other the rest of the waveform, it will be equal to supply, as you can see. This part of mode, the of mode. The restor have the same voltage as a supply. And here during the off mode till the next re angle, all of this is off mode. So if you remember correctly that V supply from QVL, if you apply KVL to the outer loop, you will find that supply is equal to VS restor plus VO. So if you take the out wave form, and add it to the VSO restor, you will get V supply. Okay? So when we have V equal V supply, VR restor is zero. When we have V equal to zero, V restor equal to V supply. Very easy. So we have this one, this one, and this one. Now, what about the current waveform? For the current waveform? So we said that the current here, we assume it is continuous. Okay? So let's first understand if it is not continuous. If it is not continuous, it will be like this. During let's draw it. So before Alpha, let's say I O as a function of Omega T. Alpha before Alpha starts conduction or before Alpha is supplied to the systor and istor starts conduction. Before this, the current will be equal to zero, right, like this from Alpha. And when the firing angle exists, it will be like this. I will start like this. Okay, keeps induction or the current will go from the supply through istor through the lute, right? Now, during this period, during this period, what will happen is that from Alpha, the current will start increasing till Pi till Pi. Now, what will happen after this? After this, this soistor is off, right. And the energy stored inside the inductor will start providing it through the free wheeling dide like this. So the current will start decaying like this until angle Peter. That's what we explained before. Then the current will be zero, all of the energy stored ended inside the inductor. So this free wheeling dite will be turned off. So both of those are restored and free wheeling diet are turned off. So all of this will be turned off until Alpha. Let's just delete this. Here's this one Okay, till here, which is Alpha plus two Pi, which is the neckst firing angle. Then the current will start increasing. Then goes down to P plus two Pi. Now, as you can see here, if you look at this waveform, zero, then the current increases, decreases, zero, increase, decrease. As you can see, this is a waveform, which is discontinuous. The current is not continuous. It goes down to zero and remains zero. Okay? So that is the first case which we explained a lot in previous cases. Now, what I would like to do is that I will assume that the current is continuous. What I mean by continuous, it doesn't go to zero. So how can I achieve this? I achieve this using a high inductance by increasing the inductance, L by increasing the inductance, the variation or change in current will be very, very slow. So the current will be continuous. As this waveform, if you look at the I out here, when we assume it is continuous, you can see that the current is continuous, it doesn't go to zero. Now, let me explain for you where did we get this one. Okay? So let's say we started the circuit for the first time like this, I out as a function of Omega t. Okay? Now, we started with Alpha again with Alpha. Current was equal to zero. So the current will start increasing. Okay, till Pi then it will start from Pi start decaying. It will start decaying. So if you look at here, it will start decaying like this. But instead of decaying at beta, it will not go to zero at Beta. It will go to zero at very, very large angle. Now, before it reaches zero here, let's say it reaches zero at this point at Alpha plus two Pi here, We start giving this iristor a pulse, right? So what will happen is that when this cyst, turn it off again, turn it on again like this, the current will be provided from the supply and go to the loot. So what will happen is that this current will start increasing again. So what you can see is that at the beginning, we had the recurrent. From Alpha, we start giving current from the supply to the inductance. So all of this is charged. Then until Pi, it will start decaying, starting decaying since it will start providing its energy from inductance to free wheeling dit. However, since the inductance is very large, this decay in current is very, very slow. So until Alpha plus two Pi, it didn't reach zero. I didn't reach zero, same as this case. So when we have Alpha plus two Bi, we start increasing the current once more because the sstors turned on. So this will start increasing again. Then it will go down until Pi. Then it will go down to Alpha plus four Pi and then increase again like this. So it will never reach zero. And as you can see here, the current is continuous. So this waveform, which is, let's say, from here, from here, from two Pi, like this. Starting frame, we neglected the first cycle and we start drawing from the steady state, which is from two Pi, like this, goes down, increase, go down, increase, as you can see. So you can see goes down, increase, goes down, increase, and so on, like this. So what you can see that the current here is continuous. From Alpha, it will start increasing. Then from Pi, start decreasing or even before Pi, the peak before Pi, whatever it is, it will start decaying, then increase, then decaying, then increase, and so on. So what you can see it does not reach zero. Why? Because the inductance in this case is very high. If the inductance is very low and the current is assumed to be discontinuous, the waveform will be like this one. From Alpha to bet instead from Alpha to Pi, then from Pi to beta, it will be from Alpha to Pi, then from Pi two pi to two pi plus Alpha, then increase again and so on. I hope this is clear for you. Now, when does the cister conduct? It conducts from Alpha? Two Pi. So the current, if you look at this current I out, and this is I diet and I supply. So as you can see, from Alpha Phi Alpha two Pi, which is the conduction of i restor. So I out from KCL at this point is equal to I diete plus Iorstor. Okay, so the outt current in the first period during the conduction of istorT one is turned off. So I out is equal to ICstor equal to I supply. So as you can see, IO, which is this part from here to here, this part is a waveform of our sirestor like this from here to here, Isistor or I supply, they are similar to each other. Then it repeats itself every two Pi, like this during the conduction of the sirest. Now, what about the second one, which is the diet. Okay? So when this soy restor is turned off, the current of IO is similar to I diet. Iot goes from inductance to diet like this in the same direction. So I out will be equal to I diet as this soistor is off, so I so restor is equal to zero. Okay, so during the other part, which is this sport from here to here, the decaying part and from here to here, all of this is our diet. You can see here this sport is similar to this free wheeling diet. And this sport is similar to the san. Okay. And the cycle repeats itself. Okay? So if you can see this two waveform, if you get the summation of these two waveforms at any instant, this one plus, this one or this one plus, this one, this one plus, this one, all of this will give you the I out. So I hope that the waveforms are clear for you and now you understand where did we get all of this, Okay? Now, as you can see here, this is very important because we will use this when we get the wave, get the equations of the current. So as you can see here four at Alpha, at Alpha, the current is not zero. It has a certain value. This value is called I node. So we have our current in general. The general current is called I node. I small node. This is the general current. All of this waveform, as you can see here. When I say I capital node, this means the current at Alpha at this point. Okay, the point at which we will start charging the inductance again or increasing the current again, which is similar to this point, similar to this point. All of these three every two Pi are the same value of current I node. Great. Then we have another current at Pi at which we will start this charging at I node one. So we have I capital node one. This is the current at Pi. So this is a general current value of current at Alpha value of current at Pi. Why this is important? Because we will use this as our boundaries in order to get the constant A and other values. Now let's look at this equation. We have here the first mode. First mode, which is the conduction of the restore mode number one from Alpha to Pi or from here to here. Remember, the initial current here I node till I node one. So we will express the current like this for the conduction mode. The voltage equation will be like this. This one will become a short circuit to the supply. So if we apply KVL in this loop, you will find that V supply, which is V max sine omega t is equal to R, R multi blood by I plus L di over dt, xs, so this is our equation during the induction mood. Now, if you would like to solve this equation, it will be I node will be equal to V max over the sin omegative minus y plus A E minus R over L T. Now, this one is the same equation that we talked about before. Now, if you are not familiar, what does this mean? You can see E to the power negative T over tau. Okay. And since the TO is L over R, T divided by L over R, it will be E to power negative T over L. Okay, you can see negative R, T over L R over L to blood by T, which is RT over L. Okay? So this equation, similar to this one, similar to this one to avoid any confusion. This is similar as negative OMI get E over OIG L. Okay? All of these are similar to each other. So as you can see, it's consisting of two parts. The current is consisting of the forced plus natural, or we can say it's consisting of the four, the steady state value plus the transient response. All of this you will find in the course of electric circuits. Okay, so now let's continue. Uh huh. What I would like to get is the value of A. So how can I get the constant A by using the initial conditions? So as you can see, e at Alpha at Omega E equal to Alpha. The value of current is equal to Ital node, right from here. So at Omegat equal Alpha, I node will be I node or you can say that at time equal to Alpha divided by Omega, all of this similar to each other. Okay, like this. So what we are going to do is that I'm going to say I node equal to I capital and take this time and substitute it here and take this omega equal Alpha and substitute it here. You will get the value of A, which will be like this. Then if I take this A and substitute it again here, you will get I node will be this equation, this large equation as a function of I node, which is initial kind. Okay. Great. Let's read this. So this is representing what this is representing our equation of the current from here to here, which is the same as the equation of the current of ISI restor or eye supply from here to here. Okay? This is from Alpha to Pi. Okay. Now for the second mode of operation, which is a free wheeling mode, you can see free wheeling mode, why we call it free wheeling mode? Because our free wheeling tide is conducting during this period instead of the S restor. So it will start from Pi to two Pi plus Alpha. You can see here from Pi to two Pi plus Alpha, from here to here, right? Okay. So from here to here, what is the current equation? The current equation will be like this will be R node plus LD RDT equal to zero, or it will be like this I node will be A, it is about negative L T. Which means it is a decaying exponential. At time equal to infinity, the current will be equal to zero. At a very, very long time, the current will finally will go to zero. That's why we say zero equal to this function. After a very long time. Now, what we would like to get is a constant A, again, for this decaying exponential. So we have the initial condition at time at omega T. Equal to Pi. At this point, we have current equal to I node one, so I node, equal to I node one, right? So when we substitute this here like this, as you can see, you will get value of A, which will be like this. So we will substitute this one here. You will get the function of the current. This is a function of the current from here to here, which represents also the value of the free wing diet from here to here. Okay? Okay, now, the question is, how can we how can we get all node one? How can I get I node one? We can get I node one, which is this one here from the previous equation. So you can see that this is the first equation, right? This is the first one for conduction mode. Conduction mode using the restore. And from here to here, number two, using the free wheeling diet. So as you can see, between these two, there is a common point, I node one. So in order to get I node one required in this equation, what I'm going to do is that I will go to this equation for this sport and substitute with Omega t equal to Pi and this equation, it will give us I node will give us the value of I node one, I node one. So we get I node one, and we can use this in this equation for the second mode of operation. Now, what if I need the average? Now, all of this, we obtained the value of current. Voltage is very easy. One of our T, integration from zero to T 0-2 Pi. However, our function exist from Alpha to Pi only. So it will be from Alpha to Pi, V max sine omega T. So it will be the same equation that we have obtained various times before, right? Okay, so let's see the leaders. Then we need the average current. Average current will be this average voltage divided by R, we out average over R. Okay, because we said before that the average voltage across the inductor is equal to zero. So we out average will be voltage across R. So the average over R will give us I average. Now if we need the RMS it will be very easy root square 12 Pi, integration from Alpha two Pi for the function square, and if we need the root mean square current, it will be VRMS divided by Z or the total imbedance for the equation, or you can simply get the square root of 1/2 pi for the current equation all square. But remember that you will divide this into two integrations because we have from here to here and from here to here, two different functions. Okay. Now, let's continue. The last thing I would like to say in this long lesson is that if you have a highly inductive lute, what does highly inductive mean or heavily inductive loud? It means N loud with a very large inductance. So if we have a loud with a large inductance, large inductance, the variation in current or Didity will be very small. So the current will be continuous like this wavef you can see it is continuous, because it doesn't go to zero, however, if this inductance becomes very, very large, which we call heavily inductive load or highly inductive, very large inductance, then the current will be treated as constant. It will not have this variation. It will be like this. I but will be just a straight line due to the presence of large inductance. This is similar to what's similar to the capacitance. So if you have a capacitance parallel to R loud or RL loud, whatever it is, the larger the capacitance, the lower the variation of voltage. So if we have a large capacitance, the voltage can be treated as constant. The out will be almost constant due to large capacitances. Now, the same idea when we have a very large inductance, the current can be treated as a constant value. Okay, so I hope this lesson was clear and you gain no more knowledge about the continuous and discontinuous mode of operations. 37. Example 8: Now let's have an example on the three phase fully controlled bridge. I have a phase bridge fully controlled directivi connected to an R up 25 Ms. The source is 380 volt. And since we say 380 volt, as we learn it a lot and we said before, line to line, RMS, sect hertz. The phase control angle is 80 degrees, which means here, Alpha, is equal to what is greater than greater than 60 degrees, or equal to 80 degrees in this example. So since Alpha is greater than 60 degrees and we have a pure resistive lute, then we will treat it as if it is highly inductive lute with a free wheeling diet. They are exactly the same. So find let's delete this find average load voltage, average ode current, average allude bower, average thyrostor current, and B criers voltage. This is very easy and straightforward. Number one, draw the wave forms as we learn it for a resistive loud with Alpha greater than 60 degrees, we will not have any negative part, and we will have mode off mode like this. Okay, so the average allude voltage will be equal to what we learned from before. It will be integration from Alpha for AV from Alpha plus 3,250 degrees. Like, Alpha plus 3,250 degrees for the function, right? V maxiline. We have 380 volt, which is V line to line or mas. So we will take this value omtolite by root two, like, root two metabolite by 380 degrees. And we converted Alpha from degrees to radiant by multiplying by Pi over 180. In order to convert from radiance to two degrees. Let's say this if you don't know, to convert from radiance two degrees. What are you going to do? You take radiance and multiply Pi 180 over Pi. If you would like to convert from degrees to radiance, take degrees, multiply by Poi over 180. That's what we exactly did. We took the Pi over 180 degrees and multiplied it by 80 degrees. So it will give us 120 volt as an average voltage. Great. Average lute current will be 120 divided by resistance, 25, right, 4.8 and pairs. Average louder, since we say average loud power, it will be V average out with average, multiplate by I out with average or I with average square multiplied by R, right like this, square multiplied by R. For average thyrestor current, we said that we have a formula for Tirestor I will conduct two times from here to here from Alpha plus 30 to 150. So it will be this equation that we took about before, so it will be Alpha 80/180, multi blot by Pi and radians, so it will give us 1.07 and pairs. For peak reverse voltage, we said that in a bridge rectifier controlled controlled or whatever it is, half controlled that we will discuss. All of this is VMAX line, right? So it will be VMAX lin root two, multilat by 180, which is 537.4 volt. 38. Example 9: Now let's have another example on the three phase bridge controlled rectifier. We have this bridge controlled 41 to 385 and so on exactly the same circuit. But this time our load is a resistance of ten oms. You can see inductance L equal to one hinary and there is a packing math 200 volt. Now look carefully here. Since our inductance equal to one hinary, one hinary is a very, very large value, which means that the current will be a constant value or this load is a highly inductive load. Okay? So the first step, the input supply is 400 volt, 550 hertz for 100 volt, line to line RMS, as we always know, find the firing angle if the average output current is ten and pairs. Find the input power factor in this case and assume that the inductance is large, as you can see one henary large enough to ensure constant current in the load, which means highly inductive load. Number one, we have a fully controlled bridge, and we have a highly inductive load. So do we have any free wheeling light? No. So if Alpha, if Alpha greater than 60 degrees or Alpha less than 60 degrees, it doesn't matter. Greater than the same equation. As you can see here, you can see continuous, right? This is if Alpha less than 60 degrees, if it's greater than six degrees, then it will go to the negative part as we learned before. Now let's assume both of them are the same equations, right? They will conduct from where V AVC will conduct from Alpha plus 30 until the next firing angle, which is at Alpha plus 90 degrees. So it'll be like V average will be 6/2 Pi, integration from Pi over six plus Alpha, which is 30 plus Alpha until Pi over two plus Alpha, which is 90 plus Alpha. Difference between them is 60 degrees. For what for V max line, sine omegty plus five or six, which is this equation for this case and this one. V average. Okay? So we will subsiute VMAX line will be 400 volt, multiplied by root two. Now what about average? So we have VMAX line 400 volt. It will be 400 volt, multiplied by root two, right for VMAX line. Now, what about so we need Alpha, right? We have this and we need V average. Now, V out average. Look carefully here, VO average. But we have Iout average, which is ten and pair, we have out, which is constant value equal to ten pairs, right? Okay, so this is our I average. Okay. Now, we would like V average. How can I get V average? If you look at here, you will find that V average, V out average is equal to what equal to I average, multiplied by resistance, plus the back in MF. Remember, average voltage across any inductance is equal to zero. So VO average will be I average, multiplied by resistance, plus packing in M from KVL here. So it will be like this. V average will be Iot average by R plus E. Or we can say that if you don't remember this, we said before that, if we have a backing F, it will I average will be VO average minus E over R. That we learned before. B, we have E, we have R given in the problem, we have It average ten amps so you can get Vout average. As you can see here, we average minus this equation, we can get average equal to 300 volts. Take that 300 volt and substitute it here, you can get the firing angle Alpha like this. Equal 2.982 radiants. We max line 400 multiplied by root two as we jet. We need the input power factor. How can I get power factor if you don't remember the same as we did before? Power factor equal to power, active power, divod apparent power. Now, what do active power here? Active power is the power will be equal to, which is B input will be equal to power output, reaching our load. Now, if you look at our load here, we have a resistance, and we have back in Math. We have DC current. So power output will be Iout average. Multibloid by resistance, plus I output average square, multi blod by resistance, plus E, multi blot by I abut average. Why do we use average equations? Because, B. Number one, our output is a constant value like this, right? Constant line. So it is DC. So to get the power, we will use average equations. So power consumed by resistance is I square or I average square, multi blood by resistance plus power absorbed by packing F. It will be E, the DC value, multi blood by average current. This will give us power absorbed by the resistance. Then we need the supply power. So in order to get S or the apparent power, it will be we have two options root three, multiplied by V line to line RMS, multiplied by I line to line RMS or three V phase RMS, I phase RMS. V line to line RMS is 400 volt given in the problem. Right? I line to line RMS. If you don't remember, it is exactly in this circuit, the supply line to line supply, root 2/3 multiplied by abut RMS. I abut RMS is a 10:00 A.M. Pair because 10:00 A.M. Pair is a constant current, constant current, which means I abut equal to I average, equal to RMS, equal to 10:00 A.M. Pairs. By using these equations, you will get perfect. So let's look at number one I average square R plus EI output average gives us 3,000 to, and I supply root 2/3 of ten pairs will give us this. You can get or apparent power root three, supply RMS, 400 volt given, and I suppliMs which is this one. The same steps that I have shown you in the previous slide. By dividing these two ratios, you will get power factor 0.53 and it is lagging power factor. Why lagging power factor? Because the current is lagging from the current is lagging from the voltage. Okay? 39. Half-Wave Controlled Rectifiers – RLE Load: Hi, and welcome everyone. In this lesson, we will raise the level a little bit and discuss another circuit. But this time, we have a half wave control dictifier with an RLE load. Instead of R, we have RL and not RL only, but RLE. So as you can see here, we have our supply. We have our CI restor, which is controlled, and we have our load which is RLE, resistance, inductance and BE math, which is considered as a battery. Now, where does the back in math come from? This backing math can be presented inside a load, such as a battery, or it can be represented as a DC motor. Okay, both of them are correct. Since we are converting AC into DC, which we will be connected to DC motor or can be connected to a battery to be charged. So what will happen in this case, let's understand what will be the effect of having back EMF on our circuit. So let's look carefully at the supplies. We have supply, Max, sinnO meg E, as we discussed before, and look carefully we have EMF E. So first, let's consider that E does not exist, as if it does not exist. Now, when does this sistor turn on? When does as turn on? We have two conditions. Number one, VT must be greater than zero, or the Cistor must be forward the highest. Number two, we need to provide the Alpha or the firing angle or the delay angle. Okay? Now let's look at the first condition when we don't have E. So starting from zero, any value of V supply greater than zero will make Vsistor greater than Z right, neglect this one. So when V supply, let's say 1 volt, it will be here plus -1 volt. So this sister will be forward biased, right? Since we don't have any since we don't have any other element except the supply. So positive and negative. Okay. So starting from here, if we give the firing angle at any instant, it will start induction, right? Now, let's say we have BEMth. What will happen in this case? Let's look carefully at the sec. Let's say the back EMF is equal two plus minus like this. Let's say five volt. Okay, it's a value of five volt. Now we have a supply like this, providing voltage in this direction or providing current in this direction, and we have our supplies that will provide a current in the opposite direction. So these two supplies max on Omegaty during the postive cycle during this part or the postive cycle is opposite to the BE math. So in order to have VSI restor or let's look at VSI restor voltage across the ester will be V supply minus E, voltage coming from E and voltage coming from supply, are opposing each other. Now, if the V supply less than less than E, let's say V supply is two volt and E is five volt, it means that the voltage across the cistor is negative three volt, which means that our crestor is reverse biased. So even if we provide even if we provide a firing angle, the cistor will not be turned on. Why? Because the voltage across it is negative. It is reverse pised. So that's why if you look carefully here at this curve, you can see E, which is a constant value, and we have supply. So you can see there is an instant. This point the intersection between here, intersection between E and V max sine omegaty a certain instant, a certain angle called seta one or Sta one or Sta one. And we have here Stat two angles at which the sine omegatyVMx and omegaty intersect with E. So these two points are VMX sine Omega E equal to its value of the supply will be equal to the value E. So we need to provide firing angle Alpha must be greater than seta one. Now, why is this? Because if you provide before Sta one, what will happen in this case? In this case, you will find that supply is less than E. So the cistor is reverse biased. So we have to provide the firing angle after S one or Alpha greater than S one. Okay? As you can see here, Alpha provided at an instant after Seda one, okay? Okay, so that's why it's important to identify Seda one in order to provide the firing angle must be greater than this value in order to as a crestor or in order to operate the crestor or turn on the Crestor, okay? So you can see that E forces us to be operating starting from a later point, okay? Later, not just early at any time as we would like, but we have to provide the firing angle after when the supply is greater than E. Okay, so that the Visa arrestor becomes a postive value. Okay? So that is the first thing. So we have V supply, we have E, and we provide the firing angle at the instant when V supply greater than E. So the iistor will be turned on. Now let's look at VO. So ciistor we provide the firing angle here at this instant. So VO will go from zero to this value and start conduction like this. So you can see like this, it goes from zero to this value and start like this. However, it doesn't start from zero as I just said, starts from E. Y is this okay. If you look at VO here, VO is equal to IR, plus L di over dt plus E, right? Plus E has the similarity as V. O is voltage drop across R plus L dI over dt plus E or the PACF, right? So when this so restor is off, Okay. When it is off, what is the value of the current in the circuit current will be equal to zero. So this one will be equal to zero. This one will be also zero, so Vout will be equal to the BEMF. That's why if you look carefully here at before firing angle at this instant, you will find that Vout is equal to value of E. Here, when the Cyrus stor is off. Okay? Now, if we continue, so it starts from here. When we give the firing angle, it will start going from E to V supply, like so it goes from here to V supply. Now, it will continue to conduct right as before. Okay? Even if we have this point at which the Sirtor is reverse biased. So after S the two, S the two here, what will happen? It should be turned off, right, because supply is now less than E. So Vistor is reverse pised. However, due to the stored energy inside the inductor, it will start giving its current to the cistor and keep it in the conduction mode. So it will extend like this until the extinction angle bit. So you can see it should at this point, say that too, it should be turned off because the pistor must be or the istor in this case is reverse piased. That's why due to the presence of energy stored, it will continue to conduct. So it will continue like this until the extinction angle beta or Peter. Okay? Then what will happen? I will go. The sores will be turned off. After this inductance finishes its energy stored, it will be turned off and VO will return back to value of E, so it will be turned off and go up to value of E. Until the next firing angle, it will be equal to V out, till the extinction angle and turn it off and so on. So when it is turned off, VO equal to E, the Pike MF and when it is turned off, VO will be equal to V supply. Turn it on off and off. Now, what about current? Current will be as I will start from Alpha to the extinction angle beta. You can see Alpha like this until Peter, from here to here, from Alpha to Beta. So what about the cystor waveform? If we look carefully when cystor is conducting from here to here from this point to this point, it will become a short circuit, right? That's why from here to here, it is a short circuit. Okay? Okay. Now let's continue. When it is turned off, let's look at it. So from KVL, when the story is off, it will be an open circuit. So if we apply KVL, then negative supply, negative supply, plus VSI restor, VT, and go like this, plus E, would be equal to zero because R plus LD over dt is equal to zero, so we neglect, so there is no volt drop here. Okay? So what about Vsstor? So VSI restor value will be supply minus E. V supply minus E. So during this part when it's turned off from here to here, from here, its value will B, which is this waveform is V supply minus E or V max sine omega E minus E. V max sine, omegati minus E. Now, at the instant of the firing angle here, Vmax sine Alpha minus E because we substitute with Alpha or Omega T equal to Alpha. Okay? Now, let's continue like this until it's turned off. Now let's look here. When it goes like this down till the sport, so let's write it. V Cistor will be equal to this value, which is the worst value. This value? What is it? It is negative V max, right? Minus E, minus E from here from this equation. So this representing the highest value in the opposite direction, which is V max, we can say negative V max plus E. So the peak reverse value, big reverse value, applied on the side restor, which is the worst reverse value is V max plus E. And the worst value, big forward value will be Vmax. Minus E, because at this point, supply will be Vmax, then minus E will give us voltage across the sistor. The peak forward voltage on this Cirstor is V max minus E, and the peak reverse voltage is V max plus E. Now, how can you do this? Just take V max line omegaty and subtract from E. You will have this wave form like this and this waveform like this. VMX plus E, worst value. And here you can see it is much lower because we have VMX sin omegty is a postive value, minus E, which is a postive value gives us a lower postive value. Here we have negative value of VMX sine omegatE minus E, which means negative, plus another negative gives us more negative. Okay, so I hope this clear for you. Okay? Now, we have to give the firing angle. As we said, when the supply value exceeds EMFE value, how can we do this? We have to get say the one by saying VMAX sine omegaty or equalizing or making VMAX sine omegaty equal to the BMF E. So during the on state, what will happen from Alpha to Beta? From Alpha to Beta, this will become a short circuit like this. So V output from QVL will be equal to V supply, which is V Maxine omegaty. Now, during the off state from Beta to two Pi plus Alpha from here to here, during the off state, the value will be equal to E because the voltage drop across R and L is equal to zero. So only the PackmF exists, so Vput will be equal to E. Now, what if I would like to get the load voltage V average, average, which we can also get the RMS using the same equation, however, with a square, as we learned before. So average is one over T, 1/2 Pi, integration from zero until two Pi until this point, two Pi, like this. Okay? So this from here to here representing one complete cycle. You can also you can also do the integration from Alpha. To Alpha plus two Pi. This period is also one complete cycle, one complete cycle of two Pi. So we can say this is much easier because we will just divide it into two integrations, integration from Alpha to Beta, from Alpha to Beta, here to here, for Vmax sine Omegaty like this from Alpha to Beta, Vmax, sine Omegaty and from this point, which is Bta to two Pi from Beda to two Pi plus Alpha, two Pi plus Alpha, not two Pi because we said one complete cycle as you see here, Beta to two Pi plus Alpha for the constant value of E for the supply E, D omegaty. So this is what we have learned and what we have done in all of the previous lessons. Now, let's assume if you have a Cistor voltage drop a constant value 0.7, volt when it is in the on state, then what we are going to do is that from QVL you can see that V O is equal to V supply minus Vs rest, right? Okay, so supply minus V Sisto. So you can see that if I would like to get, which will be available from here to here, right? So when I do the integration from Alpha Beta, from Alpha to Beta for the function, supply Vmax, sine Omega t minus V saresto D omegaty, right? So instead of having this, I divided it into two integrations, Alphabeta, Max, sine omigaty which is this one. Minus integration from Alpha to Beta four Vss, minus integration from Alpha to Beta, VSI. Okay. So these two when they are combined together will give us the value of V supply value of out in this region. Now when you are going to do the RMS, you have to do it using this don't divide it. You will have to do this Vmax and Omegaty square, 1/2 pi, and all under the square root. And of course, don't forget this part plus E square from here to here, okay? This is for the root mini square. Okay, let's just delete all of this. Now, for the current, now this current is very important current relation during conduction from here to here, right from Alpha to Beta, same as we did before. So we have when this one is conducting like this, and we assume, of course, here, let's get back this part when we have non ideal cistor when we consider that the crestor has a volt drop, if it does not have a volt drop, then this part will be equal to zero. Okay? This is just for your own knowledge. Now, let's say we are talking about an ideal circuit. Here, this one will be a short circuit like this. Okay. Now we have V Maxine omega t, and we have RLE like this. If we apply KVL, then supply, V maxine omegaty will be equal to R plus L di over dt, L di over dt plus the back math right from KVL. Now, as you can see, in order to write the current equation, we need two elements, the steady state current and the transient response. Or the forced response and natural response. Now, let's talk about first the steady state response. Now, we have two supplies that will force the presence of current. These two are V Maxine omega ty, and we have E. So we would like to get the effect of each of these carats. Okay. So how can we do this? Just wait. Let's write using this pencil. Okay. So we have here two supplies. So I steady state like this will be equal two. The two response, one from due to the supply Vmax omegaty, so it will be VMX Sine Omega T minus Seda or Phi, whatever the symbol you are going to use for the phase shift, divided by the right, as a magnitude, V max sine omega t E, which is a supply voltage minus the phase shift because current is lagging from voltage divided by Z or the total impedance. Now, this is the effect of just this supply, the effect of supply of supply, V max sine omegat. Now we need the effect of the other supply E. Now, as you see in this figure that E provide this current of the two IO. So if you assume that if you would like to get effect of E, we will just neglect Maxine omety as if it does not exist as if it is a short circuit, deactivate it. Okay? So we will have E and the resistance, right, because D E is a DC source. So inductance as if it is become a short circuit, right? When we have a DC source, inductance becomes a short circuit. So we have E, and we have R, right, and E provides current over the two I out. So the current ICD state due to the presence of the supply will be negative E over R. So we will add it here negative, EOR. Now again, why did we neglect inductance? Because inductance becomes a short circuit in the DC supply and negative sign because the supply gives the current of of the two I output. That's why we added a negative sign. However, our supply VMX sign omegaty gives us current in the same direction. In the same direction, that's why we made the sign as positive. So let's look at the steady set response, as you can see here. This part here, max over Z, sin omegat minus Y or Seda, whatever the simply are going to use, both of them representing ph shift, minus EOR. So this part representing what the steady state response. Now, what about the transient? Transient as we said before, it will be like this. It will be like this constant E to the power negative Omega T over Omega tau, right? This is a natural response when the supply NDE does not exist. Okay. So how can I get the natural response? So our circuit will be the steady state plus this one, okay? So let's write the equation and see how can we get it? Let's delete all of this and use the pencil again. That's right here. So we have our current equal to this value I Omega T equal to V max over Z, sine Omega T minus Y minus E overR. Plus A E minus omega T over omega t, right, steady state and transient. Now, we would like to get the constant A. So as you can see from this figure here or from the current equation, we will use the boundary. Okay? At omega t equal to Alpha at omegat equal to Alpha. The current will be equal to zero, right? So we will say zero will be equal to Vmax over z, Sine Omega T minus Y minus E over R, and plus A, A to Z power negative Alpha over Omega t. Okay? Okay. Now, as you can see, now, we would like to get the constant A. So from this equation, you can see that here like this. That's right here. Okay. So negative V max. So A will be E to the power Alpha over Omega T like this, multiploid by negative V max over Z, sine Omega T minus Y plus E overR. So what I did exactly is that I took this part to the other side, so it will be negative E max over that, sin omegaty minus oi not omegaty. Sorry, I will be Alpha, sine Alpha minus o. Alpha minus oi and negative E over R will go to the other side and become plus E overR we have A E minus Alpha over Omegaty. So when we divide by the other side, take this to the other side, and negative will become positive tocmeE overElpha over omegat. Now, let's delete the sport here. Okay, all of this. So just so we can understand. Okay, this is a final equation, okay? So I'm just explaining where did you get this? Okay, we can keep it like this and use the pencil okay from here. Okay, let's go. So I Omega T will be this part as it is. V max over the sine omega t. This is a steady state. It will be like this, and we remember that we have plus A E to Zibo negative Omega over omega L. Now, let's substitute this A here. Okay, so it will be plus A, which is E to Z power Alpha over Omega t like this multiplied by negative V max over Z, Omega T sine Alpha Alpha minus pi plus E over R. This is a constant multiplied by the exponential. Here this exponential, negative Omega T over Omega t. Now, as you can see that this part negative V max over that, sine Alpha minus Y, this part here plus ER, minus ER. Now what we did is that we take a negative as a common factor outside. So this one will be positive and this one will be negative, like this take negative as a common factor, positive and negative. Okay? Now, we have two exponentials multiplied by each other. EO Z power Alpha or omegata and E over negative omegat Omegato. So if we would like to combine them together, it will be E to Z power, Omegata as it is. Alpha minus Omegaty Alpha, minus omega T. Now, if we take a negative as a common factor like this, it will be omegati minus Omega t minus Alpha, divided by Omega tau. Now let's look at the equations that we have. You can see negative Omega t minus f divided by omega Omega as it is here, and the tau is L over R. So it will be L over R, like this. L over. So I think now you understand where did I get this long equation. You can see when we added the E, it became more complex than before. Okay? So let's read this. Okay. So finally, we have the steady state and the transient response. So this is the current equation for this circuit from here to here. Okay. Now, what we would like to get, again, we need also Beta, the value of Beta. We know that at Omega T equal to Beta or Peta, the value of current will be equal to zero, right? So by using a trial and error or using our calculator, we will get Beta required in the circuit. Now why we need Beta because we need the integration of the output from here to here. As you remember here, okay? That's why we need the current equation in order to get this one. Okay. Now we to average put current. We know that average voltage across the inductance is equal to zero, right? So with average, V with average will be equal to the average here, which is I average multiplod by resistance. Average volt drop across the inductance plus average voltage across E, since E is a constant value, so its average is the same value which is E. So if I would like I average, it will be V average minus E divided by R, as you can see here. Okay? Now, what about the cyst ratings that we have talked before? We said that the maximum forward value, peak forward value is V max minus E and B inverse value is V max plus E. All of these weeks blend before. So I hope in this lesson, you understand how the RLE circuit works and how we are going to use in our analysis, okay? 40. Example 10: Now let's have the first example on the three phase, half controlled bridge rict fire. We have this circuit. We have 120 volt supply RMS line to line, gives an average of current of 10:00 A.M. Pair to allowed resistance of five arms. Now, determine the firing angle of the circuit. Number one, we have a half controlled right. Now, we would like to know average. Now, is a highly inductive load or the loud, will it change anything? Will the load change anything or being resistive or highly inductive? No, it will be exactly out average. So VO average will not change. Okay. Why will not it change? Because if you remember that at Alpha, less than 60 degrees and Alpha greater than 60 degrees, we had the same V average. So let's get back. You can see here, this is our V average win, highly inductive load greater than 60. If you get back here, the same equation for Alpha less than 60 degrees. And you can see we don't have any negative part, and here and here, we don't have any negative part. So the resistance or the output average in resistance will be exactly the same as highly inductive root. Okay? So we can use the same equation. Okay, let's lead. So we have average. What will be the average?V average will be I average multiplied by R very easy. Vmax line will be V supply RMS multiplied by root two, ten multi blade by five for V average and three V max line, root 220 volt, root 220 volt. So by using this, you will get cosine Alpha and you will get firing angle equal to 112 degrees. Okay? 41. Example 11: Now let's have another example. In this example, we have a DCU requires a control of its voltage from the maximum value to a quarter of it. So we would like to make average max, drop the two quarter of it by using this half controlled the phase bridge. How can I do this by controlling the firing angle Alpha? Now, we would like the rating required for the free wheeling dt. Here, if the lute current is a constant current of 20 ambers. So what we learn from this number one, we have a highly inductive loot, right? Since the loot current is constant, so Iout, equal to Iout average, equal to Iout RMS, equal to 20 and bears, right? Okay. Now, I know that we didn't discuss the free wing dt with a free winning dt for half control but you will find that even if you don't know, you can do it. You can draw it and you can know the effect of the freewing dit. Okay? Now, first, we have average in our circuit with a free wheeling diet and without a free wheeling diet. We don't have any negative part. So when does the free wheeling diet operate? Remember, remember. That when Alpha greater than 60 degrees, when Alpha greater than 60 degrees. We had a portion of the wave in which we have zero voltage. This portion of the wave instead of t1d4 or t3d6, t5d2 causing the short circuit, freewheeling dot will act instead. It will go instead and start operating instead of them. Okay. So anyway, we don't know Alpha, we use the general equation to get Alpha. The average will be equal to what? Three V max line over two pi one plus cosine Alpha. Remember, we need to control our voltage from maximum value to quarter of it. So we have average value equal to this. Now, my question to you, what is the maximum value from this equation? If you say three V max line over two Pi, then you are wrong. Why is this? Because Alpha Alpha can be 0-180 degrees. If you substitute with Alpha equal to zero, what will happen when Alpha becomes zero, cosine zero equal to one. So one plus one gives us two. So our maximum value will be three Vmaxoline over buy, right? So maximum value occurs at firing angle equal to zero degrees. V max three Vmaxolin over. Now we would like to control this value of the abut and make it quarter of it, 1/4 of this value. So what I'm going to do is that I will say 1/43 V maxiline over Y equal to this equation, right? This is our general equation, V average, and I would like to make it equal quarter of maximum value. That is what we need. So from this equation, we will get Alpha. 120 degrees. So what we learn from this when Alpha rater than 60 degrees like this, so we will have the wave form like this. Mm hmm, zero. Mm hmm. Like this, right? So during this period, the free wheeling diet will operate. Now, first, the rating of free wheeling diet. Number one, peak reverse volt peak forward volte equal B, V maxoline. However, we don't have max line given in the problem, so we'll keep it like this. We need I free wheeling diet average. Okay, you don't know this, but let's see the wave form first. We have this waveform like this, then zero, then like this, then zero like this. During this zero portion, instead of having T one and DT one d four that will form the free wheeling dt action, when we have one dite it will be the one that will operate. So during this zero voltages free wheeling dt, free wheeling dt, free weling dt, Okay, now, how can we get the voltage or how to get the average or whatever? Now, first step, let's delete this. Okay? Now let's look at here. So let's start from Alpha plus 30 until this point, which is 210. So from Alpha plus 30 until 210. Now, the freewheeling diet will start from here from this point, which is 210 degrees, right, at which Dfour will conduct or should have been conducted without freewheeling diet. So 210 Okay, until this point, this point is 270, right? Now, why 270 because it will be Alpha. Plus 150 degrees. Remember, here we have Alpha plus 30. After 120 degrees, we have T three, T one, T three. Alpha plus 30 plus 200 plus 120 Alpha plus 30 plus 120 gives us what gives us Alpha Alpha plus 30 plus 120 Alpha plus 150 degrees. So we have Alpha plus 150 degrees and we have 210. This period. Now, how many times our free wheeling it will conduct in one cycle three times one, two, and three, right? Because this short circuit happens this zero voltage happens three times in one cycle. So it will be like this. I free wheeling dt, okay? Average will be three times of Alpha plus 15, 150, sorry, minus the beginning which is 210, right? 210, like this. Divided by the whole period, multiplied by I with average, right? Okay, now, 150 -110 gives us negative 60 degrees. Okay? Negative the spot, negative 60 degrees, which is exactly equal to what equal to Pi over three. Okay. Now we have Alpha minus Pi verse, and we have 3/360. Three son 60 equal three or three ozonO over 120. Okay? Whatever it is, you can keep it as it is. You can write it like this. Three Alpha minus Pi divided by two pi in ingredients, okay, instead of degrees, I out. So you can this equation three, three multiplied by Alpha plus Pi over six, minus Pi over two. Pi over six minus Pi over two. By over 630 degrees, minus Pi over two, 90 degrees, gives us negative 60, which is exactly equal to negative Pi over three. Okay, so it is exactly the same equation like this one. Okay. However, instead of using this part from here to here, I have used when I wrote this equation, I have used from here to here. It is the same. Anyone will give you the three minds. Now let's look at this equation. We have three Alpha minus Pi over three to Pi. Now, what I'm going to need is the rating of the diet, free winning dot. I would like to find the worst case. When is the worst case, the worst case will be when Alpha is highest or maximum. Maximum value of Alpha is Pi. Right? So if I substitute with Pi here, it will be equal to three Pi minus Pi versary divided by two Pi. Now, Pi minus bi versit is two pi versary. Multi blood by three, three will go three, so we'll have two Pi over two pi, which means unity, which means I average will be I abut and even if you take the square root of one, it will be one, so it will be also I output. The same equation, but under the square root. Okay? You have to know that I in this example, take this. Where did I obtain this from this region? This region, which is here. This written one, Okay, a smiley face. Wow. For this one root, where did we get this from here, from here, from here to here, which is 150, which is Alpha Alpha plus 30 degrees, Alpha plus 30 degrees -90 or Pi over two. Why 150? Because Alpha 120, 120 plus 30 gives us 150. That's all. Okay. I hope this example was clear for you and you understand now what will happen if we have a free wheeling guide in our half bridge rectifier. 42. Full-Wave Bridge Uncontrolled Rectifier – R Load: Hi, and welcome everyone. In this lesson, we will to pet the full wave, bridge uncontrolled rectifier an R with an R load. So in the previous lessons, we discussed the half wave rectifiers, all of the types of the half wave rectifiers. Now, this time, we need to know what is the full wave rectifier. And specifically in this lesson, the bridge uncontrolled rectifier. So starting here with the circuit, you can see we have our AC supply and we would like to convert this AC into DC at the but across an R loot. Now, since we are saying uncontrolled rectifier, it means that this circuit only only consists of dites. So as you can see here, in order to have a full wave rectifier, we have one, two, three, and four dites. This circuit, as you can see here, is a positive terminal or the first terminal is connected to the first two dites, as you can see here. Second terminal of the supply is connected to the middle of these two other dites. So as you can see here, we have this too, right? So by using these four dites, we will be able to convert AC into DC. Now, first, let's understand what does full wave mean? So all of the previous lessons, all of the previous lessons regarding half wave, it means that we were only we were only utilizing half of this wave. So if you remember in the half wave rectifier, we had wave like this VO as a function of Omega T was like this. Then zero, then like this. So only during this sine wave, we only may use or utilize it half of the wave, only half of the wave. Now, the full wave rectifier idea is to convert both of the positive and negative cycle into rectification or rectify both of the postive and negative cycles. So as you can see, instead of having AC positive negative, we will have the outs like the positive another positive. So this will lead to utilization of both of the positive and negative cycles, and at the same time, we will be able to increase VDC or the average output voltage of the circuit. Let's delete all of this. So this is what it means Pi full wave full wave, it means that we are utilizing all of the signal. In the half wave we only utilizes half of this signal. Number two, the bridge, the bridge rectifier, it means that we are talking about this circuit, which consisting of four dites. Now, we will learn in another lesson about the center tab transformer circuit, which is a full wave center tab. In this circuit, we only use two dites in order to convert or to have full wave uncontrolled circuit. Now let's understand how does this circuit work? So first, the bridge rectifier is one of the most commonly used circuit for DC power supplies. Now, as you can see, it consisting of four dies, D one, D two, D three and D four. They all of them form a bridge. So during the positive input half cycle, when this supply has this polarity, during this positive part like this in this region, you'll find the blarty is positive, negative, right. So what you can see that the current will go like this out from the supply like this. So of course, due to this polarity, this D one will be forward paced, and the die D two will also be forward biased. So the current will go like this. I will go from here like this. And do you think it will go upward or downward? It can only go upward like this? Okay, so it goes like this. Then it goes like this. This will be for a reverse biased, so it will continue like this and go through the loud in this direction. Remember, this direction is very, very important. Now, it continues like this and go like this. Now, it goes from positive like this because it would like to go to the negative polarity. So as you can see, the negative polarity is connected here. Like this. So it will go like this and go through D two like this and through the supply like this. Okay. So what you can see during the post of cycle, it will go through D one through the rule loud and go back through D two and back to the negative terminal of the supply. So during the postive half cycle, D one and D two will be forward piased, D three and D four will be reverse pis. So it will go through D one and D two, and these three will D four will not allow any current. So as you can see, when this one, you can see from KVL, you will find that VO is equal to V supply, assuming, of course, we are dealing with ideal dials. So as you can see, V supply and VO during the first part, which is D one and D two on Now, what about the negative cycle? For the negative cycle, the reversal will happen. Let's delete all of this. Now, of course, you can see the current will flow reducing a volt drop across the loud resistance out. Now, during the negative cycle, D three and D four. Now, how it will look like, the polarity of the supply will be like this, will be negative. So the positive will be here and negative will be here. So how does the current will go? It will go out from the supply like this. In this direction I supply will go like this through this terminal, like this. Now, it will go down or up down, it can't go down. Why? Because this die does not allow the current except going like this in this direction. So this one will be reverse pised. So the current will only can go through these three like this and go like this through the out. The first thing that you will notice here is that the current is also in the same direction. So I out in the positive cycle and in negative cycle has the same direction. So you can see that I out in the positive and I out in the negative has the same value with the same direction, the same direction, similar to VO o also positive and positive. Now let's continue. So it will go like this, through R three, now it will go like this. Now, this is the terminal it came from a postive terminal. I would like to go to this negative terminal. So it can go through D one or D four. Now, of course, it will go through D four like this, go like this, through D four and get back to this supply like this. So what you can see is that D one is also reverse pased. So only D three and D four are forward pased and D one and D two are reverse biased. So what we can see here is that supply like this in the positive direction like this, I supply in the positive direction, and I supply in negative direction. That's why this current, as you can see here, supply like this and I supply has positive direction and negative direction. Why? Because as you can see, positive and then during negative cycle, it will reverse its direction like this. That's why it is in negative portion of the supply. Now, number two, let's delete all of this. Okay. Now, I O, Iout has the same direction positive and postive because during positive cycle in this direction like this and during the negative cycle, it will go through these three and go through the same direction. That's why the waveform, as you can see here, positive positive. What about V O? If you apply KVL during positive cycle and during the negative cycle, you will find that VO will be equal to like this and like this. Similar to supply, but with a positive sign. If you apply VL like this through D one, like this during the post of cycle, like this, you will find that V out will be equal to V supply during post of cycle. And during the negative cycle, you will find like this. It will go through here, it will go through here like this, through D three, Like this and continue like this through D four. If you apply KVL in this big loop, you'll find that V out is equal to negative V supply. That's why the polarity is reversed as you can see here. So what you can see is that we rectified two signals, the out is positive positive. There is no negative part. Okay. Now, what else as you can see, this is the explanation for all of these waveforms. Now the current will flow, as you can see, through the loud resistance in the same direction during positive AC and negative both AC input half cycles, positive and negative. Now, if you don't understand this KVL, you can do another thing, which is Vuts equal to I output multiplied by R. And since the output current has the same direction during positive cycle and negative cycle, then the voltage will also have positive cycle and positive cycle. Okay, now let's understand more about this circuit. So you can see this previous one, this one which we have seen in the previous lesson, this one in the previous slide, D one, D two, D three and D four. Now, the same idea, you will find it here. This shape is similar to this one. There is no difference between them. For example, you can see here during postive cycle, it will go through D one, then through the loot, then through D two, then back to the negative supply. So as you can see D one, supply resistance, then D two, then back to the negative. So the direction of current is in this direction. Now, during the negative cycle, it will go like this through D three, loud, D four, and back to the supply. What you can see, again, the same direction of current. So this circuit is exactly the same circuit here. Okay. Now, supply, V max, somgaT supply, VO will be always rectified. You can see pulsating DC, pulsating DC waveform, post positive postive postive. There is no negative here. Now, what about the voltage across dites D one and D two. So when it is conducting, so during the postive cycle, it is conducting, it is a short circuit during postive cycle. I will go like this. Then go through the resistance, then D two, Okay. And this two will be open circuit. These two are open circuit. Now, what you can see here is that D one, short circuit and D two short circuit during the post of cycle, D one and D two short circuit, zero voltage. Okay? Again, during post of cycle, zero voltage. For D three and D four during the negative they will be short circuit during negative cycle. During the negative cycle here and here, they will be short circuit and short circuit. Now, we would like to get during when D one and D two, the two dites are reverse biased. Let's first look at D three and D four before D one and D two, all of them are the same idea. Now, during the positive portion, D one and D two short circuit, as you can see here, and D three and D four are open circuit. So it will be positive negative, and this one is a short circuit like this. Now, V output is equal to what equal to V supply. Now, if we apply KVL in this small loop, you will find that negative V d three. Like this, minus V out equal to zero. And this is small loop here. So you'll find that V D three, the voltage across the dot is negative V out, right? That's why you will see that when during the post of cycle, this part, this is V out, right? So voltage voltage across D three and D four, D three and D four, they are the similar. If you apply EVL here, you will get the same equation, okay? Now, during the positive cycle here, D three and D four are open circuit or reverse pies. So the voltage across them is negative V out. So we have V out like this. So their voltage will be like this, the opposite. How did we get it again from this KVL here? By putting the correctularity and doing QVL, you'll find that V across the dite is negative Vo. If you do this here or here, it will be the same VD four equal to VD three when they are reversibas equal to negative Vo. Now, what about during negative cycle? During negative cycle, current will go from here like this through these three and through the resistance, then D four, then to the lute. Now let's look at V out, and let's look at the volt. So D one and D two are reversifized, right? Let's put the polarity. Here the direction goes downward like this and from here to here, current like this, so it will be positive, negative, the same polarity of the of the diet. So D two let's apply QVL in this loop in this loop here like this. So negative V out, negative V out, and if we go like this, minus VD two equal to zero. So the voltage of the diet, again, is equal to negative V out. So during this part at which it is reverse pause during the negative portion, you will find it will be equal to V out like this, so it will be reversed like this. Similar to VD three and VD four. Okay? And as you can see, the negative part, the maximum value V max, Okay, V max. This is a peak reverse voltage applied on each dite. V max of the supply. Okay, great. Now, if you would like to understand it in a different way, instead of this KVL, I will do another one. So during negative cycle, it will be like this, like this, and like this going to the supply, right? Now, we have the supply polarity, original paraty like this, V supply. And this one is reverse pies, right? Positive, negative. Now we can get VD two applying KVL in this loop in this loop like this. So this one is a short circuit, zero voltage plus VD two, D two. And if you go down like this, negative V supply, negative. V supply equal to zero. So VD two will be equal to V supply. So let's look at the supply during negative part. This is supply during the negative part. When D two and D one are reverse biased, you will find that voltage of the supply is similar to voltage of D D one and D two, as you can see, same waveform from KVL. You can do KVL here or QVL in the outer loop. I hope you now understand where did the voltages came from with their own polarity, of course. Now, Ibut will be similar to Vut same polarity. But the difference is that but will be Vput over R. The same because it is a pure resistive lute. Now, what about ID one and ID three and ID four and I supply. Now this is very easy. How it is easy, you will see like this. Here, if you would like I supply, it will be like this. I supply from KCL at this point, you see D one, I d one, and you can see here I D four. You can see I supply entering, ID four entering, ID one leaving. So ID one from KCL, leaving current equal to entering current, I supply plus I d four. Where did we get this from KCL at this node here. So I supply from this equation will be equal to ID four minus or ID one, ID one, ID one, minus ID four. So let's look at this. We have ID one, and we have ID four, ID one, and ID four. So if you subtract ID one minus ID four, it will be the same as ID one, right. Then at this point, ID one, zero and ID four positive part. So zero minus a positive part, it means that this part will be inverse, like this here, this one minus, this one, positive, this one minus, this one will be negative. This is another way of looking at how we can get I supply. Now, where did we get ID one, ID one or the current in general, I output, is equal to ID one plus ID three. Let's look at here KCL at this point. You can see ID one and ID three entering the node here and I Out going out. So I Out is equal to ID one plus I d three. Okay. Of course, not all of them are working at the same time. So when this one is zero, then IO equal to ID one. When ID one is zero, then Iout will be equal to ID three. So as you can see here, I abut equal to ID one, plus ID three, ID one, plus ID three, ID one, plus I three, and so on. So what you can see here is that the submission of these two waves is I output or Iout will be equal to ID one when did conduct, it will be equal to ID three and ID four when they conduct and so on. So these are the waveforms for this circuit. Okay. Now, we would like to get number one output or the average Out waveform. So as we know that in the output is equal to one over t, integration from zero to T, V max, sine omega t, D omegaty. Now, this is very, very important because you will do a common mistake here. Number one, T is Pi, right? Okay? No problem at all. Now, integration from zero to T. Now, this is very, very important. As you can see, this waveform is V max sine omigaty. If you do this integration 0-2 Pi, it means that the output will be the average of this wave. However, why? Because this one is VMX sine omigaty. However, as you can see, this one is rectified. So it is divided into two parts, this one and this one, which are similar to each other. So what I'm going to do is that I'm going to integrate from here to here, from zero to Pi, and multiply the integration by two. So it will be integration from zero to Pi and multiply this integration by two in order to get the average of this one plus this one. If you integrate 0-2 Pi, it means you are integrating this big wave form, the original one. However, I would like to get average of this one, not this one. I will get this part only from zero to Pi, V max and omegat and multiply it by two. So from zero to Pi and multiplied by two. So two will go with this two. So it will be one over Pi, integration from zero to Pi, one overpi integration from zero to Pi. V maxine omegaty D omegaty. So it will be two Vmax over Pi. So I hope you understand the difference between them. If you integrate 0-2 Pi, you integrating this one, not this one because this waveform is not Vmaxine omegat. This one, VMXine omgaty is a positive part, and VMXine omegaty positive part from here to here. I hope the idea is clear for you. Now what about IOT will be V average over R, same function, but divided by R. Now what about root mean square? Same idea. Root one over Pi, integration from 02 by four Vmax omegaty all squared. You'll find that it will be Vmax over root two. Now what about I root mean square? It will be VRMs divided by R. As we learned it before, same voltage divided by R. So I root mini square will be also max over root two. Similar to Vmax over root. Now Vmax over root two, or as you can see here, one over root two is 0.707, which is one over root two. Now, Vmax over R is Imax. Okay? This equation is similar to this one except that we replaced Vmax over R with Imax. Now, what powers of power power consumed, as we learned before, is a root mean squared current multiplod by the resistance, IRs squared multiplod by R. Now, as you can see here in this figure here that we have a pulse setting, DC like this pulse setting. Now, what I would like to do is that I would like to. Instead of having the average like this, going up, down, up, down this pulseating DC wave form, I would like it to be more close to more close to DC. So what I'm going to do is that I will add barrelel to this fluid acapastans capacitor C. What does the capacitor do? Capacitor reduce the riples inside the voltage. The inductance reduce the ples inside the current. As the capacitors increase ripples inside, the voltage start decreasing. As the inductance increase, the ripples inside the current will start decaying. Okay? So capacitance for voltage, inductance for current. So you will see right now if we add a capacitor parallel to the R load or RL load, it will lead to reduce the ripples. So as you can see here, this is the original waveform going up, down, up, down like this. So instead of having a resistance like this, like this, we will connect parallel to it capastans like this, parallel to it. So this will lead instead of having this waveform, it will be like this. From this point, going down, then charging discharging, like this, charging, discharging. So our waveform will be like this. Like this. So what you can see here is that this one is becoming more and more closer to DC, right? Compared to this pulse setting DC form. Now, let's see another one. As you can see, this is the original AC form, and this is a pulse setting DC like this. Now, if we add a capacitor, it will be going like this, then it will be discharging. When it goes down to zero, the capacitor will start providing its energy to the lute resistance. So it will decrease like this. Then it will start. When the voltage becomes positive again and greater than the capacitor, it will start charging, then discharging, charging, discharging, and so on. Basically, this capacitor start charging from supply parallel to the resistance, and sometimes it will provide its stored energy to the resistance to keep the voltage from going down. That's why in the end, you will have this waveform like this. Okay, which is closer to the constant DC value. If you'd like to reduce the ripples, just increase the capacitance once more, and you will have much more smoother waveform like this. Okay? Me smoother than this one. Now, last thing about this lesson, what are the advantages of using the full wave rectifier compared to the half wave rectifier? Number one, prejctifier such as a circuit is widely used for various appliances. For example, they can convert high AC voltage into low low DC voltage. Number two, they can be used to provide or supply a polarized voltage in welding means voltage with a certain parity, a specific party either positive or negative. The full wave rectifiers also have higher rectifying efficiency than half wave rectifier. You will find that in half wave rectifier, as I remember, as I remember, half wave rectifier efficiency was around 40%. If you remember from the very very beginning in this section of the rectifiers. However, when we go to the center tab and we apply the rule of the efficiency, you will find that efficiency is more than 80%, which is much, much higher than halfway rectifiers. Why? Because we are utilizing both of the waves, positive and negative cycles. Of course, they have also low power losses because no volt signal is wasted in the rectification unlike the half wave. In the half wave we only converted half of the wave and the other is completely wasted. Here we are converting every part of the wave. Now, what are the disadvantages of using full wave rectifier? Number one, it is of course, more expensive. Here, for example, because it needs, of course, a lot of space. Why? Because as you can see here in the bridge rectifier, you can see we need four dits. One, two, three, four, four dites compared to the half wave which only required one diet. Now, if you go to the center tab, which is in the next or the next lessons, you'll find that this center tab is consisting only of two dites instead of four dites. So anyway, two dites or four dites, all of this is greater than the half wave rectifier, which needs only one dite. However, remember that this rectifier wastes or plogshalf of this signal. The half wave rectifier. That's why, of course, the full wave rectifier is widely used and most commonly used circuit. So I hope in this lesson, you understand now what is the meaning of full wave and what is a full wave, bridge uncontrolled rectifier circuit, and how it will help us in rectification process. 43. Example 1: Now let's have the first example on the full wave rectifiers, specifically the bridge rectifier. In this example which you'll see in this figure, we have a single phase, full wave bridge rectifier circuit that is required to supply a pure resistive load of one kilo at 220 volt DC. So we have a resistance of one kilo, and look at this sentence. This is very important. 220 volt DC. So we need the circuit to supply an outbut of 220 volt. What value DC, which means that the average Vout average, which is a DC outbut is equal to 220 volt, right? So we have resistance one kilo, and the output is 220 volt. We don't have the supply yet. What we need from this problem is that we need to determine the root mean square value of the input supply, the total load current drawn from the supply, the average load current passed by each light, and the total DC power dissipated by the resistance. Look at DC power, not the total power, but DC only. Assume ideal diet characteristics, which means that our diets doesn't have any voltage drop. They are ideal diet, which means when they are turned on, they become a short sect. The first step is that we need the root square of the input supply. Supply root mean square. How can I get this by using the V average the relation of the V average. As you remember that in the previous lesson, we said that average of the output is equal to this integration, which gives us two V max over Pi. Now, V max is the maximum value of the supply. So average is 220 volt. To Vmax, which we don't know divided by pi. So by using this relation, we can get that. By substituting, we can get V max equal to 210 Pi. However, what we need is the root mean square. So the root mean square is simply maximum value divided by root two, as we know from AC circuits, V max over root two. So this will give us the supply root mean square value is 244 volt. This is our supply as an RMS value. So this is the first requirement. Number two, we need to find the total loud current. Look at loud current, which means I out. Drawing from the supply since we are saying total loot current, I am talking about what I'm talking about average loot current. Okay. So we need average lute current. So simply average lute current will be out average divided by R, right. From this relation, VO average divided by R. So it will be 220/1000 oms of one kiloohm. So it will be 0.20 2:00 A.M. Pair. This is the current or the output current, for the lot the average lot current. So if you remember the waveform was like this during positive, during negative, during positive, during negative. Okay? Now, this is I out, omegaty now what we need second requirement is that we need the average luid current posed by each diet. Now, this is the Iout. What is the Iot or the current for the diet? Let's say D one or D two or D, D four. Let's say, for example, D one and D two, D one and D two. Omega T. During the postive half cycle, this two will be conducting, right conducting like this, I dite equal to I out. And during the negative cycle, they will be reverse biased. D one and D two will be reverse parts will be short circuit like this. Then or zero, not short circuit, zero. Then postive then zero and so on. So as you can see, this is I out, right? Okay. And this one is our I diet. Now, I O is simply equal to what equal to. These two we forms, right? I out average is equal to average of this waveform plus the average of this waveform or two waveforms. However, for the diet, we have only one. So what's the benefit of this is that if I need average olute current for each dite, it will be basically half of the I out. So each dite passes only half of the lute current, as you can see here, half of this total current. So the average olute current average olute current for the diet, it will be this current divided by two, because it works only for half of the psyche. Now, if you don't understand this is very simple, okay? All what you have to do is that I abut is equal to 1/2 Pi. Okay. Integration from zero to Pi. Integration from zero to Pi. Four, V max sine Omega, Omega divided by R, This is this is for the dit. Now for the loot current, it will be I out will be one over Pi, right? Integration from zero to Pi. For the same function, maxinmt over R. Now, this is similar to the voltage. They have the same waveform, except it is divided by R. Now, if you look at I average and I average for the dite let's say D for the dite, you will see that 1/2 Pi, one over pi. So the average loot current for the dite or average current going through the dite is half of the average of the average loot current or the loot current we have. So it's really clear. All you can see here, this one is half of the total loot current during one complete cycle. Then we need the total DC power dissipated by the loot. Now, since we took DC, it means VDC, multiplied by IDC, so V average, multiplied by I average, or I average square multiplied by R, or V average square divided by R. All of them will give the same solution. So as you can see, I average square 0.22 average, square multiplied by the resistance R gives us 48.4 what?o? 44. Full-Wave Bridge Uncontrolled Rectifier – RL Load: Welcome everyone to this losson. In this losson we will took a poet, the full wave bridge uncontrolled direct fire. But this time, instead of RL loud instead of R loud or a resistance, we will have an RL loud. So let's start. So we have here R load. So what's the difference? You will see right now. First, what you will have is that V O will be similar to exactly similar to the resistance case. So let's understand how does this circuit work. So during the postep half cycle, we will go like this from D one through the loud and getting through D two and get back to the supply, same as the first circuit, right? So D one and D two are turned on, and the inductance itself is charging is charging with current. Now let's look carefully here. As you can see here, the inductance, let's draw it like this during post of cycle. The current will start increasing and charging, right? Now, when the allude current starts to go down, the current itself starts to go down too, like this like this going down. Now, as you can see that at the instant to Pi, at the angle Pi, what will happen? In the half wave rectifier, this inductance started providing energy to keep the half wave rectifier in operation until if you remember, it was extended like this until extinction angle Peter and the V out was going to the negative half cycle. However, you will see that when we reach Pi, what will happen is that D one and D two are turned off and D three and D four are turned on, so the supply is switched. However, why does the inductance do similar to what we did in the half wave rectifier? In this circuit, the inductance will give current in the same direction. Remember, the current does not reverse its direction in this circuit. It is providing current in the same direction. So when D one, D three and D four, or switch it on like this goes like this, through D three inductance and go like this through D four to the negative of the supply. You can see that the inductance does not do anything. The current is just continuous as you can see here. Add the instant poi, the polarity of the supply is reversed and it will start charging this inductance once more. The current will start increasing again until the big value and then start decreasing until two poi so starting from two Pi, what should happen? At two Pi, it should continue to the negative or it should continue to zero. However, D one and D two are turned on once more, so it will go like this, increase again and go down, increase again and do it down. So you can see that the final or the steady state value, the current will be like this, state wave form like this. So as you can see, like this. So it starts increasing. Then start discharging because the VO salts decaying. So it gives a small of its energy back to the supply, then start charging again. Then discharging, charging, discharging, and so on. So what basically happens here is that you will see that the current is continuous, okay? In the and the pure resistive load it was like this goes up from zero to the peak, then goes down to zero when the supply goes to zero, and then it goes up like this, like this. However, due to the stored energy of the inductance, the current will be more continuous. It will not go to zero, it will be it goes like this and B have more energy, so it will not go to zero, then it go up and go down like this, similar to this waveform. Because the inductance or the inductance here delays or reduces the change in current, reduces the Di validity. So you can see that when supply goes to zero, the current does not go to zero. It still have delay due to the inductance. And then at Pi it will switch to D four and D three, and the inductance energy will start increasing again and decrease and increase and so. What you can see here is that during post of half cycle, D one and D two, as you can see, from this post of cycle, D one and D two, during negative cycle, D three and D four, like this and so on. D one, D two, for half of the cycle, half of the cycle, half of the cycle, half of the cycle. Now, for I supply, as we remember that from this KCL here, I supply equal to ID one minus ID four. So ID one, minus ID four. It will give us this waveform for the supply, which you can see here. Now, why does this supply have this shape similar to ID four and IO due to the presence of inductance. That's why the shape of the current due to RL is different from the out waveform O. Okay, for supply, as you can see here, postive negative postive negative. Again, same as the previous circuit, it is completely rectified, from zero to Pi. Now, now, this is something which is very important. Now, when we go to the control rectifier, you'll find this case is different. You will find that VO will go to negative part, and you'll understand then why does this happen? This does not happen here, why? Because it starts from zero to Pi and add Pi at which it should go to the negative the dour and d three switch on. Due to the reverse of the polarity of the supply. So from Pi to two Pi, D three and D four will operate, and again, it will give us positive, preventing it from going to negative part. Okay. Now for the pred circuit, the current is transferred from one pair of dites to the other pair of the dites when the source changes its polarity, as you have seen, as you have seen right now. And as you can see the current is continuous so the inductance does not have any problem here. The problem of the inductance if this current suddenly goes down to zero or switch it to plarty. However, this does not happen because the current always has the same direction, has the same direction going downward from here to here, when D one and D two operates or even when D three and D four operates. So the inductance will always has the same current in the same direction. Now, the voltage across RLE is a full wave rectified and so it's similar to the resistive load. You can see that the waveform here is different is no different from the previous one. Now let's talk about another thing here in the same lesson. Let's say we have this one. This one is a continuous waveform due to the presence of inductance. Now, when the inductance becomes very large, as we learned before, specifically when L effect is much higher than inductance or much higher than resistance, L becomes a very large combo to resistance. In this case, we have the heavily inductive load or the highly inductive load. Now, in this case, instead of having this waveform, we will have constant lines. I output let's delay this. So I Oh be like this. Should be constant line, right? So as you can see here, for V output, same as before, rectified sine wave. Due to the higher inductance, the current waveform is the only one which will change. So what you can see, I output is like this. Now, D one and DD two will operate from zero to Pi. So only this part will be our di D one and D two. You can see D one and D two like this from zero to Pi, as you can see here. Now, from Pi two Pi, the other two dites will take this rectangle like this D three and D four, then D one and D two and d3d4 and so on. Now, I supply, as we remember, is ID one minus D four, ID one minus ID four. This way four minus this way form will give us positive. T zero minus positive gives us negative, positive minus zero gives us positive, zero minus positive gives us negative. You can see it will be like this a square or rectangular waveform, positive negative, positive negative, as you see here. Similar to this one, you can see it is positive, negative, positive, negative. However, it is distorted waveform, not a rectangular waveform, as you can see here. Let's delete this. Get the pencil. Now, what about the waveform? For VO is two VMAX over Pi, one over Pi integration from zero to Pi for VMX nom, same equation for the R load. Nothing change it at all. Now for IO, it will be also the same without average over R. The average account, o average over R. Okay. Now, what about the root mean square? For the root means square Vrms. For this, it will be root one over pi integration, same equation as also the R load. This value is similar to the R load exactly. Now, what about the RMS RMS? Now, as you can see here that the root means square current is approximately equal to I about equal to the average value. This is just an approximation. If you look at this circuit, you can see that here, I put will be a DC value, constant value. So this DC value is I average. The DC value is I average. So I average is approximately equal to the root mean square value, I root mean square is approximately equal to this average value equal to the maximum value of the circuit. Now, of course, this relation here, which is root mean square is approximately equal to the average current which we obtained from here. This relation or for this part only. All of this are suitable for here and here. Now, what about this for the root mean square current? For the root means square current here, it can be easily equal to the root mean square VMs. Divided by what? Divided by Z or the total imbedanceRroot or square plus Omega L square. Here, we don't need to get the current equation. Why? Because we don't have extinction angle bit. So we don't need to get the current equation at all. However, when we go to the controlled circuit, when we have extinction angle, as you will see, in this case, you will you need to get the current equation for this circuit. Okay? 45. Example 2: Now let's have another example, example number two. In example number two, we have a full wave rectifier with an oral lute connected to 120 volt source. The allude resistance is ten ms inductance is much greater than resistance. Find number one, the average lute current, average lute voltage, average lute current, maximum lute current, RMSE value, average current in each diode, ors current in each diode, and power supplied to the root. This example is very, very simple, more than you thing. Okay? So let's take it step by step. Number one, we need the average load voltage, right, average loud vote. So we know that out average is equal to two Vmax over Pi, right? From here. V out average equal to V max over Pi from what we learned before in the RL and the pure resistive root. So V max is simply equal to 120 root two. So it would be like, two V max over Pi two multiplied by 120 root two divided by Pi gives us 108 volt. Now you can see that since L is much greater than resistance or a highly inductive loud, so you can see that all of them are rectangular wave form, as you can see here. So this is the average lute volte. What about average ute current? Very easy. Just take this voltage and divide it by the resistance. Like this, but will be out of r2v max or PR, you can see R here. So we take this value and just divide it by R. So 108/10 OMs. It will give us 10.8 am pairs. This are the two first requirements. Now let's continue. Now, what is the maximum loot current? Now, remember, the circuit like this, Iout will be a constant value Iot, equal to I average. Equal to RMS, equal to the value which we obtain 10.8. So we obtained I average V average of which is 10.8. This value is the average value, is the root mean square value is the maximum value. All of them are similar to each other. So you can see that RMS approximately equal to I output, approximately equal to IDC. So I average equal to IRMs, equal to Imax, equal to 10.8 am pairs. So this one and this one, 10.8 am pairs. Now, we need the average current in each diet. Now, as you can see here, we said that for this one, for one complete cycle, one over Pi. For the diet, it will be 1/2 Pi, half of the average current, half of the average current. So it will be this value divided by two. I average divided by 5.4 amperes. Okay, since it only provides current for half of the cycle. Now the next one is tricky. So the root mean square in each dit. Root means square of this one or this one. Now, as you can see, it works only for half of the cycle. However, remember, we have root 1/2 Pi, digron from zero to Pi to from zero to Pi for this part for Vmax, sine Omega T divided by R, all square. Or for the current. This can be like this or we can say, instead of this one, it will be I out. I out since it is a constant value. Now, what you can see that here, root 1/2 pi, right? Integration from zero to pi for the current square. Now, if we take root 1/2, so it will be one over root two. Take this part, one over half, take it outside the square root. It will be one over root two, multiply it by root. Integration. One over pi, integration from zero to Pi for the current square. Now, this part this integration from zero to Pi for a constant value, square, it will be I square, multiplied Pi Pi. Since the current is a constant value, it'll be I squared, multiplied Pi Pi. All of this under the square root, okay? Multiplied Pi, one over Pi. So this one will go with this one. So I squared gives us I. So what you can see is that the current root mini square of the diet is I divided by root two. Again, so what we did is that in order to get the current root mean square of the dit, it will be root one over T, which is a complete period, 1/2 Pi. Integration from zero to Pi. Why from zero to Pi? Because it's from here to here from pi to two pi is equal to zero. So we integrate from zero to Pi for the current square. And remember that the current is a constant value. Constant value, right? I square. So you can see I square for this part, from zero to Pi from zero to Pi, and 1/2 Pi. We took this 1/2 and put it outside. So it will be one over root two. Now, root one over Pi integration from zero to Pi for I square. This part is I square multiplod by Pi, which is this part. And multi ploid by one over Pi. This will go with this, all of this under the square root. So root I square will give us I and don't forget root two, so it will be I divided by root two. So when I get average, when I get average, it will be I average over two. When I get root Mini square, it will be I divided by root two. I hope this D is clear right now for you, as you can see here. Now, what about the power supply to the loud power supply to this load. So as you remember, power supplied is root mean square multiplied by R here we're talking about the total power consumed by the loud. Not the DC, but the total power consumed. So it will be RMS square multiplied by R. As you can see here, RMS squared multiplied by R, which is 10.8 square multiplied by ten, which is 1166 0.4 46. Full Wave Center-Tapped Uncontrolled Rectifier: Welcome everyone. In this lesson, we will talk about another circuit regarding the rectifiers, which is called the full wave center tab uncontrolled rectifier. We have uncontrolled means that we are dealing with diodes only center tab this is another circuit which is different from the bridge rectifier that we took about before. So let's start. So the full wave sent full wave, it means that we will have the output like this, or it will not like this always all the time. But full wave means that we are utilizing both of the positive cycle and the negative cycle for the outut. Now let's look at this circuit. What you can see this circuit is similar exactly to this circuit, no difference between them. So what you can see is that we have a transformer which can be found not only in the center tab, but also in other circuits, such as the previous ones, the bridge rectifiers, the half wave rectifiers, and so on. So the function of this transformer is that it steps down the voltage for a level required for the output. That is the first function. Second function is that here in this type, which is a center tab, we must have a transformer because we have the primary and secondary. On the secondary, we take a point from the center or the center tab for this transformer. So if you don't know about transformers, you can go to our course for electrical machines and understand the rectifiers or no rectifiers. Understand that transformers. Okay? So the center tab here, what does it do? You have the primary, right? So let's say primary V one, Okay. And secondary, all of the secondary neglect the center tap for now, V two. So the relation between V one and V two, if you remember V one over V two is equal to number of turns of the primary winding divided by number of turns of the secondary winding. So by controlling number of urns, we can control the output voltage. Okay? That's the first thing. Number two, if you look at here, we have V one and V two. Now, since we have a center tab, the voltage will be divided into part of it for the upper half and another part for the lower half. The upper part will be V 2/2, and this part will be V 2/2. So the total of them their submission will be V two or the output. So the center tap transformer just take half of the input voltage, half of the input V two. Okay? That is the first thing. Okay. Now, second thing that you will notice in this circuit is that we have only two dites two dites compared to the uncontrolled tire bridge rectifier, which had four dites. So what will happen here exactly this circuit is similar to this one. So let's understand how does this circuit work? This type of rectifier uses a two dites and a transformer with a center taped secondary winding. As a full wave rectifier canverte both of the half cycle since it is a full wave of the EC signal into pulsating DC. Now, during the postive half, so we have two dites D one and d2d1 works during the postive half cycle, and D two works during the negative half cycle. During the postive half cycle, we will have positive, negative, positive, negative right. Since the voltage on the secondary is positive. So as you can see, the positive goes here and negative, we can say goes till here, right? And this one will be connected the negative and this positive terminal, this one will be connected right here. So what you can see here is that this did is forward pased and this one is reverse pased. Okay. Now, as you can see D one forward pissed and current will flow through it. So if you think about it logically, you will find that during postive cycle, positive negative, current will go outside here. So this dite will allow the current to flow in the same direction. So this dite becomes a short circuit, assuming ideal dit. So the current will go like this and go through the R lute or the resistive loot. Now, this is during the postive cycle and this one is reverse pised or turn it off. Now, during the negative cycle, the reversal will happen. We will have positive, negative, positive, negative. So what you can see, negative connected to this one and the positive connected here, here, positive and negative connected here. So what you can see here is that this one is reverse piased and this one is forward pass. So the current will go like this from the transformer through D two in the same direction at which this did will allow the current to flow, so it will go like this and through the resistive loot. So we have now two dites. D one operates during the positive cycle and D two during the negative cycle. And you will see that both of the currents positive or negative goes through this terminal. From here like this, during post of cycle or like this during negative cycle. The current has the same direction, so the output voltage will be the same as the full wave rectifier or the bridge rectifier. The lot current will go through Dt during the negative week and you'll find that the lute current, as we have just explained, that its direction does not change even when the polarity of the supply changes. Now let's see this. So as you can see, during positive cycle, during negative cycle. As you can see, this is this one. This circuit is exactly similar to this one. Similar to also this one, all of them are the same, but different shapes. Now, let's see this one so you can understand that they are the same. During post of cycle, what you can see here is that this is a center tab connected to the ground and resistance connected to the ground. Or you can think of it like this point, connected to this point, both of the ground connected together. So as if the resistance will be connected like this. Same as here, right? Okay. Now, during the post of cycle, during the post of cycle of the supply, this D one will conduct, so the current will go like this, through the resistance to the ground, which will also like this as if it goes to the negative terminal of the supply. Now, as you can see here, here, current goes like this, from here through the loot from the ground like this and loop between them. These two are connected together like this. Okay, so V supply, V output will be equal to V supply. Okay? Now, during the negative cycle, the reversal will happen. D two will operate and D one is off, so the current will go downwards like this, go from here like this through the resistance, again, the same direction. So V output will have the same waveform as the positive half cycle, okay? Now let's understand more about the circuit. The same circuit that we talked about in one and then two supply one and supply two. Now remember, supply one is half of the total voltage here, half of V two, ok? Now, forget about V two. We will assume that both of these are separate supplies, okay? So us here VMX representing what representing the maximum value of V supply one, okay? Supply one. Okay, so we have the waveform like this, V max sine omgato like this. Okay? VO will be positive, positive, positive, and so on. Now, what about D one? During the positive half cycle, this one will become a short circuit, so its voltage will be zero. Now, during the negative half cycle, V two is short circuit. Then short circuit for this one, and then for this one. Cycle repeats itself. Now, during the negative half cycle, when D W D two, during the negative cycle, when D two is turned on and it goes like this, D two is a short circuit, right? And D one is an open circuit. So what we need to get is a value of D one during negative half cycle from here to here, right? We know from here to here odd circuit. Now we need from here to here. How can I get it very easy? All you have to do is KVL again, KVL in the slope. Now we have positive d one, so it will be plus V d one. Then we go like this clockwise and go like this positive out plus V output. Then we go like this negative supply, negative V supply equal to zero, right. So died will be equal to V supply minus V output, right? Now, let's look at our signal. We have V supply from here to here, negative cycle. Okay, minus minus what other value V output. So minus VO which is from here to here. So negative V output, it will be like this, right? So negative VO as if we say plus negative Vo. So inverse of VO will be like this. So add this inverse to V supply to this one. So if you add this one to this one, you will have a large negative signal of negative, negative to VM. Why? Because we have VM here and we have VM here. So when we add them together, we will have a negative week of negative to VM, okay? Now V max here representing the maximum value of supply S one or S two, not the total voltage, but S one or S two. This is very important as you will use this in selection of dites. For example, for the diet, we need to know the peak inverse voltage in order to select a suitable diet for this application. Now, the same idea for D two when D one is conducting and D two is turned off, it will be also negative cycle negative to VM, as you see here. Now what about but I output, since we have a pure resistive load, it will have the same waveform of voltage except divided by R. So Ibut will be equal to Vbut divided by R. Now, ID one will conduct during postive cycle here and from here to here, as you can see, ID two will conduct in the negative cycle here and the negative cycle here, which is here and here. It will be here like this and like this. Now what about I supply? Where is the supply current? In order to get the supply current, we have current which go like this, I supply like this. During boost of cycle, it will go through D one. It will be equal to D one and during negative cycle, it will be equal to D two. However, remember, I supply here, I supply, its direction is like this, okay? Positive direction. Now, during the negative cycle, it will go like this. So its value will be negative of the I two. Okay? So as you can see, during postive cycle, during this postive cycle, current will go in the same direction, similar as D one and similar as the polarity of the supply. So I supply equal to ID one, now during negative cycle, it will go in the reversal direction. So it means that the current is reversed like this, reverse of ID two. Remember, this polarity is important. So I supply like this going from the positive terminal here. If the current is reversed from going downward like this, it means it is a negative value. Now, the current of the diet is in the same direction, positive in the same direction and positive in the same direction. That's why ID one positive and ID two positive, okay? So now we understand all of these waveforms. Now what we need also to get is the values. So first, we need the average out voltage, V O average. You will find that it is exactly the same as the bridge rectifier, two V max over Pi without even seeing it. You can see here two V max over Pi. Integration one over T, which is 1/2 Pi. And since we say we integrate from here to here, it will be one over Pi, integration from zero to Pi, V max sine omegaty. That will give us two V max of R Pi. Since the period here, Pi, T over two which is also Pi, okay? T two Pi and T over two is Pi. So we have now DC the average voltage. Now, what about the current current will be I average over R, right? I average over R very easy two VM. Over multiblod bye bye. We don't need to say it again because we already know this from the previous lessons. Now, the RMS value of the voltage is the same as before. It will be Vmax of a root two, if I remember correctly or 0.707 VMAX. Let's see VMAX of a root two, as I exactly said. You can see VRMS and average for the center tab is exactly the same values as the bridge rectifier. Now, what about the bel here? This is important and I didn't mention this for the ripples because it is important as we will compare these values with the half wave rectifar so you can see the difference between them. For the ripple factor, if you remember it, Gamma is equal to root VRMS over VDC all squared. So we have VDC and we have VRMS substituted here. You will get Ripple factor 0.48. Now remember this, we will need it. Or you don't even remember this, I will show you a table. For the efficiency of the full wave rectifar which is power of the DCM divided by the EC input power. As you can see, power DC over power DC. Here's this one power DC over EC, not DC, it is A, C. Power of DC over AC, it will give us this formula which we talked about in the very, very beginning of the course, if you remember, the very beginning of the rectifiers. So it will be equal to this value. And if we assume our load is very large, or the resistance of the rectifier very small, its efficiency will be 81%, which is much higher than the half wave rectifier. Now, what about form effector form factor, if you remember, we are mas over VDC I will give us 1.11, which shows us how close are RS and VDC close to each other. Now, the big inverse voltage for the center tabbed full wave rectifier is two V max, as you can see here, maximum value, negative two VM, so this is a peak inverse voltage that allied must withstand. Now in the bridge rectifier, the maximum value was V max. Not to V max, but only V max. Now, if you don't remember this, I will show you quickly here. Where exactly I think here. Okay. So as you can see VD one and VD two, voltage of the die D one or D two, you can see maximum negative value negative VMX. That's why the peak inverse for each dite is the maximum value of supply. Okay? Okay. Now, let's go to what point exactly. Okay, here, I forgot to mention this. You see here I O of R, we average V of R, DC to Vmax over or multi plod by R. And we said before that IRMS will be VRMS over R. Since we have a pure resistive root, this VMAX 0.707 is similar as VMAX over root two. They are exactly the same. Okay. Now, let's continue big factor of four wave rectifier. Big factor is a ratio between V max and root mean square. So it will be root two. Now let's compare between half wave and full wave rectifiers, from the number of dits and other factors. As you can see here for half wave and full wave. Remember, in half wave we need only one dite for rectification. For the full wave rectifier, we need two in the center tab and we need four in the bridge rectifier. So we need more dites here to rectify or convert EC into DC. For the maximum efficiency, as you can see, for 40%, and we talked about this in the very very beginning. But you can see that full wave much higher than half wave, which is 81%. Peak inverse voltage, V max in the halfway ctifer. In the full wave, it is V max in bridge and two V max, which is a peak inverse voltage in the center taped. As we just explained, now, average Vmax over Pi. This one is two V max over y, we are getting more and more closer to DC because here average here is double the V average of the halfway ctifer. So we are getting more and more close to DC value. For the root mean square will be Vmax over two, here, Vmax over root two. For the ripples, you can see 1.21 and here 0.48, you can see that riples here in the full wave is much lower than the ripples inside the half wave because we are getting more closer to the AC form to the DC wave form. For the foam factor 1.5 and here 1.11, which means RMS is getting more close to the average value. They are getting close to each other. Now for the frequency here, frequency, it repeats itself every F, like this. Postive then zero, then repeats then zero, like this. So it repeats itself very every one complete cycle, every frequency, right. However, in the full wave, it is like this. Like this. This waveform repeats itself every pi or every half of the period T over two, which means the frequency here is double because the periodic time is half, it means that the frequency is double. It repeats itself at a higher frequency than the half wave rectifier. Okay. Now, when do we use the center tabbed transformers and when do we use the bridge rectifiers? Now, for the bridge, the lower peak dite voltage in the bridge makes it more suitable for high voltage application. Each of these dites when we select them, it withstand VMAX maximum value of the supply. So we can increase the value of the supply, which lead to, of course, increasing the rating of the diet. However, however, if you use a center tab, you will need diet with very high voltage rating, more than two VMAX so if you look at the center tab, and this is a bridge for center tab, as you can see here to VMAX. So if the voltage is high here, it means that we will need diet that withstand double this voltage, which means higher cost. That's why this circuit is not suitable for a high voltage application. This one is more suitable for high voltage application because it will lead to less expensive diets. That's the first thing. Second thing is that this one, center tab is more suitable for low voltage, as I have just said and high current applications. Remember that the power is the same. Power is the same, which is voltage multiplied by current. If we have a low voltage, it means that we have higher current. That's why this one low voltage, high current application, this one is high voltage low current applications. So I hope in this lesson you understand now the difference between bridge and sensor tab, you can see the, if you look at the waveforms here, here, you will find that they are exactly the same. Nothing changed. Even if you have an RL load, it will behave the same. Okay? If you have an RL, it will behave the same. That's why you can see, and you have seen that VRMS average current, everything is the same as the bridge rectifier. Okay? How the only difference is this two Vmax, VMAX and two V max. Other than that, everything is similar to each other. Now, why is this important? Because we are not going to have any solvit examples on the center tapped. Why? Because it is the same idea exactly with the same values with everything. So I'm not going to have any examples on the R or the PU resist or even the RL loads. We're going to go to the next one, which is a controlled circuit center tapped controlled and the bridge controlled, and also we will talk about the half controlled, half controlled bridges or center tapped. 47. Full-Wave Controlled Rectifier – R Load: Hi and welcome everyone to this lesson regarding the rectifiers. In this lesson, we will took at the full wave controlled rectifiers. In the previous lessons, we took the uncontrolled rectifiers. In these lessons or in these upcoming lessons, we took full wave controlled rectifier in this one full wave controlled rectifier with an R load. And if you think about this one, full wave controlled, what does controlled mean? It means that our circuit is consisting of all of the circuits consisting of thigh restors. So in uncontrolled, we had only diets in controlled circuits, we will have thigh restors. As you can see here, in a fol controlled projectifi use this for thy restors to control the average load voltage. As you can see in the circuit, we had our supply and we have four soy restors instead of four diets. With our load, which in this case is our load or a pure resistive load. So we have four soy restors. These four thistors will be switched on by using the firing angle Alpha. So let's look at here so si restors a T one and T two must be far simultaneously during the positive half cycle of the source voltage VS to allow the conduction of current. So in the positive half cycle, T one and T two will be switched on. And during the negative half cycle, T three and T four will be switched on. Let's look at what will happen here. As you can see here, we have to ensure simultaneous firing. Both of these estors must be turned on at the same time same time. We use the same firing signal. And alternatively, thy restors T three and T four will be also fired simultaneously during the negative half cycle or half wave of the source voltage. So let's look at here. So you can see here, this is a circuit for the controlled rectifier, full wave control bridge. And this one, which is similar for control dictifier but this time with a center tabbed transformer. As we remember, from previous lessons, we said in the bridge rectifier, we use four sstors or four dides for uncontrolled circuit, and for center taped transformer, we use two s restor. Both of these circuits will produce the same output waveform. For the center taped transformer rectifier, T one is for biased. When supply is positive, this one will start during the positive cycle and T two during the positive cycle. But remember, these two resors will not conduct unless they receive a gate signal Alpha. Let's look at the waveforms for this circuit to understand how does this work? So you can see here by logic in the previous lesson or in the previous lessons for the uncontrolled fires, we said that uncontrolled, it starts from zero to Pi zero to Pi for these two switches and from Pi two Pi for D three and D four. Now instead of starting from zero to Pi and from Pi two Pi, positive cycle, negative cycle. This time we will start from Alpha tel Pi and from Alpha plus Pi, tel Pi. It will be smaller, depending on the firing angle selected. So as you can see here, starting from Alpha, we have our but with V but equal to V supply, and at Alpha plus Pi here Alpha for the first two I restors and for third and fourth y restor, it will be at Alpha plus Pi. I will conduct. And again, Alpha plus two Pi, Alpha plus three Pi, and so on. What you can see here instead of starting from zero to Pi or from Pi to two Pi, we have from Alpha Pi, Alpha plus pi to two pi. This is for the conduction. During the posto off cycle, T one will conduct, so the current will go like this, similar to the circuit for uncontrolled rectifiers. Go like this through T two and back to the supply. During a negative cycle, it will go like this. From here like this through T three, through the loud back to like this, through T four and get back supply. So it is the same circuit exactly like the uncontrolled except we are starting from Alpha instead of starting from zero. Okay, so for 81 and 82, these two restore T one and T two, that's delete all of this. T one and T two will conduct during the positive half cycle for the circuit. So during this conduction, the current will go through it I ight one and 82, current will go through it to the loud. This current is equal to V supply divided by the resistance. Or to be more specific, not V supply resistance, it will be V output divided by the resistance. So if you remember I output in the load circuit will be equal to V output divided by the resistance. Now, I Obut is equal to IT one and also equal to IT three from KCL at this point, I but is equal to I three plus I one. During the postive half cycle, I will be the only one exhausting. And this one, the Cistor T three will not exist. So T one current of T one will be equal to I outputs. That's why I t one equal to I. Now, during the negative half cycle, the sport, T four and T three will conduct, the current of T three and T four will be like this, right? Now, what? Let's delete all of this. What about the supply current O supply. From TCL at this point, I supply is equal to I t one minus I t four. Okay. Right? So we will take this wave form for 81 and subtract from it 84 like this. So this subtracted from this will give us positive, as you can see here, the sport. Then here zero minus the sport, the sport here, so it will become negative. So this is our supply. So if you remember in the previous lesson for uncontrolled rectifiers, the supply was like this, right? Positive negative. However, instead of having this wave form, it started from Alpha and here from Alpha plus Pi, as you can see here. So I think if you already understand the circuit for uncontrolled rectifiers, you will be already understanding this one. There is no difference between them except that instead of starting from zero, we started from Pi not Pi. We started from Alpha or the firing angle. Now what we would like to do next is to get the equations or the equations for this circuit. So first, as you can see here, average or VO average will be as we did before for this way four. This is VO for one complete cycle. So what we can do is say 1/2 Pi, Integration from zero to Pi, V max sine Omega T or Omega T, D Omega t, right. This is the integration for VO. Vout is equal to supply. However, if you look carefully here, you can see that our waveform starts from Alpha until Pi. And again, from Alpha plus Pi to Pi. So we can say that the integration will be Alpha to Pi, Alpha, to Pi, but multiplied by two. Again, from Alpha to Pi, the sport, but multiplied by two because from here to here is one complete cycle. So we can say that the integration will be from Alpha to Pi, but multiplied by two. So from Alpha to Pi multiplied by two, so two will go with two. So we'll have one over Pi, integration from Alpha to Pi. As you can see, one over Pi, integration from Alpha two Pi, Vmax and omegaty. So it will give us VMAX of Pi one plus cosine Alpha. That's for Vout average. Let's say I would like to get the average Obit current, it will be simply V average divided by R right if you would like average because it is a pure resistive loud. Average over R so this wave so this equation divided by R. If I would like to get the root meaning square, it will be simply the same equation here, this equation about square of this function. Square V max sine omegat square all under the square root. You can see same integration, same integration as this one. However, we square of the function under the square root. So it will give us Vmax root one minus alpha art two Pi plus sine two alpha t four Pi. Now, what about the root mean square root mean square of the output current? It will be simply VRMS divided by R, as we already discussed hundreds of times before. So you can see the same equation but divided by R, as you can see here. Okay? For the power, as we said before, power consumed inside the resistance will be root mean square current multiplied by R. Here we are talking about all of the power consumed inside the resistor. So I think what you have seen in this circuit and all of the previous ones is that they have the same concept except that they are instead of being a half wave, they are full wave circuits. So I think if you already understand all of the previous circuits, everything or every upcoming circuit will be a piece of cake for you. Okay? So let's start in the next lesson in having an example on a full wave controlled rectifier. 48. Example 3: Welcome everyone. In this lesson, we will have example number three about the full controlled circuits. Full wave controlled bridge rectifier, we have here a full wave controlled bridge rectifier, full wave controlled means we have thi restors with an EC input of 120 volt. This is, of course, the root mini square value at a 60 Hertz and 20 load resistor. The delay angle here is 40 degrees. Determine the average current in the load, the power absorbed by the load, the apparent power of the supply and the power factor. Let's start step by step. First, we need to get the average current in the root. In order to get the average current, we need the average voltage. We will use the previous equations that we talked about before. The average equal V max over pi one plus cosine Alpha. Vmax will be the root mini square value, multiploPy root two. Divided by Pi one plus cosine Alpha Alpha here is 40 degrees, so it will be cosine 40 degrees. So this will give us average will be 95.4 volt. Now, if I would like to get the average current in the uid, simply will take this value and divide it by the value of resistance 20 ohms. So it'll be 95/20, gives us 4.7 7:00 A.M. Pairs. Now we would like also the power absorbed by the root. In order to get power absorbed by the loot, we need the root mini square current. And in order to get the root mini square current, we need the root mini square voltage. First, we will get VRMs by using the equation, VRMs is equal to Vmax, root of minus Alpha over two Pi plus sine two Alpha over four pi. This is what we explained in the previous lesson and avoid it by the resistance. Vmax root 220 resistance 20 Alpha. 40 degrees sine to 40 degrees, but you have to convert this, of course, to radiance. Outside in angle, we use only radiance, not degrees. As you can see here, 0.698, this is our Alpha in radians. By substituting in this equation, we will get 5.8 and pairs. Now we would like to get that absorbed. It will be root mean square, square multiplied by R, as we learned a lot before, 5.8 squared multiplied by resistance, gives us 673 what. Now what if I would like the apparent power of the supply? Now, this is very easy. The apparent power will be equal to for the supply, VRMSs of supply multiplied by IR mass, right for the supply, VRMss of the supply will be 120. This is the voltage for the supply, 120. Now, what about the current? Now, look carefully here. We have the root mean square for the current for the bit current. Now, what does the but waveform looks like? O wave form will be like this. Look carefully. It will be I but is equal to 81 plus 83. So we have 81 and we have t three. So we just add these two waveforms. So it will be like this zero until Alpha, then goes up and like this, then this part like this. Exist. So what you can see is that this is the output wave form. The root mean square of this waveform is 5.8. Now, what about oil supply? This is oil supply. You can see it is exactly, exactly, similar to the root mean square count of the out. Why? Because we have positive part and negative part. Similar to here, positive positive, but they have exact waveform. This waveforms are similar to these waveforms. Now remember the root mean square current does not care about the sign. Even if it is positive or negative, they will have the same root mean square. If you get the root mean square four this way form, it will be exactly equal to 5.8. That's why the root mean square four thus apply will be equal to 5.8. So what you can see here is S or apparent power, VRMS RMS, 100 weighting of the supply, multiplied by the root mean square for the output current. It will give us 69, six volt and bear. Now, what about the power factor? For power factor, power factor is simply equal to real power consumed, divded by the apparent power. Here we're talking about power factor of the supply. So the BOR factor will be B over S. Real power consumed is the only power consumed here. All of the power consumed is inside this resistor. So it will be 673 what? Divded by apparent power of the supply, which is 696. So it will give us a power factor of 0.967. Now, as you can see here, the power factor is much higher than the half wave rectifier circuit. If you remember, it was approximately 0.607, something like this. So the power factor in half wave was much lower than the power factor in full wave controlled rectifiers. Okay? Now, POS here, power factor here representing power factor of the supply. P input divided by S input. S input 696p input power active power input from this supply is actually equal to power consumed by lute. That's why here 673673, note 672. This was another example that will explain how to deal with full wave controlled rectifier circuits. 49. Full-Wave Controlled Rectifier – RL Load – Discontinuous Mode: Good evening, everyone. And this lesson, we will took a full wave controlled rectifiers. But this time with an RLude and the next two lessons we will too about the lute with a discontinuous mode of operation and with a continuous mode of operation. And this lesson, we will took about discontinuous mode of operation. We have the same circuit for the full wave control direct fire for thy restors or lute. The stem is an Rude resistance with an inductance. Let's look at the waveform for the circuit. As you know in the lute, we have an extinction for the induction mood. Instead of starting from Alpha and dying at Pi, instead of starting V out from Alpha till Pi due to the presence of an oral load, this will give more electrical current that will go through T two and get back through the supply, T one, and go back to the load. So it will continue to conduct despite going to the negative direction. So it will continue to conduct until it finishes all of its stored energy. So it will start conduction from Alpha to Beta. That's why, as you can see here, it will start from Alpha, and VO will continue having a negative part till the extinction angle beta. Right as we learned before from the presence of an RL lute. Now here we are talking about the discontinuous mode. It means that it starts from Alpha and ends at Beta, which is less than Pi plus Alpha. The condition for having a discontinuous current in full wave control circuit is that Beta or the extinction angle becomes less than Pi plus Alpha. If Beta reaches Pi plus Alpha, it means that the current will be continuous. It will be like this. Instead of having from Alpha going like this, it will extend like this until B plus Alpha, right. So this is will make the current like this. Okay, continuous. So the condition for having discontinuous current in the full wave control dictifR is that Beta or Beta becomes less than Bi plus Alpha. That is our condition for having discontinuous current mode. Okay? So what you can see is that from Alpha to Beta, we have the extinction angle. So V Out will be equal to supply. So as you can see here, V O is exactly equal to V supply. After the conduction mode ends, it will go from Beta and it goes to zero. Instead of going from V out equal to V supply, it will go to zero. Then it will be zero until the next firing angle Alpha plus poet, which T three and T four will conduct, it will be equal to V supply or negative V supply this time and continue like this until the extinction angle Beta plus two Pi, then it goes to zero and so. So this is exact circuit that we tube up in the bridge rectifier or even even in the half wave rectifier circuits. However, this time, I'm just explaining the difference between discontinuous and continuous modes of operation, so you can understand exactly how these circuits work. Or the work. So if you remember the current equations that we took up hundreds of times. I put equal to V max over Z, sine omegaty minus theta, minus sine Alpha minus theta, E to the pow negative Omega T, minus Alpha over Omega t, from Alpha to bit. This is the current equation from here to here. If you don't remember it, that is very easy. Remember that the current is equal to steady set current. Plus transient. We talked about this point hundreds of times. I steady state is V max over Z, V max, over Z, sine Omega t minus set. This is a steady state current or the forced response for the circuit. For the transient, we said before, A, which is a certain constant, E to the power negative omega E over omega L. This is the current equation. In order to get the constant A, we simply say at O at Omega T equal to Alpha, Omega T equal to Alpha. The current will be equal to zero from this figure, starting from Alpha. So using this, we will get a value of A and substitute it again in this equation, you will get this final equation. This is exactly the same equation that we talked about in the half wave rectifiers. So from Alpha to Beta. Okay. Now, what we would like to get is the extinction angle Beta or Beta. How can I get this simply at Omega t equal to Beta. The current value will be equal to zero. And by solving the equation numerically as we did in the half wave rectifiers, you will get value of the extinction angle beta. Now, these are, of course, the value that we talked about before. Let's first get V average. So average value, let's start from here for one complete cycle. So here we started from Alpha. So from Alpha to two Pi plus Alpha, from here to here representing one complete cycle. So in order to get the average, it will be this part all of this equal to the sport, right? So we can say one over Pi, integration from Alpha to Beta, from Alpha to beta, max sine omegaty. So instead of integrating the sports, then this part, they are equal to each other. So we say instead of 1/2 Pi 0-2 Pi, we say one over by integration of just one of these wave forms. This will give us V max over pi cosine Alpha minus cosine beta. This is our average. The next one is that we would like the average current. Average current will be average divided by R, as we learned before, because the average voltage across the inductor is equal to zero. So average across the load is equal to V average across the resistor. So it will be average over R will give us this equation, right. Now, the nucs step, we need root mean square current. So our I average also can be gotten by another way, which is 1/2 Pi for the whole cycle, for the whole cycle, integration from Alpha to Beta, from Alpha to Beta. Okay. This is for the average. However, however, there's something here which is very important. As you can see, for one complete cycle, it starts from here until Alpha plus two Pi. So we have from Alpha to Beta, from Alpha to Beta and from Pi plus Alpha to Beta plus two Pi, right? So the difference between them or Beta plus Pi. So the difference between them is that we take one integration and multiply it by two. So here, it should be one over Pi or take this integration and multiply it by two. Okay? So here, two is missing or just remove this one, okay? So as if we took one integration and multiplied by two. For the root mean squared, the same equation, V max omegat d omega T, or this square under the square root. One over by integration for same exact equation. However, under the square root. Here there is no DT. I don't know why I added this dt. Okay? The reference added this dt Domgat Domgat there is an extra T here. So V max omegatty D omigaty same as before. So here is the same integration but square of the function. That will give us this final waveform. Now for the current, you can get this function and square it and under the square root or use VRMs and void by R. Take the function as you have seen here, square of the function and integrate it 0-2 Pi again or from Alpha to Beta and multiplied by two. So here again, this one does not exist. And this one does not exist, okay? Square of the function all under the square root. Okay, or you can get VRMS and void by R, this will by Z by Z not R, the total imbedance and so we have a resistance and inductance. 50. Example 4: Now let's have example number four on the full wave bridge rectifier with an Rude in the discontinuous mode. In this example, we have a full wave controlled rectifier with 120 volt asurce, 60 rts resistance to Ms, inductance to milli henry and Alpha equal 60 degrees. Find an expression for the lot current, the average lot current, and the power absorbed by the lot. First, we will write the expression for lot current. First, we have our values V max, which is our supply 120 root two, taking the root mean square value and multiplying it by root two gives us 169.7 the imbedans for the circuit, 12.5 Theta and Omega tau, Alpha, all of these values that we talked about before. First, the expression for the alot current will be the equation that we took up in the previous lawson V max over this, I Omega tau, and so on. So Pi substituting with these values, we will get the value or the expression for current like this. Number two, average lute current. You can get average lute current by integrating this function from Alpha to Beta and divide it by one pi or by getting V average and then divided by the resistance R. So first, of course, in order to integrate, we need beta. So in order to get beta, we need to substitute with Omegata equal to Beta or Beta and equates us with zero, as we learned before. So we will get the value of beta. So Beta will be 3.78, which will help us know if this circuit is continuous or discontinuous. So as you can see here, this value is less than Pi plus Alpha. If you take the value of Alpha, which is 1.047 and add it to 3.14 representing one pi, you will find that their submission is greater than value of Beta, which means this Beta is before Pi plus Alpha, which means this circuit or this current is discontinuous. Okay, so Pi plus Alpha as you just add these two values together, you will get approximately 4.19, which is greater than beta. Now, we need the average current or the average lute current. So I average will be the average divided by R or by integrating one by integration from Alpha to Beta for the expression of the lute current. Or we average divided by R, which is much easier. So we average divided by R, we have Alpha, we have Beta. We will get the value of the current. Same value. But this time, again, I average will be just one over Pi because we have two way forms one and two, so it should be one over Pi. Integration from Alpha to Beta for the current or by using this value, it will give us finally 7.07 pairs. Now for the root mean squared current because we would like a power absorbed by the loud, so it will be RMS squared divded by R or RMS squared, multi blood by R. Let's first get the root mean square voltage, which is using this equation, one over p integration from Alpha to Beta, the same one which we discussed in the previous law. So it will give us this expression by substituting with the values that we obtained. In this example, we will get VRMS equal to, uh, VRMS will be equal to a certain value and take it and divide it by Z or impedance. So as you can see here, impedance, this one, 12.5. So take this one, which is this expression. Divide it by 12.5, you will get 8.8 am pairs. So this is the root mean squared current. To get the power absorbed by the loud, simply square this value 8.8 square and multiply it by R to give us 774.4 what? 51. Full-Wave Controlled Rectifier – RL Load – Continuous Mode: Welcome everyone in this lossm. We will cap out full wave, control, direct R or ode as the previous loon, except that we will have now instead of discontinuous mode, we will cap continuous mode. The same circuit as you see in this figure. Now since our current is continuous, you will find that this waveform is like this. Starting from Alpha, the current starts increasing, then it starts decaying and don't go to zero at Pi plus Alpha. At Pi plus Alpha, the two other sistors will start conducting leading to increasing again in the value of current. And this process keeps repeating as you see here. This waveform is, if you remember correctly, this is similar to the uncontrolled rectifier, full wave uncontrolled rectifier. But instead of starting from zero to Pi, we started from zero from zero to Pi. Instead from zero to Pi, we starting now from Alpha to Pi plus Alpha. So let's look carefully here. We start at Alpha, it starts conducting until Pi. It should be switched off these two sirstors. However, due to the stored energy inside the inductor, it will lead to conduction or continuous conduction for current through this two until from here. Until Pi plus Alpha. You can see here in this figure until Pi plus Alpha, at Pi plus Alpha, these two other switches will take the firing angle leading for the two will start conducting. Again, the voltage will return to the postive value once more. So I hope this is clear for you. Now, this is four continuous, you can see here starts at the very first cycle, zero, then at Alpha will start conduction and then it will repeat itself. So at steady state, it will be like this. Having a negative port, then Alpha start conducting, then Alpha plus Pi and so on. This is in this figure representing the steady state operation. Now, let's talk about the condition, the condition for continuous operation. So if you remember from the previous lessons, when we were talking about the discontinuous mode of operation, we said that Beta is less than Alpha plus Poi which means that the current will decay at Beta, right? So at Beta, the current will decay to zero. Now, in order to have a continuous current, we should not at least at least Beta should be at least adhere, right? So let's say, for example, it goes from Alpha, starts again until Pi plus Alpha, then start again. Decay and start again, so it will be continuous. So the minimum value for Beta in order to have a continuous waveform is that Beta must be at least equal to Alpha plus pi. Or we would like a value of current for a continuous waveform. We would like value of current at Pi plus Alpha, greater than or equal zero, so that the waveform is continuous. So as you can see here, I Pi plus Alpha, greater than or equal to zero, 00. Okay. So let's remember again, our equation, V max of Z, sun omega t minus theta, minus San Alpha minus Theta, E to the paw negative Omega t minus Alpha or Omegata. This is from Alpha to Beta, right? This equation here which you see right now is the equation for discontinuous mode. Now let's think about this with logic. If I would like to transfer this discontinuous waveform or this discontinuous current equation into continuous mode, what should I do? By logic simply Beta, we will assume that Beta or Peta at least at least equal to Pi plus Alpha, right? The minimum value for Beta for continuous mode of operation. So what I'm going to do is that I'm going to substitute with Pi plus Alpha in this equation. And see the condition required for having discontinuous waveform, a continuous waveform. So first current will be equal to V max over Z sine. Now remember, Beta here or Pita is Alpha plus Pi, the minimum value. Let's say at Alpha plus Pi like this minus eta minus sine. Alpha minus eta, E to the power negative here Omega T, which is Alpha plus Pi minus Alpha will give us Pi, so it will be Pi over Omega tau, right? So this here representing the equation after substituting with Beta equal to Alpha plus Pi. Now, what I want in order to have a continuous current, I need at this value I would like the current to be greater than zero, right? So I want this equation to be greater than zero. So what does this mean? It means that this function must be positive, right? Okay. Now look carefully here. Sine Alpha plus Pi minus eta. Sine Alpha plus Pi minus it. This one, this part is exactly equal to sine eta minus Alpha. So where did I get this? If you remember from mathematics is that if we have an angle, let's say, Pi plus Pi, this will be equal to, let's say, sine negative Pi from trigonometry, from trigonometry, sine angle plus Pi or 180 degrees is equal to sine negative the angle. So if you look here, we have Pi. And we have the angle. The angle here is Alpha minus set. So in order to remove the boy, we will just reverse Sta minus Theta minus Alpha. Okay. So now let's look carefully. I would like, again, if you remember, my own goal is to transform this into a posted value. So the condition for a posted value is that eta must be greater than Alpha to have this part becoming positive, Theta greater than Alpha. For this part, again, the same condition. Theta greater than Alpha gives us negative value multiplied by negative gives us another positive value. Again, positive Theta, greater than Alpha gives us a positive value. Here, Theta greater than Alpha gives us negative multiplied by another negative gives us positive. In general, the result for all of this is a positive value. Simply, the condition for continuous current is that theta must be greater than Alpha. Now let's see this. So if Theta is greater than Alpha, then this is a trigonometric equation or trigonometric rule that I have just used. Here, we said that theta must be greater than Alpha or equal. If it is equal, then at Pi plus Alpha, current will go to zero if Alpha equal to Theta. So eta, as we remember from previous lessons is ten minus one OmegLR, right? So Alpha must be ten or equal ten minus one omegalR four continuous cart. Now, how does this will help us? This will help us to control the circuit to be continuous by either controlling the firing angle to be less than ten minus one omic L over R, we have L and R or by controlling ratio between inductance and resistance, we can control that waveform for current or the mode of operation for the circuit being continuous or discontinuous current waveform. Okay, this is for designing the circuits to help you understand how to design these circuits. If I would like to make it continuous or discontinuous. Now, if I would like to get our usual values, average root meni squares, this is very easy as you can see here. Let's start for one complete cycle we have from Alpha, until two Pi plus Alpha two Pi plus Alpha, this from here to here representing one complete cycle. During this complete cycle, we can integrate from Alpha to Phi plus Alpha, this period here and multiply by two, we say one over Pi, similar as we did in the previous lessons, one over Bi, integration from Alpha to Pi. V Maxine omegaty. This integration, since we are saying one pi, it means we multiply the original equation or the original integration by two. If you remember the original integration, 1/2 pi, since we have two parts here, it will be one bi. This will give us two V max over Pi cosine Alpha. That is the first equation. For average, it will be the same. It will be average over R, which will be two Vmax over Pi R cosine Alpha. Now for the root mean squared, the same function but square of this function and all of this under the square root. So it will give us Vmax over root two. For the root mean square current, we divide by Z or the total imbedance like this. As you can see in this lesson and all of the other lessons, we are repeating the same equations, different circuits, but they are having the same concept, okay? Now let's say our circuit is highly inductive lud. We have an inductance with a very, very large value compared to the resistance. In this case, we will convert this pulsating DC wave form into a constant line, right. So let's look at the circut. Here this is our circuit from Alpha to Pi plus Alpha, then the other two conduct, the supply will go through here. Again, just for making sure that the explanation is clear for you from Alpha to Pi plus Alpha, T one and T two will conduct, and at Pi plus Alpha, T one and T two will switch off, and T three and T four will switch on. Again, the value of the output will return from negative to positive one more time due to the firing angles applied, okay? As you can see here. Now, Out becomes a short circuit like this ID representing the loot current. As you can see here, ID loot current is a straight line or a constant value. Now, for the current here from here, to here or from Alpha to Pi plus Alpha during this period, T one and T two or T one and T two will conduct. As you can see here, from Alpha to Alpha pas pi, T one and T two and from Alpha pas Pi to two Pi plus Alpha from here to here, the restors T three and T four will start conducting. Now, during this period, T three and T four will also conduct. Which is similar to this sport here. The sport, similar to this part. Okay? Okay. What you can see here is this, this waveform or this shape is considered as a square waveform, square waveform, not rectangular, but square waveform. Okay? I have to correct this mistake. This one is a square waveform because the period of conduction is equal to the period of non conduction or period at which the sstors are turned off. Okay? That's why it's called a square wave. If they are not equal to each other, then in this case, they will be it will be a rectangular wave form, okay? For the supply current, you can get it from here or from here, right? I supply equal to. We have I four I one and I supply. By applying ECL at this point, you can see I four entering, I supply entering, I one leaving. So I supply will be equal to I one minus I four. So we have I one minus I four. Thus minus this gives us this waveform. Okay. Now, the same idea can apply it from here. Eye supply goes like this and through the loud, then get back through here, right? We have here I supply two returning to negative of the supply. And we have I three like this, pointing award, and we have I two like this. So if we apply KCL here, I two entering, I reliving and eye supply leaving. So if I would like eye supply, it will be I two, minus I two minus three, which is this waveform, minus this one, which will give us the same solution. So either way, they will lead to the same answer, okay? Okay. So we have here the average will be the same, nothing change it at all. We average one Pi from Alpha to Pi plus Alpha, exactly the same. For the current, it will be also I average divided by R. But remember here in this circuit, RMS will be the same and IRMS this is very important. Now remember, since our output is a constant value, it means that I root means square, as equal to I out equal to I average, right? Equal to the value of I average because it is DC waveform, DC output. That's why I average, which we obtained here is the value of root mean square and value of the constant line. So IRMS equal to I average, equal to I output. Remember, IDC or I average. Remember, IRMS is not equal to RMS divided by Z. Why? Because this one is a constant or a DC value. That's why IRMS is equal to average, equal to constant value. Okay? This is very important, similar as we talked about this in the previous lessons. Now this is very important, and it will help you get it will help you on the half controlled in the half controlled rectifiers. It will help you when selecting the ratings for solstors and diets. What I would like to do is I would like to get the root mean square current rating and the current rating for thyristors, for example, in this video. What I'm going to do is that current rating for power electronic devices. For I average, it will be conduction time during one period, divided by two pi multiplied by output. For the root means square, it will be root conduction time, divided by two pi, multiplied by I out. What does this mean? Let's say I out here, which is a constant value or I average. Let's say it's 15 and pair. So we will take this 15 and pier and put it here. That is the first thing. Second thing, which is I average, it will be conduction time divided by two pi. So let's apply this rule to Pistors one and two. As you can see, they conduct from Alpha to Pi plus Alpha right so this period is Pi plus Alpha minus Alpha. This width Pi plus Alpha minus Alpha, which means they are conducting four Pi, right Pi. So what I'm going to, if I would like I average for iristors it will be simply equal to Pi, which is the conduction time divided by two Pi. Which is half, equal to half on the blood by I output, which is 15 and B gives us 7.5. Let's apply the same rule for the root mini square. It will be root conduction ti, B divided by two Pi multiplied by I out, so it will be I out, divided by root two. So it will be 15 and pairs divided by root two. We will see this and double in the next examples, okay? This is a very easy rule that will help you in solving problems related to the ser restors or related to selection of ratings for power electronic devices, and it will help you also in the half controlled rectifiers. When we go to half controlled, you will understand what I exactly mean by using this rule, okay? 52. Example 5: Hey, everyone. In this lesson, we will have example number five. In this example, we have, as you can see here in this figure, we have fully controlled rectifier. But in this example, we use center taped transformer, which uses only two thigh restors. That two pulse midpoint connection biphase half wave circuit is supplied at 120 volt line to neutron, which means voltage here, this voltage, 120 volt as a root mean square line to neutron. Two pulse midpoint connection is exactly the same as center tubed transformer, and biphase is exactly similar to center tubed transformer. So all of these definitions are talking about the same device. Okay, so what we would like to get is number one, determine the mean load voltage for firing, delay angle Alpha of zero degrees, 30 degrees, 60 degrees, 90 degrees. And this is very important. Assuming a constant voltage drop of 1.5 volt across each OSI restor when conducting, okay? Number two, determine the required rest of ratings, given that the lot current is continuous, current is continuous and level with a value of 15 and pairs. Okay? Okay, so what I'm going to do number one or the first step. Number one, here the current is continuous. Okay, continuous waveform. I will conduct from Alpha to Alpha plus Pi from Alpha to Alpha plus Pi, then second soil restore will conduct from Alpha plus Pi to two Pi plus Alpha. As we explained, exactly the same waveform as the bridge controlled rectifier. Let's look at the waveform from Alpha, to Alpha plus Pi f t one, instead of having in the bridge control directive we had T one and T two. Here we have only T one and during the other half from Alpha plus Pi to Alpha plus two Pi, we will have second Cirston. Now, I have here, continuous current and the level here level, what does level mean it means constant value. It will be a straight line like this of 15 and pairs. Now T one will conduct from Alpha to Alpha plus Pi exactly as we explained for the bridge control fire, and T two will conduct from Alpha plus Pi to Alpha plus two Pi. Now remember, here, this is T one, right? This is T two from here to here. So as you can see, for T one before it, this part is conduction for tit. Here also conduction for tit. That's why you will see a part of the wave before T one. This is the induction for T two. This is what we have done also in the previous lessons. Look carefully at this waveforms, and if we get back here, you can see here T one and DT two in the bridge from Alpha to Alpha plus Poi and you can see same part which I'm talking about regarding the i restores. And here from Alpha plus poi to two Alpha plus. So it is exactly the same waveforms. Now, what about this one? This representing voltage across T one, and we learned it voltage of the restor or diet. When we talked about uncontrolled center tab, we said that the voltage here, applied across it can reach up to double the supply by applying ivial, right? So during conduction from here to here, it will become a short circuit or to be more specific and providing the correct values, it should be 1.5 volt. It should be a little bit like this 1.5 volt during the conduction. However, since the 1.5 volt is very, very small compared to the supply value of 120. That's why it is almost very, very close to zero, but you should put it a little bit higher like this, not zero, but a little bit higher. Okay? This is during the induction. Now, when it is off, you can see from here, VT one from here to here, from here to here, and from here to here, the voltage of light on it will be double the value, right? It will be two supply, as you can see here and as you can see here. We learned about this before in the H in the center taped transfer Okay. Now, what we need exactly is number one, determine the mean load voltage for a firing delay angle of zero, 30, 60, and 90 degrees. So first, we have the values for our circuit V max 120 root two to convert it to maximum value, I load 15 and purs voltage of the restor is equal to 1.5 volt voltage drop right. So I would average will be one over by integration of Alpha to Pi plus Alpha, Vmax, nomgat minus VT, d Omega T, right. So it will be equal to Y VMX nomialty because during induction from here, to here all of this period, right? We had Cyst T one conducting from here to here and CistorT two conducting from here to here. So one Cistor all of the time is conducting, right? That's why it will be V max omegty minus VT because inducting all of the time. So it will be two Vmax over B cosine Alpha, which is the original equations that we obtain. Let's get back here to Vmax, here, as you can see, V cosine Alpha, but the difference is minus Vt because it's conducting all of the time it will be -1.5. If it is conducting for a shorter period, it will be multiplied by conduction time divided by two pi, and we will see this in the half controlled sects. Don't worry about this point. I will show you an example when we go to the half controlled sects. Suprise substituting with zero, 30, 60, 90, you will get the average as these corresponding values. Okay? Okay, so we obtained average lute voltage. Number two, determine the required systo ratings. Given that the lot count is continuous, blah blah blah, blah, blah, blah. So we need required stor ratings. What do we need? We need big reverse voltage and big forward voltage, big reverse voltage, maximum voltage can be two vs. And the maximum negative voltage will be also two vs. So it will be double this value, right? This is the worst case. We select the iristors ratings depending on the worst case. So number one, big reverse b42v max, which is 140, double 120 root two. First thing. Number two, we need the average current and the average current and second one, which is the root mini square current. So as we know, I average will be 15/2, 7.5, and I root mean squared, 15 over root two, 10.6 and bear. Where did we get this? Someone will ask me, Where did you get the values? Okay, this is very easy. All you have to do is to look at a period of conduction. You can see it's conducting from here to here, which is Alpha plus Pi minus Alpha, which is Pi. So again, if you remember the previous rule, this one, Mm hmm, this one here, these two rules. Conduction time, which is Pi divided by two Pi gives us half, multiplied by 15, gives us 7.5. And for root men square root, half which is one over root two. That's why these values are like this. 15/2, 15 over root two. Okay? 53. Example 6: Now let's have another example, example number six. In this example, we have a single phase, fully controlled bridge rectifier circuit, loaded by a resistive load of 20 ms. Us apply is 240 volt 60 rts for Alpha equal to 40 degrees determine. Number one, big load voltage and current. Number two, average load voltage and current. Number three, the root mean square load current. Okay, so this is a fully controlled circuit, right, with a resistive load. Resistive load. Okay. So what is the period of conduction? It will conduct from Alpha till Pi, right? Each sours T one and T two will conduct from Alpha until Pi, and T three and T four will conduct from Alpha plus Pi till two Pi. Why is this Because we have only a resistance. We don't have any inductance. So it will be like this, as you can see, from Alpha, two Alpha Alpha Alpha two Pi and Alpha plus pi to two pi. This one is zero because we don't have any inductans. T one and T two during the first half cycle, and T three and T four during the negative half cycle. As you can see here. Now, Iot will be similar to VO, except it will be divided by R, Vout over R because it is a pure resistive lute. So same wave form, except it is divided by R from Alpha to Pi, Alpha plus Pi to two Pi, okay? Current of T one and T two, T one and T two will conducted during this period. I but will be equal to currents of T one and T two or T one, which is equal to T two. I T one and T two, which are equal to each other will be Alpha two pi, same way form as Ibut. Now, during negative cycle, T three and T four. I output will be equal to current of T three, which will be equal to T four. So I output, same current as T three and T four. Other than that, it will be equal to zero, zero. What about I supply? I supply here, as we said before and said hundreds of times T one minus T four. This we form minus this we form, so we have positive minus zero positive, zero minus positive gives us negative. So this is our supply count. Okay? Now, what is the peak load voltage and current peak load voltage is very easy. Peak voltage is V max of the supply, which is Vmax, root two, 240 root two, right? What about current? P current will be peak load voltage divided by R, which is 20 ms like this value, divided by resistance 20 oms. What about the average load voltage? Very easy. It will be integration from Alpha to Pi, which will be one over Pi, integration from Alpha to Pi, right? Like this one overbi integration from Alpha two Pi period of induction, V max sinomity. Again, since it is repeat two times, it will be one overpi If it is half wave rectifier, which means it operates only here and here is zero, it will be 1/2 Pi. Okay? Okay. Now, let's after integrating this, you will get this function, which is 120,190.8 volt. V average. Okay. What about loud current average loud current. It will be average divided by 20 ms or the resistance, right? Okay. What about the root mean square loud current? I would like the root mean square current. It will be VRMS square divided by R can be obtained like this VRMS square divided by R. Or you can do another thing, which is taking V maxin omegati divided by R or square. Under the square root. This or this. As you can see here, we took the second solution, which is one by integration of Alpha two pi. This V max omega, the waveform divided by R gives us this I but or but current waveform. By doing this integration under square root, you will get 11.6 bars. 54. Example 7: Now let's have another one example number seven. In example number seven, we have a single phase. Again, the same values as the previous example, except that instead of having a pure resistive load, we have a highly inductive load with the same resistance, same supply, same frequency, same delay angle, everything, and the same requirements. So what will be the difference since we have a highly inductive load? It means current is continuous. So since the current is continuous, each si restor T one and T two will conduct from Alpha to Alpha plus Pi and T three and T four will conduct from Alpha plus Pi to two Pi plus Alpha, as you can see here. So as you can see, V out from here to here, first two si restors and from here to here, the second two pi restors. So the first two istors will be from Alpha two Pi plus Alpha and second sisters from Pi plus Alpha to two Pi plus Alpha. And I supply will be this with form minus this waveform. It will be likes. Okay. Another thing, I output highly inductive fluid becomes straight line like this. Now, what about VT one and VT two? We don't need this, but I'm just going to write them for VT one and V two, we said that maximum reverse voltage is V supply as we learn it in the uncontrolled circuit. So as you can see, like this, maximum value, Vmax, not to Vmax in the center tab, but here, maximum value is Vmax, okay? Now VT one, they conduct from here to here, so it will become a short circuit. Other than that, it will be equal to V supply. Now, what is the big load voltage and the current? Exactly. Big load voltage is V max, which is 240 root two, as we did before. What about big loud current? Don't be don't be trigged by this sentence. Peak load current is not equal to, not equal to, equal to, not equal to. Peak load voltage divided by resistance. Don't do this. So what is the big loud current Bg out current. Is equal to average loot current, equal to root mean squared loot current, equal to V average output average divided by R, it will be like this. We average number one, it conducts from Alpha to Pi plus Alpha two times, so we will divide by one overbi. So it will be equal to 165.5 take this value and divide it by the resistance. Okay. So as you can see, I load, equal I average, equal IB, equal root mean squared equal to V average over R. Now, again, why resistance only Because this average across RL is equal to V average across the resistance only. V average across the inductor is equal to zero. Okay? Okay. And so this is for the currents. What about average load voltage? We already obtained it. So all of this is obtained, only one remaining is power absorbed by the load. Power absorbed by the load is the root mean square current square, multi bled by R, or I average square multi plod by R because all of them are similar to each other. So it will be 8.27 square multi blood by the resistance. So this current square multiblod resistance gives us the value of power consumed by the resistive fluid, 1.37 kilo watt. 55. Half-Controlled Bridge Rectifier – Type 1: Hi, and welcome everyone to this lesson, which is half controlled bridge rectifier. So in the previous lessons, we talked up uncontrolled rectifiers, fully controlled bridge dictifers. Now what we would like to learn is half controlled. What does half control mean? So in the fully controlled, we have si restors forming our bridge. In the uncontrolled, we had dites forming our bridge. Now, in the half controlled, we will have a mixture between these two. It will be half controlled. Half of the bridge will be diets, and the other half will be si restores. So we have two types for half controlled or two configurations. In this lesson, we will too about the first type which is type number one. Now let's look at the circuit. If you remember in the fully controlled, we had four pi restors t1t2, T three, T four. In the half controlled, we will replace half of the sistors. We have two sirstors as they are. And we replace other two crestors with dites. That's why it's called half controlled because half of the bridge is consisting of si restors. You have two configurations. Type one, it will be like the sistors and two dites. In type two, it will be like this. Two CI restors like this and two dites in type two. We will explain each of these circuits in separate videos. Starting with type number one, T one, T two, D three and D four. Now let's look at the waveform. Now remember, here we will discuss them with brasons of highly inductive load, okay? Highly inductive allude. So it means that our loud here is consisting of an RL lute, RL lute Okay. And inductance is very high, leading to having an output current with a straight line like this, constantive value. Okay? Now, let's look carefully how does this circuit work? Okay? We will start again from the firing angle Alpha, Neglect what is previous, what is behind Alpha. Okay? So we will start from Alpha when the firing angle Alpha is obliged to restore T one, this one will become a short circuit during the postive cycle. During postive cycle, this one will become a short circuit. The current will go like this, go like this, through the lute and get back. We have going out from the postive terminal. Of the supply and it would like to return to the negative terminal for the supply. How it will do this by going like this, going like this and through D four because it would like to return back to the negative of the supply, like this. So during the first conduction mode between Alpha and B, let's start from Alpha to Pi from here to here. This will be the mode of operation. T one will be on and D four will be on. So as you can see, between here and here, T one and D four. Now, if we apply KVL to the circuit, you can see here by KVL here, you'll find that output is exactly equal to supply, right? During this eval using this civial, right? So from Alpha to Pi, out will be equal to V supply. As you can see, from zero goes up, Out becomes exactly equal to V supply until Pi, okay? Okay, very good. Now what about from Pi? Remember, T two will not operate until Pi plus Alpha. This is very important. T two will start operating here. So T two is already turned off, right? Remember this T two is already turned off, and T one is already turned on and D four on, right? Okay. Now, when we start going to the negative part of the supply here, going to this negative part. Now remember, and if this is a pure resistive load, if it is an R, it should be at this point, it will be turned off. T one will be turned off, right, and V will be equal to zero like this. Because it is a pure resistive luid. However, since we have a highly inductive load, RL, this will provide more current or inductance will provide stored energy to keep the current in existence. The current will go like this in the same direction. But let's look carefully how does the current will behave? This is very important. Now look carefully here during this part, this supply is reversed polarity like this, positive, negative. So this supply would like to provide current in the opposite direction like this, right through this terminal. Like this. However, remember that we have a highly inductive load with a large stored energy. The current will remain in existence. Here, this current from this suplight would like to go like this and through the loud light, but it can't go like this because D four is turned off, it will be reverse biased and it cannot go through T two because T two is turned off. I didn't get its firing angle. So the supply cannot provide any current. However, it provides a reverse voltage, a voltage that opposes the current flow. Let me explain this. Here, we would like the current due to the inductance, it should be remaining, right? It should be existing. So it will go like this. It has two options. Look at this. This is very important. It has two options. Number one, it can go like this through D four, through the supply like this, supply through T one and go back to the loot, right? Like this, go like this, through D four and phase the supply and get back through the loot, right? This is the first option for the current. The second option is moving through this through D three like this. And T one and get back to the loot. So again, the current, due to the highly inductive load, it will provide more current. This current can go through two passes. Number one, it can go through D four, then supply then T one, or it can go directly through D three and T one and get back. Which one is easier for the current? Of course, the second pass, which is the way through D three and T one because it doesn't have to face the supply. That's why you will find that. The current will go through here, go through, T one and get back to the loot. That's why from here to the next firing angle Pi plus Alpha during this period, T one and D three will operate. Why? Because the current does not have to face the supply. Now, what will happen when the current goes like this, like this. Okay. You will find that T one and D three will form a short circuit leading to V out becoming zero, right? Short circuit, parallel to V. That's why here from Pi to Alpha plus Pi, V output is equal to zero, as you can see here, zero. Okay? Why due to the short circuit, parallel to the RL loot. Okay? So T one and D three will operate in this part, okay? Now, what about from here to here? Now, when starting from biplus Alpha, T two will get the firing angle, right? So it will turn it on like this. And remember, this is during the negative cycle, positive, negative. So the current will go like this through T two, get like this, like this. Okay. And through the three because it would like to go to the negative terminal, like this. T two and D three will be on during negative cycles. T two and D three, two, cybersor and one bite. From Alpha plus Pi until two Pi. Now, again, the same thing will happen here for the lute, starting from two Pi until two Pi plus Alpha. This supply will reverse its sign again here. It will become positive again, as you can see here, positive, negative. Since this one is a highly inductive load, it would like to provide current in the same direction, the same direction like this. It will go like this and T three, remember, T one is turned off. T one is turned off during negative cycle. It has two options. It can go through these three, it can't go through these three. It has only one option. It can go through the three like two facing the supply through positive terminal, then get back like this through T two and get to the loot. Okay. That's the first option. Second solution is to go like this from the loud, go through D four, T two and get back. So which one is easier again, T two and D four. So T two and D four will form a short circuit parallel to the loud. That's why in this part, it will be zero. Okay? Okay. Then T one and D four will repeat itself and so on. Okay? Okay. Now, as you can see, IT conducts from T one, as you can see, conducts from. Look at this values from Alpha until this point, which is Pi plus Alpha, from Alpha to Pi plus Alpha. T two will conduct from here, Pi plus Alpha until two Pi plus alpha. You can see all of this T two, like this. Now for the die, this is very important. They are not similar to each other. Look at D one and D four. D four, let's look at here. Go down here. D four. D four conducts here and end is here. So it conducts from Alpha until Pi. Then if you look at D four, it will start conducting again at this point, which is here to Pi. Until here this point, which is three boy. So actually, if you look carefully here, you will find that D four also conducts here. Now, when you get this, you can look at it very easy. Look at T one and D four, starting from here. You can see there is an extra part here compared to T one, right? So there will be an extra part here compared to T one. Okay? Okay, so it will conduct here, let's lead this. Four supply current oil supply, again, from here, oil supply will be T one minus D three, or supply will be T two minus D four. Okay, T two minus D four, right? Here supply going, this one going, this one is living to entering, living and living D four minus T two, D four minus T two. Both of them will give you the same solution. Okay. So we have our waveforms here I one minus ID three, this one minus T one gives us this waveform, as you can see. Now, what we would like to do is, again, number one, V average, average here is V max of Pi one plus cosine alpha this part here, this one representing integration, exactly equal to one over boy. Integration from Alpha to Pi for Vmax, sine Omegaty, right? D omigaty, of course. One over Pi because it repeats itself two times one and two. Integrating from Alpha to Pi, you will get this equation, V max of Pi one plus cosine F which we have explained lots of time before. Now, the difference here is that if we neglect voltage drop on thi restors and dites, then this part will not exist. If we consider the presence of voltage drop on thirestors and dites, then we have to add them. Now, of course, it will be to be more sepsi it will be minus visor Multiplod boy. Period of conduction. Let's say, PT with respect to the total period. Period of conduction for sister with respect to the whole as period minus D, multiploid boy, period for diet of conduction of diet, divided by the total period. This rule which we have used in I average, if you remember, in selection of ratings for Cistors and diet or power electronic device in general. Okay, so period of conduction. So our Out exist from here in one complete cycle, of course, from here to Alpha Alpha until here. Okay? Okay, one complete cycle like this. So let's look at TI restores. You can see T one, T one, T two, T two. So during one complete cycle, during one, at least one CI restore is operating, TI restaurant diety restaurant diet, si restaurant diet. Okay, Ti restaurant diet. So in every part of this cycle, T one always conduct. So period of conduction for Ty restore will be two Pi, one complete cycle, and the period is two Pi. What about die? Four die, you can see here, D four, D, D three and D four. Also it operates at least one dite is operating through the whole cycle. So this will be also two Pi period of conduction for diet, divided by two Pi. So this will go this, this will go this. So we'll have negative Visor, negative V dite, as you can see here. Now, when we look at a different circuit or with a free wheeling dite, you will see the difference between this and the other one. Now what about Is root squares? The root mean square of the supply current. It will be simply from here by subtracting T one and D three. So it will be this waveform as we have talked before. Now let's look at I supply. It operates between Pi from Alpha to Pi. This buid is Pi minus Alpha, and the negative part is also two Pi minus Pi plus Alpha, it will be Pi minus Alpha. So in order to get the root mini square current, it is very easy. Again, the same rule for ratings. It will be period of conduction. We have Pi minus Alpha plus Pi minus Alpha, so it will be two Pi minus Alpha divided by the whole period, two Pi. Whole cycle two times Pi minus sulfa repeats itself two times in one complete cycle. Pi minosalf. So if we take this with this, you will have Pi minus sulpha divided by Pi, as you can see here. Multi ploid boy, of course, I since it is a constant value. For the pistor ratings, number one, big reverse voltage equal to the peak forward voltage, equal to V max of the supply. We talked about this before, and we said that the maximum voltage on a pystor or a diet is V max, right? Now, what about Pistor RMS or the root mean square of the pistor. Root mean square, it will be again, very easy root, period of conduction, period of conduction, divided by the total period, multiplied by I output. Period of conduction, it conducts from Alpha to Pi plus Alpha. This period Pi plus Alpha minus Alpha gives Pi. So during one complete cycle from Alpha to two Pi plus Alpha, during this complete cycle, it operates only for Pi divided by the total period to Pi. This will go all this. It will be I output divided by root two, as you can see here. Now, what about the I diet? Four died ratings, number one, big reverse Volge again, equal to big forward vulge equal to V max, nothing changing here. For the root mean square, currently you will find it operates ID three. For example, it operates from here at this point, from Pi until two Pi D three, for example, from Pi to two Pi. This period is equal to Pi or period of induction is Pi. It will be Pi divided by two pi which is half. Under the square root will give I out of a root two exactly the same value as this. So these are the ratings for stores, we average everything regarding the half controlled Tai is with a highly inductive load. Now, what if we added a free wheeling diet? If we add a free wheeling diet here, what do you think will change? I will tell you, instead of having these three here, Look carefully here. Instead of having T one and D three during this negative part or T two and D four during this negative part or positive part here, or during short circuit here or here, we will have our free wheeling diet we conduct here, and the rest will be the same. During this free wheeling diet period, we will use a diet instead of Cistors and D three or T two and D three and D four. Look carefully here. Okay, same circuit, diet. Let's look. You can see here. At this point, the diet will provide the free wiling diet process here. And here and here. And instead of, let's look here instead of T one and three, T two and D four. The rest, T one and D four, T two and D three. Lo carefully. T one and D four, T two and D three, everything as it is except that the freehing diet will operate during the extra energy provided by the inductance or preventing the diet from going or preventing the output from going to the negative portion of the cycle. Okay, what next let's draw the wave forms. So IT one will operate from here to here, I one, Okay, from here to here, Pi minus Alpha, and I T two, I t two, which is this one will operate from here to here, from here to here, which is Pi plus Alpha until two Pi. Until here from here to here, Pi plus Alpha until two Pi. And when we go to the negative cycle, the free willing diet operates. So Pi plus Alpha two Pi. Okay. Now, what about the free willing die? The free willing diet will take the rest, the sport, and this part between these two and here you can see here and here and here, which is this part, this part, this part. Okay, what about I supply? I supply again, D one minus D three, which is this way for. Subtract this one from this one like this, you will get the supply current. Okay? We average V max over by one blast cosine Alpha, this is average without considering any kind of voltage drop. F from integration from Alpha to Pi. Now, what about the voltage drop produced due to the thstors and diet free weighting diet. What we are going to do is that I will say minus like this, V ist minus V diet, minus V three wing diet. Like this. Then what are you going to do? Number two, take two Pi here, divided by the whole period, divided by the whole period, like this. Great. Number two, we need or number three, we need to add the period of conduction for each of these devices. Now let's look carefully at this circuit. T one, conduct one, let's look at the current. It conducts from Alpha to Pi. Period of conduction during one complete cycle from here to here, one complete cycle. It operates for this period only Pi minus Alpha difference between these angles, Pi minus Alpha, it will be Pi minus Alpha, right? Okay, Pi minus Alpha for t one. Now what about yeso number two? Sison number two bridge from here to here, which is two Pi minus Pi plus mel Alpha plus Alpha, it will be Pi minus Alpha. The subtraction of these two values will give us Pi minus Alpha. So during one complete cycle from here to here, from here to here, from here to here. During this complete cycle, we have two cyruses conducting T one and T two. Each one conduct for Pi minus alf. So that total volt drop or total period of conduction, it will be multiplod by two. Because we have two thy restores, Ty restor number one, Ti restor number two. Pi minus Alpha, Pi minus Alpha. So the total period of induction is two Pi minus alpha. Four diet diet, as you can see, operate from here, during this period, D four and D three operates during this period. As you can see, T one, D four, T two, and D three, so they have the same conduction period. T one, thstors and dives have the same conduction period. So it will be also two Pi minus L like this. Okay? That's why you will see here that if you take this one with this one, this one with this one, it will be Pi minus Alpha over Pi, Pi minus Alpha over Pi, VT plus VD, and negative sign at the common factor. So VT VD, because they have the same period of conduction, we add them together and multiplied by period of conduction. What about the free willing diet? I diet or I free willing diet? It operates between in one complete cycle, from zero to Pi or from Alpha two Pi plus Alpha. Whatever it is, it is the same solution. We have here from zero to Alpha zero to Alpha, it means that the period here is Alpha. Alpha minus zero gives us Alpha. And we have here Pi plus Alpha minus Pi, which is also again Alpha. And nothing else between the complete cycle here, it conducts four Alpha plus Alpha, which means two Alpha, right, this period of conduction and this period of conduction. So it will be two Alpha over two Pi. So take this with this, it will be Alpha divided by Pi, as you can see. Okay. So this is the explanation of where did we get the free wheeling dite and Ti restores and dies values. Okay? Now, if you would like the root mean square supply current, it will be the same as before from here to here, which is Alpha, Pymnus Alpha divided by Pi because Pi minus Alpha operates a conduction of the supply current two times one and two. So it will be Polpha over Pi not two Pi. Okay, or you can simply if you don't understand exactly if you would like to make sure that this equation is correct, simply you integrate from Alpha to Pi for constant value plus integration of Pi plus Alpha to two Pi for the square of the current all under the square root. We are getting the root mini square. Remember this. So we'll get the same equation. This is just a short cut. For the soy restandit ratings, again, big reverse voltage equal folded voltag equal V max, for the current. This is very important too. Now look carefully here. This is representing the period of induction in general, right? For soy restor and diet. Now, we thistor and we diet or cysto and it period of conduction depends on what here depends on the value of Alpha, right? So Alpha is between zero and boy. So where is the worst case? Or when does the value of conduction becomes the highest? When Alpha is minimum, right, which is Alpha equal to zero, right? When Alpha equal to zero, it will be bidivded by boy, which means one, right? I exist. When Pi equal to zero, it will be one. However, look carefully at the equation here. As you can see, root Pi minus Alpha divided by two Pi. Now why is this? Because you forget something, you forgot something which is important. This one for induction period for T one plus T two, right? That's why we divided by Pi. However, we are getting the ratings for each device. So let's say we are considering T one only. So it will be root period of conduction of T one only, not the total period T one only. It will be from here to here or by minus f. Divided by two Pi. This is for one Crestor. Here, when we are talking about voltage drop and when we are getting V average, we are talking about the effect of all of the devices. VisIristors, all of the ciistors, all of the dites. Here I'm getting the ratings for each cistordtes two Pi minus Pi plus Alpha will give you Pi minus Alpha over two pi. So they have the same ratings, same conduction period. Root Pi minus Alpha. The lowest value is zero, so it will be root Pi over two, which is one over root two. The ratings of the sstor should be I output divided by root two, which is the worst case when Alpha equal to zero. Okay. For free wheeling diet, again, big reverse volts equal Bk forward, equal V max exactly same as the s verstors and dites. For the current reading, look careful without even doing anything. For diet for free wheeling diet, it operates Alpha two times in one period. So it will be two Alpha divided by two Pi, which is Alpha over Pi, like this, which is this one, right? Now, Alpha can be from zero to Pi. So the worst case is Alpha becoming equal to Pi. If Alpha equal to Pi, it means that value will be equal to one. Blood by I output like this. The worst root mean square current, as you can see here, root Alpha over Pi when Alpha becoming equal to Pi, there a root mean square current will be I output at the worst case. Where did you get this from this period of conduction, divided by the whole period, gives us two Alpha divided by two Pi, gives us Alpha over Pi. All under the square root and selecting the worst case, which is Alpha equa Pi, then root mean square will be equal to I. I know that I explained a lot of details in this lesson, but I hope it's clear for you. I try to explain every single detail inside this circuit. I hope it's clear and you understand everything. 56. Half-Controlled Bridge Rectifier – Type 2: Hi, and welcome everyone. In this lesson, we will too about half controlled the bridge rectifier, again, similar to the previous lesson, except that we will too about type number two. So in type number two, as I have explained before, instead of having cysorT one and T two and D three and D four, this configuration will be changed. It will be T one and T two on one side, and D three and D four on the other side. And if you look carefully at this circuit, you will see something which is very interesting. You will see that D three and D and D three and D four are exactly providing the same function as a free wheeling diet, right. And if we have one free wheeling like this goes from the loud through D three and D four and get back like this. So D three and D four in this circuit acts as a free wheeling diet, and you will see this effect right now. So again, during postive cycle from Alpha to Pi from Alpha two Pi, D four and T one will conduct. So the current will go like this through supply and get back like this right as we explained for T one and D four, T one and D four from Alpha here until this point, T one and D four. Now in the negative half cycle from Alpha plus Pi here until two Pi, like this, during this period, D and T two will conduct, right, T two, D, T two and D exactly the same as before. But the difference is that during the period here from Pi two Pi plus Alpha and between two Pi and two Pi plus Alpha. During this period, this period and even here from zero to Alpha during all of this period, D three and D four will conduct, right, d3d4. D3d4, d3d4. D three and D four are operating exactly at C. They are a free wheeling dit, one free wheeling dit, okay? Now let's look at the currents. Since we are talking about a highly inductive loot, IO will be a straight line as we explained before for cirrus or T one, I will conduct from here to here and in the next cycle from here to here. As you can see, from Alpha two Pi and Pi plus Alpha, tell three Pi. Now for Cysto number two, IT two, it will conduct from here to here, as we learned before, from here to here, and even before a Tito only here, here, during the negative cycle, Alpha plus Pi to two Pi, Alpha plus Pi until two Pi. For the diet, you will see that diet. Let's look carefully. D four conduct during this period, and the three conduct is here and here, and here and here. Let's look at it like this. During free wheeling diet from here to here, from here to here, from here to here, free wheeling period, these three and D four. Let's look at D three and D four. Free wheeling, here and here, right? Freewheeling this part, if you go down here, this part and this part. Free wheeling dt here, this part and free wheeling communication or not communication. Commutation here and here is operating. These three and D four. For the rest, D four operates during post of cycle, and these three operates during negative cycle. You can see here, these three operates during negative cycle. And therefore during postive cycle from here to here. Okay, all of this is obtained from this graph. D three here, d three here, d three here, and here, you will find this and all of this. So by to simplify this explanation, all you have to do is draw this waveform and identify in which period each of these restores dites operate and using this figure, you will be able to draw these waveforms. Okay. I supply again will be here at this KCL at this point. It will be T one carant of restor number one minus carant of iRestor number two. I one minus I two by applying KCL at this point. So if you subtract this from this, you will get the supply. Okay? Now, let's get out average. Again, V max over piu plus cosine Alpha, this we average without considering any presence of a voltage drop on dine. Now, I would like you to really concentrate with me, because it is very important. Now, VSI restor, we are talking about SI restor. All of the voltage drop of every I restor. T one operates from here to here, right, and T two operates from here to here in one complete cycle, right? Okay. So from here, let's say from here, Alpha and two Pi plus Alpha, this is a complete cycle, at which we will provide these values. So during this period, T one operates from Pi from Alpha to Pi, right? So it will be Bi minus Alpha, period of conduction of T one, T two conducts also from Alpha plus Pi til two Pi, which means if you subtract these values, it will be Pi minus Alpha. So the total induction period of the Cirestors will be two Pi minus Alpha, right? So if you take the total induction period and divided by two Pi like this, you will get Pi minus Alpha divided by Pi, like this Pi minus Alpha divided by Pi, representing induction period with respect to total period of the whole period. What about the diet? Now, this is very important, I would like you to really, really concentrate with me because this is a little bit difficult, okay? A little bit. Alpha to tell Alpha plus two Pi. This is a complete period, right? Okay. Now, let's take each diet individually. D4d4 conducts from here. To here, D four from Alpha to Alpha plus Pi through this representing what representing this point, Alpha plus Pi minus Alpha gives us Pi. You can get it from here or you can get it from these figures. It is the same. D four conduct from here to here, D four, don't forget it also conducted during this small period from two Pi to two Pi plus Alpha, which representing this small part is Alpha. Okay, so it will be plus Alpha. This is the period of conduction of D four. Okay? What about What about D three? We need also D three because we are considering voltage drop of all of the lights and their period of conduction. The three conducts from here to here, right, which is Alpha. This part is Alpha. Plus from this also here from this point to this point, from here to here, this period, which is Alpha plus two Pi, Alpha plus Pi. Their difference or this width is equal to Pi. You can simply look at it in a different way D three, d three, d three. So all of this is d three from Pit two Pi plus Alpha. So their difference is Pi plus Alpha. Okay? So this the submission representing the total conduction for dite which will be equal two, two Pi, Two Pi plus Alpha, right. Now, divide this by two Pi, which is the total period, right? So two will go with two, so it will be Pi plus Alpha divided by Pi. So it will be Pi plus Alpha divided by Pi, right? Now you can keep it like this. Or if you divide this Pi divided by Pi, gives us one plus Alpha divided by Pi. Which is exactly this equation. Okay? So what I did it simply because we are considering the effect of each of these dites or the voltage drop of all of the dites during induction mode. So for example, to make it clear, during this period, we will have voltage drop of two VD at this instant, or in this period, okay? But I'm getting average for whole period, average, it will be like this, as I just explained. Now for the supply root mean square current after subtracting, you will get root Pi minus Alpha over Pi, this period, which is from Alpha to Pi. The difference is Pi minus Alpha and from here, which is Pi plus Alpha two Pi, their difference is Pi minus Alpha. So it will be two Pi minus Alpha divided by pi divided by two Pi, it will give us in the end Pi minus Alpha divided by exactly the same value as the type number one in bridge and half control the bridge. For thistor ratings, again, big reverse voltage equal big four voltage in bridge rectifiers, equal to V max. For the thistor root mean square current. Let's look at it. Each thi restor here, we are talking about each thistorvi individually. So each one, as you can see here, each one conducts four Pi minus Alpha, Pi minus Alpha gives us this width, which is Pi minus Alpha, with respect to the whole period to Pi. So root period of conduction divided by the whole period to Pi. So at the worst case, when Alpha minimum at zero, root Pi over two Pi gives us one over root two. This representing our design or rating of the thyrus Number two, did rating. If you look at any diet before we look at anything, look at diet, for example, diet number four. Diet number four conducts through one complete cycle, conducts from here from zero until this point, which is Pi plus Alpha, here, D 4d4d4, this period from Pi plus Alpha, all of this, so it will be root Pi, plus Alpha divided by two Pi, right? Period of conduction, divided by two Pi. Now, if I would like to get highest current, Alpha will be maximum because we have plus Alpha. The maximum value of Alpha will be Pi. So it will be root two Pi over two Pi, which will be equal to one, like this. Look at this, peak reverse volta equal to peak forward voltage of V max. Here root Pi plus Alpha over two Pi, the same exact equation. I output at the worst case, it will be Alpha equal Pi, so it will be I output. Okay. Now what if we have a free wheeling diet? Instead of D three and D four doing these two and instead of D three and D four, acting as a free wheeling diet, the free wheeling diet will do this process. So D three and D four, instead of the three and defour it will be D, and here it will be D and here will be D. So it will be like this. Look carefully, D D, D as a free wheeling diet or provide the service of free wheeling process. Okay, I add, as we have seen and the same waveform, except D three and D four replaced by T. So here T one conducts from Alpha to Pi T two from Pi plus Alpha to two Pi, free wheeling diet operates at Alpha at Pi two Alpha plus Pi Pi Pi plus Alpha, and operates here at two Pi and two Pi plus Alpha here, two Pi and two Pi plus Alpha, as you can see, and supply is the subtraction of IT one minus I t two, same as before. Now let's look at the average you will see that max over pion plus cosine Alpha, same average without considering any voltage drop. By adding voltage drop of cistor and dites, you will see T one and D four, T two and D three. So both of the ci restores and dites have the same induction period T one Pi minus Alpha, here, Pi minus Alpha, so it will be two Pi minus Alpha. For diet, it will be same induction period, same induction period. So it will be also two Pi minus Alpha. So both of them are similar to each other. So it will be VT plus VD, multiplied by just one of them, two Pi minus Alpha divided by whole period two Pi. So this will go this and we will be left with Pi minus Alpha divided ppi. The same what we have done in the previous slide and the previous lesson. For the free willing diet, again, Alpha and Alpha equal two Alpha over two Pi gives us Alpha over Pi. I supply root mean square, again, root 0.1 alpha over pic same form and same same equation. Si restor rating. Again, big reverse, equal big forward, equal V max, I restor root, Pi Alpha over two pi, exact as before. This is a period of conduction Pius Alpha over two pi for each i restor. That was the case Alpha equal to zero exactly similar as previous lesson and the previous slide. Hydrating same values, I diet again, it will be exactly similar to the sirestor because they have the same conduction period, right? For the I diet or free wheeling does this one will have this one will have the worst value of I output at Alpha equal to Pi. Remember, root two Alpha over two Pi gives us Alpha over Pi and the highest Alpha is Pi, so it will be root one multiblod by I output, which will give us this value. Okay? 57. Example 8: Welcome, everyone. In this lesson, we will have the example number eight on the half controlled circuits. So in this example, as you can see, we have type one of half bridge control dictiferT two, D three and D four. And these are the information or this is the information regarding this circuit. We have a highly inductive loud with a resistance of 20 oms. Thus apply is 240 volt, 60 ortiz, and firing angle is equal to 40 degrees. Determine number one, B load voltage and B load current, average lute voltage and current, the root means square loud current, the average ist current and power absorbed by the loot. So let's start step by step. First step is drawing the waveform for the butt. If you remember in Ty T one and D four will conduct during positive half cycle, from Alpha to Pi, and D two and D three will conduct from Alpha plus Pi to two Pi during negative cycle. And as a free wheeling diet, process D three and T one, if you remember, and T two and D four in the other half, this is a waveform that we talked about a lot before. Now, first, in order to get B load voltage and Blud current, first, we need B cloud voltage is simply V max, V max, which is 240 root two. This is straightforward, as we learned it before for Bclud current and thence we have a highly inductive loud, it means that I mass is equal to IP. Equal to I average, which is the constant value of the outt. So in order to get I average, we need V average. And if we remember, we average equal one verbi integration from Alpha to buy VMAX sine Omega t from here to here two times that's why we have one verb Pi. So it will give us VMAX pi one plus cosine Alpha, and Alpha is given as 40 degrees. So average will be equal to 100 under 90 1 volt, right? Here we don't have voltage drop, we neglected voltage drop across the presors and dites. In order to get I average, it will be V average, divided by resistance R, since it is a heavily inductive loot, divided by 20 oms, we will get 9.50 5:00 A.M. Pairs. So that's representing the average current, average output current, Obit root mean squared current and maximum or peak output current. So we obtained B cloud voltage, current average loud voltage, current with mini squared current. Now we need first the average systor currant, I average for thistor. Any of these cistor conducted from Alpha to Pi plus Alpha or for this one Alpha plus this period, which is two Pi plus Alpha minus Pi plus Alpha, which will be B plus alpha. So T two conducts for Pi plus Alpha in one complete cycle, and T one conducts also for Pi plus Alpha minus Alpha, which is equal to Pi, right? Okay, so it conducts for Pi Pi plus Alpha minus Alpha. So it will be equal to, I conducts for Pi. I average, as you said before induction period, divided by two Pi, which is the total bio gives us half. That's why half multiplied by I put gives us the average current for cysto. Now for sirestu number two, here it is also Pi not Pi plus Alpha, because if you look at here for one complete cycle, the cycle ends here at Pi. A two Pi. Okay? And this point is Pi plus Alpha, and this one is Alpha. So Alpha plus two Pi minus Pi minus Alpha. The total value will be Pi, or if you add it from here two Pi plus Alpha, you'll find it conducts only four Pi. So both of these Cystors conduct the same period, okay Pi. Now we need power absorbed by the load. It will be the current square multiplied by resistance, right? Nothing change it from the previous examples, okay? 58. Example 9: Now let's have another example. In this one, we will talk a poet circuit, the second type for half controlled circuits. In this one, we have D three and D two, which works at freewheeling Dot, as we learned before, right? Now what we would like to get, we would like for the rectifier circuit Jon sketch the waveforms of the uid voltage, uid current sistor and light currents for Alpha equal 60 degrees and Alpha equal nine degrees. The main goal for this example is to understand the effect of Alpha on the but waveforms. That is all for this example. Assume level load current, it means that our load is a heavily inductive load or a highly inductive load means that the output is a constant value. Now let's look at the difference between them. At Alpha equal six degrees, the waveform will be like this and at Alpha equal 90 degrees, it will be like this. Now, as you can see, T one and T two, T one and D two and D three and D two will act as a free wheeling die. If you get back here, T one, D two, during postive cycle and D three, D T four during negative cycles. If you look carefully, D one, D two, T four and D three, negative cycle, post of cycle. Four free wheeling diet portion, you'll see that D three and D two acts as a free wheeling dit, right? As you can see here, like this. Okay? Okay, so t1x from Alpha to pi, Alpha two pi, T four from biplus Alpha to two pi, Pi plus Alpha to two pi, and the D two can be obtained and eye supply. Now, let's see when the firing angle increased, you will find what? Number one, eye supply decreased. When firing angle increase, the supply current decreased, right number two. Look at sit Syritor conducted from Alpha to Pi. By increasing Alpha, let's say here, the induction period will decrease. You can see at 60 degrees, all of this at 90 degrees just the sport. So if you look at here, this is smaller than this one because Alpha increased. Similar for T one, T four will be the exact case. You'll find that um, wave form or the conduction period is decreased. As you can see, this one is greater than this part. Now, as a switch, since this period for conduction decreased, the conduction of D three and D two or the free wheeling dide process will be extended. So if you look at here at this one, this period here. Look at this between here and here. Look at here between here and here. You'll find that this part, this part is greater than or larger than this one. So it means that light D three and D two induction period increased. Okay? So what is the effect of increasing firing angle? Number one, V average decrease, right? VO average is equal to V max over by one plus cosine Alpha. As we increase Alpha, the output average will decay. Number two, induction period for si restores, T one and T four will decrease. On the other hand, D three and D two, conduction period will decrease, right? Will increase, will increase by increasing firing angle alpha. And sense T one and D T four conduction period is smaller. And I supply becomes smaller because I supply is simply equal to T one or thi restor T one minus the current of thi restor T four, okay? 59. Three-Phase Rectifiers: Hi, and welcome everyone. In this section, we will start discussing the three phase rectifiers. In the previous sections for this course, we took up the single phase rectifiers, including the half wave rectifier and the full wave rectifier. Now, these circuits that we discussed before are used for the single phase circuit or single phase supply conversion into DC. However, this section we will be discussing the three phase rectifiers that are used to convert the three phase supply ABC, which is commonly found in the electrical power system into DC voltage. Let's have an overview about three phase rectifiers and then we will learn about each of the circuits. So we have shown this map at the beginning of these sections regarding the rectifiers. We said that we have single phase ict fires. We said we have three phase ct fires. We discussed the uncontrolled single phase, half controlled, fully controlled, full wave split supply, which is the single phase with transformer with a midpoint. We talked about the bridge full wave rectifier. We talked about half wave rectifier. Now, we would like to do the same however, with a three phase supply. So we have uncontrolled circuit in three phase, uncontrolled, half wave and full bridge, uncontrolled. We have half controlled the bridge. We have fully controlled half wave and full bridge, and so on. We will learn about these circuits in this section. Okay, so why three phase rectifiers? Why do we use three phase rectifiers instead of the single phase rectifiers? Number one, single phase are used or single phase rectifiers are usually used in low power applications in a very small power applications, up to 15 kilowatt. So if you have a loud or a DC lute that I would like to supply with a rating up to 15 kilowatt, then we are going to use the single phase rectifier. Higher than this, we will start using the three phase rectifiers. About this po levers three phase rectifiers are usually employed. The three phase rectifiers have these several advantages converted to the single phase. Number one, they have a higher DC voltage and pet input power factor. Number two, they have less ripples inside the output current. It means that the current is almost or very, very close to DC outbit current. Therefore, we have a better load performance and lower size of the filter circuit because of the higher ripple frequency. Now, if you remember, this was our bridge rectifier full wave bridge rectifier in the single phase system. You can see that the output current usually output voltage or current when we have a pure resistance like this. It was like this, going from peak, going down, then up, down, up, down. This is a rectified line voltage, or rectified single phase line to neutral. Now, when we add a capacitor parallel to it, you can see that it will start going like this. Instead of this red one, it will be like this. Capacitor leads to a small decay in current, lower variation in current, going like this. Then start like this, then goes like this. You can see that rips this ripples here will be lower as we increase the capacitance. In something like this re phase rectifier, this is the first circuit that we will discuss as three phase, uncontrolled since we have all of these dives. And this circuit is also uncontrolled half wave circuit, half wave. Now, you will see that this circuit, for example, it will be like this, the Out will be like this. Not going like this to zero up, it will be fluctuating like this. To the rebels here will be much lower than the ripple of the bridctifier. Another thing that you will find that we have other circuits that will be like this. Like this. Tick tick, tick tick, tick tick, very, very small ripples like this. You will find that when we get the values of form factor, factors of the ripple factor, that transform or utilization factor, which we will understand what does this mean? And many other factors, you will find that this circuit, for example, is much better than the have the full wave bridge, single phase, rectifier. Another thing that you will find is that when we try to enhance the performance, we can enhance the performance by using instead of the half wave, we use a full bridge at three phase rectifier. All of this you'll understand in this section. Okay? So in general, what is the benefit of the three phase rectifiers? Why do we use three phase rectifiers? In instead of the single phase rectifiers, number one, three phase is commonly used in the electrical pool system. As you know that our electrical system is consisting of three phase or ABC. A red, yellow, blue or three phase system. Number two, the three phase can be used for supplying electrical power to disilludes more than 15 kilowatt. Number three, it has a higher DC voltage, higher average output voltage, lower amount of ripples, better supply power factor, lower size of filter compared to single phase, and much more. All of this we will see when we start analyzing our sects. Okay? 60. Three-Phase Half-Wave Uncontrolled Rectifier – R Load: Hey, everyone. Let's start our journey with the first one, which is the three phase, half wave uncontrolled rectifier with an allot. First, census one is the first lesson regarding the three phase, I will try to simplify simplify the definitions, the way these rectifiers work, the three phase rectifiers. If you understand the very the very beginning circuits, you will be able to understand the rest of these other sects. So we will start step by step explaining how you can draw the out wave form, how to get the values of current voltage, and so on, and why a dit operates or why does Cistor operate. As we will see in the next lessons. So let's start number one, what does a three phase with uncontrolled rect fire mean? Number one, three phase. Usually, you can have like this. What does this represent? This part here, this one, representing Delta Delta. Star transformer, Delta star transformer, delta connection, and star. This configuration, if you already taken my own course for electrical machines, you will understand that this configuration helps this delta, especially helps eliminating the triple harmonics inside the current. So these loads or this transformer supplies electrical power to rectifier with a resistive load. Now, you will find that this current is almost DC. It is not a sine wave, which is different from the electrical power system currently. The current should be like this, a pure sine wave, similar to the supply or voltage. However, due to the presence of rectifier or polyctronic circuit, the current will be like this, like this. Now, due to having this waveform, which is different from this waveform, we will have harmonics in the current here, which will be transferred to the primary. The harmonics due to the difference in shape between this and this, these harmonics will be transferred to the primer of the transformer and they can go to the electrical power system, right? So in order to prevent these harmonics from transferring into the electrical power system, we have this Delta connection. This Delta connection helps to eliminate any harmonics or to be more specific, not any harmonics, to be more specific, the triple harmonics the third harmonics, six, nine, 12, and so on like this. This will help the harmonics to be much lower in the electrical power system. That is a function of having Delta, a star connection. If it is a story star, then the harmonics will be transferred to the electrical power system, all of the harmonics. Okay, which is, of course, not something good to the electrical power system because these harmonics lead to number one, more voltage drop in the electrical power system, more of the power losses in the electrical power system and even more problems that can happen due to the presence of these harmonics in the electrical powe system. Okay, so that is the first thing. Number one, Delta star to eliminate or remove the third or the triple harmonics. Okay? Number two, a three phase A, B, A, A, B, and C. The phase, star connection with a neutral line with a neutral point, okay? Half wave. Why have wave? Because we have two types, half wave and full wave or bridge. So the half wave will only utilizes the post of cycle, and full wave utilizes both of the post of cycle and negative cycle. Now here when we are talking about three phase, when we are talking about postive cycle, it means we are just passing or allowing the maximum current or the maximum voltage, the maximum postive voltage only. So we have half wave, which allow maximum or the highest postive voltage only. In the full wave, it allows the highest, positive and highest or to be more specific, most negative part. Okay, we will understand what does this even mean when we learn about the bridge rectifier. But for now, we know that we have a one diet. Since we are talking about uncontrolled, it means our circuit is consisting of dites only. We don't have any fire restores. Now, these dites, which does not have any control, we have one diet for each phase, diet one for phase A, die two for phase B, diet three for phase C. And in the end, this common point between all of these outputs, I connected to the resistance. Since we are talking about R loot and the other point is connected to the neutral point of the three phase transformer. Okay. Okay, now let's understand first how does this circuit work? Number one, the half wave uncontrolled tire. It is one of the simplest of all the three phase rectifier topologies. Number two, it is known as a single way structure or a star connected rectifiers. Single way since we have they allow only the highest positive voltage. Each dot is conducting while others are plouged. So in this configuration, this configuration, let's say, for example, if D one conducts, then D two and these three are plugged. If D two conductors like this, then D one and these three are plouged I will explain why does this even happen? Now, a phase die conducts only 120 degrees when that phase voltage is the maximum of the three. So what does this even mean? Let's understand this. First, let's give an example, to understand how does the circuit work? Let's say remember VA, VB, and VC are three phase voltages. These three phase voltages are shifted by 120 degrees. If you remember from basics of three phase systems. Now, since they are shifted by 120 degrees, their values are different at different time or different angles. So they are different all the time. VA does not equal to VB does not equal to VC. Now, let's say, how does the circuit work? Number one, voltage across any diet. Let's say D one, voltage across it will be the difference between VA, you can see voltage A, positive, negative, right? VA, positive, and the negative is connected like this to the neutral, right? So VA, remember, D one, D two, these three are all closed right now. They are all closed. What I mean by closed, they are open circuit. They are not operating, okay? Okay, so D one, D two and D three. Now, voltage across D one, since it is an open circuit, open circuit, open circuit, then current, is there any voltage drop here? No, voltage drop equal to zero at this instant, zero at this instant. Okay, so if we apply KVL like this, like this, like this. Through all of this, you'll find that voltage across the diet will be equal to VA minus V. And since VO is equal to zero at the very beginning, then V diet is equal to VA. Similarly, you'll find that here, VA, VB, V C. Okay, VA, VB VC. Now, let's assume different values to understand. Let's say at any instant VA ten volt, at a certain instant, VV five volt. VC, let's say eight volt, whatever the values like this, ten, five, and eight, whatever these values. Now, what I would like to say is that when all of these have a voltage across them, all of these have a positive voltage. All of these dites have postive voltage. All of these dites should be forward, piaced All of them have postive voltage across them, so they should be forward pies right? However, however, what you will see right now is that let's say VA, which is the highest one will start conducting will be a short circuit like this. Remember, a short circuit like this. What will happen here? A instead of being here, this voltage ten volt will be transferred to the neutral point. To point P here, which is not the neutral point to point B here, this one, P, which is this one. So VA, so B became V A, right? Became VA. Let's became VA. Now, what will happen exactly? VA is equal to ten volt. Ten volt. Now let's look at the other dites. The other dites, you will see VB is here, five volt. And voltage of this point is equal to the voltage of this point, equal to voltage of B equal to ten volt, like ten volt transferred here. Now, similarly, what about C? C voltage here will be VC, which is eight volt. And this point will be also ten volt, right? So let's look at D two and D three. D two and D three are reverse biased right now because the voltage across it is five minus ten, so it will be negative voltage. So D two will be reverse pased. Similarly, for these three, you see we have here eight volt, ten volt, so the difference between them is negative value, so it will be also reverse biased. So what we learn from here is that since A is the highest volt then the diet connected to it will start conducting leading to increasing of the potential or the voltage of the point P to ten volt, leading to the other dites becoming reverse biased. Okay. So what we learn is that the highest, the highest voltage is the one which conduct. So we have VA, VB, VC. Highest voltage between these three is the one which will conduct. Since we have A is the highest voltage, then it will start conducting and VB or the dites D two and D three will be reverse pies. That's how this circuit work. Okay? Very simple, right? So as you can see when a die is conducting, the common casoterminal, P rises to the highest poster voltage of that phase. So you can see that A is the highest value. So it will be transferred using this die, become a short circuit, and voltage will be transferred to this point. Now, transferring to this point, making this point and this point becoming VA. And since we have VB and VC, voltage difference here and here will be reverse biased. Making the two be blocked. Now, the reversal will happen. Let's say VB is the highest, then VB will conduct D two will conduct, making this point becoming VB this point VB, VA, VB, and VC. The voltage here, VB higher than VC, so this one will be reverse biased and VV higher than VA, so it will be also reverse biased. So depending on the highest value of voltage or the highest, phase voltage. The other two tides will be two dites will be reverse pass or will block the flow of current. Okay? Okay, so let's look at this part only. Neglect the first part. It will not affect our analysis. Let's just concentrate concentrate with the second part here. So this is our second part here of the circuit. Now let's look at the waveform. Okay? So we have VA, VB, VC. Va Vb VC. Now remember, from electric circuits in the next lesson, I will use the port for writing. I'm sorry for using the mouse in this lesson. I forgot completely to use my own pencil. I will use this in the next lesson. Okay? VA will be equal to Vmax, sine Omega T, VB will be equal to Vmax, sine Omega E -120 VC will be V max sine omegaty plus 120 degrees, as we learned from the electric sucts. Now, VA VMAX sine omgtyPs shift equal to zero. So we have a sine wave starting from zero point VA, like this. Now, what about VB? VB is -120 degrees. It means it is lagging from A, lagging from A by 120 degrees, there is a phase shift between them. So I'm going to measure 120 degrees from A, legging to the right means leg. Okay? To the left, it means lead. So if we move the wave to the right, it means it is lagging. If we draw the wave to the left, it means it is leading. Okay? So we are having negative 120 lagging angle. So we will measure from here 120 degrees, 120 degrees. So from here to here, 120 degrees, this point, 120 degrees. So we will start from here, drawing our B like this VB like this. VB. Now, what about VC? VC, as you can see, plus 120 degrees, it is leading by 120 degrees, it means that from here to the other side like this. Of course, it will not be visible, but anyway like this, go back. It will be like this, VC leading VA by 120 degrees from here to here, 120 degrees. It will be drawn like this. So we have VA, V, and VC, VA, VB, and VC, and so on. So this point, this point here is exactly the same point here. You see this part here, this part here is exactly similar to this part. VC will be like this, so you can see VC, then VA, then VB. Okay? That is the shape phase shift between them. Now, we said before that what will be the out will be the highest of the three voltages. Let's start from the very beginning from here. You can see we have at this point, from here to here, you can see VC, VA and let's go down like this and VB Which one the highest is the highest voltage? VC is higher than VA, higher than VB. So VC is the highest. VC is the highest. So these three well conduct and VO will be equal to Vs, VC. So at this point, D three will conduct like this and out will be equal to Vs. So VO, as you can see here, is VC, like this, VO. Like this. Since VC is the highest. Okay? Now, starting from this point, you will see that from here to here like this during this part, VA V C and B. Which one is the highest? VA in this period, VA is the highest voltage among these voltages. So VA is the highest. So D one will conduct and all of the others will be plugged. A will conduct, VO will be equal to V like this transferred to VO. We will have V will be equal to VA like this, Like this. Neglect is a sport. It will be like this, VA. During this period, D one will conduct. Now, let's repeat this one more time here. VB, VA and VC, which one is the highest VB, it will out will be VB like this? Okay. And since we are talking about VB, then D two is the one which will conduct, D two will conduct. Again, from here to here, is the highest. So D three will conduct like this. Our waveform will be like this. Like this, similar to this one. Since we have a resistive loot, then the current will be V but divided by R, exact same exact waveform, except we divided by R, as you can see here, right? Iain here, VA, which means D one conducted, so the current IA the one which goes to the loud. Ibut will be A VA, A, VB, IB, VC, C, and so on. Okay. Now, what you will see here is that in one complete cycle, in one complete cycle, let's say we started here from this point intersection between A and C. Intersection between A and C is exactly equal to this intersection. Here, A and C, right, this point and this point, representing here two Pi or one complete cycle. Now, you will see that D one, one, two, and three. So each of these dites will conduct one over third of the whole cycle, one, two, three, one over third of the whole cycle, which is one over third 160 degrees, which means 120 degrees. So as we have said in the previous slide, each dite conducts 4120 degrees. Okay? This is where it came from here. Number two, I would like to also get. So this is our IO, which is continuous, exactly equal to the out except the divid by now, what about I supply? Remember, we have a phase three phase. Three phase, it means we have IA, IB, and IC. I supply can be any of these currents. It can be IA or IB or IC. It can't be three at the same time. It is just one of them. We say I supply is equal to IA or IB or IC. Okay, we don't add these currents together. Why? Because we use the phase current phase current, which will be used in our equations. Okay, now we have IA, IA will conduct from here to here, which is the conduction of here, IA from here to here. So it will be address, which is died D one, IA equal to D one, IB equal to current D two, IC equal to current D three. Now IA going to D one, so it will be like this from here, it will start conducting from here from zero to this value from zero to this value. So this part representing our IA, like this. Which is 120 degrees. After 360 degrees or two Pi, which is at this point, it will start conducting again, which is D one or IE. Similarly, if you would like to draw IB, it will be IB will be just like the sport and the rest will be equal to zero. If you would like to draw IC, it will be only the part and so on. Okay. So now, the next one, we have drawn now V Obut. We have drawn I Obut, we draw the current of each diet or supply current or phase current. Now, the last thing I would like to do in this lesson is to get a voltage across the diet. Remember, all of these dites are similar to each other, except that there is a phase shift between them. So these waveforms are similar, IA equal to IV, equal to IC, but there's a phase shift between them. Instead of being drawn here, it will be drawn like this. IC will be drawing here and so on. Now, what is important for me is the dite. Why? Because I would like to draw the waveform in order to see what is the peak reverse voltage or peak inverse voltage. Peak inverse voltage applied on the dit. In order to select the correct rating for the diet. So in order to do this, we need to draw this waveform for the diet. So how can I do this? Now, this is very easy. All you have to do. All you have to do is apply KVL to the loop consisting of D one, phase A, and the loot. Okay, what does this mean? If you look at here, we have VA. We need to draw this wave form for D one, right? So what I'm going to do is that I will start from neutral like this, KVL, like this, like this. Okay. So this large KVL consisting of VA, V diet, and VO. From this KVL, you will find that V diet is equal to VA minus V out like this. VD one equal to voltage A minus V out. By using this equation, we will be able to draw this wave form. Now, let me show you how can you do this? So we have Video one, VA minus V out. Let's start step by step. Number one, let's start drawing from here. Okay, neglect this part. We will start from here for simplicity, and you will get this part, which is similar to this part. This is similar to this, okay? So let's start with the easiest one. Number one, from this point until this point W D one conducts. Now, what are these points is very easy. This point is simply equal to 30 degrees, 30 degrees. Now, someone will ask me, how did you get this? How did you know that this intersection is 30 degrees? This is very easy. All you have to do is say VA equal to C because this point is intersection between phase A and phase C. To get this intersection, all you have to do is equate this equation. So we have Vmax, sine omegaty equal to Vmax, sine omegaty plus 120 degrees, 120 degrees. Now, as you can see that if you substitute with 30 degrees or by solving by using the calculator, as we learned before in the previous lessons, how to do this, sine omigaT equal to sine omgaTE plus 120 degrees, you will find that omega ty will be equal to 30 degrees. Sine 30, equal to sine 30 plus 120, which is sine 150. Okay, they are equal to each other. Okay. Great. Now, this point is 30 degrees. Now, what about this point? This point is after 120 degrees. We said that each dite conduct 420. So it will start from 30 degrees to 150 because 30 plus 120 gives us 150. So as you can see, omegaty 3,250. During this period, D one conducts. Now, remember our equation VD one, VA minus V output. Now since D one conducts D conductors, it means that V output will be equal to VA, equal to VA, right. Now, VA minus VA will be equal to zero, zero, right? So V equal to VA. So VD one volts of the tide will be equal to VA minus VA, which is zero. That's why you'll see that VD one is equal to zero from 30 degrees to 150. Okay? I think it is pretty clear from these figures and these equations. Number two, When D two conducts from here to here, D two conducts. What will be this waveform from here until here from 150, D two, remember, D two or any light conducts for 120 degrees. So starting from 152 plus 120 degrees, which is 270 degrees, this period, 120. Okay? From here to here, D two will conduct. So since D to conduct VO will be equal to V B. So from here to here, we have to take VA and subtract from it VB or the V line to line VAB, right like this, O equal to VB or VA minus VB, right? Okay, VA minus VH is a line to line volte. So from here to here, we will have a value of line to line voltage. Okay, or take VA, this waveform from here to here and subtract from it value of VB, so you will get this way form. Starting from here to here, 250 270 plus 120 degrees, it will be 100 and what? 300 and what and 90 degrees. Okay, 390 degrees. So 390 degrees. Now remember, it's 160 from here to here. It's hundred 90 from here to here. Okay? Don't mix everything together. It's round nine. 270 290, these three will conduct, which means Valbut will be equal to VC, right. Okay, this for substituting for getting this part, and subsiding at 180 to 110 to hundred 40, whatever, in order to draw this waveform, to get values at different points and then connect them together. Okay? For the last one, D three when conduct hundred 70 190, it will be like this from here to here, which is VA VA minus VC, which is a line to line voltage, right? VA minus VC, which is VAC. So here, zero, here, VAB, here, VAC, so you will have this weird figure. What you will find that is the big inverse voltage, maximum inverse, big inverse voltage. For any dite, it will be negative root three, V max phase. Remember, V max phase, phase voltage, maximum value of phase voltage. Root three, V max of the phase. Now before we finish this lesson, I will show you a small trick. If you would like to draw this much easier in a much easier way, there is a secret way. Now remember, this part is VAB this part is VAC. Now, if you draw the three phase, the three phase, VA, VV VC. If you draw the line voltages, VAB, VAC, VVC and all of these, which we will learn how to draw them later. Okay, we will learn the phase shift between all of these and how to draw them because we will need them in other lessons. But anyway for now, let's say you already know how to draw VAB and VAC, VAB, and VAC. Okay? Remember this. Now look carefully. We are looking for what points for drawing from here to here, right? Three parts, this part and this part. So we are looking for VAB 150-270, VAB, 250 to 270. VAB, let's look at VAB, VAB, like this. If we look carefully from 150, it will be from here, this point. Okay, VAB. And until you go down here until what until here. To this point. These two. This range is VAB, 150-270. And we would like to draw 270-390 VAC. So from here to here, VAC this one like this. So if you look carefully, this part is VAB, and this part is VAC. This shape is exactly similar to this part. What you have to do to draw D one, draw VAB and VAC, and from here, VAB, take this part only of VAB from here to here, and for VAC, take from here to here. This is another way to draw this waffle. In the next lesson, we will learn how to get the equations of the three phase, half controlled rectifier. 61. Equations of Three-Phase Half-Wave Uncontrolled Rectifiers: Everyone in this lesson, we will took a poet. The equations for this three phase half wave uncontrolled rectifiers. In the previous lesson, we took that poet. The three phase, half wave uncontrolled rectifier, how do they work? And now we would like to observe or clarify the equations or represent the equations for this three phase, half wave uncontrolled rectifiers. Okay, so let's start. Number one, these are the wave forms that we discussed before, right? Now, let's go step by step. Number one, which we always get. Every circuit in the previous section, Section four, the single phase rectifiers and now in three phase rectifiers. Number one, we need VDC, the DC output voltage. DC outbut voltage representing the average output voltage, as we said before. Now, what I would like to get is the average for this waveform. Remember this wavefo O I would like average. Now, if you look carefully here, you'll find that our waveform repeats itself three times, right? We have one cycle from here until here, right, one complete cycle. In this complete cycle, we have one and two. And three, right? So we have three identical waveforms, three identical wave forms. So we can just take one of these waveforms, get the integration for one of these and multiply by three. So that is what we are going to do. Three, remember average one over t integration 0-3 for one complete cycle. Now, one over t, our period is two pi, and it repeats itself, this repeat itself three times. That's why one overt multiplied by three. Now, let's type here. Let's again a pencil here. You can see that we can. We can say, number one, integration from 30 degrees to second 150 degrees for this function VA, which is V max sine OmigatyT this integration and multiply it by three. Divided by two pi. So we have this part, 150 degrees, 3,250 degrees for VMAX sine omegaty, which means we are integrating from here to exactly here, right for A, it is drawn like this. So this will give us a V average, as you can see here. Two Pi integration from 30 degrees, we use radiance. So 30 degrees is Pi over six, which is 180/6. 30 degrees and 150 is simply five Pi over six. If you don't know how to convert, this is very easy. All you have to do say, I would like to convert 150 into radians, I will say 150 degrees, multiploidP. Pi over 108. This will convert any angle into radians, Pi over 180. So 150/180 is simply equal to 5/6. So overall, we have five Pi over six. Similarly for 30 degrees, we can say 30, multiply it by Pi over 180 to convert this into radians. 30/180 is simply 1/6, which is multiplied by Pi Pi over six, similar to here. For V max phase B here means phase voltage. Since we have VA, VB, VC and later we will have in other lessons VAB VAC and so on. So here we are using phase voltage. Phase voltage, maximum value of phase voltage, sine omegaty. So it would give us this function three root three of our two Pi Vmax phase. Now remember that in order to convert phase voltage into line voltage, simply we multiply Pi root three. So we can take the sport and say 3/2 Pi, like this, multiplied by V maximum line, which is this part. Root three multiplied by V max phase, converts it into V max line. Now, this is the first method. Let's do another one. We can say 3/2 Pi once more, integration. Now, let's say I would like to do get the average using VB how can I do this? It will be from here from 150 degrees 270 degrees, right, this point here, this one and this one. So we can say, if we go down here like this, here, 150 to 270 degrees for VMAX phase, sine, omegaty -120 degrees. Now, this, which is integrating VB VB 150-270270. Now toblod by three root two Pi, this will give you the same value for this integration, which is three root 3/2 Pi VMAX phase. Now we can do the same for VC which is here. Degrade it 270-390 for VC, which is V max, sinomgaT plus 120 degrees. All of them will lead in the end for the same value, which is this one. Okay? Okay, great. So all of them lead to the same one. So for simplification, I always use VA or phase voltage of A. Okay? Now, let's use the mouse once more like this. Now, B is a maximum phase voltage. VML is a maximum line to line voltage. Now, if I would like I average, here we have a resistive load. That's why the current has the same waveform, similar to the voltage. This except that divided by R. I average or IDC, as you remember, from every lesson, I average is equal to V average over R. So take this value here and divide by R. So it will be three root 3/2 Pi R, V max phase, like this. Now we max phase over R maximum value of voltage divided by resistance, gives us I MIB, which is maximum phase current, like, which is this point. Maximum voltage divided by R gives us maximum phase current. We just take this two together and form one current equation or one simple. Okay. Now, what is next V output RMS exactly as we learned before, RMS is simply equal to the function square integration for the square of the function all under the square root. So if you look at this, is exactly this one, except that the function is squared and under the square root exactly as we learned from the previous section. So Pi solving this you will get 0.484 068v max phase. Of course, I don't memorize any of these values. For example, when I had exams in the past, I usually use the integrations. I look at the waveform and integrate in the exam. I don't memorize any equations. Okay? So this is because we are going why do I put these values in order to use this in the next equations? Because we will use this to prove something, which you will see right now. Now, as you can see, form effect, form factor. What does a form factor equal to V output RMS, divided by V average. The closer these values to one, there the more DC is our output. So we try to make form factor as close as one. So ORMS over VDC is this value divided by this one, not this one, this one, so VMB will go a VMB and 0.8 4/3 root 3/2 pi, gives us 1.0 165. Now, the next one, which is the ripple ripples, how much ripples we have in our output wave form, this one for voltage. In order to get triples, if you remember it was equal to root, let's type it. It was equal to root form factor square minus one. If you don't remember from the previous lessons, okay? So it will be root square of this value minus one gives us 0.18. Now, I would like I output RMS RMS of the out current. It will be simply since it is a pure resistive load, it will be VO RMS, V RMS, divided by what divided by R or our resistance. Similar to I average, average over R. So RMS RMS divided by R, like this. So it will give us 0.84 068 I maximum phase. This part and this part. Okay? What about the output power Abu power here when we say average Albo power. I'm looking for average output power. So since I'm saying average albo power, it means we are talking about the average voltage, multiploid by average current. So it will be VDC, multiplied by IDC, VDC which is the average out voltage, this value multiplied by I average which is simply equal to this value. These two together like this. Their multiplication will give us this value. You will understand why I'm using this or why I'm getting this value right now. Now for power AC, which means general power, all of the power consumed by the ut, not just the average value, but overall power consumed by the resistive lute. So it will be RMS, IRMs, AC, then we use the AC rules. DC, use the average or DC lutes. Vrms, RMS, this value multiplied by ORMs, and this value. So it will give us this one. Now, if you remember, if you remember, previously, we talked about the rectification efficiency at the very beginning of the course. The higher this efficiency, the more we are closer to the DC, right? So what does rectification efficiency equal to power of DC, divided by power of AC DC, divided by AC. It will be 69 6%. This is the rectification efficiency, right? Ratio between DC and EC. Now, let's collect all of these values. By using these values, you will find that we have this large values, right. Now, what I'm going to do is I would like to know how efficient is a three phase, how uncontrolled rectifar compared to the previous circuits for the single phase. So I brought the older table that we talked about before, half wave, full wave rectifier with these values. Let's convert them together. Number one, number of neglected number of dines for now, maximum efficiency, number one, maximum efficiency. The half wave maximum efficiency was 40%. In full wave rectifier, maximum efficiency was 81%. Here, efficiency of rectification is 96%. Now let's look at average voltage, average outt voltage here Vmax over Pi, here to V max over Pi. Here it is 0.8 V max phase. VMAX phase is similar as VMAX. Here you'll find average value is higher. Now, what about ripples here, 1.21 high ripples in full wave, 0.48 lower ripples. In the three phase half wave uncontrolled, you will see that triples is 0.182. So this is lower than this value and lower than this value. Okay, so the ripples inside, the three phase half wave uncontrolled is much lower than the previous circuits. Form factor how close our DC voltage compared to root mini square. So you can see the closer this value to one, the better. So you can see that for half wave, 1.5, for full wave, 1.11 here, it is closer to 1.0 165, which is much better than form factor for half wave and full wave. Now for the frequency frequency F, if you remember, let's just draw this. If you remember in the half wave in the half wave we had like this in one cycle like this, then zero. So it repeats itself every frequency, every one cycle. However, for the full wave rectifier, it was like this one, two. So it repeats itself two times in one cycle of the original waveform. That's why the frequency here is double. Two F. Now, in the three phase three phase, half wave or three phase, half, one, two, and three. In one cycle from here to here, this repeats itself three times. So it will be three F. Now remember, the higher the frequency, the higher the frequency, the lower the ripples. Okay? So it's better to have a higher frequency out waveform. Okay? So what we learned from this lesson is that the three phase, half wave, uncontrolled dict fire is much better than the single phase socit. Now we would like to add another thing regarding the three phase, half wave uncontrolled rectifier, another definition which we haven't discussed before. Power DC, if you remember from the previous slide, we obtained it like this, and the supply current RMS is equal to this function. Remember that the supply is equal to IA or IP or IC, either A, either P or C, not the three of them at the same time, but just one supply, one phase. So if we look at IA, here, if you would like to apeat itself every two Pi. So in one cycle, I supply only appeared once from here to here. So what I'm going to do is that if I would like the RMS because we will need this, it will be root 1/2 Pi. It repeats itself only once in a complete cycle from here to here, right for IA. Now, this integration from here 30 degrees to 150, which is 5/6 55/6. This function is our IO or IA and the current I is simply equal to. Let's type it here IA simply equal to VO or V supply, VO of V supply. Both of them are the same, Vmax, sine omega ty, which is VA divided by R. This is our current waveform, this form. Now, you can see that we did this VMAX sine omegat over R all squared, and we are getting the RMS. So it will be like this, this value. Okay. Now what we would like to introduce in this lesson is transformer utilization fact. What does this represent? Now, our goal is to deliver deliver and output power or output DC power to the resistive lute, right? So I would like to know much how much power I have utilized from the transformer. So in order to do this, I will see the output average DC power delivered to the lute compared to the rating of the transformer. So we have average power DC power delivered to the lute, which is this value divided by transformer volt and pair. Now, how can you do this? Remember, the rating of the transformer or power delivered by the transformer will three, since we have a phase, V max phase, multiploid boy, I max phase. That is one way to get that three phase power. Another way is to say root three, multiplod boy, V max, line, I max, max line. Both of them will lead to the same solution, right? Okay, so let's see what we have here. As you can see, we have used 0.68 39. This is the average output power, DC power, and three I supply. I supply RMS phase, supply RMS phase. So I supply here, this is a mistake, okay? Good thing that I have written it, okay? It should be V supply. Phase, V supply, supply, and supply. Now, not maximum value. Of course, it is RMS effective value. RMS, not maximum value. So it will be three IRMS RMS of the output current for phase, multilod by voltage, RMS phase, or root three, I supply, RMS line, V supply RMS line. Now, I supply RMS phase, the one which I got right now, 0.4 at y four IMV, this part. Now, what about V supply RMS phase? For V supply RMS phase, remember that V supply is simply because I would like to remove these two. V supply is equal to Vmax divided by root two. Right, if you don't remember, simply if you remember, V max, phase equal to root two, V phase, ors. Root two, multi blood by the RMS of the phase. So if you would like to convert RMS into maximum, we will take this one into the other side. So it will be Vmax phase, divided by root two. Now, why did I do all of this to get this with this, this with this, to have one value, which is 0.664 or six 6%. So what does this value mean? It means that only we only utilized 66% of the transformer. 62. Three-Phase Half-Wave Uncontrolled Rectifiers - HIL: Hey, everyone. In this lesson, we will too about the three phase, half wave and control dictifier. However, this time, instead of the highly inductive lute, instead of the resistive lute, we are going to use the highly inductive lute. Okay, great. Let's start. So remember this is our waveform in the pure resistive luidRmember our waveforms are continuous, right? There is no negative part. There is no negative part, and the waveform is continuous. Current is continuous. So what does a highly inductive lud will do in this circuit? Nothing. Really nothing or what it will do. It will simply convert Iout from being reflect weating into one constant line. And instead of having this waveform, we will have a constant lines like this. As you can see, I abut. Same instead of having this pulsating DC, it will be a straight line like this. Instead of having this square wave form or not square this waveform, which is close to the sine wave, it will be like this square wave, like this. That's all what will happen. So that is the effect of the highly inductive loot, I just converted the current from being similar to the voltage waveform divided by R, it will be just a constant line. And this will happen a lot during the three phase circuits, okay? So that's why you will find that R load are very close to highly inductive load from equations perspective. So let's see the equations, okay? As you can see, nothing will change except that the current will be a constant value. As you can see that the output current equal to RMS, equal to average, equal to average over R. If you remember in the highly inductive load, we said that this constant value is equal to V average over R. And since we have a constant value, this average value will be the RMS, will be the constant value. Very easy, right. Another thing that this waveform will not change at all. That's why the output will be average over R. Similar as before, right? Okay, great. What are we going to do next? Another one I average, I diet average. Average diet current, remember I diet is equal to I supply, equal to IA. When I'm getting average diet current is getting average supply current, is getting also the diet current. Now, what is the average value? Now, remember that this one conducts 420 degrees. Let's use just our lovely board here. Remember that D one, for example, only conduct 420 degrees, right? Okay. Now, what I would like to do is that I would like to know the average I diet average. How can I get it? Since we have a highly inductive load, it will be I out the constant value here, this one, metabloitPT Pi, which is one complete cycle. And here we will have our induction period, which is 120 degrees, 120 degrees here representing how much of Pi it will be 3/2, no, 2/3 Pi 2/3, so it will be 2/3 Pi, 120 induction period compared to the whole cycle. This will be equal to Pi will go with Pi, two will go with two, so it will be I output over three, which is this equation. Now, what if I would like the RMS? The same idea. IRMs for dt will be root I out Mm hmm. Two Pi and conduction period over three Pi, conduction period divided by two pi. This is similar to the rating of as a current rating for dits and thistors which we have said before in the single phase. So Pi will go with Pi, so it will be again I out over root three. Like this. So you can see, again, ID RMS, I supply RMS, I output over root three. Now, let's all of this. This one like this. Okay. What else? I supply RMS simply equal to root four Pi over 6/2 Pi. Now, this is equal to this, right? But I'm just showing you how did I get it. Root induction period four Pi over six is exactly similar to two Pi over three, which is 120 degrees. Divided by two Pi gives us root 1/3. This value, which is 0.577. Rectification efficiency in this case for BDC over AC, it will be DC, IDC, Vout RMs, put RMs, so it will be equal to V average. This value. Oput DC. Remember, DC is a constant value, I. Okay. Power E ORMs, I RMS. Remember, I RMS is exactly equal to DC, equal to or V average over R. So this equal to this, right? V RMS is the same equation. As we took it before, 0.840, this from the previous loss. The exact values. RMS here is similar as the previous loss. In the end after removing this and this and doing this division, you will get 98%. This 98% rectification efficiency is higher than our loot case. In the load it was 96%. So what did we learn from this lesson? What we learned is that the high inductive loot is exactly the same as resistive loot you'll find this happening a lot. In these examples, you will see or in these circuits that you will see, this will happen a lot. The difference between highly inductive lute and resistive loud is usually that number one. I output is a constant value in the highly inductive lute, and the current is square waves. In the resistive lute we had a pulsating DC. Find that in highly inductive load and resistive V average is the same, V output RMS is the same, and the other equations is much easier. Even here for this one, I average will be in resistive loot. It will be one over third but average. For the RMS, it will be one over root three, out RMS. Except that here in the highly inductive load, both of them are equal to each other. In resistive load, I average is different from the RMS value. 63. Example 1: Hey, everyone. In this lesson, we will have example number one in direct fires three phase fires section. Example number one will be on the three phase, uncontrolled half wave iced fire, and this example will help us how to apply the previous equations in different problems. So we have this circuit for the half wave three phase uncontrolled direct fire. Number one, we would like to, and this one has a supply of 480 volt and remember. They given similar to the single phase. Previously in the single phase circuit, when we say any type of voltage, it was the ORMs right? And now when we have a three phase system, any value given is a root mean square line to line voltage by default. So any value given inside the problem will be RMS line to line voltage. For example, 180 volta representing VAB, line to line voltage, root mean square value. Number one, determine the following when the loot is a pure resistive root of 25 OMs, and when the load is a highly inductive root, we need what? Number one, peak outwod voltage and the current, average outt voltage and the current, peak and average dite currents and peak inverse voltage of the dite. Now, let's start step by step. Step number one in any given problem. Number one, draw the wave forms. Number one, like this. For a Bre resistive load, we have the waveform that we discussed before. Now, the first thing we need big outut voltage and current. Big outut voltage here is what is equal to VMAX phase, right? This is a maximum value VMAX phase. Now, given what our giving is 480 volt RMS line to line. So we need to convert this value into VMAX phase. How can I do this? Simply you will find that. That V max phase will be take the root mini square value like this, multiply it by root two. Y root two in order to convert the value from RMS into maximum value. Okay? In order to convert from line to line to phase, we need to divide by root three. So root two to convert from root mean square to maximum value and root three divided by root three to convert this value into phase value. Okay? So the peak will be like this 480 root two multiplied by root three, a root two divided by root three gives us 191.19. Okay. Now, the second step we would like to get is this is our peak hood voltage. Now we need peak current. They have the same form. Current and voltage have same way form, except that it is divided by R. So the maximum phase current will be this value divided by the resistance R, like this which will give us 15.67 pairs. Now, the third requirement is average outwut voltage and average out current. Average outut voltage, we obtained it before in the previous lesson. We said that in this circuit, the output average will be three root three of our two Pi VMAX phase, or 0.83 V max phase by integration from here to here and molting it by three times. So it will be 325. This is the average or DC output voltage or average output voltage. For average but current, it will be simply this value divided by 25, like this. So it will be 13.0 11. Okay. Now what we need or else we need peak and the average diet current. Peak diet current, remember, this is the phase current and also the diet current. Peak value of current is exactly the big value for the outbut current. So the big current will be simply 15.67 and pair, right? They have the same waveform. And for average diet current, each diet conducts 4120 degrees. And we said that average diet current will be I average. Let's type it. It will be it will be like this I diet. Average will be 120 degrees the period of conduction, divided by 160, multiplod by I output average right to get average. This is exactly equal to 1/3 as we obtained in the previous lesson, right? So it will be I output average, divided by three. Okay, peak and average. What about the root mean square root mean square or mas? Not the root mean square, average light current, 4.337, which is the average here this value. I O average divided by three, as we just explained here. 4.337. What about big inverse voltage? If you remember when we took it before, that the big inverse voltage, as you see here, is equal to root three Vmax phase. Now, root three V max phase is exactly equal to V max line to line. Because if you multiply root three, by a phase value, you convert it into line to line. Now we have V max line to line. Now, 480, which is given value here is V V V line to line, RMS, and I would like to make it V max line to line. So what I'm going to do is simply, I will take this value and multiply it by root two to convert from RMS into max, max like this. Okay. So the answer will be root three, multiplied by VMAX peak, which is 678.7 mine. This value, remember, we obtained VMX phase 191. So in order to convert it to line to line, you can take this and multiplied by root three, or you can take 480 volt and multiplied by root two. It will give you also the same value. All of these ways lead to Rome, Okay? In Italy. Okay. Now, we would like to deal with highly inductive lute instead of instead of a resistance or L or a highly inductive lute. So what we are going to do, we are going to draw again the wave form number two, big output voltage will be the same as before, right? Big value is exactly the same as the previous slide. Now, big output current. What it will be big output current will be equal to average output current, right? So what it will be, it will be average over R because I bit is just a constant value. So big value, root mean square value, average value, all of them are equal to each other. So all of them will be equal to V average over R, right? So this is the first one exactly similar as here, 391, okay? For V average exactly also the same as the previous slide. For I would be average will be V average over R, which is 13.0 11, this constant value. Now, what about peak and the average diet current? Diet current is 13 and pairs also because it is a constant value, right? Average diet current will be this value, which is the average current, divided by three, because it works only four thirds of the cycle. Like this. I average will be I output average of three, so it will be 4.337. Now for the peak inverse voltage, again, the same answer. It will be root three V max phase, which is again 678.7 min. To this simple example shows you that there is no trick in solving these circuits. All you have to do is draw first the waveform and using the waveforms, you will get all of these requirements. 64. Three-Phase Mid-Point 6-Pulse Diode Rectifier: Hi, and welcome everyone. In this lesson, we took a pet the three phase, midpoint, six pulse died rectifier. So in the previous lessons, we took the poet, the three phase, uncontrolled half wave rectifier. Now, what's the difference between this circuit and the previous one? The difference is that you will find that number one from previous lessons that the performance of a three phase three pulse directifre. And if you remember, three pulse because we had like this, A, then B, then C, and then it repeats itself in one cycle. So we have three pulses, three pulse directfres much better than the single phase two pulse did, which is full wave rectifier, like this, right? So what we learned from this is that it is logical that when we increase the number of pulses in each cycle, it will lead to an overall improved performance. So what we did here in the phase, midpoint six pulse is that we took each phase, each phase and split it into halves. And the midpoint of the three secondary windings are connected to form the neutral. Okay, so what does this mean? Let's use our pencil. So remember that we had this A phase A, right. And we had phase B. Now, what we did is number one. Phase A is divided into two halves. So I bought here a point at which we will split A. Okay? So we have an upper half and lower half for A. Now, B will be the same. B is like this, right? So I will split into half and put it here. So it will be like this. Okay, like this. Okay, what about C? C is the same idea, if you remember, C was drawn like this. Like this, right? Mm hmm. No. Okay, let's do it here at midpoint, here, at this point. I will draw C like this, go here and draw C like this with the midpoint and then the secondary one. So we have A, then we have B, and we have C, right? Okay, A, B, and C. Now, what you can see is that we divided them into two halves. Each phase is split into two halves. The midpoint is connected all together to form This is our neutral. Okay? That is what we did exactly. We have A. Remember in the previous section, it was A, B, and C. Now we divided each into half, so we will have A and one on the other side, B, another one, C, and another one. Okay. Now, this configuration which you see here is called the three phase, midpoint six pulse died rectifier. So let's look at them. This was what we had in the previous lessons. The three phase half wave rectifier. So when we divided each into half half and half and connected them together like this, you will get this wavefur. You can see this is exactly the same as this one. A, A, you can see A, B, and C, B, C, and A. Now, I know that one will ask me, why did we do all of this? We did all of this in order to have instead of having three phase, as if we have six phase, right? Six phase. Yeah, one, two, three, four, five, and six, D one, D two, D three, D, four d, five, and D six. So this will lead to as if we have a three phase with three pulses like this, one, two, three. When we use six phase like this, we will have one, two, three, four, five, six, right? That's all we did using the three phase midpoint six pulse dite rectifier. Now, let's understand more about this and the phase shift between all of this. Okay? So let's first type it first if you remember, A, is V max, sine Omega t, right? Now, we have A in this side. So A two, which is this coil will be logically will be Vmax, sine Omega t -180 degrees, right? Fish shift, 180 degrees because it is exactly opposite to coil A on the other side. Now, similarly, for B it is V max sine Omega t -120 degrees. So B two will be the same value, but subtracted from 880 or add 180. I will subtract 180, so it will be -120 -180 will be -300. VMX sine Omega t -300. What about this coil? This is the original 1c1v max sine Omega E plus 120 degrees. The other one here will be subtracted from this 180, subtract 180. So by subtracting 180, this one will be VMX sine Omega t, 120 -180 will give us negative 60 degrees. So what you have seen right now is A, V max sine omega T C two, VMX -60 degrees, B one -120. A two -180. C one can be also instead of plus 180, we can say, subtract subtract 160 degrees. It will be 240. So it can be also exactly as negative to 140. Both of them are the same, okay? There is no difference between them. Here -300. What we learned from here is that we have how many phase one, two, three, four, five, six? We can say that's 160 degrees. Divided by six phases, it will give us 60 degree phase shift. That's why you will see that phase shift between these two and these two, and these two and the rest will be 60 degrees phase shift. Now, in the previous one in the previous one, the three phase, it was 360/3 phase. So each one was shifted by 120 degrees, which is A one, B one, and C one. So it will be like this. The same exact one, VE one phase shift between them. So you can see exactly as I have just said, VMAX sine omigaty, VMAX sine omigaty, this one -60 degrees, -120 degrees, -180, -240 -300, right? Okay. So why did we do this? Because we will need to draw all of these six phases. So you have two options, either to draw them directly or you can say that. You can draw A one, B one, C one, and the others will be opposite to them. What I mean by this, VA two is simply negative value of VA one. VC one is negative VC two. VB two is negative V one. So let's draw them and see what will happen. This is a phase shift in the vector form between the six phases, okay? Okay, so let's see our waveform. You can see here we have VA one. Okay, and Vb one. Let's use this one again. You can see Va one here and Vb one, which is here and VC one, which is here. The phase shift between these waveforms or this, um a waves of A, A one B one and C one are shifted by 120 degrees, right? Now, if I would like to draw, let's say I would like to draw, we have VA one to draw VA two, simply, it will be negative of VA one. You can see that this is Va one, right. Now, VA two will be exactly opposite to it, negative, which is this one. Va two, like this. Similarly, VB two will be opposite to it, V VB Vb one, which is this one. Opposite to it, VB two, which is this one negative Vb one. VC one. Let's say this one here, VC one. You can see exactly opposite to it, this one, which is VC two. Or you can simply say draw one wave form, one way form, similar identical waveforms except that they are shifted by 60 degrees. You can see from here to the next 160 degrees, next one, 60 degrees, 60 degrees, 60, 60, 60, and so on. Okay. You can draw them like this VA one. Then the next one will be VA one, this one, this one, then next one will be VC two. You can see A one, C two, then B one, then B one, then A two, A two, then C one like this, then B two. All of these are different methods to draw the waveform, okay? Now, since we have six phase, instead, and when we had three phase, the fish shift between them 120 degrees. And each diet conducted 420 degrees. Now when we have six phase, the fish shift is 60 degrees. So the conduction period for each dite will be 60 degrees two. So you can see here for A one, which is related to D one, conducts for 60 degrees, D 260 degrees, d three, d four, d five, d six, each for 60 degrees. Each dite related to what, again, the same concept. VO will be the highest voltage between these six phases. So the highest voltage will be, let's say, look at here, at this part, you'll see that V two is the highest voltage. V two is the highest voltage, which means V would be V two, V two like this. V two is related to which one VB two related to D six. That's why D six like this. A VA one is the highest voltage between here and here. So A one related to D. D one will conduct. So VO will be like this. So let's use this pencil like this. Okay, here to here, we will have we will have VC two VC two. This one, VC two is related to D two. If you go here, VC two and D two, and so on, you can see the abut will be the highest, the highest voltage or phase voltage at any instant. Each one conducts for how much for 60 degrees. So you can see A one conducts from 60 degrees until 120 degrees from here to here. Now, of course, out will be exactly equal to I supply, right? I Out will be equal to I supply, nothing changing. Here, like this, it will be V supply out, sorry, V divided by R. Same form except divided by R. For I supply, I supply equal to IA one or IA two or whatever. Each one, just take one of the phases. So IA one will be this part only, right? IA one, I supply equal to ID one. Okay. Now, what about D one? Again, the same problem, how to get D one. Simply if you look at here at this one D one. By applying KVL like this, let's do the KVL. If we do a QVL like this from neutral going through D one through out like this and go like this and get back to neutral. You will see VD one equal to V supply or V to be more specific minus VO exactly the same as the previous lessons. VD one, VA one minus V out. In order to draw this wave form, let's see V out. So you can see it will be divided into different regions. The first one, which is V out equal to VA one here, this part from 60 degrees 220 degrees. In this region, you will see that Vout is equal to V. So Va one minus Vout, which is VA one, have a zero because the dolt, of course, is conducting, so it will become a short circuit like this, very easy. Starting from here, from 122 plus 60 degrees, it will be until 180 from here to here. From here, 120 to 180. Which one is conducting from here to here, VC two. So V diet will be V one minus VC two Ls. VA one minus Vs two. Okay? So we substitute with two values 120-180, and we obtain this function 100 5,180. We had these two values which were used to draw this line. Okay. Now, what about it from 182 plus 120 plus 60 degrees will be 240 from here to here. Or if you look at the figure from here to here, right? From here to here, VB one. So it will be VA one minus VB one. Subtracting the two signals, you will get this function like this. Okay? Okay. Now, what about the last one? Next one, which is 200-40 at 60 degrees 200 from here to here, from here to here, VA VA one minus VA two. So that will give us this part. And by continuing for VC one, VB two, and so on you will be able to draw the rest of this function. What is most important for me is this one, which is peak reverse voltage. Now, what you will find here in this circuit is that the peak reverse voltage here is two V max phase, two times V max phase. Two times V max phase. Okay, big reversivT maximum value of the phase, two times this value. Okay? That is when we use this when we select the rating for dint. Okay, now we will start now getting our equations. Now, number one, how many pulses we have? We have in one cycle. Let's say from here, from here until here, right? This one complete cycle. One, two, three, four, five, six. So we have how many cycles, how many pulses, we have six pulses. Right? Okay, six pulses, divided by the whole period two Pi. Now, we are going to take just one wave, which let's say for example, Va one, which is Vmax, sine omgaty V max phase, sine omegatyPhase, of course, okay? Now, integration from here, which is 60 degrees until 120 degrees. 60 degrees to convert it into radiance, it will be Pi over 180. So it will be Pi over three. To convert 120 to radians, it will be 120 by Pi over 180 will give us two buy over three. You can see here the integration from 60 degrees Pi over three until 120 degrees, which is two Pi over three. VMAX phase sine Omegty. It will give us three over by VMAX phase. Okay. Okay. What about I average? I average in any problem here or even in highly inductive load, it will be average, divided by the resistance. V average divided by resistance, it will be the same function. Divided by R gives us three ver Pi I max phase. V max phase, divide bir gives us the maximum phase current. What else? We would like also the root mean square voltage out RMS. It will be the same function as we learned it before except it will be square of the function like this. I give us 0.95 58 Vmax phase. What about form factor form factor here, O mas divided by average. You will see 1.0 009, which is here, in this case, compared to the previous one for the three phase, halfwave uncontrolled, you will see that the form factor is much closer to one or much closer to unity, which means it is much better, better compared to the three phase, halfwave uncontrolled. Number two, ripple factor root square of this -10.043, which is also better than the previous one because more pulses here means better performance or better reaching to DC value or DC steady state value. For I put RMS, it will be VRMS divided by R. It will be Vmax phase, I max phase. Power of DC or average but power will be average I average. This multi blood by this will give this value. Then power of EC, we output RMS, I output RMS, multiplying them together, we will get this. By using these values, we will get the rectification efficiency. BDC of BEC gives us 99.82%, and as you can see here, this rectification efficiency is also much higher than the efficiency of the single phase or three phase half wave uncontrolled rectifier, which we discussed before. Okay, so we have these values. Now we would like to get why I'm getting the supply. I'm getting the supply because I would like to get the transformer utilization factor. Okay? So here I'm getting the supply current. So I'm going to take one of these signals only or get the supply current. We have six phase. So I'm going to get the current for one phase. RMS of one phase to get the utilization factor. So supply supply from 60 degrees to 120. It will be root 1/2 by it repeats itself once every cycle. Current supply, IA or IB or IC or whatever. So 1/2 Pi, integration from Pi over three to two Pi over three, 60 to 120 for the current square, value of current is equal to what equal to V max sine omigaty for VA, V max, phase, sine omgaty divided by R. And remember, remember, max phase, because we are going to see in the next circuits that we will have VMAX line. So be careful that we are talking about phase because I'm talking about VA one, which is phase voltage. Out phase voltage. So by doing this and square, we will get this supply current. Now transformer utilization factor will be Out DC power divided by the transformer, volt and pair rating. Here it will be 0.9 12v max phase, Imax phase like this, which is power of DC, divided by the transformer volt and pair. Remember that we have here, how many phase, we have 60 phases, right? Each phase multiplied by power of one phase. Power of one phase will be V phase RMs, multiplied by phase RMS. V phase RMS, which is a phase voltage of one RMS value of one supply, it will be Vphase max divided by root two. This representing the root mean square value, maximum divided by root two. I phase RMS at this value, 0.39 max phase. Supply current RMS. Now, by using this, you will find that the utilization factor 55.1%. So what does this mean? It means that we are only utilizing we are only utilizing 55% of the transformer, which is much lower than the previous case of the three phase half wave uncontrolled circuit. Now, why does this even happen? Because the current supply current only conducts for a short period, 60 degrees. This led to lower utilization of the transformer. Okay, so how can I solve this problem by using other circuits that you'll see in the next lessons. Now, before we continue, there is something which is very, very important, very important. You will see that then current supply current here, positive, then zero, positive then zero. Now, we will have a current only positive current. This positive current, which is in the secondary of the transformer will lead to EPECFy in the primary of the transformer. Current will lead to induced EMF in the primer of the transformer. And since we don't have any negative one that will cancel this effect, it will lead to saturation of the transformer. Now why I'm saying this because you will see in the full bridge three phase rectifier, you will find that we will have positive part and negative part. These two will cancel the effect of each other, preventing the saturation of the transformer. Now, why do we have saturation? Because the magnetization here, we have only positive current causing magnetization and we don't have any negative part that will cancel this effect. Okay? 65. Multiphase Diode Rectifier: Now let's talk about the multiphase rectifier. Now, we took about single phase, phase, half wave uncontrolled with three pulses. We took about six pulses, three phase, midpoint, six pulse rectifier. Now, what if I have multiphase? Let's say, 12, whatever the number. I would like to get a general formula without, analyzing my circuit every time. So in order to get a general formula, we think about it or we analyze the previous equations, and we found that. Number one, for a three phase, half wave rectifier. Each phase conducts for two or 3 radians in one cycle, which is exactly equal two. Here, let's type it. This is similar as what similar as. Two pi divided by three phase, right? 360, divided by three gives us 120 degrees. Four phase. This number of phases. Okay? Now, when we look at, which is, as you can see here, one, two, and three in one cycle. Now in the three phase six rectifier, this is considered as a 60 phase. As you have seen, I have considered it actually as a six phase because each phase is shifted by 120 60 degrees. So if you remember, it is in order to get the induction period, it will be two pi divided by number of phase, which is two pi, which is under and 60/6, gives us two pi over six, which is this one or 60 degrees. So what we can learn is that we can say in general, in general, in order to get induction period for a multiphase, it will be two pi over or number of phases, right? So in general an M phase with direct fire, each phase on dite will conduct for two pi divided by how many phases we have. And you will see that number of phases will be equal to actually equal to number of pulses. So if you look at here, we had ABC, three phases produced one, two, three, three pulses. Here we have one, two, three, four, five, six, produced one, two, three, four, five, six. So it means that number of phases will lead to a number of pulses and number or the induction period will be two divided by number of phases, right? Okay. Now, for M more than three, Mhas did would have a Delta connected primary and secondary would have mid taped over two windings. So what does this mean? Here we this is exactly talking about the midpoint, when we divide each winding by two. Now, number of dits will be equal to M or number of phases. If you remember that in the three phase, half wave uncontrolled, we had A, B, C, three dites equal to three phases. In the three phase midpoint, six pulse rectifier, we had six phases, which will lead to six dites, right. Now, each phase conducts for two boy over M, and this is a general formula. If you look at it, you will see this shape. Now, these are each one conducts for how much two pi over B. Now at zero instant, at zero instant, we can say we have this function Po B and pob negative pob now, why do we use this? You will see right now. So we can, in order to get general formula, we can say that VDC or V average. For a multiphase, I don't know number of phases. Well, B B, which is number of pulses, divided by two Pi. This is exactly what we did. In the three phase, the first circuit, it was three. In the second one for the six pulse, it was six pulses, right? So number of pulses divided by two Pi, integration from let's say we are integrating this one negative PiRP until PiRP. Okay, from here to here, for a cosine function. Now, why cosine because as you can see, here, zero instant. At this point, we start as a cosine with four. Maximum value at zero instant, at zero instant, here. Cosine function. Since we are integrating from here to here, you will get this function Max phase, cosine omegaty. By using this, you will get this average value and I average will be average of R, and Vout mes will be exactly the same equation except square, and I diet average average of each diet will be the output current average divided by number of pulses, right? In the first circuit, when we had three, three phase. Each diet conducts for 120 degrees or third of the average or third of the period. That's why I average will be average voided by P, and if you'd like root mini square, it will be one of a root B like this. All of these are exactly the same as before, except that it is a general formula for a multi phase. For but RMS, it will be voltage divided by R square, the same integration except divided by R to get the current wave form. And Ibut RMs will be output RMS over R. Now, before we continue to the next lesson, I would like to explain this because I know that some will be confused why cosine wave. Okay. Now, if you get back here, let's get back here. Like this here. Okay. Let's look at this waveforms. Now, let's say I would like to get V average, okay? For this out, let's say we're looking like here, for this. Now, let me ask you if I choose VA and integrate from here to here, or V B and integrate from here to here or VC integrate from here to here to get average value. Would it actually change anything? All of them will give the same answer, right? Now, let's assume that we shift it this way form and we put it here. Like this. Let me draw it like this, like this. Okay, so we have this one, Vmax, sine, Omegaty. Now, we shift it to the left side by 90 degrees. By how much 90 degrees. So it will be like this. Okay? A peak instead of peak having it at Pi over two, the peak will be at there. Okay? Okay, now we have integration from, let's say from here to here. Okay? It will be exactly integration from here to here for V max sine omegatty. If you look at this function, it is equal to V max sine, omegaty plus 90 degrees. Y plus 90 degrees because it is to the left leading. So this function is leading, V max sine omegaty by 90 degrees. So it will be V max sine omegaty plus 90. This is from trigonometry is exactly equal to cosine omegaty. Okay, so we took this cosine omega t, and we said, this will conduct a four to Pi over P, exactly similar here. This one will conduct 42 pi over P. Now, this part will be equal to this part will be equal to Pi over P, this point negative Pi over P, and this point Pi over P. So we integrated this cosine function, which is exactly equal to sine omega t plus 90 from here to here. Okay. Now, this integration, if we substitute for pulses equal to three, you will get the same value exactly as if you integrate VA from here to here. Okay? That's the whole idea regarding this one. Why did we use this one? This one. Instead of this one because it is much easier to get a general formula. Now, someone will ask me, what if I would like to get using VA Vmax and omegat? How can I do this? That is the whole idea. They shifted to the left in order to make it easier in getting this formula, which is this one, like here, this is exactly equal to this one equal to anyone. Pi, negative PRB to PRP. This makes the integration or getting general forma much easier. Okay? 66. Example 2: Hey, everyone in this los on, we will have the second example on the three phase rectifiers. In this example, we have a step down Delta star transformer, Delta star transformer with a pre phase tons ratio of five. This transformer is fit from a three phase 1,100 volt, 50 Hertz source. The second ray of this transformer through a rectifier feeds a load of our equal ten. We need number one to find the average value of output voltage, average and RMS values of diet current bow delivered to the luid in case the rectifier is a three phase, three pulse type, a three phase, six type. Number one, we have our supply Delta star transformer, right? Now, this transformer we started from Delta. Input voltage, 1,100 volt. This representing the line to line voltage, 1,100 volt, line to line voltage, and RMS now the ton of ratio between this transformer is what is equal to five. Okay? Now, we have the secondary is connected to a resistive load of ten oms. Okay? So let's type this. Okay, number one. We need first to convert 1,100 to the secondary. Now, from transformer, we know that V two over V one equal to N two over N one. Now, since our transformer is a step down transformer with a ratio of five, it means that this voltage, which is the line to line voltage, and at the same time, since it is a delta connection, it will be phase voltage. If you remember from a connection of transformer, Delta connection has a line to line voltage equal to phase voltage. Now, this ratio for phase voltages, so V two will be equal to voltage phase of secondary will be equal to V one phase of primary, multiply it by turns ratio. V one phase is equal to phase voltage of the primary 1,100 volt, and we multiply it by the turns ratio. Now, is our transformer a step up or step down. It is a step down transformer, it will be 1/5, as it will lead to reduction in our volt. So what you can see here is that secondary RMS phase phase voltage, phase and phase, phase voltage, secondary RMS will be V phase primary divided by turns ratio gives us 220 volt. Okay. Number two, we need average value of output voltage and the average and all of these. Number one, in order to get average value, we will use the equations for average. So if you remember average using the general formula for pulses, Remember that we average B over two pi integration from negative B over Pi to B over By V max phase, cosine omegaty. Here, this is a mistake. I have to change this after finishing this lesson should be cosine omigaty. As we learned in the previous lesson for the general formula for multiphase, we use a cosine function instead of a sine function. To do this integration. As you can see here, I already added goes on here correctly. Now, how many phase we have a three phase, three pulse, number of pulses will be three, three, and three xs. V max phase. This is our phase, root mini square value. In order to convert this into a peak value or maximum value, we will simply multiply this by root two xs. To become a maximum phase voltage. By doing this integration, you will get this value for V average or VDC. Now, what about the average average and R mess of diet current? Number one, in order to get diet current, we need first to get average lud current, I average output or I output average. Which will be equal to V average divided by the loud resistance of a ten Ms like this I average, we average over R, gives us 25.73. Now, what's the next step? If we remember, since we have three pulses, I average, I average four dived will be simply I, I would average divided by what P or number of pulses. So it will be like this I output average over P, which will be 25/3 gives us 8.57. What about the RMS exactly the same as this function, I RMS of dt will be It RMS divided by root B. To get I outbut RMS, we will do the integration for function like this I put RMs, root P over two P, same integration like this one. As you can see, we put it as cosine, but divided by R to get the current equation. By doing this integration and substituting, you will get 2,060.15. Now divide this by root P number of pulses, which is three, like this, 15.1, you'll get the diet RMS current. Now the last requirement in this problem is getting power delivered to the lot. In this case of the three phase three plus type, we need power delivered, power delivered here, since it does not say average power delivered, it says power delivered in general. Since it says in general power delivered, then we are going to use the root mean square rules. Like this power reaching the loot will be I mass squared multiplied by resistance. This value squared multiplied by ten OMs. It will give us this value. Now, if we say average power delivered, it means it will be power of DC, which means I average, multiplied by V average or I average squared multiplied by R. Now, let's apply the same equations for three phase six type. So number one, we need V secondary phase. Remember that we said V secondary phase in the previous, one, 220 volt. Now remember, remember this phase triples, which was like this. Let's write it like this. Like this. Now, this is very important. All of this the total 220 volt phase voltage, RMS for secondary. Remember that in the three phase M six type or the midpoint type, what did we do? We took half of the urns. We divided the urns by half, like this. Write A one and A two. What does this mean for me? This means that the output voltage will be also divided by two. Now, why is this Because remember that voltage voltage is directly proportional to the number of turns. As number of turns decreased to half, voltage voltage for the phase will also decrease to half. That's why, in this case, the phase voltage will be 220/2, like this. Because our tons for each phase is now reduced to half. Okay. Now, what the second requirement average value of out voltage. Again, the same equation, but we are going to use six pulses. You can see Vmax cosine Omega T here Vmax 110 root two. Pulses here, 66 type, which has six pulses. So it will be six, six, six, like this. Us 148.5 55 volt. Average and mass of die current, we need average lot current. It will be this value divided by ten ms to get average current, we will have to divide this by three B six, not three, B six because it will be I average, divided by number of pulses, which is six pulses, it will be 2.47 58. For RMS, we first need the RMS current, which will be root, voltage divided by R like this. Same equations for the previous slide, except that we replaced the pulses with six instead of three. This will give us 14.868. Now, if I would like the RMS of diet, it will be IRMS divided by root P, which is root six, like this. The same equations that we talked about in previous slide and in previous lesson of the multiphase. You can use these rules for multi phase, or you can also use the rules that we discussed for each type as before. Okay? The lost requirement is getting power delivered. It will be out current square multiplied by resistance like this. In this example, we learned how to apply the multiphase rules in order to get values of average, root RMS current and power delivered in a three phase repulse and three phase M six type. 67. Example 3: Welcome, everyone. In this lesson, we will have another example on the previous lessons for Half Wave diet. This example, we have a half wave rectifier with a highly inductive load. The output current is a constant value of 96 pairs. Great. Find the root mean square and average value of the diet current when six dites are used and when three dites are used. So what do we need? We need RMS of the diet, average of the dit when we have six dites and three dites. How can I solve something like this? Number one, highly inductive load, constant current, it means that I output is equal to I average. Equal to RMS, equal to mind six and pairs. Now, the general rule for diet I diet, average will be I average, which is mind six divided by P for diet. RMS it will be mine six divided by root p96 over P and 96 over root P. So one time six dites diets means six pulses. The dites means three pulses. So it will be like this for six dites. I average 96 divided by pulses or P six gives us 16 and pairs. For IRS it will be the same current divided by root P. L. Min six divided by root six, 39.19 a pairs. For the second decays for three dites, the same two equations except instead of six, it will be three pulses. Mine six over root three, 96/3. Like, very easy, right? Very easy and straightforward example on applying the multiphase rules. So I hope the circuits for the half wave pulse three phase, three pulse, and the six phase midpoint waveform half wave, uncontrolled requir is clear for you. 68. Three-Phase Half-Wave Controlled Rectifier: Hi, and welcome everyone. In this loss on, we will to APA the three phase, half wave controlled rectifier. Similarly to the previous circuit, the first circuit that we discussed Apooin the three phase rectifiers, which contained three diets, we are going to do the same circuit, but controlled. It means that we replaced the three diads with three ci restors. So first, we will look at our circuit. You can see the exact same circuit, A, P and C, three phase with a neutral line connected to our lute. We have three istors T one, T two and T three. Now we would like to analyze this circuit in the presence of a highly inductive load. This is easier way than resistive load. We will start with HIL or highly inductive load. So let's look at our circuit. In order to conduct any of the iristor, if you remember, we need two conditions. Number one. The pistor must be forward biased. For example, T one must be forward wide, voltage across it, voltage across T one, must be positive Number two, T one must get its firing angle, right? And T one or voltage, VA must be the highest voltage. So in the uncontrolled circuit, we had two conditions or one condition. Number one, voltage across it must be the highest. VA must be higher than VB, must than VC in order to conduct, D one did one. Here we will add another condition which is having the firing angle alpha. So like this being the highest post phase voltage, and firing angle is already applied to thus restore. So let's look at our circuit. So the same idea. We have three phase. VA, VB, VC shifted by 120 degrees. Now, VA, VB, VC. Now, of course, VA here starts from zero, so it should be like this. This is just a mistake in drawing, so this should be like this. Starting from here, not from here, but from here from. So we have VA, then be shifted by 120 degrees and VC shifted by 120 degrees. Okay? Now, how does these ciistors conduct or how does this circuit work? Now, look carefully here. When we had diets, dietes are basically siistors. So this is very important for you. Dietes are simply simply Ti restors with a firing angle or Alpha equal to zero degrees. Okay? So diodes can be treated as thigh restors with zero degrees or zero firing angle. Now let's look at here when we had zero firing angle or dites, let's just delete this. When we had diet, the output we forms like the slight VA, then VB, then VC, VA. This is our out we form when did when the dites we're conducting, when the circuit is uncontrolled. Now, in order to add a firing angle, we start measuring our firing angle from this reference from here, as you can see here. So the firing angle Alpha is measured from here. From the intersection point between VAN and Vcn. This point exactly is equal to 30 degrees. This is very important, as you'll see this a lot in the three phase circuits, specifically the controlled circuits. 30 degrees is very important. 30 degrees is the intersection point between VN and Vcn. Now, 30 degrees is a point at which I measure my firing angle alpha. Firing angle Alpha is not from here, not from zero, but it is measured from here. So as you can see here, we measured firing angle from this reference, Pi over six, which is 30 degrees and after a certain time, it will be this point at which VAN will be fired will be Alpha plus Pi over six. That is a point in time. Angle omega ty, at which the psorT one will start conducting. So let's look at here. So we have how many restors we have in this circuit, three thistors, right? Now, each one will conduct for 120 degrees, right? So it means that the phase shift between these istors is 120 degrees. So what does this mean? Let's say this one will be fired at Alpha. Then T two will be fired at Alpha -120 degrees. This one will be fired at Alpha -240 degrees. So each sistor will be fired after 120 degrees. So if you look at this shape, you will see first firing angle for T one after 120 degrees. Here, from here to here, from here to here, you'll find the firing angle for thirestornumber two, after 120 degrees for V B. Then we have another 120 degrees you will find the next firing angle for C. Then we have after 120 degrees, we will get back to A. Okay? So what you can see here is that each thirestor conducts for 120 degrees, and the phase shift between these sirestors is equal to 120 degrees and their firing angles. Okay, so let's see here. Number one, this intersection point at which you measure firing angle Pi over six, 30 degrees, and this firing angle first will be Pi over six plus Alpha. Now, let's look at this figure. 21 gets its firing angle. Okay? And at the same time, VA is the highest voltage right at this instant at here. So VO will be VAN right it will be like this. Okay. So let's delete all of these crowdy drawings. So we will start from here. Neglect this part. You will understand where did we get. So at this point, the firing angle for AN until the next firing angle for T two, right? So it will start from here and it keeps continuing like this, like this until the next firing angle. Now, this firing angle is for T two or VB. And if you look at this instant here, you will find that VB is the highest voltage, which means that T one will be turned off and T two will start conducting. So it will go from here from VN and go to VBN. Then it will keep conducting until the firing angle 43. So you can see firing angle is at this instant, like this. Okay. So at this point, it will go like this like this to Vcn since a VC is the highest voltage, right, and it will continue like this. Until VN will start conducting again as the highest voltage with the firing angle and like this. So this part is exactly similar to here, like this. Go down here and draw it like this. Okay? So our waveform will be like this, as you can see here. Now, of course, if you look carefully here, you will find that we have a negative part, right? So when VN goes like this to negative, it means that this store T one should be reverse biased, since we are entering the negative region. However, due to the presence of highly inductive lute, it will provide energy like this back to A and keep T one in conduction mode. So it will never turn off until the next firing angle for P. So it will go like this continuous waveform as you see here. So the negative portion exist due to the presence of the highly inductive load. Now, if you look at this is our out wave form. Okay. Now what about I will be constant current due to highly inductive load. What about IAA is our supply current, or IB supply current or IC? Not all of them together, but just phase current one of the three phases. So IA it will conduct from here until here, right? This is a period at which T one conducts. So it will be this period, this part of the current, right? This period starts from Pi over six plus Alf, right? 30 degrees plus the firing angle. Pi over six plus lf. Now, this will conduct for how much for 120 degrees. So we will add to this 180 120 degrees. So we will have Pi over six plus Alpha, add to it how much, 2/3 Pi, which is 120 degrees to get this point. Now, 2/3 is equal to 4/6 pi. Exactly the same, right? So we have Pi over six plus Alpha and four Pi over six. Four Pi over six plus five Pi over six will give us five Pi over six plus Alpha. Okay? So this here is the induction period for IA. By using this waveform, you can get IA, average current of si register, average current of out average current from here out average will be average over R and so on. Okay, so let's look at our equations. Number one, VDC, average. Now, how many pulses do we have? We have three diets, which means we have three phases, and at the same time we have three pulses. We have one, two, three, in one cycle. So it will be 3/2 Pi for one function, we will integrate just one function. For example, we will take VN as an example, which is V Maxine omegaty. We will integrate it from here to here, which is Pi over six plus Alpha until five Pi over six plus Alpha four VN. Since this is our VO will be equal to VN during this period, like this. Okay. Now remember, if you decide to choose VBN, it will be V max sine omega t -120, but you will integrate from this angle, which is five Pi over six plus Alpha until the next firing angle, which is after 120 degrees. All of them will give you the same answer which is three root 3/2 Pi, Vmax, cosine Alpha. Now, if you look carefully at this equation, if you substitute with Alpha equal to zero, which means we have three dites instead of as three thi restors, it will give us three root three of our two Pi V max, which is exactly the same function that we obtained in the first lessons for the uncontrolled rectifier. Now, if I would like I average, it will be simply V average divided by R, right? I put average, equal to IRMs, equal to V average over R. Which is this constant value. The same value here divided by R, right? Okay. Now, what if I would like RMS exactly the same equation except it is a square like this, as we always do in the previous lessons. What about I but RMS? It is exactly I average, okay? Not the but RMS over R. Since we have DC, but RMS is exactly equal to I average, okay? Now, remember this is very important, okay? When Alpha, when the firing angle is less than 90 degrees, and you will see this in the next slide. When Alpha is less than 90 degrees, we will have rectification mode. You'll find that out Average will be greater than. Now, positive output voltage. Now, if firing angle Alpha equal to 90 degrees, average by substituting here with Alpha equal 90 degrees because I 90 gives us zero, which means average will be zero when Alpha equal to 90 degrees. You will find that the positive portion of the waveform will be equal to the negative portion of the waveform. Now, when we start increasing Alpha beyond 90 degrees, greater than nine degrees, you'll find that V A average will become negative or inverted becomes negative. Okay? Now, this is important. Y. So you can produce a negative voltage Pi using a high firing angle or firing angle greater than 90. Another thing that you will find in electrical power system is that when we use when we use a firing angle Alpha greater than 90 degrees, we have inversion mode. Inversion means inverters. So we can absorb or deliver power from the lot back to the supply. By using firing angle greater than 90 degrees. Now, how is this possible if our loud contains a PMF, E or a DC DC motor, it will give us a power back to the supply. If the firing angle greater than 90 degrees. Okay? That's why you will understand in the next sections for the power electronics scores, we have inverters that convert from AC from DC voltage into AC voltage, inverters. Here we took about rectifiers that convert AC into DC. So don't worry, this is just a small note that you have to remember for yourself. That's the firing angle less than 90 degrees. It will be rectification. Normally, if you exceed 90, out will be negative or it means that the power will be back to the supply or the load will start providing electrical power to the supply or invert DC power into AC back again. Okay, now let's look at the effect of Alpha or the firing angle. You can see here this is a small firing angle, 30 degrees. This is a point at which we measure Alpha 30 degrees and Alpha 30 degrees, so it will be 60 degrees this point. You will see this is our out wave form. Less than 90 degrees, all of the out wave form is positive, right? Now, what will happen when we increase Alpha? You will find that when we increase Alpha, you'll find that we have more negative part and positive negative part and postive part, negative part, and positive part. Okay, at Alpha equal to what? Equal to 60 degrees. So 30 plus 60 gives us 90 degrees. Now, if you increase Alpha to 90 degrees, what you will see that is a negative part started increasing and becoming equal to the positive part of the waveform. So you can see this part equal to this part. Increasing the firing angle Alpha leads to increasing the negative part. And will lead in the end if we increase beyond 90 degrees, it will lead to inversion mood or even negative output voltage. Okay? Now, let's understand what will happen if we have an R loot. Okay, if we have an R loud, it will be exactly the same. However, remember, if the voltage goes to zero, if the voltage goes to zero, it means that output waveform will be zero or output voltage will be zero. Let's see this. So we have Alpha equal to zero, 15, 30 degrees and 60 degrees. Look carefully at this. At zero, we have diet like this. At 15 degrees, it will be like this, goes like this, all of it positive, so nothing changed at all from the highly inductive lute. Now, going to 30 degrees, Alpha equal to 30 degrees. You look carefully. Positive like this until 180 60 plus 120 gives us 180, 120 100. So as you can see, nothing changes, continuous waveform, exactly similar to highly inductive lute. Now, when Alpha, in this case, in the R lot becomes greater than 30 degrees, this will lead to zero voltage. Okay, let's see this. So you can see that here. Alpha becomes at what at 60 degrees, greater than 30. So it will be Alpha equal to 60. So you can see it starts from here, Alpha. Okay? Let's look at here. So this is a firing angle 90 degrees. The next one will be at 210, 120 degrees and 120 degrees. So this is for T one, T is for T two, and this is for T three, okay? Now, when it conducts here at 90 degrees, it will go to V T one will conduct, and will be equal to VA right like this. Mm hmm. Now remember, firing angle is here. So what will happen in the highly inductive lute, it was going like this, going to the negative portion, let's just delete it. Let's write it with pencil. So it will be like this, going like this. In the highly inductive lute, it was going to the negative part and then goes like this and then goes to the negative part and like this. Now, if you look carefully here, we have a resistive lute. So when it goes to the negative part, if it goes beyond zero, T one, will be reverse biased. Like this reverse Mm hmm. Let's tie with reverse pies, right? Why? Because we are going to negative part of the waveform. Now, we don't have here any highly inductive load or inductance that will give current to T one and keep it in conduction. We have only resistance. So when we go to the negative part, it will be reverse piac and T one will be turned off. So we will have this we will have zero part here until then it cyst to firing angle, and the same will happen here and again. Okay? So it doesn't go to negative like this. If you get back here. Look at Alpha equal to 60 degrees in high inductive fluid, it goes, Okay, no problem to the negative part. However, due to the presence of resistive fluid, which does not have any storage energy due to the non existence of inductance, this will lead to clipping of the negative part. Okay? So it will just conduct until this point and say and completely die, until the next to firing angle. Now, of course, none of the other cirestors will conduct unless they take a firing angle. Remember this, okay? Like this. Okay, so how does this will help us? This will help us when we solve different equations. So fund that we have two cases. Number one, when Alpha less than 30 degrees, that wave form is continuous like this, this part Alpha, less than or equal to 30 degrees, this one, or even or even equal to 30 degrees. The same. You can see continuous, no problem, continuous, continuous. So it's from Alpha until Alpha plus 120 degrees, right? Normally exactly the same equation. This is the same equation as the highly inductive lute. The average and the term is exactly the same, no problem at all. However, the change will appear when we have Alpha greater than 30 degrees. When Alpha greater than 30 degrees, we will have xs. If we look at VN, it will start from Alpha until Pi from Alpha to Pi. This is a general equation. From this point, which is Alpha plus Pi over six until Pi. So it will be as average will be Pi over 60 plus Alpha exactly the same here. However, the end point will be Pi or 180 degrees instead of five Pi over six plus Alpha. Why? Because after this, it will die completely, die completely. So it will be from here until here. Okay? That is a difference between this and this case. So it will be 3/2 Pi, Vmax, phase one plus cosine, Alpha plus Pi over six. Okay. Now what about V ORMs exactly the same equation this one except square. I average pus cases in both cases, it will be V average over R. And IRMs remember, RMS will be VRS over R in puse cases. Okay? So I hope you now understand how does half controlled a half wave controlled, not half controlled, half wave controlled rectifier circuit works in the presence of lute and in the presence of highly inductive lot. 69. Example 4: Now let's have an example in order to understand how to apply the previous equations for controlled circuit. So we have the circuit that we discussed in the previous lesson, and we have half wave, three phase, fully controlled, half wave control, fully controlled, three thistors. Connected to a resistance of 5.12, has an outwit average current of ten amperes. So it connected to a resistance, resistive load, and average output current will be 10:00 A.M. Pairs, and supply 120 volt. Of course, 120 volt. What does this mean? It means RMS line to line. We need one thing to get in the circuit, firing angle of the circut. Okay, now, how can we get the firing angle? Now, remember, since we have a pure resistive loud a pure resistive lute, it means we have two cases. Number one, how to get the firing angle first. Firing angle can be obtained from the output average, right? The question is, how can I get VO average? It will be simply. We have out average, right, and we have the resistance. We can say V out average will be. The resistance or multibloid I out average. Okay. Now using VO average, we can get firing angle Alpha. However, remember that we said in the previous lesson that pure resistive load, pure resistive load. It means that we have two different cases. Number one, we have first case when Alpha less than 30 degrees. We have an equation for VO average. And we have Alpha greater than 30 degrees, we have another equation, right? So how can I know if it is this one or this one, if Alpha is than 30 or greater than 30? Now, you can know this. You have to do a trial and error. First, you have to assume that Alpha is less than 30 degrees, for example, and get value of Alpha using VO average of Alpha. And if Alpha is actually less than 30 degrees, it means that this is a correct solution. If Alpha became greater than 30, it means that this one is incorrect. So what I'm going to do, I'm going to use this one and solve the problem. If Alpha actually came greater than 30 degrees, it means that this one is correct. If not, it means it is less than 30 degrees, and I will have to use the equation of Alpha less than 30 degrees. So it is a trial and error. You have to do this and this and your luck will appear, okay? Okay, let's delete this. Okay, great. So we will assume first Alpha greater than cert, okay? I will assume this. So what I'm going to do, I will use the equation for Alpha greater than cty that we obtain. We average got 3/2 Pi VMX phase, one plus cosine Alpha plus Pi over six, and VMX phase. We have Vmax phase. We have this, which is line to line, RMs, three phase, right. So in order to convert it into phase, we will divide by root to convert it into max from root mini square, we multiply by root two Ls. So let's look. We have V average. As I said before, I average multi blood by resistance gives us this value, 51.2 volt, and this will be equal to this equation over two Pi, 120, root two over root three. One plus cosine Alpha plus five or six. By solving this equation inside the calculator, you will get Alpha. If you don't remember how we explained how to use the calculator in order to get value of Alpha or firing angle. Or even if you don't remember Pita or Beta in the first section of the course. So what did I do after solving this? I found that Alpha equal to 54 degrees, which is actually greater than 30 degrees. What does this mean? It means that my assumption is correct. Okay? If I found that after solving that Alpha, let's say, less than 30 degrees, then it means that this answer is wrong, and I will have to repeat the same steps about use V average for Alpha, less than 30 and get value of Alpha. Okay? 70. Evolution of the Three-Phase Bridge-Wave Uncontrolled Rectifier: Hey, everyone. In this lesson, we will start or in this lesson and the next ones, we will start a new part or a new chapter regarding the three phase rectifiers. We will start now discussing the three phase bridge wave uncontrolled rectifier, then we will go to controlled rectifiers. So what we are going to learn instead of half wave, we will start from now for bridge wave uncontrolled, controlled and half controlled. So we would like to know the evolution of three phase bridge wave uncontrolled fire. Where did we get it? Okay, starting with the first circuit that we discussed before the half wave, right? This circuit, I'm going to use the pencil once more. So we have this circuit for the half wave. If you remember that we said before, since the dites are in this direction, it means that the diet will conduct when it has the highest phase voltage, right, and Vout will be positive, right? Highest, phase voltage or highest, positive phase voltage, positive value here, positive, negative, right? Okay. That's why, if you look carefully here, this is the highest voltage. That's why V Out would be equal to VA, then VB, then VC, and so on. That is what we discussed exactly, okay? Now, let me ask you another question. What if we reverse the direction of diets. And instead of this direction, we will draw it like this. So let's say, like this. Like this. Okay? And this one like this. And this one like this. What do you think will happen? In this case, the diet with the highest negative voltage voltage will be conducting or it will start conducting when it has the highest negative voltage, like this. We have D one, D two, D three, d4d5, d six, we reverse the direction, right? So what will happen is that the abut will be number one, will be operating at the diode will operate at the highest negative value, right? Highest negative. So let's say if VA is the highest negative value, then D four will conduct and abut will be negative value with respect to the neutral. Okay, negative value. Here, look carefully. Positive value, positive value, highest positive value. Here, it will be negative value. This is important because we will use this when forming our bridge. So as you can see, negative, positive, this is a value of the abut with respect to the neutral. Here it is positive with respect to the neutral. So in this case, if you look at the waveform, it will be like this. I will operate at the most negative part like this. Here, this is the most negative, then this one, then this one, then this one, so you can see, like this. This one is VB, most negative, then VC, most negative, VA, VB, and so on. So you can see we have a positive, most positive value, and most negative value. Right using by inverting the direction of diet. Now, what I'm going to do is that I'm going to combine or form a combination or combine these two circuits together. I'm going to combine this with this to take advantage of positive and negative. Now, how I'm going to do this now look carefully. We have this circuit, first one, second one. Here it's connected to positive and the other one to neutral right here. This one is positive and the neutral negative. Okay. This one here Q is negative, and the neutral is positive. So here, negative compared to the neutral. So what I'm going to do is that I'm going to combine these two neutrals together like this. So we have number one A, P and C, D one, D two, three, D one, d2d3. They are connected to positive Mm hmm. And the other terminal connected to the neutral, like this other terminal neutral here to the negative, negative, neutral, like this, right? So this part, lewd here, resistance or whatever, this part representing this circuit, right? For this one, we have A, P and C with d4d5, d six, A, P, C, D four, D five, and D six. All of them are connected to Q, a terminal Q, which is negative, okay? Q, which is negative, and the other terminal, which is a neutral positive, right? The neutral positive. So what we did is that we combined these two neutrals together. Okay, as if we connected this here, take this. Let's draw it from here, take this neutral and connect it to this one. So when you combine them together, you will have this circuit, right? You will see that the first one will form a positive voltage between P, P, and neutral, right. Also for the second circuit, it will produce a positive positive output between neutral and. Right, you can see here negative, positive, right? So it means that neutral has positive, converted to Q. Neutral has positive converted to Q, right? So you will see that these two when they are combined together in this form, we will have positive, negative, positive negative, which means that our output will be submission of these two, right? If we cancel the neutral completely like this, then Out can be considered as this voltage plus this voltage, right? Okay. Let's delete all of this. I hope it's clear we just combine these two with their signs, positive negative, positive, negative, as you can see, positive negative exactly like this. And here, neutral, positive, and Q negative, neutral, positive, and Q negative. When we add this like this, we will be able to combine them together. Very clear. Number two, we will delete this neutral and this neutral. We don't need it at all. Okay. That's the first thing. Second thing that you will see that we have phase A, phase A, phase B, phase B, phase, and C. So what I'm going to do is that I'm going to say A like this, connect it here and here, right, and say P, connect it here and here. And then C will be connected like this, here and here. So let's look at our circuit. It will be like this. A, B, C, A connected to D one, A connected to D four, like this exactly as we did. P, D two and D five, C, D six and D three, the neutral is completely removed and we have two loads in series. So we combine them together to form one loud positive and negative, right? So this is considered as our bridge. Now we can rewrite or redraw that circuit in another form like this. A, B, C, A connected to D one. You can see A connected to D one and d4d4. This is exactly similar to this. B connected to d2d5, d2d5, C connected to diode three and diode six likes us. Okay. Great. And we have two terminals. B positive connected to all of the above dites D one, d2d3, like this, positive and the Q or negative connected to the other terminals or D 4d5d6, like this Q, connected to D 4d5d6. This is exactly this, exactly this is exactly the two separate circuits. Now remember, remember. These are these diles operate when at highest, highest positive voltage, right? Voltage voltage like this. And these dites here, these dites operate at highest negative voltage right Okay. So here, as you remember, D one, d2d3, which operate at the highest voltage, and D 4d5d6 operate at highest negative. Okay? Now, this is very important, as we will use this when we draw the output voltage, okay? Great. Now, we can lead from here. We can rewrite the numbering of these dites. Instead of D one, D two, D three, it will be dd3d5, D 4d6d2. I know that someone will ask me right now, why did you do this? Why Did you change this numbering or this numbering of these dites? Why did you write it or rewrite it in this form? You have to know that in this circuit, what will happen is that D one, the sequence of switching on here, it will be D one, D two, then D two, D three, then D three, D four, then D five, D four, d five, then D five, D six, then D six and D one. Okay? So D one, D two here in this circuit, in this one, not in this one. Okay? So it will start like this. D one, D two. Then D two and D three. Then D three and D four. Then D four and D five. That's why we have rewrite this numbering due to this sequence. Here, it will be a little bit difficult to write. Here, it will be D one, then D one, D six, then D six, D two, then D two, D four, very confusing. This is much easier to understand. D one, D two, D two, D three, d four, D four, d five, D five, six, D six, d one, and so. Okay, we will see this right now in the next slide. Okay? So now we understand our bridge, and we have VA, VB, VC. Okay, great. Number one, VA VB, VC. Now, D one, D three and D five operate this upper part operates at highest positive voltage voltage. This one at highest negative voltage. So it would be a access. So when we combine two of the three pulse rectifiers that we combined in the previous slide, we obtained what we call here the three phase. Six pulse. Since we have six diles, it will be six pulse or it's called a three phase bridge rectifier. So this is our three phase bridge wave uncontrolled rectifier. Okay. Now, dites D one, d3d5 operate at the highest positive voltage. And D 4d6d2 operates at the highest negative voltage. Okay? Great. Now, let's look at here. So this is the highest. Let's start from C. Okay? Let's start from here. So C is the highest, positive voltage in this region, right? Let's just divide them like this. Okay. First, let's look at the highest positive voltage. So VC is the highest. So let's look at VC. So C is connected to D five and D two. D five will operate. Why? Because at this instant VC is the highest positive voltage. So which one will conduct D one or D three or D five, D five? Because it is connected to the highest positive voltage. So here we will have D five. Okay? Now, in this part, VA is the highest, right? So let's look at here, VA, where exactly? VA. VA is the highest positive voltage. So which one will conduct D? Right, D one, connected to highest positive voltage. So in this region, D will conduct. Now what about VB? Let's look at B, connect it to D three. And in this region, B is the highest. So it will be D three. And VC again will be D five, then D one, and so on. Okay? That is for the highest positive voltage. D one, d3d5, that is a sequence of their switching on and off. Okay? Now, let's continue. What about the negative part? For the most negative, this most negative here, most negative, most negative, most negative, and so on. So here let's start from here from here to here, from here to here. This part is, what is this waveform? This way form is VB. VB is the highest negative value, negative value. So if you look at B, here, connected to, highest negative, D six, connected to, highest and negative, which is B. So in this region, let's get back. In this region, D D six will conduct. Now, from here to here, V C, VC is the highest negative. So let's go to C, C, highest negative, connected to D two. So here, D two will conduct. Now here, this region A, A connect to D four. So it will be D four. Then again, B, which is D six, then D two. Okay? Now, let's look at our waveform. So you can see that our waveform is divided into several regions. Like this, like this. We will see the final result when we finish this. You can see here we have D one in this part, in this part, D one, and D two. In this part D three and D two. In this part D three and D four. In this part D five and D four. D five, D six, D six here, d6d1, D one, D two, so you can see, D one, D two, D two, d three, d three, d four, d five, d six, and so on. The same sequence that I have explained in the previous slide. Okay? Okay, now what how can I draw the out waveform? This is very easy. Number one. Let's look at here. So we have said the same sequence. You can see from here to here d one, from here to here D one, from here to here D two, from here to here D two, this risk, you can see the sequence for the positive group and sequence for the negative group. Okay? Okay. Now what? How can I draw this bit waveform? This is our Obit waveform. How can I draw it? Now, let's start from here, from here to here, D one and D six, right? From here to here D one and D six. So let's look at here. D one Let's type it here, D one and d6d1 positive, D six negative, right? So D one is connected to what connected to VA, right? VA. And the six connected to what D six connected to B, right? V B. So if you add, if you look carefully here, you will find that we have VA and VB. Okay? So if you look at if you apply QVL in this loop, you have this one as a short circuit. And D six as a short circuit like this. So what you can see that we have A and B, right. So A and B going like this, like this, like this, right? So VO will be V A, B, right? Positive and negative, VAB, the line to line voltage. That's why you will find that in this region from here to here to here, D one, D six, you'll find that VO is equal to VAB. Okay. Now let's look at the next one. Let's delete all of this first, like this. From here, from here to here, it will be D one and D two, D one and D two. So it will be VA, V A C. Line to line voltage is our output. So you can see here in this region VC. From here to here, D three and D two, D three, and two D three connected to B, D two connected to C. So it will be VBC like this and so on. So you can see VAB, VAC. The outut here in the bridge, not phase voltage, not maximum phase voltage. But this time due to the presence of bridge rectifier, we have an outt of line to line voltage. Okay, so great, line to line voltage. Now, here you can see maximum value is root three V max phase, which is line to line max or V max line to line, maximum value of the line to line voltage. Okay? So this is how you get the output wave form using the bridge rectifier. What are the seps that you have to follow? Number one, draw the three phase, normal three phase A, B, C, Okay, draw draw ABC, and look at the highest terminal highest voltages and look at the dies that will conduct D one or D three or D five. And for the negative part, look at the most negative and write it down. Then you will combine them together, D one, D six, d3d2, and so on. That is the first method. Another way, another way. Another way to get it is by drawing the line to line voltages, all of these line to line volte, VAB, VAC, VBC, VBA, and so on. And the highest line to line voltage is the one which will conduct. So you can see here from here to here, VAB is the highest line to line voltage. That's why from here to here, it is the one that will conduct. From here to here, AC is the highest volte, it is the one that will conduct. Now, how can we draw this waveforms? We will learn this in the next lesson, and we will learn about the equations for the circuit. 71. Three-Phase Bridge-Wave Uncontrolled Rectifier: Okay, so let's talk about the three phase, bridge wave, uncontrolled directifier, and the equations or how to draw the waveform. Okay, so we have an R load an R loud. First, we'll understand how does an R load work and we will learn what's the difference between it and the highly inductive loud. So we have our circuit here, Delta store and as we said before, Delta store is used the two trap or eliminate the triple harmonics due to the presence of DC loud and the rectifier on the other side. So we have D one, d3d5, D 4d6d2, and the three phase A, B, C, and our loud, which is exactly the same circuit as the previous las. Okay, so let's look at our circuit. So we have this circuit and the waveform here. Okay? VAB, VAC, and we learned how to get this values or how to get the out waveform, output voltage, okay? Okay, now let's look at the rest of these waveforms, and we will learn how to draw this. Okay. I will show you exactly how to draw this. So first, let's get I supply, okay? I supply before we learn how to draw, let's learn about I supply. So the supply current can be IA or IB or IC. So IA is a supply current, right? Supply current. So supply current can be equal to what? C equal to from here from KCL exactly similar to the bridge rectifier. Remember the bridge rectifier of the single phase, exactly the same idea. So we have ID one, going in this direction, and I defer going in this direction. And I supply entering. So from KCL, I supply will be equal to ID one minus ID four, right? So ID one, the one which flow through it is IA, right, A. So any current of IA, any current going out of IA or entering IA is considered as phase A. Okay, let's see this. So first, look at anyway forms involving A. You can see VAB. V, this one, this one, VVC nu, VAs, visas, VCB no, VAB, yes, VSS. What we see is that VAB. What does this mean? It means current going out from A going to B, right? Going out from A, going like this out of A, it will go through D one, IA will be positive, since it's going from A to B. So Vot is equal to IAB current going from A like this, through D one and we get back through D six back to B, like this, right? IAB. So IAB I abut will be IAB. Now, since we have a resistive loot, IAB will have the same form as AB, except that it is divided by R. IAB will be VAB divided by R or the resistance. VAC IAC will be VAC divided by R, IVC will be VBC divided by R, and so on the same we form except divided by R. Okay. Now look at IAF A to B positive, from A to C positive. So we have our output like this, AAB, IAC. Okay? So before this, we don't have any current, zero, zero, right? Here we don't have any current zero. Here we have V BA. What does BA mean? It goes from B to A. So it will go from B like this, from B like this, to A, it will go like this through D four and back to A. So you can see that the direction of current is negative. I A equal to negative. It's going to the phase, not going out, but entering the phase. That's why VBA from A B going to A or from C going to A, it will be negative, like this. Okay, so going from A out of A to B or out of A to C, it is a positive value. Going from any other phase to A, it will be negative value. Now, there is something which is very interesting here. You will find that we have a positive part and a negative part. This will not cause any saturation inside the transformer because we have a magnetization due to a positive current and magnetization due to negative current, so they will cancel each other. If you remember in the previous lessons, we said that the current was like this for the half wave, it was like this, like this. There was no negative part. That's why this one, the half wave caused saturation in transformer. This one will not cause saturation due to magnetization positive will cancel the effect of magnetization of negative or vice verse. Okay, so we learned how to draw these waveforms. So what you can see also is that I supply average will be equal to zero, positive part, equal to negative part, so I average will be equal to zero. Okay? For the D one, it will conduct during this period, right? IAB and IA this part, right? D one from here to here. Okay, ID one. Okay, now, before we go to the WayFom of died, I would like to explain this part, how to draw this wavefom. This is very easy. Okay? If you understand what are you going to do, you'll find that it is a piece of cake. Number one, all you have to draw is number one, VA VB and VC. If you remember from the phase shift, VA, VB VC, all are shifted by 120 degrees. So what you will find that here angle here, 120 degrees, this 120 degrees, and this 120 degrees. Right? Okay, great. What I would like to draw or before, what I would like to draw, you have to draw VA and negative VA. Negative VA means the same vector, but in the opposite direction. VB negative VB in the opposite direction, V negative VC in the opposite direction. So we have the vector and it's negative. I will understand now why I am doing this. Number one, I would like to draw VAB. VAB will be equal to what VAB will be equal to from vector, VA minus VB, or you can say VA plus negative VV, right? It is exactly the same. Now let's look at VA. This is our VA. Right? And whereas negative VB, negative VB is here. So what I'm going to do is that take this vector and add it to the tip of A, like this. Take this parallel to it. So this is exactly similar to this. This negative VB, VA minus VB gives us VAB. Okay. So we draw in our VAB using this VA minus VB, parallel to it, this vector, take it and put it at the end of the vector, and you will get VAB. So we have any vectors. If you'd like to add any vectors, it will be like this, let's say, VA And VB, you will put it at the end, the beginning of the second vector here, and continue like this. VB like this. If you would like to add, these vectors take the beginning and the end and connect them together like this. So we have VAB, like this. So that's what we did is that we too we have VA, and we took this vector, put it here. With the same perl like this, connect beginning with end. You will have VAB. Now, similarly, you will do this for others. VAC, let's type it here. VAC will be VA minus VC or VA plus negative VC. So what we did is that we took VA and negative VC. We took V negative VC and add here, negative VC. So we obtained VAC. By doing this, you will get VAB, VAC, VB VVC, VV A, VCA, VCB and so on. Okay. Okay, that is the first way, which is very difficult, and you'll find all of the angles here. VAB shifted from VA by 30 degrees, leading by 30 degrees and VAC between it and VA 60 degrees between this and this 60 degrees and so on. That is one way in order to draw these waveforms with their phase shift. Is there any other easier method to get these vectors? Yes. Okay, so first, what you have to do is that you will see V A P. Okay, equal to V max sine Omega t plus 30 degrees. Okay. Remember this VAB, sinot plus 30 degrees we have here. Let's look at it. VAB. At this point, 30 degrees from Zero because it is leading, right? Then look carefully here. We have VAB. Next one, start from the end, then the next letter C, then start from the end, VC, then the next letter, which is A. So we have VAB, VBC, and VCA. Now, the phase shift between each of these vectors between VAB and VVC, VVC and VCA, VCA, and then again, VAB, phase shift between them is 120 degrees. So you can draw VAB like this. VAB. Then let's say from here, after 120 degrees, here, you will get VBC. After 120 degrees, you will get VCA, like this. You can see this and this one and this one, all of them are shifted by 120 degrees from here to here and from here, to here and from here to the next one, which is VAB again, and so on. So what's the phase shift between them 180 degrees, 120 degrees, okay? VAB, VVC, VCA, VAB, and so on. Okay, this sequence. Now, similarly, in order to get VAB, let's see where exactly here. VAB, V BA and VCB, and VAC. All you have to do is take this waveform and invert it. So VBA is equal to negative VAB. VBC is equal to or VCP, negative VBC, VAC equal to negative VCA. Okay? Someone will ask me, how to do this. You see this one VAB, right? Positive. So I'm going to do like this, draw it from here like this. This will be V A. Then you continue like this, continue like this. You will get VBA. If you look at here, let's delete all of this. So we have VAB like this, it will be from here to here like this, opposite to it, PA. You can see that if you continue the waveform like this, like this, you'll get VBA, VBA, opposite to VA, B like this. This is VAB, and this is VBA. Similarly, VBC in order to draw VBC, In order to draw VBC, take VCP, or draw VCP, it will be inverse of VBC. It will be like this, take this and invert it like this and then continue like this, you'll get VCP. So on. So by drawing VAB VAB, VBC, VCA, VAB, by drawing this and drawing the inverse of these waveforms, you will get this shape. Okay? Okay. Now what about our died voltage? Here, this waveform. How to get V dite? You can get it from here. You can see here we have died one died one, I would like its voltage. Positive, negative. So we can apply QVL like this. It can be, let's say, from here to here from here to here. From here to here, you can see that D one is conducting, right? So VOO VD will be short circuit, which means zero voltage, right? Zero voltage. Okay? What about from here to here. From here to here, look at, since we are talking about D one, look at the other positive ones, D three or D five. So from here to here, from here to here, this three is conducting, it becomes a short circuit. So in order to get VD one, you can apply QVL like this, KVL, like this through VD and through these three and back to the supply like this. So you can see that VD one is equal to VAB from this KVL, right? So in this period here this part from here to here is VAB. So let's set carefully here. You can see VAB goes to the negative side like this until here, right? So this part is this part. Okay? Okay, let's read this starting from 270 from here, from here to here, from here to here. You can see what you can see that this one which is conducting D D five conducting short circuit. And I would like D one, so I can give you like this from here through D one and through D five, through D five, and back to the new like this. It will be VD one will be VAC. Since it goes through D one, through D five and like this, short sect, I'm looking for the one with a short sect. Okay? So it will be VAC from here to here, VA. If you look carefully here at this one, VAC this like this. So this is our VS, so we'll take from here to here, this part. Okay. Now, what we would like to get is our equations. So we'll find that since we have six dies, we will have six pulses. And as you can see, one, two, three, and four and five and six in one cycle from here to here one complete cycle, right? One complete cycle. Containing six pulses. I can use any one to get average, it will be 6/2 Pi. I will take AB, AB, remember, AB is VMX line, Vmax line not phase line since AB is line to line voltage. Sine omigaty plus 30 degrees. 30 degrees is Pi over six, right? Now, I would like to integrate from this point to this point. Remember, this intersection here the angle is exactly the same angle here, omegate here equal to this, and this point is 30 degrees, so this 0.30 degrees. So this point will be Pi over six until the next insert section, which is this point, or you can say it about it in a different way. We have how many pulses, one, two, three, and four and five and six. Between each of these pulses, we have six pulses. In one cycle, right? So each pulse will be 60 degrees right or by over three. It will be by over three. Okay? So this is our angle. So from here, to here, the duration is Pi over three. So from here, this one is Pi over six. So this point will be Pi over six plus Pi over three, which is which will be half or Pi over two, 90 degrees. So we started at 30. After 60 degrees, it will be 90, which is Pi over two. By doing this integration, you'll get three over Pi Vmax line. Now, what about I average? I average in any circuit, it will be V average over R. Take this one and voted by R. So you will get Vmax line over R, which is Imax line. Now, what about RMS exactly the same equation except we will square it under the square root. It will be 0.95 58 Vmax line. For the current, it will be output RMs, will be Vout RMS over R. Now, first, let's get the form factor form factor V RMS, divided by V. You can see 1.00, 09, very, very close to one, which means we are getting more closer to the DC. Ripple factor, root FF squared -10.043. I think that this number is similar to the six pole, if I remember, right? These two numbers. But you'll find something interesting in the next points. I output RMS is Vout RMS over R, this over R, which will be Imax line or multi blood by 0.95 58. Nothing changed it from the previous equations except the integration itself with Vmax line instead of Vmax phase and sine omega E plus 30 degrees. Now, power of DC will be V average DC or V average or multi blood by average. This multiblod by this will give us 0.912 Vmax line, Imax line. Now, what about AC will be Vout RMS, multi blood by put RMS. This multi blood by this will give us this value. What you can see here is that in this circuit for the three phase bridge, you can see that the DC average power is very, very close to AC average power. They are becoming very, very close to each other. Right, very close. That's why you will find that the rectification efficiency for the three phase bridge, almost 99.8%, 99.82%. We are very, very close to reaching 100%. The highest one which we reached in the previous circuits was 98%. So 99.82% is a very great achievement. Now let's look at the rest. Number one, we need the supply RMS. Look carefully at supply average zero. What about RMS? It will be integration of IEB multiplied by four, y four. You can see we have one and two, one, two, three, and four, right? They are different function, IEB, IAC, IBA, ICA. So we can integrate just one and multiply it by how much, multiply it by four, right. So it would be root 4/2 by four waveforms and integrate from here to here. From 30 degrees until Pi over two. Remember, from here to here, 30 degrees B over six to Pi over two. For what for the function? IIB if you don't remember, IB is equal to VAB, divided by R. So VAB, which is VMX line, omegaty plus B over six, divided by resistance, all square since we are getting the RMs, it will be 0.7 804. Now, what about transformer utilization factor? Now, look carefully at the utilization factor compared to the previous circuits. Bar DC 0.912 Vmaxoline Imaxolin. Transformerol, volt and Byrating. Remember, Transformer can be rating of transformer or power of the transformer S, root three, V, line to line, ms, multiplied by, I, line to line RMs, or three, VMX V phase, ms, multiplied by, I phase, rms. Okay, so let's use the second equation because we have line to line MX. Okay? Remember, MX line. So it would be root three, V line to line RMs, I line to line RMS. So root, I supply, RMS line, supply RMS line. So root three, I supply RMS line. We have 0.7 804 Imaxoline. This is our supply line two line RMS, line to line RMS. Okay? So we put er 0.7 804 Imaxoline. Okay? Not Imaxoline. It is not related to this equation. Imaxolinez 0.78 of Imaxoline gives us the RMS. Okay, so we substitute with RMS line to line. V supply RMS line to line will be Vmax line over root two, right? Vmax, line over root two. Now, why did I do this? Because we have here Vmax line Imax line, so I would like to cancel them with each other. So when we take this with this and this with this, and substitute will get 95.4%. So we are utilizing 95% of the volta and rating of the transformer. Now, this is a very large value compared to the three phase, the three phase, half wave. In the three phase, half wave, we had six 6%, if I remember, and 51%. You can see there's a huge advantage of using the three phase bridge compared to the other sect, right? So I hope now you understand the three phase bridge wave uncontrolled dictifier, how to get the equations, how to draw them. I hope it's clear for you and see you in the next lesson. 72. Example 5: Hey, everyone, in this lesson, we will have example five, which will be on the three phase, bridge rectifier. Let's look at this example. We have the same circuit and we have our lute. What we learned in the previous lesson is presence of a pure resistive lute, right? Now I will show you that even if it is highly inductive load, you will be able to solve it without knowing the equations. So in this example, we have a three phase, full wave rectifier. Full wave means bridge rectifier, similar to the single phase with a highly inductive load. So our load here is an RL load with a very high inductance leading to o. Let's write it out. Or It not be out IOut will be constant value, right? Instead of pulsating in the resistive load, it will be a constant straight line. Now, the outut power reaching the loot which will be consumed by the resistance will be 18 kilowatt. And the average of the outut is 400 volt. Okay? So what do we need? Number one, we need the peak reverse voltage of the dites. These dites, I would like to know the peak reverse voltage of their the value of peak reverse voltage and average value of dite current and EC input voltage. Okay, so the first step problem in any problem is drawing the waveforms. What you can see, it is exactly the same waveform as the pure resistive load. Remember here, VAB, VAC, and so on. This is exactly the same in the resistive load and in highly inductive load. The presence of highly inductive mode does not affect the voltage waveform at all because it is continuous. Now what about current current, if you remember it was like this. It was like this. If you remember it was like this, like this. And for supply like this, like this right this part, this one, and this part, this one. However, due to the presence of a highly inductive load, instead of having this pulsating, DC, it will be a straight line like this. Like straight line, like this. Okay. Square waveforms. Great. That is the first thing. So let's read all of this like this. Okay. Now, what's the first thing that we need? Number one, peak reverse voltage. As we remember from the previous lesson, we said that the maximum maximum reverse voltage applied on any dite will be this value, which is root three, V max phase or V max line phase, Vmax line. Okay. Now, what's the problem here? The problem is that we don't have Vmax line. We don't have the input supply. It is required in the problem. So how can I get at simply using V average? We know average in this circuit is equal to like this. V average equal to three overpi VMAX line. So we average of output 400 volt, so we can get Vmax line, right? VMAX phase is equal to 241.84, or you can get Vmax line, and it will be weak reverse voltage. So let's say VMAX phase, okay? Back reverse voltage will be roots re multiplied by this value like this, so it will be 418 thus representing V max line. Okay, now, what about the average value of the diet current? Average value of diet cart. Now, how to get at number one, remember that when we have a highly inductive loot, I out the constant value is equal to I out RMs equal to Iout average, which will be Vout average over R, right? Okay. So what we need in order to get average value of dot current is to get I put average, right? So how can I get even I output average? It will be average, which is 400 divided by resistance. However, in this problem, it didn't give you any resistance, right? It give you another thing, which is the output power. As you remember that output power is equal to V DC, multibloODC or V average, multiblo by average. Why average Because because the current is DC. So power will be DC current, multiplied by DC voltage. So I average, multiplied by V average. We average 400 power consumed by resistance 18 kilowatt, so we can get the IDC or I average like this. It will be 45. Okay? Now, what's the next step? We need average diet current. So I diet. Average will be IO. Constant value I at average will be IRMS will be 45 mpirs. So I at 45 amps. Multiploid by two over a conduction period, divided by Tupi or conduction period of course, in radians divided by Tupi or induction period in degrees divided by zero hundred and 60 degrees. Okay? Both of them are the same. So let's say I will say through 160 degrees. Now let's see how long any diet conduct in one period. So in one period from here until here. Look at D one, D one only conducts here from here to here, right from here to here, how long for how long if you go down here from here to here, starting from 30 degrees until 150 degrees, right? So the induction period of any diet is 120 degrees. So we will say 120. Let's go here, 120 degrees, divided by through 160, multiblod by output. As you can see, 120 is one of our third, multiplied by 45, which will be 15 and pairs. L. Okay? Es input supply is simply given as VMAX phase 241, 0.84 or the line to line voltage for 118.87. 73. Example 6: Now let's have another example on that three phase bridge rectifier. Now, this time we will have three phase bridge uncontrolled rectifier with a resistive load. DC at a DC voltage of 400 volta. This is our VO. Out. So our resistance here, O average will be 400 volt. Our resistance, ten Ms right or equal to ten ms for our loot. What do we need? Number one, ratings of the dite and of the three phase transformer. So let's go step by step. Number one, to get the ratings of the diet, number one, we need to get peak reverse voltage, peak forward voltage, and we need I diet, RMS, I diet, average. Right. Okay, so number one, in order to get it, we have what we need dit RMS, I diet average. In resistance, IbutRMS is different from Ibut average, right? So the first step is that we will draw the waveforms like this, right? This is our waveform. Okay. Now we need first the supply voltage, right? We have 400 volt and resistance of t 400 volt. From it, we can get our supply voltage, right? So if you remember that average in this circuit is equal to three Pi Vmaxoline. The same equation that we discussed before, which gave us three over Pi V Mxoline. Integration from Pi over six from 30 degrees to 90 degrees, right, for any waveform, VAB omega T plus 30 degrees. So from here, we can get V Mxoline which will be 418.87 volt. Now, this is our line voltage. Now, this value is simply equal to the peak reverse voltage for the diet, right? Maximum V maxoline is the worst value applied to worst value of voltage applied to the diet. This is our Peak reverse voltage equal to peak forward voltage. That is the first thing. Number two, we need V supply phase, RMS and you will understand why do we need it later? So we have Vmax line to convert this into phase, divided by divided by t divided by root three, to convert max to RMS divided by root two. So this is our value of supply as an RMS phase. Now, as we said before, peak reverse volts equal big four equal 418.87. Now, we need Idit RMs, I diet average. So in order to get this, we need current. So our current I average I average will be average divided by resistance, right? Number one, I with average, average, divided by resistance 400, divided by t gives us 40 and pairs. Now, how long does our diet conduct? It conducts four from here to here or 120 degrees. So as we said before, I diet average will be 120/160, multiplied by I at average, which is 40. So it will be one of cert, multiplied by 40 and pairs. Like this. Then we need I died RMS. In order to get I dit RMs, we need output RMs, I output RMS. So in order to get I output RMs, it will be V output RMs divided by R. So in order to get Vout rms, we will do the root square of this integration, like this. It will give us 0.95 58v max line, which we all obtained before, this and this. Take this and divide it by resistance, you will get Out RMs, like this, 40 a pairs. Now, as you can see that Iout average is exactly equal to I out RMS because this rectifier Bridge octavar has a high efficiency, 99.8%. That's why average is very close or IO average, very close to Iout RMS. Now, you have to understand that this value here, this one is actually 400 this multiplication, 400.3 volt. So when we divide by ten, this will be 40.03 ampere. So there is a very, very small difference between it and average, very small difference due to the high efficiency of this rectifier. So we have I abut RMS. We need I dete RMS will be equal to root induction period, 120 degrees, divided by 160 degrees, multiplt by the value 40 amps, which is but RMS. Like this. This root, where exactly root, root 120/360 is equal to root one over third, which is equal to root 3/3 or one of our roots. All of them are the same. One of our root ster, okay? Okay, so we have dt rms, and we have IARms and every thing. Now, another thing you can do is that this this let's go here, equal two. We can do it like this root, root two Pi a set of 362/3 Pi in radiance. This will go this and this will go this will be root one over thirdls. So you can use radiance, or you can use degrees in this ratio. It will differ or it will not make any difference. Okay? Okay. Now the next thing that we would like to get is the rating of the three phase transformer. So rating of transformer means the volt and pair rating of the transformer, right? Now, we have here also diet big. This is something that I forgot to mention. The maximum current that the diet will suffer. This is also included in the ratings of the diet. Maximum current is V max line divided by R, right, this maximum point, VA B max divided by line by resistance or Vmax line divided by resistance. So 41.8 is the maximum stress or maximum current that can be applied to the diet. Okay? Now, going to the three phase power of the transformer, the rating of transformer, how much power S can be equal to three. V phase, RMS, multiploid by I phase, RMS, or root three, V line to line, RMS multiplied by I line to line RMS. As of these equations are similar to each other. Okay? So root series, I supplMS line to line, supply line to line. Both of them will lead to the same answer. Okay? So why do we use I supply line to line and line to line? Because we have everything here in line to line. We have currents. Supply current can be obtained in line to line four. So we have V supply, RMS line to line can be obtained. This is Vmax line to convert it into. RMS, just take 418, point it and divide it by root two, to get V line RMS. For I supply RMs, line to line, we can get it using the integration. We have how many of form one, two, three, four, right? So it can be root, four of us two Pi integration from, let's say, 30 degrees, Pi over six until this point, 150. If I want to know 150, how much it will be 150/180 multiplied Pi Pi. This will convert 150 into radiants. For VMAX here VMAX line sine Omega t plus 30 degrees Pi over six, divided by R all squared under the square root. We learned how to get this. Okay? We explained this equation before we said that we take one part and multiply it by four. Okay? That is one way to do it. Another way is to take I out RMs, and multiply it by root, how many? How much does the supply conduct with respect to a whole period from here to here, 120 degrees and another 120 degrees means it conducts four supply current presented for 240 degrees for the 360 period. So this will give you the same solution as this. So what you can see here, root of period of conduction, it doesn't matter if it is positive or negative because we get the square of function. So even if this is a negative, it will become square positive value. If we are talking about average, it will be zero. Positive equal to negative. However, RMS is not zero, zero, okay? Root 240 or so on six divided by RMS but current, which is four degrees gives us 32.66. Or you can get this root, this function, I will give you the same solution. Okay? Now, we will take this and substitute in this equation, root three current, line to line RMS, multiplied by this value divided by root two converted to RMS. So our transformer rating will be 16 kilo volt and pair, okay? 74. Example 7: Hey, everyone. In this lesson, we will have another example on the three phase bridge rectifier. So we have in this example, a three phase bridge uncontrolled rectifier connected to a highly inductive loot, highly inductive with an input supply, phase voltage. Look at this sentence. Input supply phase voltage, 120 volt. So what does this mean? This value? 120 is phase and RMs, not max ms. By default, line to line RMS. If it says anything else, it means it will be as if what he says, okay? He says it is a phase voltage, then it will be phase voltage, 120 phase ms. And our current since it is highly inductive load, 60 and pairs, which means that 60 Iout, equal to I abut average, equal to I abut RMs, equal to 60 and pairs. Everything equal to each other. What do we need? We need ratings of the dit. Number one, what are the ratings of the dit? Peak reverse voltage. Peak forward voltage, peak current. I I diet, RMS, I diet, average. What else? Peak reverse volt nothing else, I think. Okay, so first, we draw the wave form that we have seen in the previous lessons. Then we would like each of these. Number one, peak reverse voltage and peak forward voltage. Peak reverse volte of diet is equal to the peak forward voltage equal to VMAX line, right? So we have here phase RMS, 120 to convert phase, let's go up here to convert phase line to line, multiply by root three, to convert RMS into max, then multiply by root two. That is what I did exactly. Root two. Root 320 gives us 200 and min 3.9, big reverse voltage, big forward voltage. Now, what is the current P current oblite on the diet is equal to this peak value, which is 60 and pairs, P current, sit and pairs. Now, we need I diet OMS and DiDt average Idit Average will be conduction period, 120 degrees divided by total period. This conducts for 120 degrees D one, D one, 120 with respect to the whole cycle. Multiplied by I output, which is 60 and pairs. I diet RMS will be exactly the same as this one, but under the square root, right? Multiplied by 60 again. Like this. I dot average, one set this one set, Multi wide by 60 gives us 20 amperes, and IRMs will be 60 am pairs, multiplied by root, this which is root one cert, like this. And the P current, as we said before, equal to 60 ampers which is I output. Now, that is all for the three phase, Bridge and control. I hope it's clear for you. The equations are clear, and the examples also are clear for you. 75. Three-Phase Twelve-Pulse Rectifier: Hi, and welcome everyone. In the previous lessons, we talked about this phase, bridge rectifier, and three phase half bridge or not half bridge, half wave rectifier. Now in this lesson, which will be a special lesson regarding the rephaseRctifier, we will talk about another one called the 12 pulse rectifier. Now, if you remember correctly, when we were choking a poet, the half wave rectifier, we had three dites, if you remember, and the half wave rectifier, right, one and two, and three. Okay, which was three pulse rectifier, half wave. Then we went to the next level and take two of these and form the six pulse, right, which is a bridge rectifier. Now in this lesson, we would like to go to the higher level, the next level, which is taking the six pulse and converting it into 12 pulse rectifier. How I'm going to do this by taking two of the bridge rectifier and combine them together. So what you can see here in this figure, we have the upper part, bridge rectifier connected in series with another bridge rectifier, two bridge rectifiers, D one, D two, three, four, five, six, d one dash, d two dash, d three dash, and so on. So what you can see here is that abut is equal to from QVL equal to the but of the first rectifier plus the abut of the second rectifier. However, you'll find that here, there is something different between these two. What is the difference? The difference is that we used here a star connection, and we used here Delta connection. Now, why did we do this? We did this in order to achieve or provide a phase shift between this part and this part. Now, let me think about it. Let's say the first one was star. And second one was also a star connection, right? So the output of the first will be like this, right? And the second will be exactly the same like this, right? So what the but will be here we have six pulse, six pulse. So the but will be also six pulse, right, but much bigger, like this. You can see this plus this higher value, higher value. Number of pulses are the same. In order to achieve or produce that lv pulse, we need to shift this one a little bit to the right or a little bit to the left. It will be a little bit to the right in order to produce at lv bolls, like this. Okay, you will see now how did we achieve this using two different connections, star and Delta connection. So let's look at the vector group, or paso diagram that will help us in producing the final waveforms. So let's talk about the first part, Delta, primary connection. The primary is a Delta connection, right? We have AP, remember the sequence. Remember the sequence, AB, then PC, then CA. You'll find that AB, BC and CA the phase shift between them, these line to line voltages and phase voltages are 120 degrees. Remember, phase voltage in Delta connection. V phase is equal to V line to line, exactly the same value and magnitude, right? Okay, so we have this vector representing our reference or the primary. Now look carefully what will happen if we take this and put it to the secondary. Our secondary is also Delta. So since this one is a Delta and Delta, so they will be exactly the same, we form with the same phase shift, right? Nothing change. You can see here. We have here A two, B two, b2c2, C two, A two, the same sequence here, you can see here, AB two, similar as AB except that the magnitude or the length of the vector will be different. Depending on the turns ratio between the primary and secondary, and we have PC two, which is similar as PC. We have Ca two, which is similar as CA. You can see they are exactly the same vector group, same phase of diagram, except different in length and simples, same shift, same phase or diagram. Now, of course, VCA two, if you would like to draw AC, it will be opposite to it, ac two negative VCA two. If you would like to draw VCP, it will be negative this vector, like this. If you would like to draw BA, it will be opposite to this one, like this. Remember that we use this positive and negative in order to draw our final waveform, right? As we did in the bridge rectifier, we had line to line voltages, right line to line voltage, AP, BC, CA, and B ACP, AC, and so on. So this is for this part, right? Now, exactly the same, exactly the same, for this one. Now remember that turn is ratio they will have the same phase angle and same phase voltage, except multiplied by number of turns. So here you can see VAB, which is this one will be exactly similar to VN or V like this parallel to this vector, except lens is different, due to what due to the urns rich. And we have P, similar as VPC, similar as this one. This one. So it'll be like this, VB, and then we have VC, similar to this one, VC. So you can see the three vectors, similar as VAB, VVC and VCA. So the line to line voltage is exactly similar to phase voltage and line to line voltage for Delta, similar as phase voltage here. So what is the difference? The difference will appear in line to line voltages. Line to line volts are shifted by 30 degrees. VAB one shifted by 30 degrees from VA now, using VAB, we have VAB, then VBC, then VCA. So AB, VBC, this one, then VCA, you'll find that the phase shift between them is 120 degrees between this 120 degrees, 120 degrees, and 120 degrees. Okay? And these vectors have their opposite values VAP VBA one, VCA, VAC, VBC, and VCB. Okay. Now, what you will see also, if you look carefully, and this is important because if you remember, out of that bridge rectifier is line to line voltage, and out of this one is line to line. So what I'm concerned with is this one line to line, this one, this one, and this one. Here too, this one, this one, and this one. So what you can see that this line to line voltage is shifted from this line to line voltage. If you look carefully and take this one here, it will be exactly here, right, V AB two. So what you will find that is VB one leading by 30 degrees VB two. And similarly, if you look at AC, you can put it here. Vac two, you will find that the phase shift between them 30 degrees, Vac one, leading Vc two by 30 degrees and similarly. This one, if we have this waveform like this, then the other waveform will be shifted by 30 degrees from it, leading in the end of achievement of the 12 balls rectifier. So let's see this in more details. So we have everything we need to know. We have what you can see here. Look at this one. We have VB one, VA PC, then VCA and the other vector. This is exactly what we learned in the previous lesson. Now look carefully here. We have VAB one starting from plus 30 degrees. So VAB one is VMX line sine, omegaty plus 30 degrees. Now, we said before that VAB one is leading VAB two by 30 degrees. So this is VAB one. So VAB two will be VMAX sine. Omega T only because this one leading by 30 degrees from VAB two. If you look at the second figure, you'll find that VAB two starts from zero. This one starts from negative 30. So VB one leading VB two by 30 degrees. Okay? Now, of course, we learned that each of these pulses conduct 460 degrees right in each rectifier this and this 60 degrees. If you don't remember, we have so 160 degrees divided by how many diets, diet six will give us 60 degrees. So each one conduct 460 degrees one, 60 degrees, two, three, four, five, and six, and then we repeat the same process. Okay. So we have VAB one and VAB two, and you can see VAB one here and VAB two shifted by 30 degrees from it. So this waveform is exactly similar to this one, this for the upper part, and this for the lower part, except that the difference is shift by 30 degrees as if we take this one and shifted it to the right by 30 degrees. You can see VB one, VB two, AC one, AC two, VVC VVC two, and so on. Shift 30 degrees between these two waveforms. Okay. Now what we would like to get, you'll see Out is equal to output voltage of one plus output voltage of two. This is out of the first rectifier. And this is out of the second bridge rectifier. Out will be submission of these two. However, due to the presence of phase shift, it will lead to the presence of 12 pulses. You can see if you combine the pluses, at any instant, you will have this final wavefall. Okay? Now remember, if there is no phase shift between these two, but will be also six pulse. However, due to this small phase shift, 30 degrees, we have now 12 pulse. Okay, great. Now, the next thing that we would like to get is the maximum value. I would like to know this maximum value or this way four. Okay, so each pulse will conduct for 30 degrees. Remember, each one of these 60 degrees, right? When we combine them together, we will have 12 pulse in one complete cycle. So each one will conduct zero hundred 60/12, gives us 30 degrees for each of these pulses. Okay, now, what about maximum value? I would like the maximum value. So let's look at any of this waveform. Let's say this this is a maximum value, right? So go like this up up up up up until here, go up up up up like until here. So what I'm going to do to get this maximum value, this maximum value, I will get value of VAB two at this instant and the value of VAB one at the same instant. Great, great. Now, I would like to know this angle. What is this angle exactly? Remember, intersection point from here to here 30 degrees, right? So this point is 30 degrees. Okay. Now, I would like to get here. So all of this is 60 degrees. So here exactly this point will be another 30 degrees. This 30 degrees and 30 degrees. So this peak value is at Omega T equal to 60 degrees, 30 plus 30. Great, great. So this point is what is 60 degrees. Now, I would like to get this point. This point is a midpoint of these two, of this part. This part is 30 degrees. So this small part alone, this very small part here is 15 degrees. Again, we have from here to here 60 degrees. So half of it 30 and 30, right. So this 0.30 degrees from here to here 30 degrees. Half of the value is 15 degrees. So we have this angle will be omgaty at which we would like to get peak value will be 30 plus another 30. So it will be 60 degrees. Plus 15, which is this small part, 15. So it will be 75 degrees. Okay? Now, what I'm going to do to get this peak value, I will substitute to 75 in this equation plus this equation. So you will get it like this. You can see V max line, sine 75 omegat 75 plus 30 degrees plus V Mxoline, sine 75, which is sine this one. It will give us 1.9 32v max line. So this is the peak value of the output voltage. Okay, I hope it's clear for you. I just divide it into regions in order to help you understand where did we get 75. Okay? Now, we have this peak value, 75 degrees, and you will find that this one, this one starts at peak value. Okay, simply like this. This peak value occurs at what occurs at 75 degrees. And this is a sine wave. It's a peak at 75. So this sine wave can be written like this. Out will be maximum value VB, which is 1.9 32v maxoline multiplied by sine. Omigat plus 15 degrees. I know someone will say, why 15 degrees. Okay, look, this one, it's a peak. Remember, any sine wave, sine wave. The peak occurs at 90 degrees, right? So here we have peak 75 degrees. So I will need to add 15 degrees to reach 90 degrees. So I will say Omigaty plus 15. Or you can simply extend this sine wave like this and like this, you'll find that this part is also 15 degrees. Both of them are the same way or the same mesod. Okay, now we would like to get our beautiful values. Average. It will be how many pulse, we have 12 pulses. So 12 or two Pi. Then integration from this point to this point. We have this function. If you remember it as I just said, VB sign Omegaty plus 15, right? So it will be like this 12/2 Pi. Let's say I'm talking about this one here. This one here, 12 or two by integration from this point to this point. This point, if you go up here, if you go up here, it will be peak of VP one. VP one, its peak occurs at 60 degrees, so it will be by over three. Until this point, each pulse remains for 30 degrees. So 30 degrees plus 60 degrees will give us 90 degrees until Pi over two. For VB, sine Omega t plus 15 degrees, right? That is one way. Okay, so that is one of the ways you can do it. Another way, like I did here. So the previous one is okay, I will give you this same output, okay? Another way is to take like this one, this way form is lagging from here. This one starts here. This 14 will start after how much after 30 degrees, right? So it will start after setting. So it is lagging p 30 degrees. So if the first function is V B sine Omega plus 15, then since it is lagging, lagging p 30 degrees from this function, it will be VB, sine, omigatE 15 -30 gives us negative 50. And we will integrate it from here. To here. This point, if you go up here is this point which is 90 degrees and at 30 degrees to 120 degrees. So you can see where it degraded from. P over two, 90 degrees, 220 degrees for snomegt -15/180 point in order to convert it into radiance. Okay, to convert this into radius. It will be the same equation. Okay? This will give you this value, or if you integrate it like I have just showing you, it will give you the same answer. If you integrate number five, lagging by another 30 degrees from here, it will be negative 45 degrees. All of them will give you the same solution. VOR Ms, the same equation but squared give you this value. Now, if we look at form factor, OR mass divided by V average, it will be 1.00 005, very, very close, very, very close to one or unit. Ripples root square of this value minus one gives us 0.0 102. So what you can see, ripples are very, very small. If you compare this with this 12 pulse, you can see it is almost DC, almost DC, almost a pure DC voltage. If we compare the pulse rectifier, the half wave, the six pulse midpoint, the six pulse bread rectifier, the 12 pulse we have here, you can see form factor, the closer to one, the more DC we have. You can see 1.01 becomes smaller, similar to each other, then very, very small here. Now look at ripples. Ripples 18%, 4.3%, 4.27, very small difference, and then we have 1%. So what does this mean? It means that the 12 pulse rectifier is the best between these rectifiers, because it produced almost DC output voltage, meaning that we will need a small filter in order to have a constant or pure DC output. Okay? 76. Three-Phase Bridge-Wave Fully-Controlled Rectifier - HIL: Hi, and welcome everyone. In this lesson, we took the three phase bridge with fully controlled rectifier with a highly inductive load. We discussed the three phase bridge uncontrolled, which consisting of diodes. In this lesson, we will replace the diodes with cistors and we will have the fully controlled rectifier. But we will start this lesson with a highly inductive lute and in the next lesson we took a about resistive lute. So let's go step by step. We have our bridge rectifier and controlled D one, d3d5, d 4d6d2. In order to have the fully controlled, just replace all of these dites with istors with the same numbering. So this is our waveform for the three phase bridge wave and control direct fire. Now, we replaced D one with D one, D three with D three, and so on. Now, what did we change? Nothing changed at all. It is the same circuit, except that we remove dials and put side restors. Now, before we understand how does this circuit work, you have to understand that the diet is exactly threstor. D one is exactly as T one here, except that with a firing angle Alpha equal to zero. When firing angle of T one equal to zero and phase shift between these soistors equal to zero. Phase shift of Alpha equal to zero in these stors or firing angle of the restor zero. It means that this circuit will become uncontrolled rectifier. Let's look at the circuit. You can see we have V A V, B, V, C V, A, and so on. We said before that we measure the firing angle from where we measure the firing angle from the intersection point between A and C, right, this point, which is 30 degrees, this intersection 0.30 degrees. Now, first, if Alpha equal to zero, it means that here this point is the first firing angle at 30 degrees for the first restore T one. So here we will have T one. Okay? Now, the phase shift between T one and T 320 degrees between T three and T 520 degrees. So you can see T one after 120 degrees T three, after another 120 degrees, T five. What about T one and T two? Remember, we have six switches. So it means that we have 360 degrees divided by six, so it will mean that 60 degrees phase shift. So between each of these switches, T one and T two, there is a phase shift of 60 degrees in the firing angle. So what you can see that from here, let's use the pencil between here and T one and T two, you can see here between them. 60 degrees between T two and T three, 60 degrees between T three and T four, 60 degrees. Okay? So we have we can say, get T one likes and after 120 degrees, T three, and after another hundred 20 degrees, T five. Then T one and T 260 degrees between them, T two. Then we measure from here 120 degrees to get T four, 120 degrees to get T six. So what we have learned between each consecutive each consecutive s restorsPhase shift between them is 60 degrees. T one and T 260 degrees, T two, T three, 60 degrees, T three and T four, 60 degrees, T 45, and so on. Between any of the two si restors in the same group same positive group or same negative group. Phase shift between them will be 120 degrees. Okay? Now, each this each thistor will conduct for how many, 120 degrees. So you can see that we have here, T one started here, so it will conduct for 120 degrees. So you can see T one and from here to here, 120 degrees. So from here, T three from here and here, 120 degrees, T five. Similarly, T two after 60 degrees, two, 120, 120, 120, and so on. Now, this is important. When you draw this, you will be able to get what to get out waveform. Okay? So at Alpha equal to zero, the out will be exactly similar to this. We will start from here to here. So if you look at this group and this group, you will see D one and T one are exactly the same. So out waveform will be like this. Let's delete all of this. Let's say you will start increasing firing angle Alpha. Alpha, instead of starting from this point, it will be after a little bit. Let's say here, Alpha. Here, T one will start. So if you go down here, you can see T one will start for how much, 420 degrees. Then what T 320 degrees, then T 520 degrees. What about T two from here 60 degrees, T two will start for 120 degrees, T 420, and so on. Now, using this shape, you will be able to draw this out with four. You will see that T one starts from here. Let's look at this. You can see at any instant, let's say from here. Here, this part, T five and T four. This two will conduct. Let's go here, T five and T four. T five connected to C, so we will have V, C, and T four connected to A, it will be a VCA. If you look at here, VCA. Now, from here, to here T five and T six, T five, which means C again, and six means P, VCP, so it will be VCP and so on. So you will have this waveform. Okay? So where do you draw it, you will see it will be like this, let's say from here. So it will be like this goes from here, VAB, like this, and then VCP like this and like this. So it'll be something like this. So it will give us this one. Okay? So first, we draw the waveforms AB, AC, BC in the same figure like here, like here. And depending on each period, which one conducts, we will make our output like this. Now, we have different cases for highly inductive load. Let's say first Alpha less than 60 degrees and let's draw it. Alpha less than 60. So if we remember, each thi restor conducts 420 degrees. Each thy restor, t1t2 to three to five. Fifth, switching between each consecutive ones, T one and T two, T two, T three, T three to four. Each phase shift will be 60 degrees, right? However, switching between two positive thi restors or two negative thi restors will be 120 degrees. Switching patterns will be similar to dites. Remember D one, D two, D two, d, d3d4, the same idea, t1t2, T two, three, and so on. And in this circuit, exactly the same as diodes. The peak reverse voltage equal peak forward voltage, equal B Vmax lie. So let's draw the out wave form. Alpha less than 60 degrees. So the first step number one, we will start measuring Alpha from here from this intersection point. You can draw this or something which is much easier for me and you. What es this intersection point here between VA and VC, which is at 30 degrees is exactly exactly the same point here between VAP and VCB. Okay, VCP and this point, This point here is exactly this one. So you can look at this figure much easier. This point is 30 degrees. Okay. From here, we measure our firing angle alpha. Let's say our Alpha is less than or equal to 60 degrees. So we measure from here 60 degrees. Less than 60 degrees I value 40 degrees, whatever it is. Okay? So we will have this point here. This is a start of what start off T one. So T one will start from here four, 120 degrees, four, 120 degrees like this. From here to here. So from here to here, we will have what we will have T one. Let's look at it from here to here 60 degrees, another 60 from here to here. Yeah. So if you look down here, I one, representing this T one, I two, is T two, I three, I four, I five, I six. So the first i restor starts from here from Alpha equal plus 30 degrees this 0.30 degrees plus firing angle Alpha. So it starts from Alpha plus 30 degrees until this will last for 120 degrees. So 120 plus Alpha plus 30 will give us this point, which is what Alpha plus 150. You can see if you go down here, Alpha plus 150 degrees. We can do something else. We can magnify it like this to make it more clear. You can see from here, Alpha plus 30 until 120 degrees this induction period of T one. Now, T two, T two will conduct after 60 degrees. Start from here another 60 degrees. 60 degrees plus Alpha plus 30 gives us alpha plus nine, start of it for 120 degrees. T three after 60 degrees, 60 plus 90, 150, 120 degrees, I four after 60 degrees, after 6660, and so on. Even you'll find that there are some parts at the beginning. You can see that we surpass one cycle, surpass 160 degrees. So a part of it here from it is found at the beginning. Okay. So if you go here like this, you will see that VCP, okay, VCP from here to here, VCP C and P, C and P. T five and T six. So you find here, T five and T six in this part from here to here, T five and T six. Okay? Okay, you can also look at it in a different way. Here you look at the highest voltage, A, which is related to T one, VB, T three, VC, T five, and the most negative VV related to T six, VC related to T two, T four, related to TA and so on. But I like this method more. If you draw this, you will be able to know the induction period for any of the cistors and you will be able to draw the out wave form. So let's understand how does the outwave form will be made let's delete this like this. Okay. Then magnify once more like this. So we have here. Let's look at here starting from here. Okay, from here, T one will conduct, and which one is already conducting T six, right? T six is already conducting T one. So if you go here, T one and T six. So T one related to A, T six related to B, so it will be VAB. So if you go like this from here to here, right from here to here, we look at the two c restores, T six and T one. If you go here, from here, VAB. So we go from here, VB until this point, VAB from here, you can see from here to here, T one, and let's go down here and T two, T one and T two. Only this conduct this during this period. T one and T two. If you go here, T A T two, C, it will be VAC. If you go like this, starting from here, VAC goes from here to AC. Mm hmm. Like this. Okay. Then from here, you'll find that VBC and so on. So you look at what si restors are conducting. You can see here in this part, T one and T six. So if you go here, T one and T six. So we look at the positive ci restor and negative ci reistor and we will get V out. Similarly here, we look at postive and negative between the two and this two and this two and so on. Now, what you can see here that VAB here and VAB here. There is a small part before it. The spot from here to here is exactly this region. If you go down here, you will find that this region from here to here, this part is exactly this part. The spot is exactly this part. So after drawing this, you'll be able to get the first part here. Okay, so this is our V O. Very easy and clear, right? IA, for example, IA, IB, IC, all of them are similar to each other except shifted by 120 degrees, right? As you know, IA, IB, IC, or the phase currents are shifted by 120 degrees. So you can see IA, then IB shifted by 120 degrees from it. Now, how to draw IA if you go here, IA is equal to I one minus I four. Like this one, minus I four. If you subtract one minus I four, this way four, you will get positive and negative, right? Exactly similar to the bridge rectifier. The three phase bridge uncontrolled and single phase. In the single phase, we did the same steps. For IB, IB will be three minus I six from Kcal. You go to minus six, subtract this from this, you will have this waif. Of course, I average equal to zero, however, IRS exist. Okay, um Okay. Now, what we need now is the waveforms. Now we need the average out RMS, RMS. I register RMS, I supply, and everything. Now, let's look at this waveform. So we have, let's say, I would like to get out. So it will be how many pulses divided by two Pi. So how many pulses we have one, two, let's say, from VAV one, two, three, four, five, six, then the cycle repeats itself. So we have six pulses over two Pi. Now, if we decide to take VAB, it will be VAB V max line. SnomgU plus 30 degrees, Pi over six. Then integration from here, this point is Alpha plus 30 degrees, Alpha plus 30. From here to here, 30 degrees plus firing angle Alpha plus 30. Okay. And each thistor conducts for how long or not each thistor between each of these pulses, 60 degrees, right? Between here and here, 60 degrees half of the conduction period of the thisto. So it will be plus 60 degrees. So we have Alpha plus 30 plus 60 degrees gives us Alpha plus 90 degrees. From here to here. Alpha plus 30, Alpha plus 90. Like this, for the function gives us three VM line over Bicusine alpha. For boot RMs, the same function except the square. Iout RM, this one is a highly inductive loot, IRMs, equal to I abut, equal to I at average, which will be V average over R. Now, each thiresor conducts for how long 120 degrees, 120 degrees we said before 120/360. I conducts for third of the cycle. So it'll be one of our third for average, one of our third of I out. For RMS, it will be this value but under the square root, one of our root three I out. What about eye supply? If we look at eye supply, we have one, two, I conducts 420 plus 120 means 240/360, which is 2/3. So for RMS, it will be root conduction period divided by total period, root 2/3, I out. Okay, this figure here is exactly the same figure here for errans, but it's more clear. You can see T one conducts for 120 degrees from 30 plus Alpha until 30 plus 90 30 plus 120 150, right, 150. 30 plus Alpha plus 120 gives us Alpha plus 150. You can see Alpha plus 150. Between each thysor 60 degrees, 30 plus Alpha, 30 plus 90 90 plus Alpha, sorry, 90 plus Alpha, another 60, 150 plus Alpha, 210 plus Alpha, 270 plus Alpha, 330 plus Alpha, and so on. Okay? Okay. What about Alpha greater than 60 degrees? What will it change? Let's look at the figures and see how this will affect us. You will see that Alpha greater than 60 degrees. You'll find that in the previous figure here, we had only positive, right? So when Alpha becomes greater than 60, let's say 90 degrees, you can see we have a negative part, right? Same conduction. You can see VAB conducts from here, 424 60 degrees. Then VC, then VVC, and so on, exactly the same as the previous one. And you can see the restores that conduct during each of these periods. However, the difference is that we will have a negative part, and despite having negative part, the thyristor will not be turned off due to the presence of what highly inductive loot highly inductive loot. That's why this out this i restor does not turn off. So you can see that current and voltage equations for VRMS VORMS, supply VRMS RMS, I output RMS, I output average, I supply RMS, everything the same as before. Okay? Nothing changed at all. Now, what if we decide to add a free wheeling dt? What do you think will happen? So we have our circuit here with this part, you can see, this is the outta this is the original one, AB, A, B, C, this is out, o? In the output here, you can see AB, AC, B, C, B A, and so on. You can see postive part and negative part. Now, if we add a free wheeling dt, it will conduct during negative part. So what will happen when it conduct, it will remove this negative part. So you can see the output will be like this VAB, VAC, V VC, and so on. You can see VAB, VAB like this, only this part. You can see only this part. And this part will be removed by the effect of the free wheeling dt. If we add the free wheeling dt, if you don't know how, like this parallel to the loot like this in the opposite direction. So it will conduct when we have a negative voltage. So we will remove this negative part. All of this is deleted and we have only this like this, like this. So you can see like this. Okay? So the free wheeling dt current will present when during this zero period, current, current, current, and so on. For the thy restore it will be not one square wave. It will be several waves. You can see 81 conducts during AB and AC, you can see here, conducts during this part only and during this part. Let's draw it during this part, and zero will have free wheeling did. We will have another one for VAC and VAB, like this. Okay? So this part and this part. If you go down here to T one, the sport, and this part. Great. Okay, so we draw free wheeling dite and Trest. Now let's look at a comparison between firing angles, highly inductive load with pF, and let's look at the equations for the Brevia circuit for the free wheeling dite. So you can see effect of E, it will not change anything in the circuit. It will not change anything except that the outt current It average will be VO average minus E over R. Okay, similar to the single phase circuit that was containing the back AMF. However, effect of B AMF is useless here due to the process of highly inductive load, that force the circuit to go in the negative part. You can see Alpha equal to zero. You can see this part. Then Alpha equal to 60 degrees, we will have this greater than zero. Then when Alpha becomes 90 degrees, we will have positive negative, positive negative as before. Then increasing beyond 90 degrees, what will happen exactly 150? You can see negative part. All of this was shifted and became negative, which means inversion mode. So this circuit can supply electrical power back to the three phase or inverted from DC to AC if Alpha greater than 90 degrees, the inversion mode that we talked about before. Okay? Now, when Alpha greater than 60 degrees, we will add a free wheeling die to remove the negative part, right? We said that. Now, I would like to know the equations. So number one, our V average will be like this. How many we form for VAB, one, then, two, three, four, five, six. Then cyclo bits, so we have 6/2 Pi. VAB, V max lines anomty plus 30 degrees. AB will conduct from here. What is this point Alpha plus 30, Alpha plus Pi over six, which is 30 degrees. Okay, now, until this point, this point, what is this point at which VAB becomes zero? So what is this point, if you remember, our function is V max line sine Omegaty plus 30 degrees, right? So when does the sine becomes zero? Sine becomes zero at sine 180 degrees, right? So in order to achieve zero voltage for AB, Omigay will be 150 because 150 plus 30 gives us 180 leading to this point at which AB is zero and at which the free weighting diet will start conducting. So it will be from here. Integration will be from Alpha plus 30 until Alpha plus until Alpha plus until 150 degrees. No Alpha, 150 degrees. You can see Alpha plus 30 until 150 degrees, 150 pi over 180 to convert this angle into radiance. So it will give us this output value. Okay, now, what about Vout RMS? The same function squared, right? What about output RMS? Since this one is a highly inductive load, I out RMs, equal to I output average, equal V average divided by R. Now, what about Icystor average? Look at the cystor. Cirestor will conduct it from here to here and from here to here. So Isrestor average will be we have two Pi Like this. And it will conduct how many times one and two, white, one and two. So it will be two times. This period is exactly a B from here to here, which is similar as our integration from 150 from Pi over six plus Alpha until 150. So we can say 150 Pi over 180 minus Alpha plus Pi over six. So this difference representing what representing this period from here to here. Okay, so you can see 150/180 is five or six Pi minus Alpha plus pi or 6/2 pi two. So it will give us this value. Okay, let's delete this. What about a free wheeling diet? So if you look at the free wheeling diet, it always conduct it conducts how many times it conducts six times in one cycle. It conduct, let's say, from AB here. This is the beginning. So we have one, two, three, four, five, and six. Then we will repeat again. So we have six of these of where is the free wheeling guys? Free wheeling diet average, six of the free wheeling diet average. And each free wheeling diet conducts from where to where. If you look here, from here 150 degrees until the next firing angle. Remember, from here to here 60 degrees, right? So 60 degrees. So this point is Alpha plus 90, right? Alpha plus 90. This point. Alpha plus 90, this point, 150. So this period is Alpha plus 90 -150. Alpha plus 90 -150. At blood by six times our divided by two, Mult blood by I out. It will give us this function for si rest or RMS, it will be the same equation except root of this value. Now, what about eye supply? Remember, I supply is simply two times root RMS, two times one of the sisters. Okay? One of the isors induction period. So it will be root two times this value. You can see here if we get back here, here, let's get back here. Again, again, here, okay, you can see 120, 120, which is one of the restored current. So it will be double the restored current. So the waveform will be two times root two times this, and the average is zero, similar as before. For I free wheeling dit, RMS, it will be root this value. So in this lesson, we explained the three phase, fully controlled the bridge rectifier with a highly inductive load, and in the presence of a free wheeling dit, I hope it's clear for you. 77. Three-Phase Bridge-Wave Fully-Controlled Rectifier – R Load: Hey, everyone. In this lesson, we took apart the three phase bridge wave fully controlled rectifier with an R lute. We took apart the highly inductive lute in the previous lesson and this one we took apart the R lute. So when Alpha or the firing angle is then or equal to six St degrees, which you have seen, it will be as you will see, it will be exactly similar to highly inductive lute. Nothing changed at all. So if you look at here, this is our circuit with a resistive lute. You'll see that VAB, VAC, and so on the line to line voltages, and it will start from here. You can see this 0.30 degrees. Let's use the pencil, as I usually forgot here, this 0.30 degrees and after firing angle Alpha, which is this point will be Alpha plus 30 degrees. You can see from Alpha plus 30 degrees, it will conduct normally until the next firing angle, then it will conduct, next firing angle, and so on. You can see here from here, which is Alpha plus 30, for how much for 60 degrees. I'll be Alpha plus 90 degrees. So you can see we have t1t2, T two, T three, T four, T three, four, T 45, t5t6, then t6t1, then t1t2, and so on. You can see here we have t16, which is exactly at this one. So brevis one will be t56, which is this one. Okay, great. You can see nothing changed at all. It is exactly similar to Alpha less than 60 degrees in highly inductive lot like this continuous. Okay, what about the equations? Equations are very easy from Alpha plus 30, Alpha plus Pi over six until Alpha plus 90, Alpha plus Pi over two. F V max lines omega ty plus 30 degrees six times one, two, three, four, five, six, exactly the same equation. ORMs exactly is a function, but square four I out average will be average divided by R, and IO rms will be V RMS over R. And I I restore here will be third of the I out average. Why? If you look carefully here, we have this one T one conducts from here to here, T one continuous from here to here. So this biode is 120 degrees. So 120 with respect to 360 degrees will be cert, one over third of I out average. Remember, we have a pure resistive lute, not highly inductive lute. For ISAs or RMS, it will be one of a root of IO RMS. Since they are not equal to each other, average are not equal to RMS due to resistive root. Force applied will be root two verse. Why? Because it conducts for how much 4240 degrees exactly the same with form that we discussed, except instead of straight lines like this. Instead of like this, this square waves like this. It will be what it will be like this one. It will be I Abbott will be like this. Okay, which is this one. So it will be VO, divided by R. Same with form since we have pure resistive lot. Now, here the change will happen. When Alpha greater than 60 degrees, we will have negative part. If you remember, VAV conducts from here, Alpha plus 30, starts conduction, so it will start conducting and before reaching the next firing angle, it will start going to the negative part. So in the highly inductive lute, it was like this, going to the negative part, then go like this, then go down like this, and then like this and like this. So we had a negative part. This negative part in the highly inductive lute was removed by using the free wheeling light. Here when we have a resistive lute, the crestor will stop conducting when we go to the negative portion. When we go to the negative portion of the cycle, the cistor will stop conducting, right? So it will conduct only from here to here exactly similar to what exactly similar to the case of highly inductive lute with free wheeling dite. Why? Because the resistive fluid and resistive fluid, voltage across systor will be negative, sorry, systors through the thistor will be negative. So the thistor will be reverse biased, the highly inductive fluid due to the presence of high inductance, it allowed the flow of current through thethystor and making it in conduction mode. So it will conduct from Alpha plus 30 until 150 exactly similar to highly inductive loduse free wheeling dit. So the equations will be exactly similar to highly inductive louds free wheeling dt from Alpha plus 3,250 degrees. For the same function AB, and for AutRMs it will be the same function square, and IO average will be average over R and IORMs will be VRMs over R, since we have a pure resistive lute now for I average I thy restor average, it will be what it will be this equation. Now, where did you get this? It is from here for IT one. You can see IO is exactly similar to what similar to a voltage, voltage as you see here. Same with form except divided by R. It one conduct this here and conduct this here. Let's delete this first. Here and here, right? So it will be like this here and here. So it will be two waveform, so it will be multiplied by any of these functions from Alpha plus 30 to 150. So it will be 150 minus alpha plus 30 to give us this period. 150 minus Alpha plus 30 gives us this conduction period. Multiplied by two is respected to two Pi. So it will give us this out, right? I out average multiplied by this division. Okay, great. For I thyrostor average. Okay, these values 110, don't worry about it. This value is from another example that we will see, okay? It should be here general formula Alpha plus 30. These values are from one of the examples that I will discuss. Now you can see IT one. Here I A -84 84 conduct is here and here. I will be 84, 84 and negative. Instead of what? If you remember, in the highly inductive load, we had them like this. We have like this and like this and like this and like this. But here since we have a pure resistive, it will be like this, similar to supply voltage. I theist RMS will be IORMS multiplied by root, this value. I supply Rs, this is Iristor I thi restor root of this I supply will be two times two times this value. Two times this one, two multiplied by this. You can see two multiplied by this. Why? Because you can see one, two. However, I supply one, two, three, four, four wave forms. So they are double thus restor current. So it will be two, multi bled by this conduction aro. This representing all of these conduction is, period, okay? 78. Example 8: Now let's have an example on the three phase fully controlled bridge. I have a phase bridge fully controlled directivi connected to an R up 25 Ms. The source is 380 volt. And since we say 380 volt, as we learn it a lot and we said before, line to line, RMS, sect hertz. The phase control angle is 80 degrees, which means here, Alpha, is equal to what is greater than greater than 60 degrees, or equal to 80 degrees in this example. So since Alpha is greater than 60 degrees and we have a pure resistive lute, then we will treat it as if it is highly inductive lute with a free wheeling diet. They are exactly the same. So find let's delete this find average load voltage, average ode current, average allude bower, average thyrostor current, and B criers voltage. This is very easy and straightforward. Number one, draw the wave forms as we learn it for a resistive loud with Alpha greater than 60 degrees, we will not have any negative part, and we will have mode off mode like this. Okay, so the average allude voltage will be equal to what we learned from before. It will be integration from Alpha for AV from Alpha plus 3,250 degrees. Like, Alpha plus 3,250 degrees for the function, right? V maxiline. We have 380 volt, which is V line to line or mas. So we will take this value omtolite by root two, like, root two metabolite by 380 degrees. And we converted Alpha from degrees to radiant by multiplying by Pi over 180. In order to convert from radiance to two degrees. Let's say this if you don't know, to convert from radiance two degrees. What are you going to do? You take radiance and multiply Pi 180 over Pi. If you would like to convert from degrees to radiance, take degrees, multiply by Poi over 180. That's what we exactly did. We took the Pi over 180 degrees and multiplied it by 80 degrees. So it will give us 120 volt as an average voltage. Great. Average lute current will be 120 divided by resistance, 25, right, 4.8 and pairs. Average louder, since we say average loud power, it will be V average out with average, multiplate by I out with average or I with average square multiplied by R, right like this, square multiplied by R. For average thyrestor current, we said that we have a formula for Tirestor I will conduct two times from here to here from Alpha plus 30 to 150. So it will be this equation that we took about before, so it will be Alpha 80/180, multi blot by Pi and radians, so it will give us 1.07 and pairs. For peak reverse voltage, we said that in a bridge rectifier controlled controlled or whatever it is, half controlled that we will discuss. All of this is VMAX line, right? So it will be VMAX lin root two, multilat by 180, which is 537.4 volt. 79. Example 9: Now let's have another example on the three phase bridge controlled rectifier. We have this bridge controlled 41 to 385 and so on exactly the same circuit. But this time our load is a resistance of ten oms. You can see inductance L equal to one hinary and there is a packing math 200 volt. Now look carefully here. Since our inductance equal to one hinary, one hinary is a very, very large value, which means that the current will be a constant value or this load is a highly inductive load. Okay? So the first step, the input supply is 400 volt, 550 hertz for 100 volt, line to line RMS, as we always know, find the firing angle if the average output current is ten and pairs. Find the input power factor in this case and assume that the inductance is large, as you can see one henary large enough to ensure constant current in the load, which means highly inductive load. Number one, we have a fully controlled bridge, and we have a highly inductive load. So do we have any free wheeling light? No. So if Alpha, if Alpha greater than 60 degrees or Alpha less than 60 degrees, it doesn't matter. Greater than the same equation. As you can see here, you can see continuous, right? This is if Alpha less than 60 degrees, if it's greater than six degrees, then it will go to the negative part as we learned before. Now let's assume both of them are the same equations, right? They will conduct from where V AVC will conduct from Alpha plus 30 until the next firing angle, which is at Alpha plus 90 degrees. So it'll be like V average will be 6/2 Pi, integration from Pi over six plus Alpha, which is 30 plus Alpha until Pi over two plus Alpha, which is 90 plus Alpha. Difference between them is 60 degrees. For what for V max line, sine omegty plus five or six, which is this equation for this case and this one. V average. Okay? So we will subsiute VMAX line will be 400 volt, multiplied by root two. Now what about average? So we have VMAX line 400 volt. It will be 400 volt, multiplied by root two, right for VMAX line. Now, what about so we need Alpha, right? We have this and we need V average. Now, V out average. Look carefully here, VO average. But we have Iout average, which is ten and pair, we have out, which is constant value equal to ten pairs, right? Okay, so this is our I average. Okay. Now, we would like V average. How can I get V average? If you look at here, you will find that V average, V out average is equal to what equal to I average, multiplied by resistance, plus the back in MF. Remember, average voltage across any inductance is equal to zero. So VO average will be I average, multiplied by resistance, plus packing in M from KVL here. So it will be like this. V average will be Iot average by R plus E. Or we can say that if you don't remember this, we said before that, if we have a backing F, it will I average will be VO average minus E over R. That we learned before. B, we have E, we have R given in the problem, we have It average ten amps so you can get Vout average. As you can see here, we average minus this equation, we can get average equal to 300 volts. Take that 300 volt and substitute it here, you can get the firing angle Alpha like this. Equal 2.982 radiants. We max line 400 multiplied by root two as we jet. We need the input power factor. How can I get power factor if you don't remember the same as we did before? Power factor equal to power, active power, divod apparent power. Now, what do active power here? Active power is the power will be equal to, which is B input will be equal to power output, reaching our load. Now, if you look at our load here, we have a resistance, and we have back in Math. We have DC current. So power output will be Iout average. Multibloid by resistance, plus I output average square, multi blod by resistance, plus E, multi blot by I abut average. Why do we use average equations? Because, B. Number one, our output is a constant value like this, right? Constant line. So it is DC. So to get the power, we will use average equations. So power consumed by resistance is I square or I average square, multi blood by resistance plus power absorbed by packing F. It will be E, the DC value, multi blood by average current. This will give us power absorbed by the resistance. Then we need the supply power. So in order to get S or the apparent power, it will be we have two options root three, multiplied by V line to line RMS, multiplied by I line to line RMS or three V phase RMS, I phase RMS. V line to line RMS is 400 volt given in the problem. Right? I line to line RMS. If you don't remember, it is exactly in this circuit, the supply line to line supply, root 2/3 multiplied by abut RMS. I abut RMS is a 10:00 A.M. Pair because 10:00 A.M. Pair is a constant current, constant current, which means I abut equal to I average, equal to RMS, equal to 10:00 A.M. Pairs. By using these equations, you will get perfect. So let's look at number one I average square R plus EI output average gives us 3,000 to, and I supply root 2/3 of ten pairs will give us this. You can get or apparent power root three, supply RMS, 400 volt given, and I suppliMs which is this one. The same steps that I have shown you in the previous slide. By dividing these two ratios, you will get power factor 0.53 and it is lagging power factor. Why lagging power factor? Because the current is lagging from the current is lagging from the voltage. Okay? 80. Three-Phase Bridge-Wave Half(Semi)-Controlled Rectifier – HIL Load: Now to the last section or the last circuit that we will discuss in this section for the three phase rectifiers. I hope you now understand three phase rectifiers and it is clear for you. In this lesson, we will discuss the phase bridge, half controlled rectifier with a highly inductive load. We took DAP uncontrolled the bridge. We took DAPA with the fully controlled the bridge, similar to the single phase, uncontrolled and fully controlled, we have also the half or semictrolled. Since we say half or semi controlled, it means that our bridge, half of it is our diets, and the other half is si restors like this. Okay, so someone will say, How can I deal with something like this? It is difficult to think about thi restors and diets. Actually, it is very, very easy, more than you think. You have remember the fully controlled, we had T four, T six and T two. Let's use the pen instead of this, instead of this one, like this. Okay, so if you remember, we had T one, T two, T three, T four, T five, T six. And you said T one conducts at Alpha plus 30 degrees, T two after 60 degrees, Alpha plus 90, T three after another 60, and T four after runs are 60, 210, after unos are 60, 270, after unos are 60, which will be 0130 degrees. Right. We had T one, T two, T three, T four, like the fully controlled. Now, we are going to replace T four with D four, right, and D six with T six with D six. And T two with D two, right? So what are we going to do? It is very, very easy. Remember that we said diodes or thyrestors with zero, zero firing angle, right? So what are you going to D two will be Alpha plus 90. No, it will be just zero spiring angle plus nine, so it will be 90 degrees. Firing angle zero. D four, zero, firing angle. Instead of Alpha plus 210, it will be 210. D six will be 330 degrees, and T T three and T five will be the same as before. That is all of our circuit. So if you look here, each cistor or diet will conduct 420 degrees and switching between each one. A positive group or positive group or negative group would be 120 degrees similar as before. And between each two will be here, it will not be 60 degrees, it will be different. The dyes are treated as if they are crestors with a zero firing angle, as I have just seen. The switching pattern will be T one, D two, D two, D, D three, d four, D four, D five, the same numbering t1d2, D two, d three, d3d4 as if you remove T and add D. Okay. And remember, peak reverse volte, we called Big forward volge In this circuit, maxline similar as all of the bridge rictfres. Now look at here. You can see that is what I have just said. You can see zero degrees. Okay? T one conducts at Alpha plus 30. So we have 30 degrees and add it to Alpha. So this angle Alpha plus 30. D two operates at 90 degrees. T three operates at what? Alpha plus 150, so you can see 120 plus Alpha plus 30, gives us all of this 150 or 120 degrees from T one. Okay? And T 520 degrees from T three. You can see D 290 degrees, D 4210, D 630 exactly as I have just said. The difference is that we put Alpha equal to 04 dites. So if you look at here, remove this, you can see as we have just said. Now we will use this in order to draw our waveforms. So look at here first. T one, let's start with T one. Okay, so let's magnify this instead of writing. Magnify this. Okay. Look at here. We start from here or from here. V here we measure our firing anglph. So let's say our firing anglpha starts here. So this will be T one. We'll start here. T one will conduct from here. For how long? 420 degrees, right? So it will conduct it from here to here, right, 120 degrees. So you can see from here to here T one, T one. Okay? From here to here for another 120 T, you can see from here to here T, T three. From here to here, T 5t5t5. So T one after 120 degrees, T three, after 120 degrees, T five. Okay? Now, that is the first part, second part D two, D four and D six. We said before, D two conducts at 90 degrees, right? At what at what angle 90 degrees. So starting from where here. This angle 90 degrees, you can see 90 degrees. If you go down here from here. So from here, the two will start from here. For how long 120 degrees. So starting from 90 degrees for 120 degrees, it will give us 210 at which D four will start from here, for 120 degrees, it will end here at 330. Then what? Then D six will start from here until the next firing angle, which is D two, right? That's why you'll find that. The part before D two is equal to d six. You can see d2d4, D six, then D two. So D two here. So before it, D six, which is this part. Now, by using this knowledge, d6d2, DT four d six, you can see from here to here, d2d2, from here, d four, d4d4, d6d6. So Pi combining thy restores and you can see t1d6, t1d2, t3d2, Pi combining these two together. Now, by using this information, we will be able to draw the A wave form. Let's start from here. So we have from here to here, t5d6. So let's go here t5d6, right? So T five is connected to what to C, D six is connected to B, so it will be VCB. Let's go here. So from here to here, VCB you can see VCB right from here to here. Now, from here to here, we have T one, D six, let's go. T one, D six, T one connected to A, D six connected to P. So it will be VAB. Go like this VAB from here to here. Starting from here, we have t1d2, go here, T one, D two, Vac, it would be VAC and so. Now, what you will find here that our pulses here are not symmetric. You can see we have this part and we have another different part. So we can say we have integration from here to here plus integration from here to here repeats itself three times, right? One, two, three, and starts again. Okay? So now you understand how we draws that we form in a half or semictrolled. Great. Now let's look at the current. We have a highly inductive load. Output is a constant current. So T one will conduct from here to here for how long 120 degrees from Alpha plus 30 to Alpha plus 150. ID tool conducted from here to here, 420 degrees and so on. So you can draw all of the waveforms. Now look at I supply. IE will be T one minus D four. Now, this is important, okay? Because there is a difference here. T one minus D four. So this minus zero will be like this, and T four, this will be negative, right? I T one minus D four. So nothing changed it from the previous equations, right? This doesn't change because we have Alpha less than 60 degrees. So this what you see here for Alpha, less than 60 degrees. Now, when we start going to Alpha greater than six degrees, you will find a difference. So I hope you now understand this circuit. Now, let's go. Let's write our equations. They are very simple. Don't worry about these integrations. They are very, very easy. Number one, we have how many of these it repeats itself three times one, two, and three. So we have 3/2 pi of what? Integration from here to here for VAB, VAB, V max lines, anomget plus 30 degrees. Here Alpha plus 30 until what until this intersection point, which is 90 degrees, this 0.90 degrees between AB and AC. So it will be from Alpha plus 30 until Pi over two. Plus this part from here to here for VAC. Here it is 90 degrees Pi over two until the next firing ang right or until this point, from here to here, this is 120 degrees. You can see here from here to here, 120 exactly as from here to here, right? So Alpha plus 30 plus 120 degrees gives us here Alpha plus 150. So you can see here Alpha plus 150. From Pi over 290 degrees until Alpha plus 150, like this. Now, what you will see here is that this function, AC. If you look at AC here, you'll find that its phase shape is lagging by 30 degrees. So it will be snomegati minus Pi over six. Okay, lagging by what? Lagging by 30 degrees, sorry, 30 degrees, not 60 degrees. 30 degrees. Because the fish 50 between AB and AC is 60 degrees. So this part is 30 degrees. If you are not sure about this, you can see here AC at this point, right? Zero angle. If you go here, this intersection point, which is 30 degrees. So it is lagging by 30 degrees minus Pi over six. This is our V average. Now, it will be like this, equal to this value. Now for RMS, take the same integration, square of this function, a square of this function like this. Very easy. Now, what about I average? I average, equal to IRMS equal two, average over R since we have a highly inductive loot. What about I thyst each thistor conducted for 120 degrees, so it will be one cere out similar to I dite. Both of the dites and thistors conduct 420 degrees. So it will be one s of the cycle or one of I out. IRMS will be one of our root root of the conduction period with respect to the total period. Okay, what about supply? Supply we said one, 240 degrees, two or three similar as before. Great. Now, what will happen if Alpha greater than 60? Here, something interesting will happen. It will change everything. Now, this is when Alpha greater than six let's magnify like this. Okay, Alpha from here, right, 30 30 degrees plus Alpha here. Alpha greater than 60. We will have positive and negative, right? Now, look carefully what will happen. Here Alpha plus search. Right, starting from here. So T one will conduct right from where to where from here. Until here, right T one, till T three and T three will conduct it from here to here, and T five from here to here, 420 degrees exactly at the previous circuit. D two will start conducting from wherehere from 90 degrees to 120 degrees, 120 degrees. D four will conduct from here to 110 until 130. D six from here exactly as the previous circuit, and you see here t5d6, d5d2, and so on. Now, what will change here? We have a highly inductive louid. Remember, Look here, look at this one. We have these divisions, right? T one, D two, right, T one and D two. So if you go here, T one, D two, VAC. Okay, so we'll start here with VAC. So from here, VAC. Okay, that is the first difference. So before in the previous circuit, we started with AB, right? At Alpha plus 30, we started here. Mm hmm at AB. However, here we started due to Alpha being greater than 60, it changed everything. We started from Alpha plus 30 at AC instead of AB. Okay? Now we started here. Now let's look D one, D two from here until here, right? So this period from here until 210, VAC. Now, let's look from here. F here, you will see T one and D four. If you look here, T one and D four. So what does this mean? So T one conductors and D four conductor. What does this mean? It means a short circuit across our lot, right? If you look here, like this, let me write it. You can see here that in this part from here to here, you can see t1d4, right? T one, D four. Both of them will conduct and they become a short circuit. So what does this mean? It means that our out is parallel to a short circuit. It means that out will equal to zero, right? That's why from here to here, you can see zero voltage. Now, why did this happen? Because because T one, T 385 when Alpha increase, they are shifted to the right. They are shifted to the right. However, D 2d4d6 are not affected or unaffected by the firing angle. They are still in their position 90210330. This shift in T one, T 385 beyond 60 degrees Alpha greater than 60 led to this phenomena When we have voltage and zero voltage, similar to highly inductive loot. Okay? Okay, now let's magnify again. Similarly, you will find that so we have one period of conduction, zero, then conduction, zero, conduction, zero, and so on. It repeats itself. Okay? Okay. Now let's look at our currents. So we have this conduction period from here to here, t1d2, right, and T one conduct. So look at T one. It one T one, T three, T five conducts all the time. D 2d4d6 conducts also, not all the time, 120, hundred 20, hundred 20, and these are the same. You can see here, IT 120, then T three after 120 degrees, it start conducting. Then T five after 120 degrees, it start conducting. Okay, four D two, D four and D six exactly. Starts at 9,210, 200 and tend and 30 and so on. Now, what you will see different from before. So what you will see different from before that I supply. If you remember I supply, we said it is IT one minus ID four, IT one minus ID four. CLC is at IT one minus ID four. Now, what's the difference here? You will find that there is a common part here, this common with this one. So they will cancel each other, and we will have this part only like this. If you look at the rest I one, then this common part removed, then we will have this part as negative. You will see that this supply current is different from the previous one. It is not true 2/3, it is different and we will see its equation right now. So do you understand why does this happen? Because I one is shifted. Now, if you draw the previous one, it will be drawn from here, let's say, from here, from this part, you can say, I one will be drawn like this. Okay, let's draw it. It will be like this. It one was drawn like this. Okay, something like this. Okay? So this will not cause an intersection with this one with ID four. So nothing will cancel each other 100 2,120. Okay? Similar to this, you can see here. This sort does not intersect with this spot. However, due to firing angle, delay increased beyond 60, this was pushed to the right. So there was an intersection region between it and ID four. Okay? So you can see here. Like this became an intersection ratio. That's why I supply became smaller. Okay? Now, let's see the equations. Okay, number one, V average. V average will be integration of our function. Which one? Let's start with Alpha plus 30 until 210. So Alpha plus 30 until 210 is 7/6 Pi, 210. Okay? How many bulls we have one, two, and three. So it will be 3/2 Pi. V max line sine omega t for the function. Remember, we are integrating AC, not AB. So AC is lagging Pi 30 degrees, so it will be minus Pi over six. Okay, we are integrating this one. Okay? So by integration, it will give us this value. And surprisingly, it is the same exact value as Alpha and 60 degrees. Okay? The ORMs, the same square, same integration square. What about Iot RMS? Since we have a highly inductive load, Iot RMS equal I average equal the average over R. What about the I thistor this Iristor will be one over cert and exactly I thiisor RMS, one of root three, and dites for average and RMS are exactly the same, equal to this value and this value because they are conducting for the same conduction period. What about eye supply? Now let's look at it. Magnify go down here like this. So we have eye supply. We would average equal to zero, right? For RMS, it will be integration from the induction period from here to here, multiplod by two. Now here, Alpha plus 30. This point, this point is what is 210, this point. So it will be this region, 210 minus Alpha plus 30, multiplod by two. So it will be like this, 210 minus Alpha plus 30 degrees divided by zero hundred and 60. Okay? This tblad by two. And this under the square root. So root induction period divided by zero hundred and 60 or too Pi. Induction period here two times. Here we have one and two, each period 210 minus Alpha plus 30. So 210 -30 degrees gives us 180 minus Alpha and two. Okay, so it will be root two, 180 minus Alpha over two Pi. Okay, or instead of two Pi, let's say two multi blood Bi, 180. Okay, take two with two. So we will have root 180 minus Alpha over 180, which is exactly similar as root Pi minus Alpha over Pi. All of them will lead to the same solution. Okay? 81. Example 10: Now let's have the first example on the three phase, half controlled bridge rict fire. We have this circuit. We have 120 volt supply RMS line to line, gives an average of current of 10:00 A.M. Pair to allowed resistance of five arms. Now, determine the firing angle of the circuit. Number one, we have a half controlled, right. Now, we would like to know average. Now, is a highly inductive load or the load will it change anything? Will the load change anything or being resistive or highly inductive? No, it will be exactly out average. So VO average will not change. Okay. Why will not it change? Because if you remember that at Alpha, less than 60 degrees and Alpha greater than 60 degrees, we had the same V average. So let's get back. You can see here, this is our V average win, highly inductive load greater than 60. If you get back here, the same equation for Alpha less than 60 degrees. And you can see we don't have any negative part, and here and here, we don't have any negative part. So the resistance or the output average in resistance will be exactly the same as highly inductive root. Okay? So we can use the same equation. Okay, let's lead. So we have average. What will be the average?V average will be I average multiplied by R, very easy. And Vmax line will be V supply RMS multiplied by root two, ten multiblod by five for V average and three Vmax line, root 220 volt, root 220 volt. By using this, you will get cosine Alpha and you will get firing angle equal to 112 degrees. Okay? 82. Example 11: Now let's have another example. In this example, we have a DCU requires a control of its voltage from the maximum value to a quarter of it. So we would like to make average max, drop the two quarter of it by using this half controlled the phase bridge. How can I do this by controlling the firing angle Alpha? Now, we would like the rating required for the free wheeling dt. Here, if the lute current is a constant current of 20 ambers. So what we learn from this number one, we have a highly inductive loot, right? Since the loot current is constant, so Iout, equal to Iout average, equal to Iout RMS, equal to 20 and bears, right? Okay. Now, I know that we didn't discuss the free wing dt with a free winning dt for half control but you will find that even if you don't know, you can do it. You can draw it and you can know the effect of the freewing dit. Okay? Now, first, we have average in our circuit with a free wheeling diet and without a free wheeling diet. We don't have any negative part. So when does the free wheeling diet operate? Remember, remember. That when Alpha greater than 60 degrees, when Alpha greater than 60 degrees. We had a portion of the wave in which we have zero voltage. This portion of the wave instead of t1d4 or t3d6, t5d2 causing the short circuit, freewheeling dot will act instead. It will go instead and start operating instead of them. Okay. So anyway, we don't know Alpha, we use the general equation to get Alpha. The average will be equal to what? Three V max line over two pi one plus cosine Alpha. Remember, we need to control our voltage from maximum value to quarter of it. So we have average value equal to this. Now, my question to you, what is the maximum value from this equation? If you say three V max line over two Pi, then you are wrong. Why is this? Because Alpha Alpha can be 0-180 degrees. If you substitute with Alpha equal to zero, what will happen when Alpha becomes zero, cosine zero equal to one. So one plus one gives us two. So our maximum value will be three Vmaxoline over buy, right? So maximum value occurs at firing angle equal to zero degrees. V max three Vmaxolin over. Now we would like to control this value of the abut and make it quarter of it, 1/4 of this value. So what I'm going to do is that I will say 1/43 V maxiline over Y equal to this equation, right? This is our general equation, V average, and I would like to make it equal quarter of maximum value. That is what we need. So from this equation, we will get Alpha. 120 degrees. So what we learn from this when Alpha rater than 60 degrees like this, so we will have the wave form like this. Mm hmm, zero. Mm hmm. Like this, right? So during this period, the free wheeling diet will operate. Now, first, the rating of free wheeling diet. Number one, peak reverse volt peak forward volte equal B, V maxoline. However, we don't have max line given in the problem, so we'll keep it like this. We need I free wheeling diet average. Okay, you don't know this, but let's see the wave form first. We have this waveform like this, then zero, then like this, then zero like this. During this zero portion, instead of having T one and DT one d four that will form the free wheeling dt action, when we have one dite it will be the one that will operate. So during this zero voltages free wheeling dt, free wheeling dt, free weling dt, Okay, now, how can we get the voltage or how to get the average or whatever? Now, first step, let's delete this. Okay? Now let's look at here. So let's start from Alpha plus 30 until this point, which is 210. So from Alpha plus 30 until 210. Now, the freewheeling diet will start from here from this point, which is 210 degrees, right, at which Dfour will conduct or should have been conducted without freewheeling diet. So 210 Okay, until this point, this point is 270, right? Now, why 270 because it will be Alpha. Plus 150 degrees. Remember, here we have Alpha plus 30. After 120 degrees, we have T three, T one, T three. Alpha plus 30 plus 200 plus 120 Alpha plus 30 plus 120 gives us what gives us Alpha Alpha plus 30 plus 120 Alpha plus 150 degrees. So we have Alpha plus 150 degrees and we have 210. This period. Now, how many times our free wheeling it will conduct in one cycle three times one, two, and three, right? Because this short circuit happens this zero voltage happens three times in one cycle. So it will be like this. I free wheeling dt, okay? Average will be three times of Alpha plus 15, 150, sorry, minus the beginning which is 210, right? 210, like this. Divided by the whole period, multiplied by I with average, right? Okay, now, 150 -110 gives us negative 60 degrees. Okay? Negative the spot, negative 60 degrees, which is exactly equal to what equal to Pi over three. Okay. Now we have Alpha minus Pi verse, and we have 3/360. Three son 60 equal three or three ozonO over 120. Okay? Whatever it is, you can keep it as it is. You can write it like this. Three Alpha minus Pi divided by two pi in ingredients, okay, instead of degrees, I out. So you can this equation three, three multiplied by Alpha plus Pi over six, minus Pi over two. Pi over six minus Pi over two. By over 630 degrees, minus Pi over two, 90 degrees, gives us negative 60, which is exactly equal to negative Pi over three. Okay, so it is exactly the same equation like this one. Okay. However, instead of using this part from here to here, I have used when I wrote this equation, I have used from here to here. It is the same. Anyone will give you the three minds. Now let's look at this equation. We have three Alpha minus Pi over three to Pi. Now, what I'm going to need is the rating of the diet, free winning dot. I would like to find the worst case. When is the worst case, the worst case will be when Alpha is highest or maximum. Maximum value of Alpha is Pi. Right? So if I substitute with Pi here, it will be equal to three Pi minus Pi versary divided by two Pi. Now, Pi minus bi versit is two pi versary. Multi blood by three, three will go three, so we'll have two Pi over two pi, which means unity, which means I average will be I abut and even if you take the square root of one, it will be one, so it will be also I output. The same equation, but under the square root. Okay? You have to know that I in this example, take this. Where did I obtain this from this region? This region, which is here. This written one, Okay, a smiley face. Wow. For this one root, where did we get this from here, from here, from here to here, which is 150, which is Alpha Alpha plus 30 degrees, Alpha plus 30 degrees -90 or Pi over two. Why 150? Because Alpha 120, 120 plus 30 gives us 150. That's all. Okay. I hope this example was clear for you and you understand now what will happen if we have a free wheeling guide in our half bridge rectifier. 83. Definition of AC Chopper: Now we would like first to understand the definition of an essential part. So what is an ISI, Chopin and EC shopper is simply this sock, OK, this circuit, what does it do? It simply converted as the input is a voltage to another output AC voltage, okay? Which changes there is a voltage input 2a EC shopper to another value of RMS. Saw was at ISI. Shopper operates as a step-down transformer. So it decreases. The input is a voltage. So it can be used to produce our variable ac voltage from 0 V supply. So this one, if it has resupply Xunzi Albert TV or what can be controlled from 0 to V supply, okay, from Z minimum value to the maximum value. So most likely it will be used as a step down as it decreases the input voltage. So we can control Z voltage by user usage of AC Chopin. So one of the applications in which we use as Zai Yi Si Shuo part. For example, when we change the Z value of Z voltage, which changes the value of Z current and the torque produced by the motor. Okay, so in case of starting our motto, we will use something which is called a softer starting z software starting means that we are decreasing or increasing Z voltage input to the motor gradually. As the voltage increases. Or ads or beginning Z, starting current is very high. So we decrease the voltage to limit the starting current. Then we start increasing our voltage two up to Z maximum value in order to reduce the maximum torque. So z is u shopper is used in software starting in motors in order to increase the voltage a gradually during the starting of Amato. So it, it converges if exit AC voltage and VS into our control, the voltage waveform at same frequency. So frequency here is the same as the frequency here in most cases except in z. Integral cycle control, which will also we are going to discuss. So what are the methods of controlling the AC voltage? We can control Z value of AC voltage of pi that transformers, Z transformers, which is used to obtain a pure sinusoidal waveform. Remember that the transformer is considered as the backbone enzyme bar system. It is used, uh, to step up the voltage or step down the voltage at the same frequency. Remember that in a transformer V1, the primary voltage over z. Secondly, secondary voltage equal to the number of turns of z, primary over z number of turns of secondary. So by controlling this ratio, we can exchange a Z output voltage. Okay? So if any two greater than N1, then it will act as a step up as v2 equal to V1 multiplied by N2 over N1. V2 equal to n0 tomato blood by V1 over N1 in the tomato blood by V1 over N1. So as we increase the number of turns of secondary or a greater Zen N1 is in V2 will be greater than V1. If in a tomb, listen N1 x_N, v2 less than v1, and it will act as a step-down transformer. So we can use as transformer in order to step up or step down the voltage. And the benefit of the transformer is that it produces also a pure sinusoidal wave. Okay? Z input is a sine omega t and the output is also a sinusoidal waveform. And the second method for controlling z voltage is AC Shopper, which we are discussing in the course. We use EC Schumpeter and order to step down the voltage by switching on and off voltage during its cycle, causing Z total RMS of the voltage to change or switching on and off periodically. So we'll understand how an essential part walks to step down the voltage in the next two lectures of z AC sharper. You will understand that when we connect an essential par with our Lord, how we can change its magnitude. Okay, how we can change is the RMS of z voltage. 84. Switching Techniques in AC Choppers: Let us discuss a z switching techniques and sides the AC sharper. So one of the ferocity techniques which we will find in the course is the phase angle control. What does it mean? It means that we control the output voltage RMS, or is the output waveform buys a usage of z. Phase angle controls the phase angle here is meaning the Boise alpha as the firing angle alpha. So by controlling busy finding the angle alpha, we can exchange a Z value of Z waveform or z a period where is our wave form is available. So this will result in is changing is the total RMS value of Z voltage. As we will see in Zen extra videos when we discuss LC circuits, you will understand how it changing Gulf War, it changes the root mean square. You will find here is that this distance which represented the presence of our wave, volume minus alpha. This period is pi minus alpha k as this period from here to here. Okay? By changing a Z value of alpha, we can change as the widths of the presence of Z-Wave. Okay, which result is interchanging Ghazi at z Albert RMS. So z power electronic switch operates once every half cycle. You will find that here. This one is a half cycle. This one is a half cycle. So we'll find here is that the first Osiris store, and we will see next videos. And the first side is store here on doctors once in z cycle. Okay? This one also on doctors once in the psych. Ok. So it conducts oneself version half cycle in the phase angle control, the integral cycle control, for example, you will find that it operates for a group of cycles in off for a group of cycles, or the pulse-width modulation, as we will see now. The problem of this method is that it has atomos harmonics, okay? Is this waveform contains lots of harmonics away from the fundamental component. Okay? If we analyze this waveform by using Fourier series, you will find that it contains lots of harmonic, which of course not good for us. The second method, which is pulse width modulation, several switching Aldabra half cycle. Here we said that we are switching for one in one cycle on Zen, it is often due to its natural commutation of the restaurant, as we will see. Then on the negative cycle and of my natural commutation here several switching our DOM perhaps cycle we will switch on and off several times. It's a problem of this. It is more complicated and expensive method for switching. Here you will find as an example. For example, here we switch it here. Okay, and then here off, here, on here of Zen on, Zen off, Zen on Xin off and so on. So finds that in z half cycle, you will find 12345. You can see five switches. 12345. We have switched at our photos dot 4-5 times in societal. Here we switched it only for one time. So as this one consists of a group of pulses, you will find this method. In case of the inverter is of course no finds that z pulse-width modulation technique is discussed dessert. Z advanced of this method is that it has a low harmonics, but it is expensive and complicated control circuit. Z lost technique here is z integral cyclic control. We operate our service or for N cycles and turn it off for M cycles. Here, you will find that here our sinusoidal is going out normally in Boise usage offsite or store or operating it on continuously. Then off for a group of period Xin ONE again, Zen or for a group of cycles and so on. Here we'll find that z, z previous frequency was from here. Here, okay? Is a frequency of z sinusoidal wave, okay? Is this frequency now z, as this pattern operates or repeats itself through this period, p from here to here. This is considered as a new periodic time or frequency. So here you will find in integral cycle control is a frequency with a change. Of course, we will discuss the z integral cycle control in z and its own video or on its own electron. This message is not recommended in motors that drive. Why do to audible noise? What is the problem? Z frequency, it change it, which causes inside Z mortar on Azara rotating magnetic fields or rotating fields, which causes Z mortar to have a reverse torque against its main torque, causing it to have this kind of noise. So the integrals over Madonna, we don't use it in water because it will cause noise and cause rotating greens against its mean-field. 85. Applications on AC Choppers: Now let's discuss some applications of the ISI shopper before we start discussing the AC shopper itself, the first thing it is used in lighting controlled as incandescent lamp control. So as an example, this is an easy shopper. Okay? And this is you should pour control disease magnitude or the RMS of z voltage. Here we have a plus minus VS, which is input 2i z circuit orders gene, the incandescent lamp. And by user usage of ISI isobar, we can control the voltage from 0 two VS. OK. So by changing the value of the voltage, z intensity of light will change. Okay? So we say that the intensity of light or the flighting of z lamp depends on the value of the input voltage. So by varying this voltage, we can change the intensity of z lighting. Okay? So by changing the z value of the voltage across Z lambda here, by sausage off a session four, we can change as the intensity of lighting. So this variation in the voltage depends on the angle alpha, okay? Z phase angle control. Also it can be used in the software starting and softer stop of z motor during the starting of induction motor, which we said before. The current is very high. So we need to limit this current. So buys or usage of Z sc support, we can exchange the input voltage from 0 to V supply to limit z starting current and to produce the soft distorting of z motor. It is used also in industrial heating control as the current depends on the voltage. So in industry, we use z variation of the voltage in order to vary the amount of current entering a circuit. So we can vary the amount of Z heat produced. Remember that for examined in resistance, a Z value of z power dissipated or the heat dissipated is I square R. So by changing or decreasing z inbuilt voltage, we decrease the input current. As an example for this here, let's all of this you will find here if we have here plus, minus VS Zimbardo sort of ISI Chapin, We decrease or changes the voltage from 0 to VS. So CSA read of heating the events on the firing angle alpha, same as before here, but here's the intensity of light. Here. Z rate of heating. Sos also was some applications in AC shoppers. 86. Types of AC Choppers: So what are the types of ISI shoppers? We have two main types in EC shall possess unidirectional AC super and z pi directional EC shopper. So the first one which is that unidirectional EC shop or how does it look? It consisting of one diode barrel to one's high-risk. So let us see the circuit that we have here, our diet parallel to Osiris. Ok, so you will find that as we remember from our course for power electronics, that's z psi restaurant can control when it is on. Okay? So diode cannot be controlled is this one is called as semi controlled because we can control when it is on, but we cannot control when it is off. As allied weak, it is uncontrolled because we cannot control it. So why it's called a unidirectional because we can control only z half positive cycle of z and z AC shopper. Okay, so all of this is called an ISI short, but all of this together see parallel is called Dizzy EC shopper. It's called an electrical equipment. Okay? So Z during the half cycle here we can control AC shopper. During the negative cycle. We cannot, it is uncontrolled, so we can control from one direction only. Controling here is done in the positive cycle. The second type is the PI directional AC. Sharpen. It consisting of two parallel Cyrus tours. And it's called a bidirectional because we can control z positive and a negative cycles. This one which is the most commonly used is the shopper. When we are talking about EC shoppers. As you will see here is that here we have ADD. Is your Schober consisting of a toast virus stores in battery. This one controls is equals to half cycle. This one controls the negative cycle. So it's called RPI directional or tool direction because we can control the positive and the negative. Here. It's called the unidirectional because we control only with z positive psych. 87. AC Chopper with R Load: Now let's discuss first circuit, which is the EC shopper, loaded with an R or a resistive load. So we have here our V supply. Let's use our Penn. We have here our resupply, the supply voltage, which is AC, and having this waveform, sine wave, okay? Z voltage as a function of omega T. Okay? So this is e supply voltage is input through Ze, EC shopper. This one is the AC shopper. And to our resistive load, okay? Z voltage here across z voltage from here to here is v out. Okay? I would voltage across the resistance R. And we have here a current or output which is the current Abbas things rosy or loot. Now, the first thing is that we have here is this is the input sine wave. Now, we would like to find is the output voltage. Okay? So the first thing we have at also restores connected embedder which represented our AC, this one and this one, which is polished directional essential part one used before. And Z control of z positive cycle ends the other one for the control of z negative cycle. Z. Second thing is that the server store T1 over it's at angle alpha. And T2 operates at angle alpha plus pi, which is alpha plus a 180 degree. Okay? So this So restore or breaks at the angle alpha. This one operates at angle alpha plus pi. So let us see that we output, remember that this point is, for example, our alpha, okay? And this point is pi plus alpha, okay? Or alpha plus a 180 degree. This point now adds up beginning at alpha, okay, before alpha or alpha plus pi is a servicer is off and this one is off. So our circuit will be like this. Okay? We have here is our resistance, okay? And the V supply. So the V output here across the resistance R is equal to 0, okay? No voltage across the resistance. So before the angle alpha, when D1 is off and T2 is off, T2 before alpha z Alberto will be equal to 0, as you see here. Starting from alpha, thyristor, T1 will start to, to conduct, okay, it will be a short circuit. So our circuit will be like this. The supply voltage. And T1 short-circuit T2 is off. Since it has no firing angle and it is also reverse biased from our first course. So it will be an open circuit and I output enzyme resistance. So from KVL here, you will find that V output is equal to the supply voltage. So at this point z, this one is the supply voltage, our sine wave. So at alpha, it will start becoming the supply voltage. So here is the value of the supply voltage. Then it will continue to conduct until point. So from alpha to pi, this I restore T1 conductors. So the output is the same as the supply voltage, okay? At this point, pi z voltage across T1 is, will be equal to 0 or it will be reverse biased. So T1 will be off, okay? So T1 and T2 are off until pi plus alpha. So during this period, z, so it is T2, T1 and T2 are off because this one does not have any firing angle. This one does not have any foreign angle. And the voltage across them, 41 is reverse biased. Now at pi plus alpha is Osiris store T2 will start to conduct. So the circuit will be like this. T2 will start to conduct likes us. So the supply voltage will be equal to the output voltage across the resistance, same as this circuit is. This one will be assault circuit, and this one will be an open circuit. So from GAVI l z output equal to the supply. So starting from pi plus alpha to two pi, Z wave form will be the same as the supply voltage. Is this one. Same as starting from here, this one. Since during this period, T2 conductors, during this period, T1 conducts. So z voltage across the thyristor, T one will be equal to, at this point, it is reversible iced or open circuit. So it will be equal to the supply voltage from here to here. It is conducting. So it is a short circuit like this. From here to here. The voltage across it, it is reverse biased. So it is an open circuit like this case. So the voltage across it is a same as V supply a starting from here, D2 conductors. So z voltage across Z t1 is the same as T2 since they are in parallel. So it will be a short circuit. So in the end, if you sum from giving ALL here, you will find that supply voltage equal to t1 plus v-out. So if we sum this waveform. With this full waveform, you will find it will be equal to supply. This plus 0 will give us z supply. This one plus 0 gives us the supply. This part, which is this part plus this part to give us supply and so on. So in the end, we have this waveforms and I, I would, is the same as the supply voltage. Why? Since the ZR and having a resistive load associate Have the same waveform. The current I out is the same as V output over the resistance R. So I, I would have the same waveform as the V output. So we'll find that by increasing, let's delete all of this by increasing Z alpha from 0 to a 180 at alpha equals 0. From here, then the circuit will be like this. We will draw and the bipolars alpha, it will be y plus 0, which is starting from point. Suppose I V output will be equal to the supply voltage at alpha equals 0. So it will be equal to VS at a 180, none of the restaurants will conduct. Since alpha pi, it will be at this point and at pi it will be at this point to point. So at this point x0, so I restore is reversible as due to z negative part and the same S for T2. So at alpha equals a hundred eighty zero, voltage output will be equal to 0. So the alpha is from 0 to pi. And to understand the now by changing the alpha, we can achieve the root mean square of our voltage. So by increasing value of alpha as we approached z by z, root-mean-squared decreases. So why do we make alpha and alpha plus a 180, alpha and alpha plus a 180 in order to make z positive Bharti year equal to z negative part here. So that V output average to be equal to 0 as here, z positive cycle equal to z negative cycle. So the average of sine wave is 0 and also the average here will be 0. And that's why we make z alpha and alpha plus a 180 as a firing angles. Now, if we look at x0 equations for Z sc SOPA was an onload, you will find that the RMS value, output voltage, your nose at the RMS is a square root of the function one over two pi integration from 0 to two pi 2Z function square, right? So the function here is V1 and this one is Z, maximum sine omega t v maximum sine omega t, which is this point. This point is root two V1. Ok? Why R2D2? Because v1 is the RMS, okay? And in order to change RMS 2Z maximum, remembers that v-max. Over V R M S equal to root two. So V maximum, which is the maximum of the sine wave, is rho V RMS root to V1. V1 is given RMSE value. So it will be root two v1 and it is our sine wave with a maximum value over to V1. All square, okay? Here, one over two pi. But you will see that since we are getting is ultimately square of this function, okay? We are getting is that root mean square of this function. So we'll integrate from alpha to pi, okay? Alpha to buy. And the plus integration from pi plus alpha to two pi. So instead of doing dizzy integration two times, because when z squared, z function will be like this to this waveform which is a negative Y squared AND gate will be positive, the same as this waveform. So instead of integrating this plus this, we can say is integration of this one multiplied by two. So here's integration should be one over too far. Okay? One over the period multiplied by two. Since we have hears this integration. And instead of this plus this, we will say that the integration of this only, so it will be multiplied by two, is integration from alpha to pi. Alpha to 0.2 will go with the 22 will be one over pi. Degradation from alpha to pi. Now's the integration of this function will give us this value. Ok? You will find that here as alpha increases, sine two alpha, and this value will increase. But at the same time this value will decrease. So you will find that the relation as alpha increases, the V output will decrease. Now is the RMS of the current RMS I output, RMS is V output RMS over r, Since it is a resistive load. Now if I would like to find the RMS of z psi restored current. So we will find that z, sorry, restored, for example, t1. Okay? And this is the output current as this where T1 opiates and this when T2 operates. So if I would like to find the T1 as a waveform, it will be this waveform like this. Okay? From alpha to pi. So the question here, I would like to find the RMS of Z current of Osiris tone. It will be square root one over two pi integration from alpha to pi, since it is only existing during the half cycle from alpha to pi, alpha to pi z, the value of z. So I restore current which is same as I output. It will be supply current over the resistance R. Okay? Is this gives us Z waveform of the output or the square since we are getting z RMS. So we'll have finally is this value of Z RMS of the cyrus took current. So in order to understand more about this EC sharper, let's go and have an example on this in order to understand how to apply these equations. 88. Example 1 on AC Chopper with R Load: So let's have an example on the EC shopper with an onload. So we have here as single-phase inverse parallel thyristor diode phase control shopper circuit. So phase control means that we are controlling the FBI is the angle alpha, and we have years at parallel Cyrus total diet. So it means that it is unidirectional function or unidirectional AC Chopin. And it supplies a load of eight ohm here. It only means that resistance R. Okay, we should make it more ghrelin, which can see are equal it on ok, because the OEM can represent a z value of the inductance or any value of z capacitance or anything, okay, as an ecstasy of course and exile. So z ac voltage source of a 120 volts, this is given as an RMS, determines z power at firing angle of 60. And the root mean square of this will restore count. So how we can solve an equation like this? So the first step is that you draw the circuit. We have our supply voltage, 220 volt. The enzyme RMS. We have add parallel thyristor diode supplies and eight ohm resistance. So the first, we draw the circuit, second sim finds the waveforms. So for this I restore. It will conduct a from alpha to pi. Okay? Z Guide is uncontrolled. Saw during the negative cycle, it will start conducting when it is forward biased. So starting from pi, it will start conducting. Or to make it simple for you, remember that in Z parallel combination we said alpha and alpha plus pi. Ok? So for XY and Z, alpha is equal to 0. So it started conducting from pi. This is another way to understand it. Sosa waveform is like this and the current waveform will have the same value but divided by R. The thyristor current which is required. It's from alpha to pi only. Z diet is conducting from pi to two pi y naught from alpha, because this is uncontrolled and I cannot control it. So as soon as the voltage across it becomes a positive, it will start to conducting from here. So the first one determines the power. So remember that z power is simply equal to z i square as a root mean square multiplied by the resistance R. Or the value of z voltage as a root mean square over the resistance. Are any of them are equal or Zr? Correct? So let's start to pi using getting dizzy VL as Oldman square. So we have the function like this, and I would like to get z square. So simply remembers as a v l as a root mean square is square root of one over z period. Period is from 0 to two pi, one over two pi. Integration from alpha to two pi since it is here continuous. So we'll say from alpha to two pi, directly, from alpha to two pi to the function. Function here is root two VS sine omega T all squared d omega t. So we can get Z in this form or substitute directly with z values V S is given as at a 120 volt x0 sine omega t Ruto alpha is given as 60 or pi over three. So you can get simply z value of v l as ultimate square. So we can say that here we got as heavy as this function and then squaring it for disempower over r. So it will be VS square. This function after removing zeros over z resistance R, R is given as eight ohm. So we can get z power as 5.46 kilowatt. Now, Z RMS of societal store current simply z t RMS assembly X0 integration from alpha to pi forward the value of the current. So square root one over the period two pi integration from alpha to pi, alpha two by four z function of z voltage over our Ruto VS sine omega t over r, all square d omega t. Now, as this function. And we have our, we have sine omega t, We have everything, okay? We have alpha equal biodiversity. So by solving this integration and under the square root, you can get that the ISO restored as a root mean square equal to 17.44 and bear. Ok. So this is another simple integration. So this was our first example in z AC support. Now in the next video, let's have another example on Z AC Schumpeter was on our loot. 89. Example 2 on AC Chopper with R Load: Let's have another example. Here. We have an ISI super circuit loaded by a two ohm resistor. So we have here anode as two ohm resistor. This is supply voltage is at a 120 volt Z firing angle is adjusted to 60, or alpha equal pi over six. Neglect as E, sorry, restore volts drop, remembers it in Z course on power electronics. So we said that z psi restores end every power electronic device reduces our voltage drop. But if we compare at two volt, as I've also Strobe2 is at 120 volt. It is a very small value, so we neglect the z, sorry, restore volts drop. Now what is our requirement? The requirement is determines the load power. Here we have an ISI super circuit. This means that we have i pi direction, like this, this one and this one. So the first step, we draw our circuit and we should draw our waveforms from alpha to pi and pi plus alpha to two pi 0 current is the same as Z load voltage, but divided by R. So restore T1 conduct is during this part and so I restore T2 on ductus during his spot. So someone will ask me now, why did z are what current in this life exists? And so I restore Tito likes us, okay? Simply during the negative cycle when z, so I restore T2 conducts in this direction. Okay? So the current will flow like this. And the current of T2 is in this direction. T2. So this direction is in the same direction of the supply. So it is a positive direction. But z, Assume the value of i in this direction, okay, in the reverse direction. And that's why you will find I output is negative as it is equal to supply over z or a z over z resistance, okay? 0 d is negative. So it has an negative direction same as the voltage. And i2. I2 or is Osiris to T2 has reversed the direction of i out. So it will be opposite of this wave form of Ireland. Now, I need a load power. We said that the load power is I square RMS R or V output Goodman square over r. So v n as a root mean square is square root, one over two pi k. And there were multiple ads or integration by two, as we said before, Xin we integrate from alpha to pi two z function root 2V as sine omega t r squared d omega t to a 120 volt is V supply alphas pi over three. Everything is given. So we can get V L as a root mean square. Zr is given as a two ohm. So we can divide z v L square over r to get z power load, which is 19.469 kilo watt. So this is another simple example on how we use a super to a change in the value of Z output voltage. And we determine the Z load power. In case of Z s You shall polluted to buy our loomed. 90. AC Chopper with L Load Part 1: Now let's discuss Z. C Shapiro is an L or an inductive load, or a pure inductive load. So we have here our supply VS. And we have here our ISI shopper and our load, which is a pure inductive yoked. So as a first single would like to draw the output waveform. So how does this circuit works? So assumes that we are starting from alpha at alpha z, sorry, restore T1, which is this one, this one is t one, and this one is T2. T1 will start conducting at alpha. So it will go here. It will be equal to V output as supply voltage or 0 volt output voltage equal to the supply voltage from KVM, since it is assault circuit. So starting from alpha two will be V, output will be equal to V supply going through some till pi. At pi, we said that T1 is reversed biased. But however, Z inductance itself was a charged Boise current. So z inductance all started this charging and reducing current in the same direction of the original current. So it bridges or current in this direction goes in the Zt1 to continue conduction. So it will be from alpha to an angle called the beta, which is the extinction nk. So similar will happen for Z T2. It will start from alpha plus pi and they'll continue to two pi. Two pi z inductance will have or be charged it in the opposite direction, causing the T2 to continue to conduct until beta plus pi. So this waveform will be repeated here. Okay, is this part is similar twos's part. So the poster from alpha to beta, thus the negative from alpha plus pi to beta plus pi. So this is point beta plus pi, This one is beta and Zen, this point will be beta minus pi. So here subtracted by, we're, we added by. So as the current, you will find that the current is charging from alpha to beta as opposed to okay, and not having disease same wave form because it is an inductive load. So at alpha z current is 0. And starting charging Zen started discharging until betas extinction angle, where it become 0. And beta is equal to two pi minus alpha. And we will prove this part. So when beta equal to pi minus alpha, you will find that this area is equal to this area, equal to this area equal to this area. So in the integration we can use one area only and multiply by four. But how did we prove this? You will find that when we broke that beta equal to pi minus alpha, you can find that this area is equal to this one to this one. Now, z current is from alpha to beta four, z positive T1, or it's a positive cycle. And for the negative cycle from y plus alpha, two pi plus beta. And this part will repeat itself here. So again, when T1 is on, this causes the current Abbas Rosa inductance and charge it. When the voltage reaches a 180 degree at pi, the thyristor T1 want is to turn off. However is the inductance itself will start to producing a current in the same direction of the original cannot causing the T1 to continue to conduct Zen. So restore T1 will be on ontology angle beta or the extinction angle. Then T1 will be off. At alpha plus pi z psi restore T2 will start to conduct. T2 will keep conducting until beta plus pi by the same effect of z inductance, but in the opposite direction or in the reverse direction. So our conduction period here extended from or increased, not buy it become pi to pita, boy to Peter. This is the extra part. And for this extra 40 year is from two pi to beta plus pi, two pi, two plus pi. Now we would like to find the z equation of Z current in order wish or which will help us in our analysis. So we'll see during busy on cycle and during the OFF cycle. So during the on cycle when T1 is conducting from alpha 2s extinction angle beta, okay? When T1 is conducting, our circuit is V output will be equal to the supply voltage V, right? So V out is equal to 0 and voltage equal tos a supply voltage. Sources supply voltage would be equal to V out. And the voltage across an inductance is equal to L d over d t. So z d i is equal to D t V supply, d t V supply, which is same as V Lord over the inductance L. Okay? Just cross multiplication. So the current is equal to one over L, one over L, xa integration of v l, which is the V supply, which is root two VS sine omega t. And gt, okay, we said the year DT, okay? But we would like to z function here of z sine in function of omega t. So would like instead of d t, we need it to be d omega t. So multiplied by omega and volunteer point omega. Ok? So one over omega n equal to the integration of this function. So the integration of this function will give us i as a function of omega t, which is the current output during the on period from R2, it will be equal to root over S over omega l cosine alpha minus cosine omega t. So why did we do this proof in order to finds extinction angle beta? We know alpha, alpha is given, which is controlled by me. And the inductance is z value, which is also put inside the circuit of zeros. Supply is controlled. And the only unknown here is extinction angle beta. So how we can nose extinction incubator? Remember that the current was conducting from alpha to beta, alpha to beta. Okay? So add alpha is the current is equal to 0 and at betas the current equal to 0. So we can see that ad is current equal to 0 at z extinction angle beta at omega T equal beta, the current will be equal to 0. So 0 will be equal to two V S omega l cosine alpha minus cosine instead of omega t, we will put beta because we would like to find CB ITER at betas, the current is 0. So this will go with z 0, of course, anything TO brackets Motorola by Chaucer equals 0. So this equal to 0 or this equal to 0. So this one is Z1 actually equal to 0. So cosine alpha equal cosine beta. So alpha equal beta, which is of course refused alpha cannot be equal to Peter or there will be no extension angle. Or this means that Z are at the same point. X0 current will be like this point, alpha equal to beta. So of course it is refused. Or your nose at cosine, we say that cosine X is the same as cosine. Two. Phi minus x, write Z are the same. This is the first quadrant and this quadrant. So x or two pi minus x is the same. So we can replace alpha, which is x by two pi minus alpha. So we can say is it's beta equal to two pi minus alpha, which is the value which is accepted and which we said before, will help us to understand that z areas are equal. So let's get back to z with four. You will find here is that this part is beta minus K, beta minus pi minus 0. So this part is beta minus pi. Let's see, this part. This part is beta my employee minus alpha, okay, is this one is pi minus alpha. Minus alpha itself is equal to beta minus y. So it is equal to beta minus two pi k, which will give us beta minus pi. Ok? So this part is equal to beta minus pi as we proved now, which is similar to beta minus pi here. If you do this here, beta minus pi, same as before. Alpha plus pi to pi minus alpha plus pi will give us beta minus pi. So this area is equal to this area equal to this area equal to this area. Okay? That is a very good thing in case of pure inductive. And these four areas are equal to each other. So remember that here, that beta equal to pi minus alpha. Now we would like to see that now is that z beta, of course, would be greater than alpha. And the Denzin acres that firing the paths to avoid the interference or the overlap between the extinction angle and z, a firing angle of t2. So to avoid over lab z beta should be less than the firing angle of t2. You knows at here z sine wave, like this until Peter. Okay? Zen, starting from pi plus alpha, we will continue by K plus alpha. So z beta should be less than this one because in case of beta greater Here, for example, Zenzele will be interference and then this one would not work. And you will see what will happen. But for now, beta should be less Xunzi firing angle of t2. So beta this m pi plus alpha. And we said that beta two pi minus alpha. So Z alpha should be greater Zan, boy, over to, okay, this is the minimum value of alpha. Now what we'd like to understand what will happen if alpha equal to pi over two or alpha list Zambia i over two. 91. AC Chopper with L Load Part 2: What will happen if the firing angle is 90 degree or pi over two? In case one, we have something called Dizzy short pulse or a single pacifying and longer balls. So in case of Z, salt balls, sort tuples here means that for z, so I residue number one will give only one pulse at alpha and arg2 give it one pulse at pi plus alpha. So it takes only one pulse. Let us see what will happen. This one starting from alpha, okay? Then it goes through here until Peter. Beta, which we said that it is two pi minus alpha. K is a extinction angle. So here, this is beta, okay? And since alpha less than pi over two, which is our case, then z pulse here as this pulse for z psi restaurant tomb was adhere for example, before z beta. So T1 during this party one is working. So when T2 gets its a pulse, it will not work because T1 is still in conduction. So here T2 or the soil astronomers will never conduct. So you will find that alpha to two pi minus alpha. This waveform is the same as, as if we have an SOI restore, okay? As if we have a circuit like this. Here we have our supply VS. And we have here are Cyrus Ok circuit which we discussed in our first course. Like this. Only once I restore existing only workers into positive scikit. So it starts from alpha z into z extinction angle beta, and off until the next firing angle, which is two pi plus alpha and so on. So that what happens in the sort of ours, in the sort of pulse Z, sorry, restaurant number two will miss. It's a firing angle. And our AC shopper will operate as a rectifier. Xis I restore Tito a mess. It's a finding y because z T1 was still in conduction. So z, sorry, restore T2, the voltage across it is 0. Okay? Since this one acts as a short circuit during conduction until beta, so the voltage across it is 0. So it will not operate when it gets the firing angle as well. Remember that z psi restore to operate should have z firing angle and it should be forward biased. But since it is a short circuit, it will not be forward biased. So let us see the other case. If we have a train of pulses or along the pulse firing, okay? In guess of z train of pulses, you will find that it operates from alpha until here, before alpha plus pi k. Then this one has a longer pulse, so it gifts balls all of this times. Okay, like this pulse, pulse, pulse spots. So adds a beginning. Let's see the first society likes us. We have an alpha conducting until here, until this point, which is Peter. T alpha until this point which is z beta. And we said before that z alpha plus pi will be less than beta here, for example, is alpha plus pi. So T2 will miss, it's a firing pulse until, so T1 will keep, keep conducting until beta and beta z. So Restore T1 will be off because z voltage across it is negative. Now at this point, x0, so I restore T2 will have, it's a firing angle. We are giving it always firing angle. So what will happen? The voltage across this Aristotle would be equal to 0 and z supply our z, our motto will be equal to supply. So it will continue to conduct likes us until peta, okay? X1, x2 beta, which is beta plus pi. Similarly, Z Cyrus told T1 will miss its defining angle, but it has a train of pulses or a longer pulse gapes, giving zip files. So when T2 wants to be off, T1 will start to operate and continue my exists. So we'll find that in the end, Z output is a sinusoidal waveform. There is no control. So we say that i, ac, shopper will go out of control as each Osiris to operates immediately after the awesome at beta T1 is off and T2 will continue to operate. So it will go from here, sine wave, and T1 is off or T2 is off, T1 will start to conduct and so on. So each one conducts after the other. So we will have no control on our soil store. So in case of Z shorter pulse, we will have psi restoring and T1 will, it will act as a rectifier. And if we have a train of pulses, Zen it will act as no control or we would have no control. Z out is the same as the supply voltage. 92. Example on AC Chopper with L Load: So let's have an example on the EC show powder was an invalid in order to understand is the idea. So we have a single phase is issue percent loaded by an inductor of one milli Henry. The supply voltage is at 120 volt, at 50 hertz, determines the RMS root voltage. And the V average over a half period at alpha equal 60 degree and alpha equals a 120 degree. If the control signal of which servicer is single shorter pulse and train of pulses, and draw the load voltage and current waveforms in each case. So let us start one by one. So the first one is the shorter pulse for the alpha equals 16 and alpha equals 60 less than pi over two. So as we remember, and this m pi over two, then in case of a sorta bolts, our soils store will act as E, as E and rectifier, okay, from alpha to extinction angle beta, beta, which is two pi minus alpha k. So z v output as a root mean square is the square root of R one over two pi integration from alpha to beta for the function squared. But you will find is that this area equal to this area. This part is pi minus alpha. And this part is point minus alpha to two pi minus alpha minus pi, pi minus alpha. So this area equal to this area. So the integration from alpha to pi multiplied by two, giving us one over pi integration from alpha to pi root two VS sine omega T all squared. By solving this integration, we will get that the root mean square value is 107.3 volt. Now with the second requirement is that z a v output average of the half cycle. So we need scrutiny RMS and the average load voltage. So in order to find the z v output average over half period. So what does it mean? You, you remember that here we said that Z, we have alpha and pi plus alpha in order to have a to equal areas and the V output average equal to 0. So if we look at the average of this wave, will be equal to 0. So in order to find a value for the average, we take the half period. We assume that our waveform repeat itself each half period. So we see that the average is equal to one over T d here for z, half of the period is what is pi? Pi is the half of the period of the sine wave. The integration from alpha to pi, this is considered as the hub of the wave function will to VS sine omega t d omega t. So by solving this integration gives us the average of Z ofs, like it is a 148.55 volt. Now we would like to see for alpha equals a 120 degree, for alpha equal a 120 degree. Remember that this one, z alpha greater Zimbabwe were two souls. Cyrus Tom will operate normally or the EC shopper will act as an ISI shopper simply. So z v l as a root mean square is one over pi integration from alpha to beta K. So we talk X0 integration. We have one over T, which is the total period. And we took the integration of Zippo stuff cycle or Zao Windsor Forest, Cyrus tours conducting Ruto VS sine omega T all squared. And multiplied by two for one for z positive cycle and one for z negative cycles, they seem as we did in z or loot. So beta is two pi minus alpha equal to 140 degree. So this integration will give us root mean square value of a 137.57 volt V average of the half cycle is of course, one over pi. And remember that the integration from alpha to pi, okay? But someone will tell me we have here at two, the reason for this, why? Let's take it back. Okay? Now we have this waveform. Okay? Remember that we said the integration from alpha to beat. Okay? We need the root mean square. Root mean square is this part and this part, this part, and this part k, this four parts. So we took the integration from alpha to beta and multiplied it by two as the four areas equal. Let us, the first thing we did. Second thing we did is that we need the average. So we said that we will get from alpha to buy or Gazi as the average for half cycle, of course, from alpha to buy and multiply it by two. Why? Because we have this area and this area for the V output average 40.5 cycle. So we took the integration of this and multiplied it by two from alpha to pi. Let's get back to our example. From alpha to pi and multiplied it by two. So this will give us an average of 99.03 volt. And here is the waveform which we discussed in case of Z pulse or Z sort tuples or longer pulse, which will be the scene. Now in case of the train of pulses for Z alpha equals 60, remember that authors to will be out of control, right? We said that alpha equals cosine 90 degree. So in case of train pulse or a long bolts, Z is equal to the supply. So V output as a written squared, is equal to the supply root mean square, which is at a 120 volt. Now we need the average of the half cycle. So we said that it is out of control like this. So the average of the half cycle is from 0 to pi, from 0 to pi and one over pi root two v r sine omega t, giving us an average of the half cycle, a 198.06 volt. Now for the alpha equals a 120 degree, we said that for long and short is the, are the same values. It does not depend on this. Xj walk more muddy in the two cases. So this was our example on Z AC Shapiro is an antidote. You have to solve this by hand in order to understand where we did or how do we find this equations? 93. AC Chopper with RL Series Load: Now in this video, we would like to discuss a z. Ac Chopin was an RLC loop. So we have here our supply, which is an indices. And that we have here our AC sharper consisting of T1 and T2. And we have here, in this case, our load is RLE load resistance series with zinc inductance. So we need to understand this circuit. So adds up beginning, you will find that this circuit is similar to Z inductive circuit. Ok? You will find that z current or Z inductance causes the current tarballs, Susie, sorry, restores and conduct until angle beta. So let's see. We have here our input V supply V maximum sine omega t. And we have here our V output. So starting from alpha at this point x0, so I restore, we'll conduct a t1. Okay? So when this one conduct z, V supply is equal to V output Z I would voltage equal to supply voltage. So it increase until maximum or key or until 2Z sine omega t At this point. Then it continues until pi. And by So restore T1 should turn off. However, since this inductance was discharged during this period. So at z angle by when z inductance, when to turn off or when z. So Aristotle warns us to turn off, z inductance will produce a current in this direction. Keeping is at T1 conducting. So you will find that z waveform will continue here until angle beta or the extension angle. So you will find the years at 40 current will find it is a charging from alpha, okay? It is 0 at alpha. Xin increases until maximum value. Then the stores to discharge until Z extension angle beta. So charging Zen, discharging to keep z square restore G1 on. Similarly, it will happen in z negative half cycle and alpha plus pi. This I restore T2 will conduct. So the current will flow here, and it charges the inductance here from plus minus. So it will be charged until z, sorry, restored wants to store, turn them off. And by or at 2 here. So it will continue to conduct, okay, boys effect of the inductance, that the inductance is it, charge it. So it will produce a current in the same direction of original current, causing T2 to keep conducting. So it will continue until another angle, which is theta plus pi. So this is the waveform of the current. This is the waveform of savvy out and here are the signal or the pulses for our Cyrus tours. Now would like to analyze our circuit and we will understand why you would like to analyze in order to find Z extension angle beta and Z required the alpha. So at the beginning we will discuss during the conduction period when T1 conductors, when this one is on, it is a short circuit. So z V output is equal to the supply voltage by applying KVL. And here in this loop, then V output equal to the supply voltage. And supply voltage is equal to root two VS sine omega t, where v is the supply input voltage in Zao root mean square value or RMS. The effective value equal to the voltage across the output. And the voltage across the element is equal to r multiplied by Z current plus L d i over d t. So R i plus L d i over d t. So this is a differential equation and its solution is z is consisting of I steady-state plus the transient component. So for steady state, okay, if we pass Z transient, then the current is simply equal to z supply voltage over z. So it will be root two VS, which is the value of this apply as a maximum over Z total impedance which are present here, which is R squared plus omega l square, of course, under 0. Now z sine this current due to the presence of inductance, this current is lagging by an angle phi, where phi is tan minus one omega L over R. So we have the steady-state value equal to the maximum of the voltage over z. Sine omega t minus z lagging angle phi, or the power factor angle. So we have the city-state for Z transient as before, z current transient is equal to I naught e power negative t over tau. Okay? This is the transient component, okay? And we will see how it changes in a graph. So that transient component or the solution for this part is I naught ebonics t over ten, where I naught is the starting value. Tau is equal to L over R and omega tau is equal to ten. Following where if we multiply omega Boys Town, so we will have omega L over R, which is the same as here. Then phi. So from this equation, ten phi is omega L over R. So omega tau is equal to ten pi. Now, by boating disease, this two components in our equation, we will have root 2V S over z sine omega t minus phi. Plus I naught IPO negative t over tau, which is omega t over omega tau, which is tan phi. This equation is this one, and this equation is this one, the eastern state. So we need to find I naught. I naught at the beginning of the current, as we remember from our here, z beginning of Z current is at alpha. So add alpha z current is equal to 0 at the current by substituting omega t equal alpha, it will be equal to 0. So I alpha equal 0. So 0 will be equal to root two VS over z sine alpha minus phi. And omega t here is replaced by alpha. So it will be I naught E bar negative alpha of r term phi, which is the same as omega tau. So from this we have all of this. In this equation is known. The only variable is the node. So I naught from this equation is equal to negative two and negative root 2V S over Zed. So an alpha minus phi e power alpha over two m phi. So our final equation is taking I naught here and substituting here in this equation. So we'll have i as a function of omega T Ruto VS over z as a common factor, sine omega t minus phi. And we have this part in negative sine alpha minus phi e power alpha t over ten phi. And here IPO negative omega t over ten. So it will be e power here. Since it's common between them. And taking negative as a common factor, we will have omega t minus alpha. And there is a negative sign here. So this represents the equation of our current buys a usage of this equation, we can find the extinction angle beta hell oil substituting at omega T equal beta z current will be equal to 0. This is z boundary condition, z final condition. So to get beta at i or omega equal beta z current equal to 0. So 0 equal to z equation, but add omega t with peta. So everything here is known except the CPD soap by doing as simplification and very simple simplification, you will find that sine beta minus phi equals sine alpha minus phi e power negative beta minus alpha over two m phi. So this equation helps us to find z extinction angle beta. Now, you will find that for proper operation, you will find that here alpha. Should be, Lizzie should be greater than phi, alpha should be greater than phi and at the same time less than pi. Okay? You know that ad by Z firing angle will start is Israel restore and at the same time the surface two will be turned off by is z, where z is thyristor will not conduct because at pi z, sorry, restore will be turned off and after a millisecond or a very small time, it will be turned off due to the negative half cycle. So pi here, not accepted, but it is z limit of alpha, z highest value of alpha. So alpha should be between phi and between by y because we would like cistern to be greater than 0. So it will be negative here. Negative and positive BOD. So this is the condition of our circuit. In case of Z a pure inductive load, as you remember, abuse inductive load having phi equal to 90 degree. So as we remember that we said that alpha should be greater than boys over two in case of Z, pure inductive load. Okay? Here, pi over two here represents a z phi or the power factor, power factor angle is 90 degree in case of a pure inductive load. In case of a resistive load, alpha should be greater than 0. For enzyme, resistive load is equal to 0 and phi in case of Z, pure inductive route is pi over two. So this is a general formula for Z alpha. Now we would like to understand why short Z alpha greater than phi and what will happen if it is less than or equal to phi? The first case where alpha is some y. So z sine alpha minus foil, which is this part, will be an active part. So this part is negative. And with another negative, this one is alpha minus phi and alpha less than foil, so it will give us a negative value and we have another negative here. So it will be, so I am beta minus phi, the Lus, this exponential form. So if we draw the function, you will find that we have Z transient current is abosDF, okay? Since it is narrative with another negative social people step. So the first one, this one is representing this bolted one, representing the, our current as a steady-state value, lagging by angle phi. So if you get back this one, let's get back here, here, here. This one. Ok, we have Ruto VS over z sine omega t minus phi is this one representing is the steady-state value. And this one with this one representing the transient devalue and exponential decaying. So in this case, this one is the steady-state with a lagging angle phi. And we have here our exponential decaying Z transient part. So in the case of alpha resume foil, Zenzele transient will be positive value, then the total current will be some measure of steady-state velocity z transient. So we'll find that the total current will be greater than the other curve. You will find here. This curve shifted aboard, ok, from alpha to beta. So we'll find that here from alpha to beta. This one is beta minus phi, beta minus alpha. This one is the beginning of the steady-state, which is fi. This one is end of the city-state, which is pi plus phi. Where did we get this simply, how did I know this point? If we draw a sine wave like this, this point is 0. This point is pi. So if I shifted this wave starting from phi y exists, then this point will be foil, and this new intersection point will be pi plus phi. So this point is phi plus phi. So we'll find that this part is equal to by the difference between sports. And from here to here is beta minus alpha. Okay? So we'll find that the conduction angle gamma, which is beta minus alpha, is greater than pi. Beta minus alpha greater than pi. So what does it mean? It means that beta Z extension angle is greater than pi plus alpha. So what does it mean? It means that the thyristor T1 will be turned it off after Z Cyrus two number to take it's a firing angle. So T2 will mess. It's defining NK. So this is similar as is the case where alpha less than pi over two in case of Z B, your inductive load. So it's a similar here. If z alpha less than fi, then beta or the extension angle will be greater than pi plus alpha. So in this case of a short pulse, you will find that z. So I restore T1 will be turned on and T2 will mess itself firing angle. So that easy short power will act as a rectifier, similar as the pure inductive load case. In case of the long pulse or train of pulses Z AC super, we'll go out of control, similar as the case of z, your inductive load when alpha less Zen while you were two. And z long pulse case. The second case here, when we have our alpha equal to phi z critical condition. So at alpha equal foil, you will find that sine alpha minus phi, which is representing is the transient component, will be equal to 0 since alpha equal to phi, so alpha minus phi equal to 0. So that transient component will be equal to 0. So the steady state waveform is the same as the transient waveform. So in this case, beta minus alpha will be equal to pi. Let's say get back in this waveform Windsor electrons and does not exist, then the total current is similar as the steady-state current, this current. So z conduction videoed gamma will be from pi to pi plus phi, which is similar as pi. So in this case, gamma will be equal to y and the gamma is beta minus alpha equal pi. So beta equal pi plus alpha. So as soon as z, sorry, restore T1 will be turned off, D2 will be turned on. So we'll have not control case similar as zinc case of Z, pure inductive load when alpha was critically equal to pi over two. So the ISI shopper will be out of control as T2 will be on or will be turned on as soon as T1 is off. Like this. Okay. This is the steady state which assess him as a z door total current as the transient equal to 0. Gets number three, which is the normal case when alpha is greater than phi. So we'll have i transient equal negative. So in this case, the extension angle or beta minus alpha, z gamma will be less than pi. Okay, listen the first curve. So in this case, our issue sub-par will work normally and will have 0 cm one forms. As we said before. Why? Because in this case, the extension angle beta will be less than the firing angle of T2. So T1 will stop and after a small time, t2 will be on normally. So this was our ISI. Schapiro is an RLC series loot. Let's have an example on it to understand it more. 94. Example on AC Chopper with Rl Series Load: Let's have an example on AC. Schapiro is an RLC series load. We have a single phase AC Shofar feeds n r eally series loot that supply voltage 220 volt as an RMS, 50 orders as a frequency. The firing angle is 75 degree, the extinction angle beta is 205 degree. First one draw the output voltage waveform and the calculate Z RMS value of the output voltage. And the average value over 1.5 cycle starting from x0 0 of the sine wave. So let's start step-by-step Z. First thing is that we have the firing angle alpha 75 and extinction angle 205 degree. So the first step I would like to see if alpha is less than z, phi or not. So we have V supply equal to a 120 volt as an RMS. Alpha equals 75 degree and beta or the extension angle equal to 105. So you will find that in this case, the waveform we will be drawing nominee, which is the case where alpha is greater than phi. Ok? The reason for this is that you will find that beta minus alpha, beta minus alpha here, 205 minus 75. You will find it is less than pi. Ok? So it means that in this case, since beta minus alpha less than pi, then it means that we are in the third case, where z beta will be less than Z next to firing angle, which is alpha plus pi. Alpha to pi is equal to 180 plus 75 is equal to 11103600080, which is z by m plus 75 will give us a 255, which is an aqueous defining angular of t2. This firing angle is greater than z m beta. So y plus alpha greater than beta. So ours, Aristotle will walk normally from alpha to beta, then from pi plus alpha to beta plus pi and so on. And here, beta minus pi, similar as Z pure inductive load case. So the first requirement is the draw the output waveform. The second requirement and creates the RMS of z, our voltage and the average value over 1.5 cycle. So let's define the average. The average that we would average for half cycle is starting from here until pi, since it's saying half cycle. So it will be one over pi. Since it is half cycle. Integration from 0 to pi for the function will do VS sine omega t d omega t. This integration where z function exists is from 0 to pi. So our integration will be from 0 to beta minus pi, 0 to beta minus pi. And again from alpha to the extension angle beta or two z pi, okay, pardon here two pi from alpha to pi. So from 0 to beta minus pi and from alpha to pi. So we have the extension angle, beta given 205. Why is a 180 degree? Alpha is given as 75 degree. And we have here R2D2 VS, VS is z supply as a root mean square 220 and invoicing is given. So by solving is a simple integration, we will get zeta V average of 40.5 cycle from 0 to pi z. Voltage here average is a 133.9457 volt. Four z root-mean-squared. Find the RMS of the output voltage, would like to find z root mean square for this function. So the square root one over t, t is the total period which is two pi multiplied by. You will find that here. For this function, we can integrate from 0 to two pi, okay? Or we can integrate from alpha. For example, two alpha plus two pi two this point. Ok? So from here to here give us a complete cycle. From here to here gives us another complete cycle. So you will use this one from alpha two beta and from pi plus alpha to beta plus pi y. Because this part from here to here give us a complete cycle. So it is easier for me to use this area and multiply it by two. Why? Because this area equal to this area from pi plus alpha to beta plus pi is similar to alpha to beat. So we said that one over T multiplied by 22 areas are equal. This area and this one too, which is represents the tools are positive and the negative borders multiplied by alpha two Peter. For the function Ruto VS sine omega T all squared d omega t. So alpha 75 beta is 205. By solving this integration, we will have a 181.401 volt as RMS value. Rms value is representing the equivalent or the effective value, or the equivalent of a DC supply of the same value. Now we will notice something again, again, again that here two multiplied by this from alpha to beta k we talk, we would like to find z. Now RMS from here to here. So we said that this area is equal to this area. So multiple which got one integration and multiplied it by two. So this was our example on the AC super within our elicit is loot. 95. AC Chopper with RL Parallel Load: In this video, we would like to discuss a z AC sharper but with an R L parallel loot. So in this case we have our supply as before, and we have here our AC sharpen, and we have a resistance parallel to an inductance L. Now in this case, we would like to draw the wave form and find the equation of Z cart. So first we will start to Boise firing angle alpha 2s extension angle beta. So at the beginning we fire our alpha from this point and we will understand why at alpha, T1 is conducting. So the voltage v output, which is the voltage across the inductance, or the voltage across the resistance is equal to V supply. So at starting from alpha, our voltage will be V supply Xin. It will keep being a supply until T1 is off at boys. But however, the inductance is the charge it, so z inductance contain a geologist so it will discharge through Zt1 Cavic in conducting until beta. Ok, the same case as AC shopper was an electrode and RLC series loot. So starting from alpha to the extension angle beta, where t1 will be off. Okay? So adds a beginning because this one is a charged pi plus minus. Then during discharging plus minus, reducing a current through T1 to keep it conducting. But our supply is producing a current in the opposite direction until a point where the inductance cannot keep z T1 on. So in this case at beta. So what happens after this point? In this case, after this point when T1 is off or starting from beta, z, inductances still have an energy stored. So in this case it will restore it discharging serosa resistance R. So add this part. It is discharging resources apply and through T1. Starting from beta, it's still have, still have stored energy, but it cannot. Boss throws a sublime. It chose Beta for itself, serosa resistance like this. So it will be soldiered through the resistance R. So this part is acting as if it is a transient RLC parallel circuit, where z voltage decays exponentially on till firing angle of T2. So in this case, we will draw our voltage decaying. Ok, this is the V out, okay? Starting from here, from the end of Zt1 conduction. At this point, you will find that the voltage here is similar to the voltage across the inductance. And the voltage across the inductance is equal to L d over d t. So the current is decaying, so the voltage is decreasing. So starting from here, it will decrease until pi plus alpha. Okay? The firing angle off T2 and firing angle of T2, z V output will increase back to resupply. So at pi plus alpha it will increase back to V supply. Okay? Because at pi plus alpha, T2 is conducting, so V output is equal to the supply voltage like this. But this part is representing z exponential decay when the inductance is discharging through the resistance R. And our third strip at alpha plus a 180 degree in the third step, the second Cyrus store T2, start receiving. It's a firing angle causing the voltage to increase again until beta plus a 180 degree. So during this period, V out equal to V supply. So starting from here, del pi plus beta, where T2 is conducting. So during this part, similar as this part, Z, sorry, restore T2 is conducting. And at B pi plus beta, it will form another exponential part here. Okay? Similar as here. Since T2 is off and T1 is off. So Walter starts to decay from beta plus pi until alpha plus a 180, likes this, or until the firing angle of T1. So it will be like this. Okay? So dividing our conduction here into two periods. The first from pi_plus alpha to pi plus beta, where T1 is a, T2 is conducting. Then from here to here Z RL parallel transient circuit. Whereas the voltage will decay. Knowing is the same here. It will be attributed to add beta minus pi. So here, as you see here, this part, at alpha plus pi to beta plus pi, it will be minus pi for alpha. And for this point it will be P2 minus pi. Since we have here beta, therefore it will be beta minus pi. Okay? And this one is pi plus alpha. Therefore, this one is alpha, and this point is alpha plus two pi. To be more specific. Okay, not pi, alpha plus two pi, and this one is beta plus pi. So subtracting two pi, it will be beta minus pi and the alpha will be alpha only. Okay? So this is our final waveform in case of Z is issue par with an R L parallel loot. So for the outward current, it will only exist when T1 or T2 is conducting. And between them. When T1 and T2 is not conducting, we have a current circulating in Z RL loot. So here it will conduct from alpha to beta and from pi plus alpha to pi plus beta here, conducting T2, here, conducting T1, here conducting T2. Okay? Same as the case of the RLC series Z L, pure inductive load condition. All of this are similar to each other. The difference is here that the waveform has an extra part, Z extra bar to when z. We have our L upon z are forming and exponential decaying voltage. So for the resistive current, it will have the same waveform of the output voltage. Okay, so I output will find you. Let's get back to the circuit. We have here, i n, i l, and you will have your IR. I would withdraw it as alpha to beta, k from alpha to beat. For Z current through the inductance and the current through the resistance would like to find the waveforms. So getting back to this point, you will find here that z resistive load or the resistive current is equal to V over R, Okay? Zr in and in the same waveform like each other, okay? Z resistance, the resistive current and the output voltage have the same waveform. So that the output is like this. And this one is I would. So V output is similar to IR, but Ir is V out divided by R. So it will have this waveform, but divided by r, which is a constant. So it will be like this. Okay? This one is same as V, And so this representing the current through the resistance. This one representing z or v out, and this one representing Z total current. Now we would like to draw the inductive current, which is the most difficult one to m. So remember that, that I out is equal to IL IR. This is during the conduction period. So at alpha and beta. The current is equal to 0, okay? Z1 and the beginning of the charging and at the end of the conduction of t1 and t0, t1, of course, or of course attitude. So here we have the current I output equal to 0. So IL will be equal to negative IR. So we have here Z waveform of IR at beta minus pi, we will have IL equal negative IR. So we will take this part. And would it here as a negative value, okay? Because IN equaled negative IR add beta minus alpha beta y plus alpha and y plus beta z boundary over the conduction of T1 and T2. So this one is a negative. This part, which is a small part here, is also a negative here. This board at beta here have a negative. So it will be here as a positive. This part here will be similar, but with a positive value. This one is negative. So it will be here at Boston. This one is negative. Here will be negative, here will be positive, here, positive will be negative, and so on. So you will find that here we talked z points which are representing ir equal negative I am okay at z boundary conditions. Now another sink by connecting this XA position point that we can get to Z inductive current. So you will find that here, this is z ir. And we have here x, x and this x and this x and x and so on. So we can connect a line exists. Okay, likes us. And you'll see that here it is. Zinc going up like this. So we'll find this waveform is like this waveform, okay? Connecting all of the points together. So at here and this point you will find here is that we have our little beak, Xin decane, this peak representing this part, representing z charging them from here to here, this a charging. Okay? So we have here are charging, Zen discharging. Ok, similar here too, charging discharging and so on. So this is how we found z current, IR, IL, and Iout. Now we would like to analyze our circuit and get the equation of Z current in order to find the boundary conditions. So we will first discuss the current equation during the conduction. During the conduction T1 is on. So it will be a short circuit, okay? Assuming is positive Bart or add the firing angle alpha. So we'll find that V supply is equal to the voltage across the resistance. Equal to voltage across the inductance and z i dot equal to iR plus I l. So during conduction period from alpha to beta or the extinction angle beta, i output equal to IL, IR Z output current. So z is equal to the resistance of current plus the inductive current, of course, uh, from KCL, Kirchoff's current law. For our node, you know that the V output here is equal to IR multiplied by the resistance, okay? Ir multiplied by the resistance. So this voltage here is equal to V supply. So IR onto blood buys a resistance equal to VS. So IR multiplied by resistance equal VS. And this apply current itself is R2D2 VS sine omega t. So I r z current through the resistance R is equal to root 2V sine omega t over r. So this representing z value of the current through the resistance R. And you'll find that the current here is in-phase with the voltage. There is no minus phi here, since this is a pure resistive load and phi equal to 0. For the node here, the voltage here across the Z load or the inductance equal to VS. So VS, which is root 2V as sine omega t, is equal to the voltage across the inductance. And you know that the voltage across the inductance is equal to L d over d t L d I L, which is the current of z inductance over d t. So you will find by getting this T DT here and integrating z from 0 to i l in the value of z inductive current. And here we have root 2V sine omega t or sine omega t. And we have d t, but we multiplied it by omega, since here it is function of omega. We multiply it by omega and divide by omega. And here get goat here and divided here binds the function. So we have the integration from 0 to L. And we have the integration here from alpha to omega t at alpha Z load current equal to 0. And at omega t, at any omega t, We would like to find the value of I l. So I l as a function of omega t is simply equal to root 2V S over omega l cosine alpha minus cosine omega t. Plus I naught i is the initial value of current. The reason for this is that here should not be equal to 0, okay? At alpha, there is a current. Okay, let's take it back. For, look at IL. For example, here at alpha. At alpha we have a value of i n, And at y plus alpha we have also a current. So we will have to boat here I naught. In order to be more specific, we can say the integration from I naught the initial value of current to any value of z i l I M, which is inductive current. So it will be i omega t minus i naught. Okay? Equal to root 2V S over omega l cosine alpha minus cosine omega t. So we dogs this and put it in the other side. So it will be plus I naught. Ok? So we'd like to find the initial value of I naught in this equation. So for xenon conduction videos from beta minus pi to alpha, okay? We have i naught equal to 0, z output current equals 0, and I eloquent negative ir beta minus pi to alpha to make you remember, we had it likes us at beta minus pi is indicating the two alpha. Okay? This is the voltage from beta minus pi to alpha z are in the transient circuit. In the our electrons Ian circuit, we have I, L equal negative IR and IR would equal 0. So IL is equal to a negative omega t over omega tail. As human minds that this is a transient circuit and it is decaying and decaying current. The current is decaying or discharging through this resistance R. So it should take this form in the RA Le Baron transient circuit. So to find z constant a, we know that at omega t equal to theta minus pi, we have IL equal negative IR. Okay, so where did we get us? Let's see, get back. At beta minus pi. This is point, ok? You will find that this component is equal to this part, right? Since IR equals negative i n, So Z value of z ir here equal to negative IR. So IR itself, IR itself is equal to VS over R root 2V as sine over our write sine omega t over r. This is the value of z resistive current. But we would like to replace omega t by beta minus pi. So we replace this u omega t year by beta minus pi. So this is the value of Z current at beta minus pi, z resistive current. So ZIM or is inductive current will be equal to negative IR. So I l equal negative root two VS sine beta minus pi. Let's delete all of this. So IN connective route to VS sine beta minus pi over r. So this part is equal to IS exponential here at beta minus pi. So z constant air will be equal to negative root two VS sine beta minus point over r e power beta minus Pi over omega. Tell. You will find here in this office you are in liberal loot. There is too much equation. Okay? Is there are too much equations, so don't worry, it is normal. You have to read this again in order to understand it well. So we have year ZA Zeno substitute resist value here. So i l as a function of omega t is equal to negative root 2V as sine beta minus pi over Are you born negative omega T minus beta minus pi over omega l. This in case of discharging period, in this case of the RL parallel transient circuit. So we would like to find the initial current i naught, which is was in the first equation. So if we get back, so we'd like to find the current I naught here. Okay? This current. So we have to substitute with alpha in the equation and phi l here and the equation of I L here. So at omega t equal alpha i l equal i naught. So in Z RL, the barrier transient inductive current, we saw system by omega t equal alpha, saw in that previous equation, this one, we replaced each omega t I alpha, which is this one. So this representing the Z value of I naught. But the problem is that we have unknown, we have beta is not known. We have every single bought only beat. So how we can get beta? We can get beta or Z extension angle at omega T equal beta I naught equals 0. Okay? Z Albert current itself is equal to 0 and extension angle beta. So rosy IN equal negative IR or I l plus ir equal to 0. So I n. In case of a z, this part, which is z conduction period, okay, is the start of the conduction period, Ruto VS over omega l cosine alpha minus cosine beta. This is representing z equation of IN. This represents a z equation of IR in case of from alpha to beta, as we remember. And we replaced here each omega t by beta. So this equation, we can get the value of z beta. I naught here is this value. Ok, we talk this part here. So we'll have a large equation by solving it, we can get VDD. Also remember that in parallel R L root, then phi is r over omega n. But in case of series tan phi is omega L over R. So this is very important in case of x0 cosine phi or defining or z power factor. Remember that tan phi here in series omega L over R, but in Barron is R over omega_0 is different because here we got Z 4pi, the currents. Here we got the Z by Z voltage phasors. Ok. So z bar, z. Purpose of all of this is that I would like to show you how to get c beta if you don't know anything about dessert. But in z example, I am going to give, I'm not going to let you find beat. But if you would like more complex examples you will find in z resources in our course, I am going to give you more example. You can sell, you can solve it by yourself. 96. Example on AC Chopper with RL Parallel Load: So let's have an example on Z AC Shapiro is an RLC parallel load. We have on our illiberal load resistance of five on omega equal ten on alpha equal ten plus Phi Beta equal to a 170 degree v supply equal to a 120 RMS. Draw the output voltage. Find the output average per half cycle and Z alpha. So how we can get this simply, we will start from here to here. The first thing, which is C alpha alpha is equal to ten plus phi then plus phi. So would like to find z phi, as we remember that we have here R L parallel loot. So z phi is equal to ten minus one over omega L. In case of parallel load. In case of series load, we have omega L over R. So the resistance is five ohms. And omega l is ten on. So damn minus1 five of our thing give us this value or function. This angle representing z phi. So alpha will be ten plus phi. So it will give assert 60 degree point 5-6. So this is the alpha. Now the V output average per half cycle, or Z I would voltage wave form will be like this, okay? Same as we did before. Z bar, z conduction, Xin, z bar z exponential decay as in continuously and so on. The conduction of T2 then decaying, Zim conduction and so on. So VR would average per half cycle is the losses. And so V out would average is one over pi. Integration of the function from 0 to pi, okay? From 0 to pi. So this is really important and you have to think about it here. Integration from 0 to beta minus pi. And another integration from alpha to beta. So here, from 0 to beta minus pi and from alpha to pi, okay, for the function rule to VS sine omega t and here Ruto VS sine omega t. This part from here to here. And from here to here. Now we have two integration, z only partner meaning is from beta minus pi to alpha, this decaying apart. So how we can represent this part simply, you know that the function root 2m VS sine omega t, right? But we are starting from a value here. I would like to find it saw a replace omega t by beta minus pi, beta minus pi. So this representing the Z initial value here at beta minus pi. So we have also an exponential. So we multiply this value, which is the starting of the exponential, exponential negative, we should say omega t. But here we are starting from what? We are starting from beta minus pi. So it should be omega T minus beta minus pi. So what does it mean? It means that it is an exponential decaying with a starting value root two VS sine beta minus pi. This starting value, and decaying exponentially as a function of omega T, uh, but it is shifted by beta minus pi. Okay? This will point is beta minus boy. So as if the exponential not starting from here, but starting from this point, we shifted the exponential. So we'll find that true to VS sine beta minus pi EPA or negative omega T minus beta minus pi over z omega l, or tail which is L over R, which is omega L over R. Okay? Remember that this part is not then phi, okay? It is one over ten file. So for starting from beta minus pi to alpha. So by solving this integration, you can find that z value of the voltage output average for half cycle is a 196.57 volt. So this was an example on the RL parallel load. The most important thing is that you will get Z of Z, principle of operation of Z R, L parallel load. And if you have any questions, don't hesitate to ask me. 97. AC Chopper with Pure Capacitive Load: In this video, we would like to discuss as CSI is, you show partner, but with App your capacitive load. So we have here our supply and that we have here our ISI shopper and our loop now is capacitive load. So a capacitive load of course, means that we have a capacitance C. In this case, we have two cases. If alpha is m pi over two, what will happen? So simply adds a beginning. We will start at this point, okay, at alpha. So add alpha, z at t1 will be charging or it will be on. So it will be a short circuit. Therefore, the voltage across the capacitance is similar to the voltage across the supply. And in this case, z supply is providing current, doozy capacitance and the charging gate. So starting from here from alpha, it is equal to V out equal to V supply. So it will be equal V supply equal V supply on ten Z maximum point here. At this point z V output is equal to VS. OK. And our capacitance is already charged by VS. OK. It a charge it from alpha until zap voltage across the capacitance V s. So in this case, z capacitance is acting as if it is a supply VS with a positive minus. And at the same time, at this point, x0 supply when I started to decrease. Okay, the star to decrease. So what happened in this case? In this case, let's see, let's delete this. So let us see the voltage across Z. So Restore t1. So T1 is plus minus. So T1 is on when z voltage here is greater than this one and at the same time giving it firing angle. So Z problem here is that T1 will be off y because z positive will be root two. Vs sine omega t and z negative potential will be 0 voltage across the capacitance in this case, which is V supply. So you will find that z negative terminal have a higher potential Zai Zheng Zhi positive terminal. The negative which is V supply, since our capacitance at this point will be charged by VS and Xr voltage here will be root two v r sine omega t. So it will be at this part. Which is lower Zan VS. So what happens in this case, t1 will be off. So this EC sharper will be of SO2 will be open circuit. The voltage across the capacitance in this case, which is V output, will remain equal to VS. So at this point, which is the maximum charging of Zika Boston's, we'll keep as if it is VS. OK, until y plus alpha at pi plus alpha, T2 will conduct, okay? So T2 will conduct when this plus minus greater than 0. So z positive terminal is connected to V supply and z negative terminal connected to the V S Ruto via sine omega t. And in z negative cycle, in this part, Z supply voltage is in this direction to plus minus. So this providing current in this direction and the combustor and supervising current in the same direction. So c and this apply our boss helping T2, T2 conduct with the firing angle z. So rest of T2 will conduct. So z voltage V output will be equal to v s. So in this point, pi plus alpha, it will change from VS to roto VS sine omega t equal to supply. Then it will keep it charging in the negative cycle until maximum negative VS and remain constant until alpha plus two pi, which is the same as here. This part is similar to this port. So add alpha, it will conduct again going to z sine wave. So z capacitor, the capacitor C is the charge advises her ploy until VC equal v maximum. Ok, which is root two VS is NC combustor will stop conduction. Or to be more specific, not z capacitor or it will not be a be conducted Boise current. G1 will be off. So z z combustor will cause the AC part to be off. And same procedure is happening in negative half cycle. So adds a beginning at alpha two z maximum z capacitance is charging until maximum value. Then it will make T1 off. So V output will be keep the as rho two V S or Z maximum value until pi plus alpha it will be, switch it to the opposite direction. This happens when alpha is less than pi over two. So what happens if alpha greater than pi over two? So simply, what will happen at alpha? Z capacitance will have this value, okay, R2D2. Vs sine alpha, okay? So when T1 is conducting at alpha, then z voltage V output will be equal to supply, which is root two at this point through to VS sine z angle, which is alpha. Then you will find that Z-Wave itself is decaying. The supply is lower. Zan wrote to VS sine alpha, which is the value, which is the charge it Susie capacitance. So this part is rho two VS sine alpha. And this one is root two VS sine omega t. This is plus minus. So z positive terminal at this point will be lower than the negative terminal which is connected to Zika Boston's roto VS sine omega t, sine alpha. So T1 will be off. So V output will be keep the as a root 2V sine alpha. So it will be root two VS sine alpha like this until alpha plus pi and alpha plus pi, it will be reverse it. So z capacitance will be charged by negative erato sine alpha or roto VS sine alpha. So also again, z voltage will start to increase. So t2 will be off. So it will be kept as it is. So I will find that here it is a charging. So sine wave is in keep the constant. But at alpha greater than y over two, it will be kept as a square wave as you see it. So Zika Buster is charged, buys a supply to a value according to our alpha rho two VS sine alpha, at which is a voltage of z capacitance will be greater Zan V supply. Starting from here, all of this. So z AC shopper will stop conduction because it will be reversed biased. In this case, z value of Z voltage of capacitance is always greater than supply voltage. So we'd like to understand the V output average in each case for half cycle. So v out average for z half cycle for Z keystone per 11 over pi integration from 0 to pi. So instead of don't exists, you will find something here. If we integrate from 0 to pi to this point, we will not have z maximum average for this hiker. Ok, in this case, when we are talking about a Zi Bu capacitive load, when we wanted to find the V output average of 4.5 cycle, we will start the forum alpha, two alpha plus pi from alpha to alpha plus pi. So from alpha to this point, which is pi over two. We will have this sine wave from alpha to pi over two is our sine wave route to VS sine omega t d omega t plus integration from here, from pi over two to pi plus alpha pi over two to pi plus alpha is equal to this waveform. This was form is DC and its value is a roto via this route to VS, from pi over two to pi plus alpha. So this integration in case of z average half cycle, we'll integrate from alpha to pi plus alpha y in order to get Z. Maximum average. Of course is a VR would average for a complete cycle equal to 0 in case of this pipe or any other type of Z. At Bjork, pure inductive load or z are in series or parallel. All of this AR and VR would averages equal sends alpha to pi plus alpha or z and conduction angles. Now, in the second case here. And this one. If we would like to find the V output average for half cycle again, Zen we'll integrate from alpha to pi plus alpha for our value. Look at the function from alpha to pi plus alpha. This value is Ruto VS sine alpha is this value DC. So we'll integrate from alpha to pi plus alpha rho to VS sine alpha value of the DC component d omega t. So this will give us an average of root two VS sine alpha. Now if we would like to find the RMS, in this case z RMS voltage, soul finds that this part is negative. So when we square it, it will be like this. And when we square this negative part, it will be like this. So this is square wave as a root mean square value is equivalent to a DC supply or fruit to VS sine alpha. So by squaring z function, you can find that it will be equal to root two via sign-on for all square, which is V out square as r square Zynga V RMS is the square root of this, which is root two VS sine alpha. Or you can simply, if you don't understand this, you can integrate from, you can say z V RMS is equal to the square root of the function one over T, which is two pi, two pi integration from 40 function, from 0 to alpha. From 0 to alpha. 40 functioning, which is square of course, negative root two VS sine alpha all square. So it will be root two VS sine alpha square. Ok? So this representing disease part for this part from alpha to pi plus integration from alpha 24 to 2 k, we are integrating for a complete cycle. For the same function would do we assign alpha? This is similar here. So by integrating this, it will give us as if it was a dc component of Ruto VS sine alpha. So this was our example on Z EC shopper with a pure capacitive load. We would like to have a numerical example on this. 98. Example on AC Chopper with Pure Capacitive Load: So let's have an example on CSI is you borrow with a pure capacitive load. So we have a single phase AC Chopin feeds are pure capacitive load for alpha 90 degree. The rosy output waveform and calculates the average of 0 and voltage over 1.5 cycle. And the RMS value, given that the supply voltage is 220 RMS and 50 hertz. So simply at alpha starting from 90 degree, it will be a square wave. As we said before, since at alpha it will be charged by Z, maximum value here, then the supply voltage will start to decrease. Saw Z output will be a square wave. Y exists at alpha equal 90 degree, will be charged by root two V s. So it will be kept constant until alpha plus pi Xin again in the negative half cycle and so on. So this z a v output as a waveform. Now we would like to find the average value of zeros voltage over 1.5 cycle. So we said that in this case, we will have one over pi integration from alpha to alpha plus point in order to get Z maximum average value. So from alpha to alpha plus pi, we have Z value of Z DC component is root two VS sine alpha. And our alpha is 90 degree. So sine 90 is one is this one. So one over pi root two VS integration from alpha to alpha plus pi is, as our subtraction of Zim senses, this one is a constant. So it will give us point. So this will give us finally boy will go as pi. Finally, Ruto VS, which is root 2220 is 711 volt. So this as an average, root to VS is z, same as Z row RMS, as you will see now. Four is the RMS value, root one over two pi integration from alpha to alpha plus pi, which are presented pie or half cycle, okay? Which is similar to alpha plus pi, two alpha plus two pi. So here we integrated at two-point from alpha to alpha plus two pi complete cycle. Or you can integrate from 0 to alpha to two pi. From 0 to two pi as you would like. Okay, at two pi is here. So instead of integrating from 0 to two pi, and the integration will be in this case, 0 to alpha, Xin from alpha to alpha plus pi Zim from alpha plus pi to two pi. We integrated from alpha to alpha plus pi Zen from alpha plus 0, alpha plus 2y. And this area squaring it as similar as this area. So we said that integration from alpha to alpha plus pi root two VS sine alpha, which is this value of the DC component, or a square from alpha to alpha plus pi and multiplied it by two. Since we have here this part and this part. So this will give us similar here, root two VS sine alpha, which is timeline t equal one. So it will be root two VS, which is similar to this part. So simply the pure capacitive load can help us to produce a square wave. Okay, so we can use up your capacitive load by controlling gate. We can produce an output n in the form of a square wave instead of a sine wave. 99. AC Chopper Loaded by Heavy Rectifier: Now let's discuss is loaded by a heavy rectifier. So assumes that we have this circuit. As you see here. We have a resistive load connected to bridge rectifier. Then this bridge rectifier is connected to a transformer, a step-down transformer. And we have here our AC sharper. So what happens here? For heavy current DC application? What does it mean? It means that if we have here and would absorbing a heavy current, a very large current. So the price of societies stores like this one in Z AC shopper requires very high cost or very expensive compared to Z diodes in zee bridge rectifier. So the restore, if we connected this AC bridge to this load and this load ticks lots of current. Therefore, Z see shoppers required or z psi restore Zi required is very high in case of cost with respect tos or diet. So what do we do? We simply take Z shopping or Z sc shopper and connected it to the high voltage side of the transformer. And the Z lawful society is connected to the DC bridge or zee bridge rectifier or uncontrolled bridge rectifier, Zen two zeros. So as you remember in the transformer, V1 over V2 equal to N1 over N2 z number of turns of z primary over z number of turns of the Secondly equal to R two over R one. So here in this application, you will find that this is the high voltage side, which means it has large number of Turner's. So V01 is having a higher value is n V2, V2 representing Zillow voltage aside. So V1 to V2 is a step-down transformer. The current here is i2 and current here I1. So from this Turner's ritual finds that v1 is greater than V2. So i1 is less than T2. So z heavy current here are absorbed by the load, which is represented in case of i2. I1 is lower current by the effect of z transformer. So Zach EC sharper is connected to the high voltage side, which means a z low value of current. So in this case we can take a, sorry, restore with low current rating. And instead of connecting resist iso Schober to this part. So the advantage of this method is using Guillot current rating. So I restore as the current is here, lower Xunzi current here. The current in high-voltage site is lower than the current in low voltage side. So if we look at the waveform, you will find that the AC shopper here takes z supply voltage, which is V supply. Supply is represented by a higher value here. Then it chops Z voltage. V1 here will be the output of the AC shopper. Starting from alpha, it will conduct a from alpha to pi. Then it will conduct from pi plus alpha to two pi. Then it repeats itself. The effect of a C sharp are loaded by a resistive load, okay? Our load here is a resistance. So conducting from alpha to pi, then from pi_plus alpha to two pi. Now this we form the upper one representing the v1. And this one is also V1. Zen, this V1 will be step-down wise effect of z transformer producing a v2. So V2 is this value or this waveform, which is Zm b2 z bridge rectifier. So this waveform from alpha to pi is in, from pi plus alpha to two pi is rectified by the effect of Bridge and z, this uncontrolled bridge, what does it do? It simply reverses z negative cycle as abosDF. So this one will be alpha to pi z m pi plus alpha to two pi is also reverse it 2Z positive direction. This circuit, which is the prejudice, is also discussed in my own first course of power electronics. So buys a usage of z transformer. The current scene Boise AC shopper is lower than the actual current drawn by the supply. So this is the case where we have AC shopper loaded by a heavy rectifier or a heavy current loot. 100. AC Chopper Loaded by an AC Motor with Sinusoidal Back Emf: Now we would like to discuss as he is super loaded Marianne AC motor with sinusodal and back in math. So we have here our supply VS, and we have here our ISI shopper. And now our load is an AC motor. And our AC motor can be represented by a resistance and back EMF, which is in the form of a sine wave. So you will find that our supply is VS equals root 2V sine omega t. And our back EMF, since it is sinusoidal back EMF, since it is an ISI mortal. Therefore it will be V back or Z back in myth of Zomato will be equal to root two V S, which is the value of z supply or can be any other value. Sine omega t minus epsilon. Ok? So this back EMF is lagging by an angle ellipsoid. This ellipsoid represent is the phase shift between the supply voltage and dizzy backing myth. And we have phi, which is the power factor angle, which are representing z phase shift between z input supply voltage and the current draw employ 0. And we have also a z alpha z firing angle of Cyrus T2, T1, and T2, which is alpha plus pi. Now, if we draw these two waveforms, the resupply you will find it is starting from 0. And this is our V supply. And if we draw our back EMF Xin, it will start to from a psi, which is the angle or the phase shift between resupply and VB. So it will start from VB here, V supply and from psi. And it can be the same value of VS or it can be Elysium VS. So it can be instead of root two V S, it can be root two Vb for general formula. So starting from upside and drawing our second waveform, which are presenting our packing left. This one representing our supply. Now what happens at the beginning from alpha to pi minus y by minus phi, representing the intersection between z V S and V B. Now at alpha two by minus phi from mere at alpha, this thyristor will be turned on. Here we have plus minus x0 Boston is connected to our V supply and the negative is connected to our back EMF. Now, from alpha to pi minus phi during this part, from alpha to pi minus phi at the intersection, you will find that the supply voltage is greater than the back EMF. For example, at this point here, V is greater than VB. Here, V S greater than V, V Here, VGS greater than VT until the intersection where z will be equal to each other. So starting from alpha to pi minus alpha v is greater than VB. So T1 will be on, and the output voltage will be equal to supply voltage. So this is our output from alpha. It will be equal to Z supply voltage, this black line, okay, on del V, Vg until the intersection here. And the intersection, you will find that after it, V back is greater than V, S, V back here greater than VSV, back greater than VS, and so on. So what we'll have in the back is connected to negative terminal of z. So I restore and the positive terminal is connected to the supply. And the negative greater than z positive. So T1 will be off. So this semester will be off. So the output voltage will be equal to what? Equal only to v back because no current will be drawn since it is an open circuit. So starting from intersection, our output will be V back, as you see here, we lack line which are representing our Albert after intersection. Until the 0.2 pi plus alpha. At pi plus alpha, we have used the positive and the negative for the other cellular stone. So z positive connected to V back ends in negative connected to VS. And you will find here during this part, all of this V back greater than V S, V back greater than V S. So t2 will be on at pi plus alpha. So Add by plus alpha z output voltage will be equal to the supply voltage, okay? At this point, okay, by plus alpha to two pi minus alpha Z intersection. Again here, the bag greater than V S T2 is on, and the output voltage here will be equal to the supply voltage. So from here, all of this equal to the supply voltage, okay? This is the wave of the supply. So starting from pi plus alpha until intersection, at intersection or finds that the supply is greater than Z V bank. So supply which is the negative, greater than z positive, which is V back. So t2 will be off. And again, T2 is off. Therefore the output will be equal to V back. So output from here will be equal to Z back voltage, this voltage, okay? And then the cycle repeats itself at alpha going into VS onto intersection and so on. So this representing our waveform for Z output for the current. In this case, we assume that our current is leading by an angle phi. So the current, this current is leading from the supply by foil. So Z current waveform, which is a sine wave, will start before z supply by an angle phi. Therefore, a sine wave here representing DC current, okay? And at angle phi. So this point, the current existence when T1 or T2 on. So T1 is on from alpha to the intersection. From alpha to the intersection. What is the value of intersection? Simply, your nose at this point is negative phi Z starting of current. And from here to here is pi. So this point will be pi minus alpha. All of this is pi half cycle. And starting from negative phi. So this point will be pi minus phi K, which are representing Z intersection here. And similarly here at pi plus alpha to two pi plus alpha, which are representing Zheng. So restore T2 from y plus alpha to two pi minus alpha. Okay? So this representing the waveform of the current. So what we would like to do in this case, or Z AC motor with a sinusoidal pecking myth. We would like to draw the wave forms and we would like to get Z value of z power factor phi, which is unknown for us. So the first thing you are going to do now is that we are going to get z phasor diagram. So at the beginning we have VS. Vs is the supply voltage, which is VS sine omega t. And we have here our output current leading by an angle Foy. Ok. And do we have here our back in math, current, vacuum if voltage lagging by an angle epsilon. So from KVL in this loop, you will find that the supply voltage equal to VB plus o multiplied by R. In case of this one is conducting. So VS equal IR multiplied by Vb or plus VB. So we have VGS equal to what? Equal to VB plus R multiplied by R. So Vb plus IR give us Z supply voltage. So you will find that this line is parallel to this line. And we have here a horizontal line, and we have yet another drawing horizontal line. So the angle here is phi and the angle here is phi, since the ZR parallel lines. So this angle can help us to divide our component for ir. So IR in this direction will be IR cosine phi. And in this direction it will be ir sine phi. So this component, ir sine phi and z, upper direction for VB. We can analyze it into two components. One at VS, which is VB cosine psi, and another one in Z, opposite direction downward, since the bone here and going here, VB sign epsilon. So we'll find here we have a component ir sine phi. And do I have here another component, downward VB sinusoid. And we have here a horizontal line here, horizontal line here. So this component equal to z component. So ir sine phi equal to negative V, V sinusoid. Why negative sense this a vector in this direction and this direction. But this vector will be in the upper direction and in the direction of x. So one in z negative y, and one in z positive y. So from this we can get sine phi equal negative VB sine psi over ir. So from cosine law, we have VB and we have sine psi given. We need to find the phi, but we don't know r multiplied by r. So simply by cosine law, you see that here we have add triangle. From this triangle we can get ir. Ir square is equal to v squared plus vb squared v squared plus v squared minus two V S VDB VSB VI cosine of the angle between the S and the VBE. The angle between V S and V B, which is epsilon. So I r square will be the S square plus V S square minus two v as v v cosine psi. This is from z equals iron low in trigonometry. Okay, so this was our explanation for z, is C sharp are loaded by an AC motor. Now would like to have an example on this. 101. Example on AC Chopper Loaded by an AC Motor with Sinusoidal Back Emf: Now let's have an example on z is c bar loaded by an AC motor with a Science Sociale back EMF. So we have a resistive load of one arm with a sinusoidal back EMF V back equal to a 100 sine omega t plus 60 degree. And then ISI salts say v equal 0 sine omega t will finds at VB here is not equal to here. Okay? So throw assign guilt phase is issue per se. The Rosa load voltage and current waveforms at firing angle alpha, called 60-degree, determines z RMS value of Z load voltage. So we have a z firing angle alpha. We have Rebecca meth, and we have the ac source. And you'll find that z value of Z maximum VB is not equal to the supply. And you will find that VB is leading our AC source. So first we will draw our load voltage and current waveforms. So we have VS equals 0 sine omega t And VB equal to 100 sine omega t plus 60. As we remember that ellipsoid, the angle between VB and VS. But remember that we said sine omega t minus epsilon minus epsilon equal to plus 60. So ellipsoid will be equal to minus 60. Why? Because as you remember, Vb, we assumed it is Vb sine omega t minus epsilon. So equating a plus 60 with minus psi, we have US Ipswich one negative 60. So by calculating z, ir will be equal to z square root of z V S square plus Vb square minus two v multiplied by Vb cosine 60. So IR would be finally equal to 264.5675 volt. Now this is the value of iR. So sine phi is equal to negative VB sign epsilon over IR minus 200 sine negative six seasons if psi equal negative 16. And VB is given as 200 IR calculated. So we can get that phi is equal to 40 degree. So we can draw our function, okay? Starting from Z, alpha, alpha is equal to 60 degree and the intersection will be at 48.8 mine. So at alpha, we have drawn here now vibing and we have drawn our visa apply, you will find that VB is lagging by an angle psi, by leading by an angle of pi. Ok, so this part is epsilon. Now Z VBE continued until the point of intersection. This point intersection now it will be fine. And z at alpha VDS greater than VB. Saw, our supply will conduct from here or our Cyrus told conduct. So VS will continually exists until alpha plus pi. And alpha plus pi, or before alpha plus pi, we have Amazon intersection here. So at this intersection will, will start to buy VV. At this point we have VB. So again, from Alpha, v is greater than VB, so T1 will conduct normally VS until the point of intersection. At intersection, we will have z part of VV, Okay? Since D1 will be of Zen at alpha plus pi, the thyristor T2 will conduct. So we will go through VS. This is our supply again until intersection going through v, b and so on here. Similarly here. Now if we draw our waveform of current, our current will conduct a from alpha like this. And going until what? Until Z intersection here. Then it will conduct again from alpha plus pi until the intersection here. This intersection is similar to this one. Okay? Current now is lagging by an angle phi, okay? The current in z, previous explanation was leading because z epsilon here was a positive value. So the current was leading. But now Z is a negative value. So the current is lagging. So this one is phi. So this point is, of course what? It will be vi plus 5x, since you are starting here from phi. And all of this is pi. So this point will be y plus the starting point, which is phi alpha plus pi until the intersection here. So now we have drawn our waveforms and we got our IR and sine phi, we got z phi in order to draw our waveform. Now we would like to find the RMSE value of zeros voltage. So as you remember that the V output RMS is equal to the square root, one over two pi or one over T. The integration from, from 0 until pi or until two pi. Ok? So we will integrate our function from 0 to pi and multiplied it by two since it is cemetery or it is a symmetric function. So finds that here multiplied by two. And then we'll integrate from 0 until by now starting from here from 0 until Z intersection point which is far away from 0 until intersection, which is fine. What is our function? Our function here is z, V supply, OK, before the intersection we have V supply. So two will be V supply, which is surrounded with sine omega t Serrano sat on sine omega T all squared. Then from phi until alpha, okay? From phi two alpha, we will have after intersection VV. So it will be 200 sine omega t plus 60, which is VB all square. Then plus from alpha to pi. From alpha to pi, we have VS surrounded and sine omega t or a square alpha two by. So when you draw this wave forms by yourself, we will understand how you can get Z RMS is Lee. Okay? So as the most important thing is that you solve as examples The by your hand so that VR would RMS will be equal to after substitution. It will be 207.45 volt. So this our, this is our example on z, is he motor loaded by acid and sodium back in math. 102. Integral Cycle Control: In this video, we would like to discuss z integral cycle control. And this method is important in z AC sharper. In this method it is different Zen z. Previous methods of controlling z essential part we have discussed is the controlling of AC short bar pi z firing angle alpha. Now is the integral cycle control, which is known as z 0 voltage switching circuit z cycle selection circuit Z on off control circuit z psi kilo converter. And we will understand the now from z, meaning of integral cycle control, why it's called the wisest names. So simply, we have this circuit similar as before. Now we have our supply, our ISI shopper control, the Buys a integral cycle control technique. And do we have here our node? Now our input is VS supply voltage VS our output. In this case, we'll have a group of cycles representing z on period Zen 0 cycle for our group of cycles, then it repeats itself. So we'll find that here. It's called on-off control circuit because we have some cycles on and some cycles of all of this together representing the new output frequency. Ok? So before is infrequency was from 0 until two pi. Now, since we have here our group of cycles on Xin off, isn't it means that this pattern will repeat itself from 0 until here. Okay? Then it will start to repeat itself again. So we have a new frequency. The frequency here is one over T, one over z videoed here from 0 until here it is year 10, whatever. So we have a group of cycles on, a group of cycles off. So this type of converters simply allows the passage of sine wave for a group of cycles. Zen Turners of orthologs the sine wave for another group of cycles. So this causes the total RMS of z voltage and the frequency of the output voltage to change. However, in firing angle alpha z, frequency of z supply was similar to the frequency of the output. But here the frequency of the supply is different from the frequency of Z. Z. Restore T1 operates at alpha equals 0. Answers to T2 operates at alpha equal 180. So in order to allow the sine wave here to buffs Rosa side story here. The first thyristor, T1 will conduct at 0. So it will conduct a from 0 until by. The second service to T2 will make it conduct at, by in order to allow the passage of the negative Bach and so on. So one at 0 and the one at a 180 degree in order to keep them producing cycles. So it will allow zing, complete sinewave to boss as if z we're diets. Because as you remember from our first course that z psi restore can be considered as diet, but with a 0 firing angle. Z, sorry, store will be on this service told conducting for a group of cycles of cold the end. And we'll be off for a group of cycles equalled m. So T1 and T2 are off for another group of cycles, so-called em block, the sine wave here. Now we would like to understand a very important concept. Okay, from here, we have a group of cycles on Zen apart of, So our new, our new frequency is from here to here. So from here to here representing our two pi. Okay? As you remember that two pi representing a complete cycle. So from here, we're representing a complete psych. And do we have an N cycles on and MSI cuts off. So each one cycle here representing how much simply each a complete cycle will be. Two pi Z, total cycle over z, N a plus M. So two pi over N a plus M and M plus m representing T. T here is not the aperiodic time, but t here representing the number of cycles. So one complete cycle, this complete cycle until here, this part is equal to two pi over t. Okay? So this half cycle is poverty only. Now for this one, another one cycle. So z it will be two pi over t multiplied by what onto blood by two. And the four talking about Z total cycles here until this point, then it will be two pi multiplied by N cycles over t. Ok? So from this our understanding we can prove the RMS of the output voltage n z power factor. V output as a function of omega dt equal root 2V sine omega t d y t. Because now we have, we don't have omega t only, but we have n plus m cycles. So our new frequency change. So it will be t omega t. So z period of conduction is from 0 to pi over T multiplied by n. Remember that two pi over T representing one cycle, one cycle inside our large cycle. So N two by N over t representing the total number of cycles of operation. So when we substitute in this equation, sine two pi n over T, it will give us sine two pi n. Okay? And from two by N over t to two pi, which is a total equal to 02 by N over t is the end of z on cycles. Now, the IV root mean square is one over z. Payload videoed here is two pi, of course. From the beginning of the cycle to the end. The integration from 0 to be a two by N over t 2Z function here, root VS sine t omega T all squared. So by doing some simple mathematics, you will find that savvy output RMS is equal to VS multiplied by root N over t, where n is the number of cycles and T is the total number of cycles. N a plus M. For z power factor. From definition, z power factor is equal to the active power over the apparent power S. So the active power, and do we have here at resistive load. So z power output is simply equal to v squared over r, v Albert RMS square over r. So this representing z output power on Z resistive load. And S z input power, or the apparent power equal to V supply as an RMS, of course, multiplied by Z current RMS. Now, instead we can see it like this, or we can say is that the power output rosa resistance is equal to V output RMS. I would RMS, since it is obscure, obscure and pure and resistance. Okay, we don't have any inductance. So ZB across a 0 resistance as similar to zs across the resistance. So V output root-mean-squared untroubled by Albert root-mean-squared over VS. root mean square multiplied by IS root-mean-squared. Here from our circuit. I would similar to I supply, okay? When T1 is conducting, then I equal to I supply. When T1 is often T2 is off, this will be equal to 0 and this one is equal to 0. So our function or our current output equal to z input current. So V output equal to input current, z equal to i is not equal to i s. So we can cancel this with this. So we'll have V out as an RMS over V S. V R would RMS over VS will equal to what road? N over t root N over t. We'll finally, we'll have our power factor as root N over t. So from this we can define that by controlling z number of cycles, we can exchanges a V RMS. And by controlling Z total T N a plus M, or the total number of cycles, we can change the frequency and we can, it changes the output root mean square. Also z power factor depends on the ratio between all 0 root of m over t. So let's have an example on integral cyclic control to understand it. 103. Example On Integral Cycle Control: So let's have an example on integral cycle control. Single-phase integrals cycle control is see sharper resistance of ten on, as our load and supply voltage offered 120 volt RMS, frequency of 50 hertz, number of cycles, 12 cycles. And the number of cycle equal eight determines z, V out as a root mean squared z power factor and the current, so I restore as root mean square. So let us start step-by-step Z. First thing is that our cycles or our Cyrus total will be on for a group of cycles, okay? N cycles, 12 cycles, and off for M cycles. So which is number of cycles, which is eight here. So V out as a root mean square. We said it is equal to the square root of n over t multiplied by VS as a root mean square. N is given as 12. M is given as eight VS supply voltage as a root mean square 220. So our output symbol b would be equal to a 170 volt as a root mean square z bar vector. We proved it before, equal to root n over t. And it's 1212 plus eight. So we'll simply have a power factor of 0.77. Now to z, our new requirement is the eye. So I restore as an RMS. Remember that here. This representing z a period where the cellular stores are on. Ok. I also restore T1 is conducting at here from 0 to pi over T and conducting here and the conducting here. So it's conducting a for n cycles. Okay? So z psi restore RMS, of course for T1 or T2 are similar to each other. For T1 starting from 0 to pi over T, representing one cycle of conduction for the Siris too. So I, so I restore as our root mean square is equal to the square root, one over the total period from here to here, which is one over two pi multiplied by z integration from 0 to pi over T. Z value of Z current here. Value of the current is the voltage wrote to VS sine t omega t divided by the resistance. All of this square. So z voltage over resistance will give us z output current, which is the current across the solar system. Now this representing z current here only, but this is repeated for n times. So we will multiply this integration by n. So we will finally have at welfare 0.05. and bear. So this is our psi restaurant current as the root mean square or effective value. We simply integrated this part and multiply it by N, since it is repeated N times in this cycle. Okay, we have one cycle, 23. So in this three cycles, in this city and silence, it will be 123. So z number of conduction is equal to number of cycles. So this was our simple example on the integral cycle control. 104. Definition and Applications of DC Choppers: So first, let us discuss a z definition of DC shopper. So what is DC? Dc shopper? Assembly and electronic equipment or an electric equipment which is used to convert the Z input, dc voltage. Dc voltages the constant DC voltage input to the decision part into a variable DC voltage. So that these usurp or provide a variable DC voltage, or varying or changing the DC voltage from a constant input dc voltage. As an example for this, if we have here a voltage input of a 12-volt DC, we can use some kinds of or some types of these usurper to produce 14 volt. For example, 14 volt DC two. Or we can use it to produce eight volt, for example, eight volt DC. Okay? So it basically changes the input DC voltage into a variable or changing a value of DC output voltage. What are the applications of decision? When we are using or when we use z? Dc shoppers. Z, we use them in xhat traction, motor control in DC traction system, in DC traction system. Or in trends, we use DC motors as driving for or drive for our equipment or our train. Okay, so we use DC motors. So DC motors require DC input voltage. So by varying z input dc voltage into a DC motor, we can produce variable is speed. Okay? One of the methods to control a DC motor is by a changing z value of the input voltage, which changes z is z traction system or the speed of z. Second application of DC shopper is that dc voltage regulation. What does it mean? It means that we can use, or we can control Z value of DC voltage by using DC shoppers. Okay? Is up by step are being this DCM book or step down this DC input. A third application which is a really important, it is used in solar energy conversions. You will find this in BV system and you'll find it in my own course for solar energy. You will find that z DC shoppers are found inside the solar charger controller, which regulates the charging of batteries. Solar charge controller used to regulate Z charging batteries. As solar charged controller simply takes a z DC voltage output from our photovoltaic or our output from our baby system. Zen. It changes the value of z DC voltage to a value suitable for our batteries or solar charger control does something called Z, maximum power point tracking. It produces the maximum possible power by varying z value or choosing a suitable value for our battery. Another application for DC shoppers is the audios, Danish electric cars, airplanes, and space ships. So from this lecture, we will understand that DC shopper simply reduces a variable DC voltage. What does it mean? Variable. Variable does not mean easing. Variable means that we can produce different values for DC voltage, okay? If we have here ten volt, then we can produce mind voltage, DC, voltage, DC, voltage, DC, and so on. And those are the applications of the DC shopper. 105. Step Down DC Chopper with R Load: Now let's discuss Z. First subtype of DC shoppers with z, step-down DC sharper with resistive load. So what is a step down DC shopper stepped down in DC Super means we are stepping down or decreasing z value of the input voltage. Okay? And our load means a resistive load. So our step-down DC short circuit will be simply like this. What we have here, our V supply, the input DC voltage. And then we have here our output voltage, V0 Albert over load, which is the resistive load. And I0 or Z output current through the lute. S is the switch, the switch which will be on and off in order to add step-downs are DC voltage. So what happens here is that Swiss as I switch S here, this is which will be closed for a certain period called T1 or conduction period will be T1. Then it will be opened for another period called T2. And the summation of this two time, T1 and T2 were represented t, which represent the Z periodic time or one over the switching frequency. Ok, because the periodic time, as you know, is one over f. Zach switching here which will be used, will be AD BD or MOSFET, or G20 or IGBT, which we discussed in our first course on power electronics. So simply, when this switch is closed, z, V supply would be equal to V Albert or V output will be equal to V supply, okay? When it is off or when it is an open circuit, V out will be equal to 0. So this process will cause a Z output voltage to be Liz Zan Zi, DC input voltage. And we'll see now hell, during the conduction period or S of the switches on, our circuit will be like this, is a switch becomes a short circuit. So V output from KVL, V output is equal to V supply from the KVL here. And the current output will be equal to z V supply. Or what is the resistance R V supply over R from KVL here, when the switches of i output is equal to 0 because we have an open circuit and the output will be equal to 0 because we don't have any. Voltage across the resistance here. So this two cases, this one during a period called the T1, this one during the period t. So if we draw the V output as a function of time and I output as a function of time. We will have v Albert during T1 will be equal to V supply. And i will be equal to VS over R. During the off period T2, we will have an output of 0. Because we have an open circuit. And the output current will be also equal to z. Zi summation of t1 velocity two or all of this period is called as aperiodic time tin. Okay? Remember that here, T1, this conduction period is equal to K, a certain constant call decay, which represented the duty cycle multiplied by t or z daughter period. Okay, and you will understand where did we get this part? So V output is the output voltage across the resistance I, R. What is the current going into the resistance R? And here you will find that i output is equal to the current coming from the supply IS duty cycle represented as the ratio between T1, 2a beauty time or the conduction period with respect to two total time. That's why. Or is that duty cycle multiplied bys aperiodic time will give us T1. So T1 represented Z portion of that conduction time or due to cycle represented z ratio between Z conduction period to Z daughter period. Ok. So k, which is the duty cycle, is equal to T1 over z period T. So T1 or the conduction part, or the conduction pi1 will be equal to k, or the duty cycle multiplied bys aperiodic time. T1 plus t2 is equal to t or the total time. And t1 is equal to K t, K T1 plus T2. Equality. So T2 will be equal to one minus k multiplied by t or the total period. So t1 finally is equal to the duty cycle multiplied by Xp aperiodic time, and z off period or the off time, one minus k multiplied by it is aperiodic time. Now we need to have or to find the z important laws here. So we have here this waveform of the output voltage. Waveform is the force, the current. Now, I would like to find the V output average or the average output voltage. So I know that the average of any waveform is equal to one over t. Integration from 0 to T or the total period. But our function here exists as from 0 to k t as t two t is equal to 0. So the average of this function will be the integration from 0 to k t for supply voltage VS. So one over t, 0 TO katy VS d t. So it will be equal to K T over T will give us k v s. So the V output average will be equal to the duty cycle, motto, blood vessel supply voltage. Now, if we would like to find z root mean square or the effective value of the output voltage. It will be equal to the square root of one over t. Integration from 0 to t. And we said that our function exists only from 0 to k t two z function square. So VS all square. So this will give us finally root k or the square root of duty cycle multiplied by CSA supply voltage. This one, the effective value represent the z output dc value, okay? As the root mean square is equal to z DC value. Now if we'd like to find is the output average, here you will find is that this function is similar to this function but divided by r. So I output average will be equal to V output average, that voltage over resistance. So it will be GAVI S, which is V out average over the resistance. For the power. It will be equal to v squared over r ij. Remember that z power is equal to i square as a written square multiplied by the resistance or the voltage square as r square, or the effective value of the resistance. Now comes here an important factor called the input resistance, seen by the source are, or is the input resistance. This represented the equivalent resistance seen by our source. Okay, so let's get back to our circuit here. So here we have our visibly here, our I supply. So if we, let's mark this one, Ben. So here, if we look at that circuit from here, this will give us our input. Okay? So this one is our input, represents a z effective resistance seen by the source. Okay? All of this part, how it looks like from the point of view over the supply or all of this as a switch. The switch and the resistance. All of this represents what as an, as an equivalent resistance from their point of view of the sublime. Okay? So this one, same as seven, is equal to V supply over I supply a V over r. Okay? Or the voltage over the current. So here you will find that the input resistance seen by the source is equal to V supply over our supply average. And we know that I supply average is equal to I, i would average. So V supply over I would average, which is caveats over r. So it will give us r over k. Okay? So at K equal one, when we don't have any switch, just as the supply voltage, always supplying a resistive load, then we will have a resistance R. When we change the duty cycle, we will, it changes the value of the equivalent input resistance. So this was our step-down DC shopper. Four is on our, our, our resistive load. Now in the next video we are going to have an example. 106. Example on Step Down DC Chopper with R Load: Now let's have an example on CSER, step-down DC shopper, who is on our load. So we have here as a step down DC coupon with a resistance of ten on. So Z load is resistance with a value of ten on the input DC voltage, or the supply voltage is equal to 220 volt. Dc. Voltage drop on the switch is two volt. So our switch, I have alters drop because our switch is not idea. Okay, from our first course in power electronics, we said that there are two types of switches. We have an ideal switch which does not cause any voltage drop is equal to 0. But actually 0 voltage drop is existing in switch, okay? Because nothing is ideal. But in real life, Z two volt here can be neglected with respect to 2a supply voltage. That's why sometimes in z problems or, or in my own other courses, you will find that I don't mention the voltage drop. The frequency of switching of our obesity or MOSFET or any switching device is one kilohertz. So the frequency of switching means that our transistor is switching on and off 1 thousand times in one, in one cycle. Okay? Z, duty cycle of switching is 50%. It means that we are switching on our switch or our switching device for 0.5 z periodic time or hover over the cycle and the other half is as a switch is off. So what is the requirement in this program? Determines z v I would average the average output voltage, RMS or root mean square value of the output voltage. The efficiency of z converter, Zillow Singh is the effective input resistance seen by the source. So simply Zafar single are gong to do is we would like to find the z average voltage. Okay, so the first thing we are going to draw our circuit. So we have here our supply, and we have here a voltage drop across our switch. And is there a v output across the resistance R? So by applying a caveat here, V supply is equal to V switch blas v-out. So z is equal to supply voltage minus voltage drop on the switch. Okay? So before in CSER equations which we discussed, we had the voltage across the switch to be 0. But here we assume that we have a value for the switch. Voltage drop on the switch. So our album from caveat will be equal to V supply minus z voltage of the switch, opens the switch. So our output wave form will be during the on period will be VS minus z voltage across the switch. And the current will be z voltage over resistance. So Zafar set requirement is V output average, which is equal to one over t or z videoed integration from 0 to Katie or until T1 to z function here, which is VS minus the voltage drop on the switch d t. So it will be k t multiplied by t over t, which will give us K V S minus V switch. So VS to a 120, a voltage of the switch. Voltage drop on the switch is equal to two volt. And said Dude cycle 0.5. So our output will be 109 volt. Our second requirement is a V root mean square, similar as before. And instead of having V S, it will be VS minus z voltage drop on the switch, all square. So it will finally give us root k VS minus voltage robots or switch. So it will give us a 154.15 volt. Now, 2Z sordid requirement is the efficiency of the converter. Efficiency is simply equal to z power output over z input power is this is the efficiency in any electrical equipment. So zap or Albert is equal to v squared over r. We said this in z laws which we discussed Sousa via what root mean square is 154 and the resistance given by ten ohm. So we'll have 2376.22. What? Now for z in both power, our input is DC supply. So for DC supply, as we discussed in our first two course for power electronics, z power is equal to in DC or DC voltages supply will be equal to V average of supply, or V supply multiplied by z average supply current. So v sub y given by 220. Nosy supply average current is equal to I, R would average. Now's the IR would average, or we say that two will be equal to V output average over r. So Z output average voltage is equal to a 100. And tonight, which we got before, a resistance of ten on all of this will give us 2798. What Sousa efficiency is equal to power Albert overpower input, which will give us nine to 9% as an efficiency of conversion. Z lost requirement in this problem is the effective input resistance, or it will be equal to z v emote, or the supply voltage over Z, output average, Oryza, supply average, okay? Supply average here is the same as I would average. So supply 220, I would average is given as, as VR would average over r. Ok? Or it will be K V S minus V switch over as a resistance R. Okay? This part represented savvy. I would average over the resistance gives us r would average. So I will have, this one goes here. Since we are dividing here, will go here. And the KV S minus V switching. And by substitution, we get at 20 ohm as an equivalent resistance. So we can, instead of doing this, we have AES 220 and I would average we got, and before as VR would average over R, 109 over ten. That's all. Okay. So this was our simple example on the step-down decision. So you have to understand where we got zeros and how does this circuit walkers and solve this example by your hand? Okay. 107. Generation of Duty Cycle: In this video, we are going to discuss a z generation of the duty cycle. How we can produce a key from 0 to one, ok. How we can produce Agile cycle of 50% or 40%. How we can reduce this values? So our circuit or zed duty cycle generation is simply very easy. How we can produce this Z for assessing all we have a comparator and op amp, which is working as a comparator. We have two signal, one on z bold step and one on z negative. If you don't know what is an op M or an operational amplifier, you can go to my own course for complete electric circuits. So z first, we have two signals here. The comparator assembly compares it between two signals or two voltages. Here we have first the voltage which is called a z, a v r or v reference or reference signal. Second one on Zan negative is called Dizzy carrier secondary. So z, if z reference signal greater Zan Zi carrier signal, the output here will be high. And if z reference signal Li Zan Zi carrier signal or z reference voltage less than the carrier voltage, then our output will be low. So Z output here representing z gate of electronics, which as an example, if we have a BJT, you remember that from our first of course is at z gate signal is z or z base current. So by changing, by putting a high IBS, we will give an and Alberto will be high. And when z equals 0 or low, then the transistor or our switch VGD is off. This is also can be used for Z transistor or z and z, sorry, restaurant service or as you know that it switches on when we give it an bots. So this can be considered as opposed to z gate of electronics, which is the gate which controls the on and off of our switching device. So z reference signal and the carrier signal can be represented by reference signal can be represented by a DC voltage. Okay, we'll find the here. This is a straight line with an or parallel to z axis. And the z carrier signal can be represented by triangular wave, okay? Or we can say it is sutras, okay? It is a so tos not. A triangular. Okay, anyway, we compare the value of z voltage here with respect to z reference signal. Okay? So you will find that when ZIM, This one is the reference signal at this point, reference signal is greater than Z carrier signal. So z Alberto will be high. At this point from here until here, you will find that z value of Z carrier signal is a straight line here is greater Zenzele reference signal. So our output will be equal to 0. Again at this point, x0 carrier signal. And this one is the reference signal. So z carrier signal is li Zeng Zi reference signal. So z reference, which is this one is higher. So our output will be high. So z Vr reference signal is this line and carrier signal as this line. Okay? And comparing between them, we can produce either both step and the negative pulses. Okay? So, and here we can control our duty cycle. This one representing z time of switching or when the switches on. And this one, this board representing z time when it is off. So this equation, or the equation of the straight line here of our carrier signal is equal to, as you remember that from mathematics, who knows at y, y2 minus y1 over x2 minus x1. So this one, for example, we have here a point is 00, this one is T and V reference. So V reference Z final value of y minus the initial value of y, which is 0, vr minus 0, Y2 minus Y1 over X2, which is here, is t minus X1, which is 0. So here we can represent this part by y two or y y1 minus y2 minus y1 over x2 minus x1. Okay, from Z mathematics. So this one represents as the equation of our carrier signal. So at any value, Y minus Y1 over x minus x1. So Y1 is 0, X1 is 0. This is the initial point. Here. We have z, y, y representing z output, okay? Z1 is the voltage and which is representing VI carrier. The carrier is varying, okay? Vicariance representing z values on this straight line over t, which is x. So we have the equation of the straight line is y, which is v carrier, is equal to Vr B over T total period multiplied by z x, which is T, y equal m x m is the slope of the line. And x is our varying, which is the time. And the y is our z voltage of savvy carrier VAR RB, which is beak reference voltage, is this signal. When we introduce it up Z maximum value, or the maximum possible value is at this point. So this point represents a Z maximum of reference signal. We can increase our signal up to here. So in order to find our due to cycle at t equal t1, at t equal t1, okay? Which is K T z voltage will be equal to f z reference value, Z carrier voltage. This line at this point is equal to 0 reference evalue. So V carrier is equal to VR. So VR, and instead of v carrier is equal to V reference B over T multiplied by KT. Kt is a time t1. So K is the ratio between the reference voltage and this one to z A-B-C value. Okay? So if we increase Z line of 0, France signal up to here, Xin will produce k equal one. So by varying the Z reference signal VR, we can exchange a Z duty cycle key. So as v r increases, K increases. As v r decreases, k will decrease, and so on. Ok? So this how we can generate our duty cycle k by having two voltages, Z, so tos. Okay, along go is z reference voltage, which is a straight line parallel to the x-axis. By combining these two values. And we can, by varying a VR, we can vary z, duty cycle k. This video is for your own knowledge, it will not change in a sink in our equations. 108. Switching Techniques: Now, let's discuss the Z switching techniques. So for your own knowledge at all, there are two switching techniques. First one is as z constant frequency operation or Z pulse width modulation. What does it mean? It means that we are having a constant T. What are constant and periodic time, or a constant switching frequency. And we are varying z t on, or a varying is at t off, but Z total period T will be constant. So we have here an example tee off entity on this one representing AZT. Okay? We can increase the T off, decreasing Tian. Okay? But the total period T will be constant. Ok? This is one of the switching techniques by varying Z widths of z pulse. Adds a same frequency or adds asymptotic time. We can produce as switching technique. Okay, this one is a switching technique. So what is Z? Advantages of this type? So being frequency is constant, which is important to prevent is a presence of harmonics. We valleys at time t1 by varying the widths of z balsa of t on, by varying Z height of the reference voltage, as we did before. We said that the VaR is oblivious video by varying z reference voltage weekend, very hour out. Or our Dude psych. The second technique is called the z variable frequency operation, also frequency modulation. In this time, which change our periodic time t ZAB beauty climb is not constant. So this one is periodic time t, t off will still be constant, OK, as an example. And we increase our Tian. So this type t here is different from t here, z t here as the front from this one. So we change our frequency or a change in our periodic time. We keep T constant. Or we can do the reverse. We can increase t off and decay constant. Ok, in the end, we will change our frequency. Z. Shoving frequency is varied z. This advantage of this time is generation of harmonics, which means that we will need to design more complex filter in order to remove the harmonics, reduce the from this time. 109. Step down DC Chopper with RLE Load Part 1: In this video, we are going to discuss a z step-down DC Shofar was an ore in the loot. So a step down. Dc shopper means that we are decreasing our input DC voltage or in the load which is a resistance inductance and the back EMF load, this RLE load represent this DC motor, okay? We can say it is a DC motor sensors. The own book is DC voltage. So in this type of load, we have two conditions. We have z, continuous current mode, and we have discontinuous current amount, okay? Due to the presence of the inductance, the current can be continuous and it can be discontinuous. Zab reasons of inductance along with AMF. Okay? Zany MF causes the current to be discontinuous and z inductance may cause also the current to be continuous or not continuous. And all see now. So the first ammonia, which is a continuous current mode. Our circuit is consisting of DC supply as switch, like BGT MOSFET or any switching device. We have here our diet. And we have here our Albert current resistance inductance and back EMF. This load represented Z motor. Okay? And we'll find here is that Z output current is not equal to supply current. So adds a beginning. We have two different cases. Z first one wins, this switches on. When the switch is on. Then our equivalent circuit will be like this, because the diode will be reverse biased. So it will be an open circuit. When z switches off, our circuit, our or our inductance have, will have a stored energy. So it will produce current in the same direction of Zan, original current. So z supply current will produce a current in this direction, which a charging dizzy inductance during busy, unmoored, during busy off mode during this circuit. And z inductance will produce a current in the same direction of z, a charging current. So it will have a current in this direction during busy off mode and becomes a short-circuit. So during this as Bart Z IVR would from KVL is equal to V supply. During busy off mod Z, the output is equal to 0, okay? Since it is parallel to z diet, which is as hot circuit, therefore the voltage here will be equal to 0. Now before drowsy V out and I out as a function of time during busy on videoed, V out is equal to V supply. And during the off period, V output is equal to 0. Here for our current, since it is a continuous current, continuous, it means that it does not go to 0. So z i Abbott, during the conduction period, when it is on the supply voltage, it charges z inductance. Okay? So from here to here during the conduction period, is the current is increasing from I1 to I2. During the off period. When z supply does not exist, Z current will start discharging. We have here during conduction. Current is charging or the inductance is charged it from here, I2, I1, z current is discharging, okay? Or the inductance itself is discharging. So we need to find z value of I1, I2, and we need to find z rebel. What does the current treble mean? The rebel is the difference between I2 minus I1, okay? Although differences between i2 and i1. So let's see our equations. So during the on period, we have our supply and our load from I1 to I2. So by applying a KVL here, in this loop, we will have V supply is equal to o multiplied by the resistance plus L d i over d t plus z back EMF e. Remember that during busy conduction, we assume that the current here is called the I1, okay? The current I1 representing z current during the conduction period. So from here to here, all of this is named as RE1 i small. I capital represented Z minimum current, and I to represent a Z maximum current. So this equation will be simplified to this one. Okay? This type of equations of z transient equations can be solved by assuming that our current will be equal to I started ICT state blas I transient. So this equation, this second ordered as a first order differential equation, can be solved by assuming that we have the current I steady-state bolus transient, steady-state when our circuit will be like this, okay? And sub dc supply with this load. Okay? So during steady-state, Z, current will be equal to VS minus Ea over R. Vs minus Ea over R because we have a DC supply. And do we know that as an inductance in case alpha dc is equal to 0 X L equals 0. So we have VS minus Ea over RT plus z transient to component which are represented as z a charging of Zan inductance, a certain constant, a z exponential negative t over tau, where tau represents L over R. Okay? So this representing transient circuit was on our envelope. Okay? So a E bonding 50 over tau representing the charging of the current. Ok? So this equation from year should be an exponential. But for simplification, we draw it as straight line. Ok? Del is equal to the ratio between the inductance. So 2a resistance. Now i1 will be equal to VS minus Ea over R plus a ebonite of r over l t r over n is one over ten. So now we would like to find z certain constant a. So how we can get it by using the initial condition. So as you see here, at equals 0, we have Z value of current i1 will be equal to i capital. So at i1 equals capital V, S1 is C over r. At time equals 0, exponential of 0 is equal to one, so plus e. So we can get a equal I1 minus VS minus Ea over R. So we will take this and substitute in this equation. So we'll have I1. Okay, let's delete all of this. We will have I1 equal VS minus Ea over RT plus z constant a, which is this one, I1 minus VS minus Ea over R, z exponential negative R over L DT. So I one as a function of time is equal to r u one z exponential of this equation. Okay? So I1 here, we exhausted, take I1 here, ebonite of R over NDT. Here, this part is this part. Okay? And we took a VS minus Ea over R as a common factor. So we'll have one minus IPO negative R over L T. Ok, just simple simplification. Now, we got z equation of Z current as a function of time i1 representing z equation here of the exponential is exponential here, which is nearly a straight line if we made a simplification. So z equation here of z exponential, we would like to find z value of current i2. So in order to get i2, we will put here z time equal to t1. So the current will be I2. So by using the final condition here, at time T1, at this point, time equal Katie. And the value of the current will be I2. So i2 will be equal to r u one e bonding the r over l t will be k t, and this t will be k t. So from this we got Z first equation, which only consisting of i2 as unknown and I1 is another unknown. Okay? So we need another equation, find I1 and I2. So during the off videos from T1 to T capital or periodic time, you will find that the current is another exponential, but can be approximated to a straight line. So at our equation here will be equal to from KVL is I2 R plus L D I by DT plus the back EMF equal to 0. So here we have a current, i2, i2 represented as z equation or z value of the current during the off period. So this one is called i2. This one is called the i2. This one is i2, and this one is I two and so on. So all of these values is called i2 as a small little. I2 capital represented Z maximum value, I1, small i capital represented Z minimum value. So this equation, another transient equation can be solved by our STD state, but also I transient steady-state value when the inductance equal to 0 because we have a DC supply. So this gives me n will be finally equal to negative e over r or negative e over r. Okay? When this one is removed after a steady state, when the current is steady-state, zz0 a2 will be equal to negative e over r. Negative Ea over R. Z transient will be another constant a. This is the front or from the other aim. Ipo negative t over tau. Tau is a lower are same as before, exactly the same steps. So we need to find the air by using the initial condition at time equal 0, z prime years 0. What does 0 mean? Because with Rosa straight line here, assuming that we at this point is 0. This point is 0, and this one is t minus K t, g minus k t. Okay? As if this solid line starts from here, okay, from 0. This is for simplification. That's all. So. And at time equals 0, the current will be i2. So i2 minus Ea over R plus a at 0 here is a. So a equals i2 plus e over r. By taking a and substitute it here, we will get Z fun. Final equation I two as a function of time would be this equation. Now, we need to find our u1 at time equal, one, t minus KT at time equal. Or Tito 1minus q all t, which is t minus KT. We will have a current of I1. So by taking this here, are y1 would be equal to i2 Ebola negative R over L. T is t minus KT, okay? And minus Ea over R, one minus R over L, t is d minus KT or one minus k multiplied by T. Okay? So, and this equation is similar to starting from Katie to teen. Okay? So they're France is that we just shifted our function. So we have here i1, i2, and the second equation between them. So this is our second equation by solving z second equation and is the first equation and the previous one here, I2 and I1. From these two equations, we can finally get i1 and i2 VS over R Ebola cases it minus1 over e bars at minus1 minus Ea over R. I2 will be similar, but with a negative sign here. Here we have as constant called dissent, where Zed is R T over N. Now in order to find the z current treble i2 minus i1. So difference between the maximum and the minimum current. So by subtracting these two equations, we will have this long equation. Okay? Now, we need to find a Z maximum rebel. So Z maximum current ripple is d z differentiation of z ribbon who's respected to our variable, which is k. K is our variable here we can change our key to reducing Z maximum current triple or Z minimum. Okay? So, and we have to take this equation, which is delta i or repel current and differentiate it with respect to k because k is our only variable here. So by differentiation this function, we get this function and the equating it with 0, we will finally get that k is equal to 0.5. So add the duty cycle of 50%, we will have z maximum amount of current rebels, okay, Z maximum difference in between them. So by taking izzy k here and substituting it here, and our equation at K equal 0.5, you will find that this equation, which is VS over R, V S over r, This one will be equivalent to two. Danish are over four f L Tanisha is Z hyperbolic tan function or tangent function. Okay, in mathematics. So, So as you did this by your hand, by substituting of K equal 0.5 here. And simplifying this equation into this form, you can replace danish with this exponentials. So Z maximum rebel is VS over R Tanisha, or over four f l. Now we will find that as that Z four multiplied by Zan frequency, which is a switching frequency which is in kilohertz. So it is a very larger value with the inductance Zomato publication of z3 factors will give us a value much greater Zenzele resistance. So this value is very small. So we can assume in mathematics is that Danish are over four. Fl is similar to all over for FAR. So delta i maximum or Z maximum ripple will be equal to VS over R, R over four FL. So we have finally Z maximum rebels VS over four f L. So this was our step-down DC short bar in case of RLE load. And we got the equations of the current i2 and i1, Z maximum current and Z minimum current and Z maximum ribbons. Now in the next video we will see discontinuous mode. 110. Step down DC Chopper with RLE Load Part 2: Now let's discuss Z. Discontinuous current mode is oblivious. Video with Cassese continuous. Continuous current mode means that I1 will not go to z value of 0. I1 will never be equal to 0. So z i discontinuous mode means that our current is 0 for aspecific period of time. So agains as him circuit here are in a lot and a step down DC. So how we can check if our I have circuit or our load is continuous or discontinuous. As remember that I1 is equal to VS over R ybar gizzard minus1 Ebola at minus1 minus Ea over R. This equation, which we proved in the previous video. So if I1, which is the minimum current greater than 0, so it means that Z minimum current greater than 0 means that our current is continuous. There is a value of Z, load current always, ok. In case of I1 equals 0, then it means that we are in the critical value. And if i one less than 0, it means that we are in discontinuous operation or in the discontinuous mode. So what does, or what are the factors which affect z? Continuous or continuity of Xen current inside our RLE load. So the first one is Z current can be discontinuous due to one OZ following reasons. Number one, there could be low switching frequency. Switching frequency is low, so our inductance store to store discharge and donors to 0. Then after some time it starts to each larger again. Okay? But z on and off period, making z rebels are very high. So z inductance, so go to or Z value of Z current goes to 0 due to the basis of low switching frequency z. Second factor or the second reason is the inductance of value Z inductance can be very small, which means that our inductance will not store much energy. So it will charge a fast and discharged fast. The value of Zach E or Z back EMF is near 2a value of supply voltage. So this effect, this z value of the charging current, making it harder for the inductance to store more energy. So these are the three factors which will affect z current. Now, we need to find a Z wave forms equations in case of discontinuous mode. So in case of Z unmanned. Again, we will have VR emoticon V supply. Now in the off barrios wins. This one is a short circuit. We have two cases here. The first one, when our inductance is having a store the energy and still we have a current. So VR would in this case, will be equal to 0 because V output is similar to Z voltage across the diode, which is 0. Now, when Z current i2 or when the current goes to 0 during the off period, I two is 0. So V output in this case will have a voltage, voltage of E. Ok? Because V output is equal to r multiplied by the current bolus LD i over d t plus z back EMF. But now when the current is 0 and the i over d t will be also 0. Therefore, V output will be the voltage only across the back EMF E. So here we had a value of 0 because we have i2 plus L d i over d t plus E. But now we have only the back EMF. Therefore, the voltage will be equal to E. So our wave form will be law exists. We have during the conduction period, the current charges from 0, ok, because you are talking about the discontinuous mode. So from 0 increasing up to the maximum value i2. Zen during busy off build. All of this is the off period. Z. Current starts to discharge. So during discharging period, output voltage is 0. Okay? Now when Z current equal to 0 until Xin Zhi necrosis switching action from here to here. And V output will be equal to E, as we discussed before in xy equations. So during the on period, we have this function where I1, where were our current sources from I1 equals 0 to i2. So what do we want to find? We need to find I2, and we need to find z time here, time t2, which is from Katie until here. Okay? We need to find this period or z pi m of discharging. Okay? So how we can find this by using the equations of current during the on period and off period as we did before. So here by applying KVL, we would have V supply equal to i multiplied by resistance plus LDIDT plus Z back EMF. Now, this equation can be equivalent to i, i1, which are representing z currently during discharging period. Will be equal to steady-state plus z transient current. Z steady-state as before, VS minus z over r plus a IPO negative t over tau was ECM equation exactly. And tau, a lover are same as before because we are talking about Osiris transient circuit as here, this RLC circuit. So RE1 VS minus Ea over R plus a IPO negative out of our entity. Now we need to find the z certain constant a. Here. It will be different. Why? Because before we said that at time equals 0, we had a value of I1. Now, our current will be equal to 0. So at time equal 0, Z current will be equal to 0 because it is a discontinuous mode. So by substituting this in Zi Men and equation here, at time equals 0, which is one and i one equals 0. So a would be equal to negative VS minus Ea over R and D. By taking this and substituting again, it's a question of I1. We will have I1 as a function of time VS minus Ea over R, one minus E0 negative R over L t. So we'll find that our time equals 0, I1 will be equal to 0. Now in order to get i2, i2 is the maximum value. We will substitute with t equal catering. And I1 will be i2. So I do will be VS minus V over R1 minus evil or negative R over L K T, where z again is r over l t. Ok, so we can substitute here, who is it? So I do will be equal to VS minus Ea over R, one minus e bar negative gaze at Zed is r over l t. Now, during the off period from T1 to T capital or xP periodic time. From here, we will have KVL. By applying KVL here. Remember that here during the off period is Z from T1 until T. And we will divide it into two parts. The first apology during discharging. Second, the Barto when the current is 0. So during this part, we will have an KVL in this loop. Okay? Our circuit will be the same as before. So I2 multiplied by the resistance R plus L d i over d t plus z backing Mf is equal to 0. So z function here. Can be solved by assuming a steady state transient part. A steady-state is at t equals 0. So i two will be negative Ea over R, negative e over r, same as before. Plus z ebonics 50 over L a which is z certain constant. Or we can assume it is be anything, okay? Is this a is different from the other EI del L over R. So I2 will be equal to negative e over r plus a negative r over Entity. Now we need to find a at time here we can say at time equals 0 because we shifted our function i two will be equal to i2 capital at time equals 0 here. Because as we said before, this function, we assumed it is shifted to the left. So time equals 0, I two equal to capital. So i2 will be equal to minus e over r plus a. So a will be equal to i2 plus e over r. So finally, by substituting in z i2 equation, we get I2 is equal to i2 IPO negative R over L t minus E over R1 minus R over L t. So now we need to find the Xen time t2. So at time T2, let's get back to see what I'm talking about. So we got this equation, which is exponential, and for simplification, it is drawn as a straight line. So we need to find T2. We said that this point is 0. So this point is T2. So by substituting at time equal t two in the equation, we get z time and the current will be 0. Current equals 0 at the end of discharging at time equal t2. So 0 will be equal to t2. And this function, by doing some simple mathematics here, we can finally get that T2 will be equal to L over R. Lend one bolus I2 R over E. So we have here z time t2. Now i2, or the current I2, we obtained it before. Okay, let's get back to it. You'll see that i2 VS minus E over R1 minus y1 Native kids it. So VS minus Ea over R minus VS minus Ea over RT bonding of Giza, which is a same as the same equation. Ok? So by simplification here too, we will finally get T2 L over R ln one plus VS minus e, one minus IPO negative kills it. So we obtained our T2, we obtained our voltage. Now we need to find the z v average. Ok? So our waveform will be like this. So z v average, V average will be equal to one over the period t. Integration from 0 TO Katie. And integration from this point to t from 0 to k t VS 0 to k t VS from this point, which is KT. And until to this point, KT plus z2, okay? This is one plus z at the added time. So it will be Katie velocity 2GM until the periodic time t for the function e. So we can finally get the output average is equal to k VS plus Ea over RT, one minus k t minus t0, t2. So, and this is a long equations. But the most important thing is that you understand the reasons for being discontinuous and continuous inside Z DC shoppers. And understanding how to get z T2 and the current I1 and I2 in this type of DC shoppers. Now in the next video, where we'll have unexamined on a step down DC shopper was on R and E loot. 111. Example 1 on Step down DC Chopper with RLE Load: Now let's have an example on Z, step-down DC short bar with an RLE load. So in this example we have a step down DC part was on r eally lot. Where's the resistance equal five Ohm. Z inductance is 7.5 millihenry Z switching frequency as one kilohertz. Z duty-cycle equals 0.5 and Z back EMF equal to 0. So here we don't have a back EMF. It is considered the ASM are in Middle Earth. So what do we need in this problem? We need i1, the minimum current I2, z, maximum current Z currently labor rebels or z delta i z average of the current, the root mean square of z Albert current, Z effective input resistance and z supply RMS current. So let's find each of them step-by-step. So the first thing we draw our circuit, we have here our switch, our diet, and are in the loot. So we need I1, which is the minimum current. We broke dizzy equation I1 equal VS over R, e bar minus one e bars at the minus1 minus Ea over R. So E equal to 0. So sister animal will be gone or B Z. Now, Z is equal to r over l t, where t is one over that frequency. So it will be one over 100 thousand Z inductance L is 7.5 millihenry, 7.5 milli, which is Dembo negative 30 resistance, which is five ohm. Now Z current. So Zed dot will be equal to 0.667. Now as the current i1 will be equal to V supply, which is at 220 volt. Okay? Xy VS is the same as the supply ends up previous example, Z resistance five on Ebola K, K is 50% or 0.5 duty cycle. Z is 0.667 E bar 0.667 minus one minus one. And this term is equal to 0, since is the back EMF equal to 0. So i one will be equal to 18.37 and bare. Now, find I2 same as I1. We will get Z equation which will we broke the before and substitute directly with z, k, z, VS, and so on. So by doing this, we will have a maximum value of 25.6 is three. So z minimum 18.37, Z maximum. 25.6 is three and bare. Now z delta i or 0 people is a difference between I2 minus I1 or Z. Maximum value minus the minimum matures 18. So it will be 7.26 and bare. Now, let's see our force requirement. We need the average of x0 Albert current. So z average, we have here only one increasing to IDO is in decreasing to i1. So, and for simplification, we draw it as a straight line. Now in order to find the average account. Okay, let's support here Zi Ben I average is the average of this current. This value, okay? At this point is called I average. The average of this current. Okay? For simplification, we can see is that average current is equal to Z maximum plus Z minimum over two. Same as getting as the average of a two component or two voltages or Docker runs or any two signals. So I1 plus I2 z maximum plus Z minimum over two, which will be equal to 22 and bare. Now our served or number e requirement is the album as a root mean square. So I R what we said that we have a charging period, Xena discharging period. Remember that here, E is equal to 0. So our current waltz continuous. So I would as a root mean square is square root of one over z period. Integration from 0 to capital T, or is aperiodic time the function of the current all square. So here we have from I1 to I2 and from IDO to i1. So we have two integrations. One over t, integration from 0 to getting to this function, which is the current charging I1 square plus integration from gating to T capital. So this one is t capital or is a periodic time for z function here, I2 r square d t. So our output ordinary square is divided into a charging part and discharging current. Now, we need to find I1 as a function of time and I two as a function of time. Here, we will not use the exponential function for simplification or we will assume it is a straight line. So in order to find the equation of line, remember that here we have four z I1. I1 is the current here as a function of time i, one, as a function of T. Okay, so I1 as a function of time, is equal to this line, okay? We assumed it is aligned. Not an exponential for simplification. Ok, because you are going to integrate a function in time containing exponential, so it will be a difficult one. Now, z first one at this point at time equals 0, X1, and the value of the current i1, which is y one. And the final value is x2 at here Katie and value of current i2. So we have X1, Y1 exit all Y2. So z function of z line assembly equal to Y equals MX plus C. Okay? The equation of a straight line. So exit all white or minus y1 over x2 minus x1. So Y2, which is I2 minus Y1, which is I1 over x2 minus x1. This is, this one is z slope of z line. Ok. Now here are u one, which is x minus and y y1, Y minus Y1 over x minus x one. So y, which is Z current i as a function of time minus Y1 over X1 over x, which is representing the z variable time minus X1, which is 0. So I1 as a function of time, is equal to I1 plus I2 minus I1 over KT multiplied by z time. So this one representing is the x, this one representing our y, this one representing our slope m, this one representing the intersection C. So I want as a function of time by substituting, we will have 18.67 plus ones are 14,520 t, okay? This is the function of Zeff crystalline. Second line here will have two points. Again, Katie, I do and T i1. By doing this ECM process as we did here for Z equations of straight line, we will have I2 equal to 25.36 minus 14,520 t minus 0.5 multiplied by ten power negative three. So by taking this equation and this equation is squaring z m n integration. We will have i out as ultimately squared. So root one over z period. Degradation from 0 until Katie, from Katie to t. This is the function i1 or the square function I2 square. By knowing on solving this integration, which is a very simple integration, we will find Eddy gets effective output current. To be 21.65 and bear Z. Final requirement is the effective input resistance. As we said before in our first example, that the effective input resistance is equal to V supply over I supply average. Ok? So V supply is 220 volt, okay? Z I supply average is I would average multiplied by k. So k is 0.5 and I would average is 22 and bare. So it will give us 20 ohm. Now we need to understand why is the supply voltage equals K i, i would average. So if we get back here, assuming that our converter is lossless converter, okay, what does our lossless mean? Means that there is no loss as z input power is equal to the output power. So in low-power cheese is VS is equal to z, v. What? I. And V output is equal to V supply multiplied by k. Ok? This proof is repeated in case off z buck converters which are going to discuss, okay, so U V supply, i0 supply, this is the input power equal to the output power, which is VS K, which are representing savvy I output. So V supply will go with VSV supply. So I supply is equal to k multiplied by i out. So we are k multiplied by i out. So we got here the effective input resistance. Now, our loss requirement is that supply root mean square value. So let's get back to z waveform. You will see that the supply is only existing during busy conduction period from I1 to I2. So I supply as a root mean square is the integration of this port square. So it will be equal to one over Z. Total videoed integration from 0 to Katie during busy conduction. Because our supply current only existed during this failed Zi current I1 square as a function of t d t. So I wonder resenting z equation of the line during the conduction. So by substituting with z equation which we obtained, we can get that supply voltage as a root mean square is equal to 15.626. You have to solve this example by your hand in order to benefit from this example. 112. Example 2 on Step down DC Chopper with RLE Load: Now let's have another example on the step-down DC short bar was on RLE load because this example is much simpler than before. So if we have a step down, DC Chopin was an RLE load and having a resistance of 0.25 on Z, supply voltage of 550 volt Z back EMF equals 0 is the average output current is equal to a 100 and bear and z frequency to a 150 hertz frequency of switching. Find is inductance required to reduce delta i maximum equal 0.1 I output average. So we need the inductance, which are called as Z. Maximum rebel to be 0.1 is the average current. So how we can solve with this simple problem for a sink. In order to solve this, we need to get delta i maximum. Okay? Z equation we will write as i delta i maximum. So delta i maximum as approximation as we did before, is equal to VS over four fl. Okay? So delta i maximum, which is VS over four f L equal to 0.1 I r would average 0.1. I would average, which is 100 MB. So Zach VS, which is 550 volt for frequency 250 hertz inductance, which is unknown equal to 0.1 multiplied by 200, which is 20. Sousa inductances required is 0.0275 Henry, or at 27.5 millihenry. Ok. So you will see that this is a very simple problem without direct substitution. 113. Step up DC Chopper with R or RL Load: Now in this video we are going to discuss a z step of DC shopper. Zach step up. Dc shopper means that we are taking our input voltage or input dc voltage and step it up, or boast z in both voltage. So Z stub abdication par, having an R or on our involute. Okay, is this is the first satire which we are going to discuss in this video. So how does the circuit looks like? We have here? At first, do we have here and inductance? We have here our switch barrel 2Z supply source with the inductance and the parallel with a diode and a Z load, which is R or R, are a little bit. Okay, so at first and we have two cases. Z first one during busy on. When this switch is on. Then our circuit will be this part which is supply voltage, the inductance and short-circuit. And this one, the other side will be Z, by which is an open circuit and V output. So two will be V output only. Now during the off period, when the switch is off, then our circuit will be V supply Z inductance, Z diode, which will be assault circuit because the tool being forward biased, then our v-out. So we have two cases here. We have our load or I l, or Z inductance, current Anansi and load current I l. Here, l means a z inductance or inductance during the on period T1, you will find that Xen current changes from I1 to I2 is the current is charging. And here, during the off period, Z store the energy and science, the inductance will go TS Rosie out. So the current will start decreasing or discharging during the off period. So how does the step up dc shall bug or zr a little Dworkin. In this type of DC Super, we use a two step up or increase the input DC voltage to higher value at the output. So this simply work is pie. At first, during the on period, we are charging our inductance, okay? Our inductance is a charged by Zach basins of DC supply. This is happening during the period when the current is charging. During the off state, the inductance starts discharging, shows a diet alone. Go is the input DC voltage. So we have here is a DC V output is equal to zag V supply. Along who is the voltage or the voltage produced from here, or Z. Current will be produced from the stored energy and Sada inductance. So as if we have here as supply voltage VS and another supply coming from our inductance. So the total output voltage will increase. So during the on period, we said that we have here the supply voltage, Z inductance and here assault circuit when the switch is on. So we need to find the equation for the current. So we know that from KVL here, you will find that negative V supply, negative V supply plus Z Vn is equal to 0 or V supply. The supply voltage is equal to the voltage drop across z inductance. So 0 voltage across the inductance is l d i over d t. Okay? And since here, our diode will be exalted in the form of linear form. Okay? We assume that Zahn, the charging, is in the form of a linear equation. So Z D i can be approximately B delta i, OK? In a time t1. So d i by d d, it means that the variation in the current, in a small DT time, there is a variation in the current and our time called the D T. It is a same as delta i in case of a time t1. So if we look at this one, you will find that during the time t1, our current it change it from I1 to I2, which means delta i. So delta I and II, which is delta i and d T will be T1. So from here we can get delta i as VS t one over L. So here you will find when we integrate this equation d i by d t, it will give us a linear equation. Okay? Not as before, because before we had an on-off mode. So r l Give us an exponential form. But here we have only inductance. Sows are charging is linear. Ok, if we integrate this function, it will give us the same delta i. But if we got it as a function of time, as a current, as a function of time, I1 as a function of time. So delta I is equal to V over L. Now, during the off period from t1 to t, our circuit will be as following. We have a supply voltage, Z inductance, and we have here our output. So give me N V output is equal to V supply plus v n. Why? Because during the charging, our polarity plus minus ok. So after orange z at time of discharging, polarity will be reverse it, okay? Because our inductance here acts as a source, which is giving a current in the same direction of original current. So again, this part we have during the charging, whilst during discharging, our polarity will be reverse it because and instead of charging told me to such forging Z and inductance will charge and discharge in the same direction. So our current i L was in this direction during the charging, during discharging tool gives the same current. So as if we have the supply voltage and Vn supplying V out. So V out is equal to V supply plus vn. We supply plus L d i, d t is a variation in the current. They'll die, which is from i2 to i1 over the small variation in time delta t zap time here, which is represented by here. This time, is called the T2, which is t minus KT. So you have the output v sublime L delta I is a variation from current from i2 to I1 in a small time called T2. So delta I was VSD one over L. So VS t one over L By zoster moving z similar terminus, we will have v-out equals V supply plus VS T1 over T2. Now we know that T1 is equal to k is a duty cycle multiplied by t is aperiodic time. And t2 is equal to one minus k t, or it will be equal to G, the total period minus KT, which is t1. Now by removing this, we will have k over one minus k. So we will have our V supply, one plus k over one minus kx. So we finally get a v output as VS over one minus k. And d k was from 0 to one. So the value of k will produce or cos z v Albert to be great or Zanzibar supply voltage. 114. Step up DC Chopper with RE Load: Now let us discuss a Z stub DC short part of B2 is an RE load. We have here. Our load is different from our previous video. Our previous load was resistive or an art a lot, a resistance. So with an inductance, here we have on our waters, are there France? So difference here is that xii can cause the current to be continuous or be discontinuous. So r e is a special case in the step of DC. So our DC shoe, but again, having a guess of z on, switch off, switch when it is on, we have a circuit like this one as before. And this one in case of it is off when our inductance is discharging Rosie diode and through our load, we will have from charging from I1 to I2, Zen from i0 to i1. Now, what are we going to do in this lecture? We are going to find the value of i1 and i2 and delta i's and if France between I1 and I2. So the first step we are going do is I'll analyze our circuit in case of zinc on estate and analyze it in case of Z off state. So Duolingo is the only state from 0 to t1. So during this hour, supply, DC supply voltage, we'll start a charging or storing energy inside our inductance. So our current will increase from I1 to i2. And from KVL here, our supply voltage is equal to z v l or load current. So V supply here will be l d i over d t. Okay? So it's oblivious case within Toronto to find I1 and I2, all what we need is to get z V output as a function of V sub blowing, Okay? So V output as a function of V supply was one over one minus 6y, okay? The case where we get that V equal to V supply over one minus k. Okay? Is that what we wanted in a previous video? But here in this case, we wanted to find I1 and I2. So in order to do this, we need to do is as same as we did in the step-down. Okay? So we have here as the current is increasing and this current is called I1, okay? Current as a function of time during charging period. So d psi is equal to V supply multiplied by d t over l. Okay? It's the same equation but cross multiplication. So by integrating this part and this part, we will have i one as a function of time equal to VS over L t. Because this is a constant and it's stimulation with respect to time is de, ballasts, certain constant a. Okay, as any integration in our life, now we need to find the z value of E. So again, as we remember or as we did in the step-down, we will say at time equals 0. Our current would be equal to I1. So at, to get a using the initial condition at t equals 0, at the initial point, our current is equal to one. So from this, this part will be equal to 0, so a will be equal to one, okay? Because I1 as function of time equals one at this point. So we finally get I1 as a function of time VS over L t, VS over l t plus a, which is I1 capital I one capital is minimum value of current. Now we need to get a z value of i2, i2 by using the final condition. At t equal katie, Our current will be I2. So we can get I2 equal to VS over L at t equal KT plus I1. Now, let's analyze our circuit in case of Z off period. So we have here I2, I1 decreasing linearly, okay? Same as before. Now by blowing a KVL, our Albert here is equal to V supply and plus v L. Or we can say VS plus z voltage, VL minus z voltage drop here plus e. So let's see. At S open to by applying KVL, we have negative V supply plus V L plus I two R plus E equal to 0. So supply voltage here is supplying all of these elements together. So VS, which is a supply voltage, is supplying Z inductance L DIE by DT plus z. Voltage drop across the resistance R y2 are bluffs Z back EMF e. Okay? So here our supply is supplying the inductance resistance and the back EMF E, OK, z d i by d does not have any negative sign. And the reason for this you'll find the Now, if we'll get back to Zim step of DC show partner was on our elute. Let's get back to z equation. In case of z, there is no problem. We have delta i as a positive value, okay? They'll die here. Is I2 minus I1, which is our Boston value. Now, in case of Z, load is off, we reverses, polarity, is reversed. A year is an utiles or basins of decaying. Ok? Or to be more specific, you'll see that the output is equal to V output here, or Z voltage. Let's drowsy equation. From KVL here, our supply voltage supplies the inductance and sublime our absolve. The supply should be equal to V output plus z voltage here, V L. Okay? So the output is the same, but V L is L, d i over d t. Ok? So d i here is when we changed it to delta I over a small time T2, delta I over small time t2. Here. Remember that delta i is i2 minus one. Ok? So delta I is I2 minus I1. But actually here, let's say get back. You will find that here. From here we are decaying. So delta i year is negative delta i. We variation in the current is I1 minus I2, okay? Since we are decaying, so I1 minus I2 is equal to negative delta i, because delta I is i2 minus i1. So here we need to add a negative sign, okay? In order for this to be I1 minus I2. So you will find that here, this voltage is now when it goes here, it will be V output equal to V supply plus this B2, which is delta i over 2m. So the reverse here in ZAB polarity or assuming here and negative, this negative or this one is due to substituting with negative Delta i. Ok? So I'll find this equation is similar as this equation. Okay? So originally when we are getting z equation for z off period, we apply KVL here saying that V supply is equal to v plus v n. And the output here plus VL is the LDIDT. Ok, this one, which is in the step up with an loot. So D I by D T, When we are simplifying it, the i is delta I, or to be more specific, negative delta i. So from this we can say V out is equal to V supply. Blas LDIDT or negative the eyeball entity which she became Boston. Okay, from here we obtain disease one. But in our case here, let's get to this equation. In this case, we said that V supply again equal to V L plus R multiplied by R plus E. So we said V supply equal to V L plus R plus E. Now, we also substituted as before, l d i2 of RDT. The i2 is the current as a function of time during Z decaying devalued. Okay, this one is i2. But in this case we didn't substitute who is delta i? If we substitute with delta i, it will be negative L delta I over delta t. Ok? But here we won't I do as a function of time. So we keep this equation as it is. And this is as a step down DC short part, where we said that this one is RL load, which it can be ICT state glosses the transient. So i2 as a function of time, is equal to the steady-state value when I by d t equal to 0. So I2 will be VS minus Ea over R, V S minus Ea over R. Or we can assume that the inductance does not exist. So I2 will be equal to VS minus Ea over R, V S minus V over R plus the transient to component a ebonite if T over tau, where tau is L over R as before. So we have your i2 and we would like to find a. So add time here equals 0, the current will be i2. Same as the step-down decision at time equals 0. The current is i2. So i2 equal VS minus z over r plus a. So a will be this value. Then we will take the a and substitute it here. So i2 as a function of time VS minus Ea over R, which is the steady-state component, plus i2 minus VS minus Ea over R, a bond negative t over tau, which is the transient component. So we'll finally get our equation. So in order to check if this equation correct or not, if we substitute with time equals 0, then we would have i2. And this one will be one minus one, which is 0. So i two at time equals 0 is equal to two. Now we need to find I1. So to get ZI1, we will use at time equal one minus k, t or T2. Okay? So difference here. The time t2 here is t minus KT. So t minus KT, we will get Z current will be equal to I1, which is the final value. So I1 will be this equation. So we have here our second equation and our first equation obtained during busy charging period by solving these two equation as same as the step-down DC shopper, we will get that i1 will be equal to this larger equation. And i two will be equal to large equation. And same as before equals teen is epidemic time t. That resistance over L. And the ripple here will be equal to I2 minus I1, which is VS over L. Katie, the difference between these two parties equal VS over lf K. Okay? So this was that both of the equations of xhat, step up DC shop in case of prisons, of RE load. 115. Example on Step up DC Chopper with RE Load: Now let's have an example. Anzac step of DC short borrow is an R e root. Ok? So our example here is that we have a step of DC converter or a DC show bar with a back EMF E equal 47 volt. Resistance equal one Ohm. Z supply voltage equal 24 volt, ands inductance equal ten milli Henry. The frequency of switching is five kilohertz and k equal 0.5 or duty cycle. Now what do we want to find Rosie waveforms of z supply current, Z battery current Z, I would voltage across R3 and lewd RE, load. Okay? And also we need to find the minimum instantaneous current, which is I1, beak to beak inductor River, which is delta i z average value of Z battery current I average the effective input resistance R i and as the supply root mean square current. So this looks are larger problem, but it is an easy one. So the first step we draw our circuit here we have the supply voltage, Z inductance. And here our switch, which is MOSFET, is in Biot Z load, which is our elude. Now's the first requirement is the waveform of the supply current. So we need to draw the IS. So if we concentrate here, you will find that this one is the ISI, which is equal to z inductance, the current. So as we remember that the inductance was in this form, we know that the inductance current is charging from I1 to I2 during busy on period and during the period from i2 to i1 discharging. So I supply is equal to the inductance current from KCL. So Duolingo on charging during off discharging. So I supply is very simple. Now is the second requirement is the output voltage. The output voltage, which is the same as battery current, Z battery current is the current bossing years Rosie, e, m, f, m f here is considered as our battery. So z battery current is same as output current. And our, our current only exists when the MOSFET is off. So if we look at the waveform and we'll find that the current from i2 to i1. And this charging, ok. This is considered as our output. Now we need to find z and output voltage is 0, voltage across R3. So during the on period, we said that this one will become a short circuit. So our circuit will be like this k, let's draw it. So during Z and on videos. We have here, our inductance likes us. Okay, V supply, this is battery. So let's extend this a little bit. And from here, and here we will have short-circuit when this one was on. But this one will be reversed biased. The why, because the back EMF produces a current in this direction. So this diode will be reverse it and it will be an open circuit. So we have heated like this and open-circuit Xin resistance ZIM back in F. Okay, like this. So this one during z on period. So we need to find out voltage. So the output is plus minus R cross z are elute. So here we don't have any current because our circuit is an open circuit. So V output will be equal to Z back EMF E. So let us see here during busy on period, our output will be the back EMF E. Let us see during the on period, during the on period will have a supply voltage, V As and inductance. Then short circuit of the diode, then our elute or E, same as we did before. Okay? R and E. So the voltage V output will be equal to the current year i2. Small, of course are not capital I two multiplied by resistance. Blas is the back EMF assembly. So here is the output is I2, R, loss is the back EMF. So we need to find the current of i2 and multiplied it by resistance adding e. So here, this one is considered the as i2, right? So i2 at this point, for example, I2 multiplied by the resistance velocity. And at this point is o to I2. Here is the current as a function of time, which is at this point is one multiplied by the resistance. So this one, by connecting these two lines, we will have our final value of V out. So again, V output is equal to e plus z current year multiplied by resistance. So multiplying z current by resistance tool be i2 or are you want, are then adding e here and here. So as if it was this curve or but shifted a little bit, K plus z back EMF, ie. So we draw our Siri and waveforms which was required. Now, our first requirement is minimum in instantaneous load current. So it needs one. This is the minimum load current. Load current here is the minimum value, which is I1. I1 is equal to this logic equation. And by getting Zed as similar as we did before, we can get I1 equal 0.88 and they're now z. Second requirement was the rebel or Z back to be beak to beak. And Dr. rebel, which is x0 delta i. So delta i, which we proved the from other equation, is equal to I2 minus I1. We can get I2 from this equation and subtracting it from I1. Or we can get it from this equation. K, this one, delta I equal VS over L, Katie. Okay? So z will be similar to each other. Like this. Getting the toy equal 0.24 and bare. Now, for our third requirement is the average value of battery current. So how we can get average battery current. Okay, let's see. So we have here is this one is a recurrent and would like to get z average. So simply z average, or average of any wave form is equal to I average is equal to one over the period. One over video to GSD integration from 0 to t. But our waveform only exists as from here to here. So it will be equal to 0 to n from Katie 2T. For our function here, which is this function or this straight line. So we can get it by using this method. Okay, the integration, we can say it, let's write it to make it more clear. Integration from gating doe t for this function, okay? One over t outside. Okay? And this equation, the straight line. So it can get Z equation Boyd, as we did before, y equals m, x plus c. By doing this, we can get Z current as a function of time and put it here. D t. We can get our battery average k or the average of battery current. This is the first method. And also an instead of writing this, we said that Why two minus one over x? Two minus x one is equal to y minus y one over x minus x one. So we can get Z, Y as a function of x or z current as a function of x and substitute in the integration. This is the first method. Second method we can get z average current of this function. Z average current is i1 plus i2 over to, okay? This is the average current. But this current only exists for this function for Z period, half of the period, okay? Because z, this one, which is KT, is equal to t minus KT. Why? Because our duty cycle is 0.5. So we can say that we have our current only for the hovels cycle. Okay? So we can say 0.05 multiplied by I1 plus I2 over two, okay? Which is half of the cycle. Or our second method is by using that power output equal to power input or power in political power output. So we SIS equal V average I average, which is our output. So our v average, we said that it is VS over one minus k. So IS equal to I average over one minus 6y, okay, from this equation. So k equal to 0.5. So our supply current will be equal to I average over 0.5, which is two i average. And we know that the supply current average, this one is I2 minus I1 or I2 plus i1 over two. So the supply average current is one and bear. So z average will be half of Z I supply. So this is the second method. Okay? So we have here as three methods. Number one, which is direct one by integration from Katie to T. Two, this function over T, we can get the battery average current. Second method is by knowing that the average current here, okay, is one over T, z average multiplied by KT, KT over T, representing z period or the time or where our average exists. Okay? Zim, sir, the method which is by assuming z input power equal to the output power, we can get that V average equal VS over one minus k. So we can get supply current equal to I average one minus K. So I supply equal w z average current, or the average battery current and supply current as an average from this curve by integrating or by getting Z maximum and minimum divided by two. We can only use this when we have z variation here existing for the whole period. Okay? Now, this equations, this equation is valid when our M k equal 0.5. If our key not equal 0.5, then we will change this part to z time where our battery exists. Okay, so this time is similar to Katie or one minus k t z are similar because z k equal 0.5. Now, let us see the force requirement here. The effective input resistance. We know that from the step-down decision per that z, effective input resistance equal resupply over I supply average the supply voltage given as 24 volt Z I supply. We got it before with one and beer by knowing as z some measure of dorm maximum and a minimum over two. So our, our input will be 24 over one, which is 24 ohm. Now, our lost requirement is supply current as a root-mean-squared. So here, this one, again, again, again, this one in case we have this period equal to this period. So I1 to i2 is equal to item two, I1 as an area, okay? As an area. But if z time or z k not equal to 0.05, we cannot use zig zag multiplication by two. What I mean by the root mean square current is equal to one over the period integration from 0 to t capital Z total period. Josie function, all square. Okay? This is the general root-mean-squared equation which we will discuss the before. But here we can say one over C period and get one integration only and multiplied it by two. Okay, from 0 to t1, this is Z charging period. And multiplied it by two, we can get the root mean square. This function is z, function of charging a period, okay? And squaring it, integrating Motorola by two will give us root mean square. So this one would xy multiplication by two, only valid when our k equal 0.5? If our gain not equal 0.5 Hz, then we will say that one over t integration from 0 to Katie to this function of Z line and Zn plus integration from t1 to t two this line, all square of course. But since our and the period is 0.5 or our K 0.5. Therefore, this area equal to this area. So when we are squaring this m, we will have similar area here, similar to this area. So we've got one area and multiplied it by two to get zirconia square. So here I1 and I2. Z function here will be I1, which is the interception with or intersection with y axis plus MX. So let's get the equation. I1 is the intersection. You notice that z y equals m x plus c. So y here is our current as a function of time. N is z delta y over delta x, or the slope of the line. So delta y is I2 minus I1 over delta x z distance which we removed all the time which we passed during transformation from I1 to I2 is t1 multiplied by z, x or t here, plus C, which is the intersection with y axis. So by integrating this part, we get iOS as min squared equal one and bear Z period is one over the frequency and T1 is KT. Ok, there is no problem in this, only this equation, this two is valid when we have k equal 0.5. In case of gay, for example, equal 0.25, then this area will not be equal to this area. Okay? So that was our example on the step up decision, but try to solve this exerted by your hand in order to understand and benefit from this example. 116. Buck Regulator Part 1: In this video, we are going to discuss as Z buck regulator in Zionists atoll videos, we also are going to discuss Z bolster regulator and z buck-boost regulator. So what is a buck regulator? Bucket regulator is step-down when DC shopper, okay. It is a step down, DC shall part. But in addition to being a step-down, it has an LSE filter adds the Albert end, which is used to reduce Z voltage rebels. Okay? So let's see here our circuit, as we remember, that's a step down DC shopper. In case of a load, for example, we had a switch and then we had our load. And we had our diet ok circuit, which we'll discuss in a step down DC shopper with an RLE load. In this case, we had z v Albert and z diet and switch. Okay. Now Z buck regulator is different from Z regulator which we discussed it before, which is z step-down DC shopper. So difference is, is the addition of an l c filter. Okay? This LSE filter is used, uh, to decrease z ripples at the output, ok, so it decreases the rebels. Now, this type of converters is used in SMB S always associate mode power supply circuits as this circuits, whereas a DC output voltage, it needs to be lower than DC voltage. As all simply, it means that we are stepping down our DC voltage. So what does ask switched mode, power supply means switched-mode ball supply is simply a power converter. This power converter, like this one, is user to utilize or utilizes a switching device such as MOSFET transistors or any switching device that have a high frequency of switching. Okay, for example, in this case, we have this switches such as the MOSFET, the transistor, which is having a high frequency. It is continuously on and off. And we have all we use energy storage devices such as the capacitors and inductors to supply power during the non conduction state of the switching device. So when the switch is on, we are supplying here our power or our Albert with Rosa inductance and Zika Buster and during off more than we have only xi1 doctrines and combustor. But in case of, for example, Z, step-down, step up, DC sharper, you will find that we inject the Bower to Z inductance two. Store energy inside it, as in yours, this power in the owning xenon conduction state of the switching device. So in certain cases, enzyme guess of z stepped down and step up or step down, which means here, buck regulator and step up. Bocce means a here is you boast regulator was other Francis between them is that we have an LSE filter which is used to decrease rebels as the output decreases the delta V, okay? Or is a variation in the voltage at the output, okay, to boy, to be more DC or direct current value. Now, Z buckling you later, which we have here. And let's see how it looks like during the honest state and during the off state. So here we have during the on-state switches on. So z by dt will be reverse biased. So we have here our switch as a short-circuit, our inductance, our capacitance, and our output. So we'll find that the voltage across the diode is equal to z, V supply. And this voltage is equal to v sub y because wins this one is assault circuit and apply KVL here. Find the voltage across Z die during the reverse bias is simply equal to so supply from KVL. Okay, so our, let's draw it so you can understand more. So higher our diode, okay, should have been exist. So this diode having a voltage across it equal to the supply voltage. Now z second sings the IL increasing. Ok, we are charging our inductance using Z supply. Okay, is this supply injected current 2Z inductance and a charging it. So i equal to i n z supply current is equal to the inductance current. Supply current is the same as the inductance current. The id m, m, m equals 0, Z diode here, current equal to 0 because it is an open circuit or because it is a reverse surprised. Of course, when we are talking about an ideal diet. I equal to i average, we assume that our current year approximately equal to I average. This is an approximation, but actually the current is increasing and decreasing, as we will see now. Z cover stanza current. The current year is equal to from KCL I L minus i, n minus i output thrown keys l, z supply current, give guarantees to Zika capacitance. And if Karen to 2Z output. In case of Berio, when z switches off, you'll find that z inductance starters providing current. Ok. So you will find here the inductance produces a current through two charges, he inductance and to our loop. Ok? And here you will find that the voltage across Z equal to 0 because it is assault circuit or because it is a forward biased, the inductance current is decreasing. Okay? Because it has a stored energy from our supply Xin, It is now discharging because there is no supply. Z IS equal to 0. Supply current equal to 0 because the switch is off. Z dot current equal to Z load current. Ok? So you will find all not Z load current is the inductance current, because here IgM equal 2s inductance current. Z would equal to or approximately equal to I average. Ok, so here we assume this is an approximation, is that our Albert is on I average and am approximately equal to I average. Do 2a less amount of rebels. And the output is nearly equal to V average due to xylose amount of rebels. But we will see now is the X1 waveform. Also our combustion. So current equal to I L minus i R would seem as here. Now, let us use our waveforms. We have here is the voltage across the diode during, on estate and off state during an honest date. Instead of our switch here, this one is reversed biased, so it will be equal to V supply here during off, it is forward biased, so it is a short circuit. So the voltage across it equal 0. So here V supply and here equals z. Now our current or the inductance, the current actually, we said that during the on state it is a charging from I1 to I2, same as a step down decision. Borrow is on RLE load during the off period, it is discharging. And the difference between i2 and i1 is delta i. Now for the supply current, supply current equals 0 during the period and during the on period equal to I l. So I l here, the inductance current, same as the supply current for Z diode current I d M. You will find that during the on state ID m equals 0 and during office state equal to Z inductance current. This Esperanza equations or the relationship which we obtained in our own state and off-state simply. Now, our output, output current here. We assume that it is nearly equal to a constant value called i average due to reasons of LSA filter which decreases the rebels. Okay? Now, Z inductance or capacitance current. What does a capacitance current equal to? You will find that IC equal to IL minus i Albert IC equal to I L minus I L. So simply or finds at this one, I L minus I output L z inductance current minus I, which is nearly I averaged due to the decrease in ribbons. So if we subtract this from this, then this curve will be shifted downwards. Okay? So it will just point will be i1 minus i average. And this one i2 minus i average. So it will be i1 minus i average, i2 minus i1 average, and this one, I1 minus i average. So as this here, you will find that in this part, that z value here of I average is greater than one, greater than the minimum value of current, and less than Z minimum as the maximum value of current. That's why this one is negative. And this one is positive. Z distance from here to here is delta I y because this one minus this one give us deltoid I2 minus I1. Okay? And I average Hugo is I average. So it will be I2 minus I1, which is delta i. So this part alone is delta i over two, and this part alone is delta i over to soul finds this one is deltoid over to this one is delta i over two. Now, for Zika stance, note that the capacitance IC is the current of the capacitance is equal to C dvc by D T, c d v by d t. So it will equal to capacitance multiplied by is the differentiation of z voltage of z combustor with respect to time. So finally from this we can get that z value of Z Tibetans as the voltage of the capacitance equal to one over C integration of Z current and eating. So you will find that here. That's the voltage of zag capacitance equal to one over C. So this one is the integration of this function. So Vc is equal to the integration of this line which is occurs. And this one integration of this line, which is another curve. Okay? And the difference between the maximum and minimum value is z delta VC, okay? So where did we get zapped, VCM or DC voltage across the capacitor by integrating f z currently function or z equation of our current or current waveform. So by integrating as a straight line, it becomes a curve, and this one becomes another curve decreasing. Now, Z most important steps for proving busy current equations in guess off our buck boost and buck-boost converters as three types which we are going to discuss. So this steps of which you are going to do is S3 or in Zen and six videos. In case of broaching in, in buck and the boost and buck-boost converters. So what are the steps? First step is number one, we apply KVL during his air conduction period or the period of our switch in order to get an equation for ripple current or delta i. Second time applying KVL during off period to get another equation for rebel current. So third step, we are going to equate xhat to equations of cerebral current, delta I and deltoid y to get Z V output average as a function of the supply voltage, the average output power, output voltage with respect to, to the supply voltage. And step number four, we will assume a lossless converter, which means that the input power is equal to the output power or the ASIS equal V average I average. The reason for this in order to get a relation between the supply current and I average. Okay? So we get the relation between supply current and a Z. I would average current. Step number five, we will get an equation for the current ripples. Okay, we had here an equation for recurrent rivers and on-site equation. Now we are going to get z equation for the current ribose. We wanted to find z rebels as a function of knowing parameters in our circuit. Is step number six, we are going to get any equation for voltage ripple on the capacitor. We need to find Delta V C, or delta v, the voltage variation or voltage ripples on Zika Boston. Number seven, we are going to get an equation for the critical inductance value. We need to find the minimum value of inductance for continuous current. We are also going to get an equation for z critical capacitance value. We need to find the lowest or the most. They're listed acquired the capacitance. In our case, it's the least amount of capacitance in order to get Z and V minimum two p z. Ok, so let's go to each of these steps in our buck converter in the next video. 117. Buck Regulator Part 2: Now let's discuss a z steps. He used it to get z values, which we would like to find. So the first step or step number one is getting an equation of Zerubbabel current during Z conduction mode or during the on period. So first, during the on period, we have our circuit like this which we discussed before. Now what are we going to do? We are going to find z current rebels. So in order to do this, we are going to apply KVL here, scurvy l from the supply 2Z output. So we will find that V supply is equal to or negative V supply bolus vn plus V out equals 0. Negative resupply plus VL plus v equal 0 or V supply equal to v n plus v out z v l is l d over d t equal VS minus V average. Ok? We said that V output is nearly equal to V average, okay? Sinces the ripples is very low. Now with z or z inductance voltage l d over d t d i, or is a variation in z current is equal to delta i in that time of charging, which is T1. Ok, as we did before, delta I over delta t, which is delta i over T1, which is the conduction time V S minus V average, V average is at v out. So from this equation, we can get delta i equals V S minus V average over L T1. So this was our first equation. Now, step number two, we are going to get an equation for z rebel current during the off period or the switching off period. So let's draw the circuit, our circuit walls like this. We will apply again KVL Rosie output. So VL is equal to V out. So V L equal to V output, which is V average. And Vn is LDIDT. So it will be L delta I over delta t. So delta t here is a time t2 zip buy-in of switching off or z non conduction period, same as we did before. So we have here delta i as a function of T1 and delta i in a function of T2. Now, step number three, we are going to equate these two equation. So step number three, getting an average voltage as a function of sublime voltage. So delta y1 equal to delta i two or z, z ripple current in the first step, or z equal to z second equation, delta i again. So from this, you will find that we can get V average equal K V S, same as z v average which we obtained, or XVI output which we obtained in step down in DC shopper. Okay, it's the same equation. Now, in step number four, we are going to get the supply current as a function of average current by assuming Z lossless conductance. So power input, power output. So VS equals MATLAB IS equal V average, which is KV S multiplied by I average. So supply current equal k i average. So from this we got z v average and we got the supply current. Now, step number five, we are going to get delta i. So how we can get delta I. Now, you know that is that the total period T is equal to T1 plus T2. T1 is the conduction time and T2 is non conduction time. So how we can get D1 and D2 from the equations which we obtained equation number one. Let's get back here. And equation number two, we can get T1 function in deltoid and D2 function in delta i. So we get here is that delta L over L multiplied by l. We assume minus V average plus delta I L over V average. So we have here delta i as a common factor. We will get delta i and some factors here. So delta i finally equal to V average V S minus V average t l VS. Now, you know that Z T or Z periodic time is equal one over the frequency. It will be one over f. And the nose at v average is equal to k V S, V average KV s. So by taking VS here as a common factor, we will have VS square one minus K. So one VS here we'll go with a VS here. So we'll finally get as an approximation, or after simplifying our equations, delta I equal to k V S one minus K over L f. So this equation is for our Rebel currents. Step number six, we would like to get an equation for the voltage ripples for Zika stock. So we need z voltage ripple, which is Delta V C. Okay? So how we can get this, we know that delta VC, or the voltage from here to here, z variation in z voltage is equal to one over C integration of the current DT. So you will see here is that z delta Vc is a variation in voltage from where from this point is the minimum value to this point, which is the maximum value. So in order to find this z delta VC, we need to integrate which current from here to this current. So how we can integrate this current is the integration of the current. Means we are talking about the area under the curve. We have here, this current, the area between this line and the x axis. So this area is simply the area of a rectangle, a triangle. So the area of the triangle is equal to half base multiplied by Z height. So the 1.5th one base, this one, we have here 1234. So this four parties, this for borders are divided all of them and forms our broad at time t. So this part is equal to this one is 4r Z period, which is our base and height, which is delta i over two, as we explained before. Now, our second triangle will be half d over four again because it is z base multiplied by the height which is delta over two. So by simplifying, we will get and delta VC equal delta I over eight FEC. And the we got z delta i before. So by substituting was delta i, we have given one minus k over LF. So finally, delta VC will be equal to GAVI Es one minus K, it F a square LC. So this our equation for the voltage ripples. Step number seven, we would like to get zx oppression for x_ critical inductance. X_ critical inductance means a minimum value of inductance two and having a continuous current. So when he Z critical case is a critical case is when our current I1 equals 0. Okay, this is a minimum value. So here, when we draw our function, it will be I2, I1. And this one is delta i i one is equal to 0. So delta i here, which is I2 minus I1, is equal to i2, or delta I0 is equal to two V average. This one is v average, is this one is v average. So delta I equal to two V average. So delta i equals I2, because I2 minus I1. And average current of the capacitor is equal to 0. Where did we obtain this? We know that z Tabasco, z average current investors is equal to 0 and average voltage in inductor equal to 0. So here, i average of the combustor is equal to i2 over two z average year. Is equal to delta i over two and delta i is equal to i2. So I average this one is equal to delta i over two, which is i2 over two. So I2 equal to i average. And we know that delta I equal to I average where because I2 similar as delta i, i do similar as delta i, delta i, which is current rebels equal to i average and delta i. We obtained it before, GAVI S1 minus K over LCF. So I average. So Zach LSE here, by equating these two equations, we can get a z value of LLC. And I average simply because we have here are resistive load, it will be V average over r. Or even in an inductive load, it will be the average of our sensor z, average voltage equal to 0 across the inductance. And the v average equals KV S, as we explained before. So Zach, critical value of inductance equal to this value. So we somebody OK, z, when we want as a critical condition or Z critical inductance, we bought I1 equals 0 and by using some equations we obtained is a critical value of inductance. Now, forwarders that continuous operation, we need to have an inductance at least greater than this value or equal step number it for our combustions or is a critical capacitance and minimum value of capacitance at all ensure zap continuous or voltage. So when this happens, when V1, well to 0, V1 is the minimum voltage value. And this one is delta VC. So delta VC equal to v average and delta VC equal to V average. Why? Because this part equal v average and this part equals V average. Now we attended before again delta VC from our proof, KV S1 minus k, it L C, F square. Now we know 2V average V average is equal. Caveats. Saw. Finally, we can get that, that critical value of the inductance or capacitance required one minus k over 16 L F square. So z, summary of our laws which were discussed, V average equal two caveats and I supply equal k pi average, same as we did in a step down. Dc SOPA delta i, which we obtained before, was GAVI S1 minus Q0 over Fn. Delta VC owns a variation or Z. Voltage ripples equal to this function. Z critical value of the inductance equal one minus k bar over two f c critical inductance is at i1 equals 0. The minimum value or require the Forza continuous current z critical value of z capacitance is equal to one minus K over 16 LF square, which is the minimum value of capacitance to ensure Zen contiguity in z voltage. So what are the advantages and disadvantages of our buck regulator? Number one, it requires only one queue. It requires only one switch, as we said before, although as we discussed and now it is simple. Efficiency is greater than 90%. Z d i by d t can be limited by the inductance L. As we remember that in z power electronics as Z inductance is used it to limit z variation in current. Z inductance does not allow sick the i over d t two and surpass a specific value. So L inside our circuit used to limit the variation in z current. One of the applications of z buck regulator, it is used in the solar photovoltaic cell, electrical and hybrid vehicles and robotics. So z or z1, which can be used in the solar charger controller. In the solar cells, you can talk to step down z voltage, which is used for making a z voltage suitable for our charging of z batteries. There are disadvantages of our buck regulator is z, input current is discontinuous and the content high rebels and so filter is required. The current input, as you remember, that's a supply only exists during Z switching on and during off period. It does not exist. And that's why when the current is discontinuous or when z waveform is available during a specific Beirut and another bureau does not, is not available. This causes high ripples and high harmonics at the output. So this causes Z, existence of filter required to remove this harmonics. The protection of electric circuit is required as a short-circuit can occur on the diet, Z diode is parallel to our supply. So if assault circuit happened on the diet Xin, What does it mean? It means that our supply will be short-circuit. So this case is very dangerous. So we need to do some protection for our diode. Can only produce output voltage was one polarity. We can only produce poster voltage. We cannot produce negative or reverses polarity. So now let's have an example on Z buck regulator and see if it is easy or not. 118. Example on Buck Regulator: Now let's have an example on Z. Buck regulator. We have here at buck regulator with a supply voltage of 12 volt z, average voltage of five volt. 0 resistance equal phi form z. Voltage ripple is 20 millivolt z frequency equal to 25 kilohertz and z delta I or Z current triple equal pointed and bear, find z due to cycle k, z inductance, capacitance, Z critical inductance, and critical capacitance. So what are we going to do? We are assembling, knows that for the first requirement that we have the relation between the average and supply voltage, we know that z v average equal K V S. So V average, which is five volt, V supply, which is 12 volt, we can get a duty cycle of 0.4167. The second requirement, which is inductance, we know that z voltage ribbon or as the current ripple is equal to k v S1 minus K over f l. And this is from Zillow, which we obtained in case of the buck regulator. Delta i is given as 0.8 and bear ZK is 0.4167 supply 12 volt ZK, 0.4167 frequency for 25 kilo vertice, z inductance is the only unknown. So we can get our inductance similar for Zika Boston's, but here we will use delta-V. Delta V, which is 20 millivolt equal to GAVI Es one minus k over eight LSE F a square, which is as same as before. This will give us an capacitance off 200 microfarad. Now we want a Z critical inductance and critical capacitance by using the Z director laws which we obtained before. In z buck regulator, we have Z critical inductance one minus q r over two F and critical capacitance one minus k over 16 square. So by substituting year and the year, we can get the minimum value of inductance and the minimum value of Tibetans, you will find here is that the capacitance He is a critical one is 0.4 and z1, which is used to provide delta VCU 20 millivolt is 200 microfarad. So this 1.24 is a critical and 200 for our problem. Here we have a critical or 58.3 stream microhenry. But here 145 microhenry soil was at critical is of course, it is Zan Zi value required in this problem. So this was a simple example on Z buck regulator. 119. Boost Regulator: In this video we are going to discuss as post regulatory. So what does Z bolster regulator? It is apples to convertor or a step up converter. So it is a step of DC sharper or these C to DC power converter. That step's OBC voltage. It increases our input voltage while stepping down our current from its input supply to its output loot. So simply our step, DC sharper increases our voltage, but at the same time it steps down z current. Why? Because as you know that z power is equal to z voltage multiplied by current. So if we have z power equal to Z output power, so if the voltage increases, the current decreases. So it is as step up DC show part. So this one is the circuit of the Postal Regulatory which causes as step up. And you'll find here a capacitance which is user to decrease z voltage ripples. And our, our output or the output current and V output is assumed to be constant with a value called average and v average. Zika bus tours user to decrease the voltage rebels is same as the buck regulator. Now lets us see is our waveforms. Here you will find you. We have two cases. When the switch is on, our circuit will be like this. Let's say get back 2s as switch. When the switch is on, you will find that this half is separated from this half and the light is open circuit. So during the on period, this one is a short-circuit. And you'll find here is that the supply current is equal to z inductance current I m equal two ends. Inductance current is increasing because our supply is charging or storing energy. And sides, the inductance IDM equals 0 because our diode is an open circuit ICU equal IN Z current through our transistor. Or a switching device is the same as the inductance current, same as the supply current. The capacitance current econ negative Ali Albert, you will find here is that our capacitance here, the current of Zika persons are assumed to be entering gate. This is an assumption. Ok, we assumed our direction for the cover stance. This direction, because this direction is for a charging our combustor. And I output is assumed in this direction because the current is going through siloed. So you will find here from KCL is at i Albert equal negative IC, I would negative IC or IC negative ion. You will find that, that savvy l equal VS voltage across the inductance is the same as the voltage of the supply. Now, you will find here during the off period when Z is switch is off and the light is conducting. You will find that here, and that's the inductance current starts stored charge. Z supply current is again equal to the inductance current. So i is equal to I, L here and l here. The diode current is the same as Z lot as the inductance current, or same as the supply current from KCL is the current here is similar to each other. Z v L, or the voltage across the inductance is equal to from KVL here, equal to VGS minus V out, which is V0 average V S minus V out, which is assumed to be V average or a corner Santa value. Iq is the current through the transistor, or our switching was equal to 0 because it is an open circuit. The current here is open circuit because 4x equals 0, because it's an open circuit. The current IC from kissy n is equal to I L minus I Albert. And I went is I average. So I L minus I, which is I average. Now, let us see the wave form. You will find here that during the on period Z inductance current or the supply current is the same as the current is charging from I1 to I2 increasing. And during this period it is decreasing from i2 to i1. Charging, discharging. Now for the light current, current here equals 0 during the on period, because it is an open circuit or reverse of ice. Here, it will be equal to Z supply current, iodide equal to supply current or the inductance current. So it will take the same waveform, IQ or Z current of Z and transistor only existence during the on period and equal to supply current. So it will it charged again, same as supply, same as inductance. So z capacitance current here, i c is equal to negative I out. And I out is I average. So it would be negative i average. During the on period. Now, during the off period, our capacitance current equal to either m or supply current minus I, I L minus I Albert or IL minus average. So it will be inductance current is this one, I2, i1 minus i average. Here you will find that the current does not go to z negative part because z and z average year is lower than the average in the case of z buck regulator is the average current is much lower as n z buck regulator. So i1 is a still larger Zan on average. Now let us see the voltage of z capacitance. Here, the capacitance is simply equal to z voltage equal z current equal to C, d v over d t Z current Rosetta buttons. The differentiation of the voltage. So the voltage is simply the integration of the current. So during the on period here, when we integrate this part, it will give us a straight line. And during gives you own videoed here, which is asked to read line by integrating this one, it'll give us a straight line. The straight line is decaying. Why? Because our capacitance has a stored energy. So it is discharging through our loop. So the voltage across it is decreasing with time during the on period. During the off period. Xe and xe capacitance is a charging. So the voltage will increase again. So during the off period it is discharging, during the on period is charging. During the on period is discharging, as you see here, capacitance discharging through zeroeth during the on period is increasing. During this period, which is Z, current is supplying the capacitance and the charging gate. Now let's see z v l, or the inductance voltage during the on period VL equal VS, VL equal VS supply current from Xie KVL. And during the off period, we said that that inductance current equal to supply minus V average. So supply minus v average and supply is greater. Of course, Li Zan Zi Zi supply is less thans I V average because as ACIS is up, boast, regulate, it increases the voltage. So V average is greater than V supply, okay? Because v average is the output voltage and it is a step up, so it increases the voltage greater than the supply. That's why is this part is negative. So what are the steps which we are going to do again is a seems SIP Sochi we did inside Z buck regulator. First we are going to get z equation of ripple current during the on period. So during the on period by applying KVL and here we have z, V supply is equal to the inductance voltage. And the inductance is l delta over delta t is a variation in current. I2 minus I1 over delta t is the time during the on period which is T1 will supply voltage. So z ripple current is equal to this equation, which is our first equation. Now we will go into get z equation of rebel current earnings or period. So here by applying KVL, we savvy Albert because we are what is known as the average. So negative VS or Zan supply voltage, or the inductance voltage, as we said before, the inductance voltage equal to VS minus the average. The inductance voltage equal V average minus VS. Or by applying KVL, we can get the same equation. Now seen. And L is the inductance voltage can be L delta I over delta t, same as before. But delta T here is T2, okay? And v average and resupply. Okay? Remember that we said before that z inductance voltage or is equal to VS minus V average. Ok, this is correct. Ok. But after we substitute here, we substituted the width L delta I. And as you remember that delta u was and negative one. So leave it because someone can have some confusions. We said that v l equals supply minus V average, right? So this one is equal to l negative because z delta I is one minus i2. Okay? Here, during the off period over at t two equal V S minus B average. So you will find that this negative caused z, v average, it'll be minus V S and V average minus VS l delta over delta two over delta t, which is T2. There is on for this negative because negative delta I is I1 minus I2. Okay? Since our current is decaying. So we go to z, second equation of delta I. Now Z sorted the step what you are going to equate z2 equations. So by equating question 12, you will find the finally, by doing some simple mathematics that the average voltage will be equal to supply voltage over one minus qi. Remember that k was between 01. But here one will be refused with cause it called z to be infinity. So whatever is the value of K between 01, this will cause one minus a small value gives us a value lower than one. So VS over a value less than one causes that V average to be greater than the supply voltage. So this was the equation of OZ step up DC. So about the same as the equation which you obtained in guess off z, step up decisional par. Before. Now step number four, we are going to get us apply current as a function of average current by equating or assuming that we have a lossless converter. Power input equals z power output. So VS IS equal V average, which is VS over one minus k i average. Supply current is greater than the output current a by one minus 6y. Okay? So I average. Here is lower than the supply current. Note as the buck regulator or we had the Z supply current equals k or u average. Ok. So here is our average. Current is lower than the average of current in case of bulk regulatory. Step number five, we're going to get z equation for z ripple current. So you have here T, T1 plus T2. And T1 can be obtained from the delta i. As a question of deltoid number one and T2 equation to all deltoid from equation number two by substituting and doing some simple mathematics as same as before. Delta I L over V S delta IL of T average minus VS. We talked delta L as a common factor. So we have one V S plus one over V average minus VS. We will have as this pi cross multiplications or some summing two factors, two fractions. So we'll have L V average over VS. V average minus VS. And we have also that x0 and delta i by going to the other side called VS. V average minus VS t divided by N Livy average. So delta i would be equal to v average here will be VS over one minus k. V average here will be V S1 minus k and a. By removing z common terms, we would have finally an equation for z ripple to be cavea over L f. Okay? This is some mathematics, simple mathematics. You can do it by yourself. Step number six, we are going to get an expression for Z. Voltage ripples are for Zika pesto. So we have here our waveform for the current of Zika Buster and is the voltage of Zika bust. And would like to get z delta VC or the variation in the voltage. We know that delta VC or 0 voltage across the capacitance is equal to one over seen integration of the current in the team. So we have here a simple one, which is this one. Okay? This part, the integration of this part representing is the area under the curve. So this period, this escape the team. And this part is negative. I average. So one over sin is the integration of the current DT here is the area under the curve which is negative i average a KT, negative I average Katie. But we assume the year We want a z magnitude only. So we remove this negative sign because we wanted the magnitude, okay? But we know that delta Vc here is decaying and delta Vc here is increasing. We need to find is the value of delta VC. So Katie overseen by average. So delta Vc will be finally k I average over c and the d is one over the frequency. So we obtained the equation for the voltage ribbons. Now we need to find is the expression for Z critical inductance. We know that again, that's a critical condition, means that I one is equal to 0. So delta I or is a variation or Z current rebels is equal to i2. Because delta I is I2 minus I1. I1 is 0. So delta i equal to i two. Now we have here an average of four. Our inductance, or Z i L is simply equal to z rebels over two, okay? Delta I over 2m. So find hears that IL average, which is the average is equal to delta i over two, which is same as i2 over two. So I2 is equal to i average. Now we have seen rebel equation delta i, which is two i average. And instead of seeing this, we can say that z average here from our curve is this part is delta i over two, and this part is delta e over two. So we can say is that the average is delta i over two, average is delta y over two. And instead of doing business, but anyway, anyway, that dy is equal to two I average deltoid was obtained before KV S over L critical frequency. And the two I average, I average is equal to V average over r. And v average is VS minus K. So by doing some simplification, we will get that the critical value of inductance in case of Z bosons regulatory scare y minus k over two F. Now, let's again say exhibition for Z critical capacitance, we assume that V2, v0, v1 here is equal to 0, is the minimum value is 0. So V average is equal to delta VS over two. So Delta V S equal to V average in this part is this one is v average and this part is V0 average. So delta VS is equal to two v average. Delta VS was obtained before the voltage rebels, K I average over v of r, f and c critical do v average and do an ozone v average is equal to i average are. So I average can be taken was I averaged. And the value of critical capacitance is equal to K over two. For nauseam, summary of laws which we discussed, V average year is equal to V S1 minus K. I is, supply is equal to I average one minus qi delta i gave us over L F z delta V S. Or is a variation in z, a voltage or the voltage rebels is K I average over fs. Ls is the critical capacity, critical inductance. And C, c is the critical capacitance. You will find here that the average is lower than the supply of sublime. The average here is the average current, which is lower than the supply current. Now, let's discuss the advantages and disadvantages. Xen advanced is that z, both the convertor can boast as in DC voltage as the POMC DC voltage, same as the step-up transformer in EC with a higher efficiencies and z buck boost converter, which we are going to discuss. Its efficiency is higher than 90%. The input current is continuous, which means lower harmonics. If you will see, is that Z waveform was the same as the inductance in praising is in decreasing, increasing and decreasing solves the inward current is continuous. A disadvantage of the boost converter. It has a high beam current in the inductance. Because you will see that during the on period, we had the inductance along with a supply. So the current year will be very high. So xy inductance should lose stand on Z high current during the on period. There, V output is sensitive whiskey. Why? Because V output is equal to V S1 minus k. So a small variation in, or any small change in k can cause a larger change in our output. That's why i what is sensitive whiskey. Since the Wii and avoiding VS over one minus k, z average current is less compared with SAP AAC. We said before that I average is equal to supply current one minus K. Okay? Is the average is o Z out is lower than, but Reebok was Z. Supply current is equal to k i average supply current was lower than the average current. So let's have an example on the boost converter. 120. Example on Boost Regulator: Now let's have an example on Z post converter. We have here at supply voltage of five volt Z output average voltage, which is of course greater Xunzi supply, is 15 volt. Z average equal 0.5 and bear. The frequency of switching is 25 kilohertz. Z inductance is on 150 microhenry, and the Z bosons 220 microfarad. So find the ZK, Z0 current rebels Z maximum current. The voltage ripples Z minimum or Z critical inductance and critical capacitance. So let us see this simple and examine Z first one, which is z k. We know that the relation between the supply and the v average is equal to v average is equal to supply over one minus K. V average is equal to V supply over one minus k. So we have supply voltage or five volt V average 15 volt. So we can get k equal to two over three Z. Second requirement is the current rebels by using xenon equations, which we proved caveats over LF, we have frequency 25 kilometers supply five volt z k equal to three, z inductance, 150 micro furrowed microhenry. So our delta equals eight over mine and bare. Now we need to find dizzy current i2. So we simply know that the current I2 is equal to Z load current velocity delta i over 2M z, sorry, z as the inductance current plus delta i over two. Where did we get this information? Your nose at x0. And that's to get back to the equations to Z wave form. Here. You will see here is that Z. And we need to find the z i2, i2, which is the maximum value. And we have here a z inductance current plus delta i over two. So let's draw it. So here we have here z average. This is average of what? Average of sublime voltage. Okay? Supply. And when we add delta i over two, we get y2. Okay? So again, I average here. And the average value of Z inductance current or the supply current. Remember, I S or I average of x0 inductance. And z delta i is z i two minus I1 here, similar two-year, similar to anywhere I2 minus I1. So add I2 minus I1, which is delta i over two, gives us this space or this distance. Okay? And so adding I average two delta i over two, we get I2. But what average I, average of the supply or the inductance. But we remember here is at x0 and inductance or the relation between Z inductance current and dizzy supply voltage as supply current. Let's see. You remember that or Z, supply current or inductance current is equal to z Albert average occurrence over one minus k. Ok, this, we broke the from's at lossless converter. So I average which is 0.51 minus k, which is two over three, give us Z IL 2p1 B15 plus delta i over two gives us 1.2944. So again, i two, we said that z delta u over two is as small as a hub of x_ variation in the current or have OZ rebels a plus the average value of Z supply. And is the average value of so supply is Albert average over one minus k. So i two will be finally this value, z force requirement is delta VC simply by substituting in x1 value which we obtained or the equation which we obtained K I average over f for C, we will have two of our citizen revolt, Z L E critical and critical capacitance simply as a equation which we proved. And the other equation which will be approved and just direct substitution. Z resistance is simply equal to V average over I average because it is a resistive load. So we finally get z critical value of inductance and the critical value of covers and exhausted by direct substitution. Okay, so this is a very simple example on Z and boost converter. 121. Buck Boost Converter: Now let's discuss as the boost converter, which is z lost the type of our converters. Z buck-boost converter has a circuit like this one. We have here our switch. We have your iron inductance. We have here a diode which is reverse at Zan Zi. Previous ones have reverse one was in the opposite direction. Z combustions. And of course, for Z ripples and our loot. Now, Z buck-boost converter, what does it do? The buck-boost converter is used, uh, to step up or step down the input voltage. So according to the value of z duty cycle, we can control if it is a step up converter or a step down converter. It is primarily used aware and negative Albert is required with respect to the ground. What does it mean? It means that previously we had here is a supply voltage plus minus, and the current goes through here. So rosy lot, so siloed previously had sign positive narrative, same as Z supply voltage in case of the buck converter, poster converter or any type of converter such as the step-up, step-down, everyone. But here we can obtain and negative why? Because you will find that the current, now instead of going through this direction, it Brian money goes through this direction and it goes through here. You'll see that the current i output is here due to the composition of our circuit. Okay? So we'll see that the polarity here is reversal due to reverse the direction of the current. So lets us see Z wave forms and z composition of our circuit. So you will find there is a switch during on period, we will have all of our component. This part alone. This part would be an open circuit and this part will be alone. And during the OFF period is S1 is off. So our circuit will be this part, which is the inductance capacitance and inductance capacitance and the output. Now, you will see that during the on period is the same as before. Z supply voltage is discharging our inductance. Z capacitance here is discharging through Z load. Okay? So you will find it from PVN is that VL is the voltage across the inductance here is the same as the supply voltage. Is inductance, current is increasing because we are charging our inductance Z supply current equal to Z inductance current 0, current through z dot equal to 0 because the right is an open circuit. Z or z capacitance current equal to negative i out IC equal to negative by Albert or equal to negative i average. Same as zinc bolster convert. Now so our, our inductance is a charged in this direction. Okay? So the current is in this form, okay, in this direction, downward. So during the off period, we would have the inductance capacitance and Z i Albert and do we have here a diode here in this direction. So z current does not go through this direction because the diode will be reverse wised. Both Z, Z inductance current goes in the same current of Z in the same direction as the original current. So I L goes here, and this direction downward. So i l will go here downward. So i n will go here, go buy exists and charges the capacitance in the opposite direction. And the charged our or go through our loop. That's why you will find that the output here is reversed from z, both to convert y due to the composition of our circuit and z inductance and produces a direction in this, in this direction, same as before. But the difference was that, and previously we had the inductance here, which will produce a current in this direction and the Goes through a light. Let's say get back so you can understand what I mean by this. Okay? No problem of repeating. Here is the inductance will reduce a current in this direction after being a charge, z current goes through here, through the diode and the ghosts Rosie capacitance and I out. But now is the inductance is here. So it produces a current density direction. And like this, so that there's a difference between z and z boast and is Epoch. Now, let us see V L, or the voltage across the inductance is equal to negative V average. Because from KVL here, z inductance voltage equal negative v average inductance current is decreasing because he wanted to charging Z supply current equal to 0 because it is an open circuit ID m equal IN Z current of z inductance is the same as IgM. Igm here is a as I, l and z capacitance current is equal to the capacitance current plus i or i average equal to the inductance current or the IBM. So Zika Boston's current equal I L minus I out. Same as the previous example or oblivious circuit of z. Both to convert. Now a 0 voltage across the inductor during the on period is equal to V. Supply during this period is equal to negative V average Z inductance current during on Beirut charging during of periodicity charging same as before. So apply current is existing during his own videos and equal to IL. During of Beirut equal to 0. So during on period equal to I, L, during a period equal to 0, IDM equal I ended during the off period, IDM iOS, or I l equal to IDM during this period. During on period it is a reverse biased. So the current will be equal to 0. Capacitance current as the capacitance current. And we said that here it can negative by Albert or negative I average. Here. We said that it is I L minus I output or I L minus I average. So I L minus I average i to minus i average and I1 minus i average. Now, Vc, again, is the voltage across the capacitance. During busy. On videoed. It is discharging. During off period it is it charge it. So during off period discharging and during on bedded charging during her own period is discharging and during off period is charging. Same as Z circuit off Z bolster converter. The same sings exactly. Now we would like to do these steps in order to find our ripples is the voltage and everything as we did before. So the first step we are going to get dizzy equation of Z ripple current during on period same as before. By applying KVL, we will have as the supply voltage equal tos inductance voltage. So V L equal VS delta over delta t equal VS and delta t equal t1. Delta I is Z one which is same as the circuit of z. Both the converter same equation. During the off period. We will have this circuit which is differences. And before. Now we will find that z inductance voltage, which is considered here as a supply, is equal to 0 V output or savvy average. So V L, which is delta r over delta t. And delta t is t2. The average here equal to z, so delta I equal to V average over LT two, same as before. So here we said that the current, or by applying KVL, assuming that no source is existing and, and it will be negative L adult oil water Delta T, which will give us a positive sign in the end, the same as we did before. We said before that VL plus V output equal to 0. Okay? If we assume that it is a voltage drop across the inductance is then we said that this one is l delta i. But since it is decaying, so it will be a negative over delta t is same as what we did in z balls to convert. You'll find that V output is equal to L delta I over delta G v equal to L delta I over delta t, same as before. Now we will equate z2 equations. We will have as the first equation of delta i, second equation of delta i. You will find that finally, it is a different equation. V average equal V supply k over one minus key. On funds that it is a mixture between Z bolster converter and the buck convertor. So we'll finds out here between as k, between 0.5, our converter actors as our buck. And if we increase K between 0.5 and the one, it will act as a boost converter. So simply by controlling z value of duty cycle, we can choose either to work as a buck or two work as opposed to convert. Now getting as a supply current as a function of African. Again, we will assume and lossless converter z v SIS equal V average, I average. And do we have V average is equal to VS K over one minus k. So I supply equal k over one minus q. I average to sort according to again value of k. We can say it is a bug or a boast. And we can say ease or dizzy, I average greater or less than the supply. So we can control the value of current by controlling ZK. Now step number five, we are going to get the expression for Z repel current delta I. So as we did before, T z aperiodic time equal z plus z off period. And do we have an equation of one of z ripple and equation two OZ Ripple as a function of T1 and function of D2. We will substitute with XAMPP here as we did before. So we'll finally, after doing some simple mathematics as at delta I or cerebral, cannot equal to GAVI S over lf. Ok? What we did here is that we take, again, they'll die as a common factor, one over V S plus one over V average. Then we have the average VS K over one minus K, v average k over one minus k. And you will find the final exists equation, z, step number six and getting the expression for Z ripple voltage delta V. So we gnaws at, again with the same integration. We have z delta Vc is equal to one over Zika Vestas is integration of Z current of z capacitance. And we have here our area which is I average gaiety and same as what we did in z balls to convert. So delta VC equal K I average over f for C. The same equation. Z value of Z critical inductance against assumption of I1 equals 0. So delta i equals I2. And here we have our average. So delta I is equal to the LTI, which is all of this is equal to two I average delta I equal to I average, same as the previous one and is the previous one buck and the boost. So delta i, which you obtain the KBS over LCF, do I average. So we have year giving us over LCF to V average over R. So we'll finally gets at a critical value of inductance equal one minus q, r over two F. Step number eight, getting the value of the critical Boston's, boy, assuming if you want to be equal to 0, so delta VC equal to V average, same as z, delta equal to i average. And we have the equation of delta VC and V average equal i average are, so Z critical capacitance equal K over two FOR The same equation, you will see that I'm not doing it the same as the Andon z, same steps as we did before. The same steps. And you'll find that it's nearly the same equations. Z, somebody of his laws, V average equal VS K over one minus k, z supply current, or Z inductance current equal i average k one minus k z rebel current equals KV S over lf. That the VS equality I average over FSC, Z critical inductance and critical capacitance. So what are the advantages and disadvantages of using this type of converters? The advantage is we can obtain negative output voltage. And not like Z, but not like zip post. There is a self or protection against default as L d i over d t. Remember that zi or zj inductance limits is e the i over d t and Zika Boston's eliminate DV over DT. So by using Z inductance, we can have a sulfur protection against the variation or the large variation in the cart. We can use the buck-boost converter two, those are two functions, a step up and step down by controlling the Z value of Z k, wisdom, minimal number of component in instead of having a buck alone or boast alone we have accompany nation but boast it's of course, convert the Zen, Zen most the types of the other inverters. Z. This advantage is z supply current is discontinuous or finds out the current. Supply current only exists as during the on period. So during the on period is equal to 0. So this means that the supply current is discontinuous. Which means we have harmonics, which means we will need a filter. Z also z, a transistor current is high, which means we need a high rating IGBT, or high rating of transistor, or high rating of our switching device. Because the transistor current is very high. Z efficiency of our buck boost converter can be bought as 60% for duty-cycle of 0.7 or 0.3 or 0.3, but has a 90% efficiency for duty cycle of 0.5. So zi or zj, duty-cycle factor, or Azi, is ratio between t1 and CTQ olds aperiodic time. This ratio or is a variation in duty cycle causes variation and our efficiency. Just so YZ probably of buck-boost is there bad efficiency. So sometimes we cannot use it when we need the efficiency of 60%. It means that 40% of our power is gone as our losses. So this was our buck boost converter. Let's have an example on Z. Buck boost converter. 122. Example on Buck Boost Converter: Examine on Z buck-boost converter. You will notice that the Xen and we have a circuit like this one which is the buck-boost converter. We have supply voltage of 12 volt Z, duty-cycle k equal 0.5. The frequency of switching 25 kilohertz Z inductance equal a 150 microhenry Z capacitance equal to a 120 microfarad. I average equal 1.25. Find the number one. It's a V average. So the average, we know that is that we averaged equal resupply k over one minus k z equation which will be approved or for our buck boost converter. V supply 12, K 0.5 and K Boyd five. So it will be V output will be 12 volt, same as the supply voltage, okay? But of course it will be reversed polarity z delta VC. We will substitute in our equation, K I average over FSC z k equal 0.5. The i average given as 1.25. Frequency 25 kilowatt-hours capacitance to honor, and 20 microfarad give us five over 44 volt. Now number C, which is x0 delta I, or is a variation in z current. The current ripples GAVI S over lf z equation, which we obtained k equal 0.5. Supply voltage at 12-volt Z inductance equal a 150 microfarad microhenry z frequency, 25 kilohertz. So delta I equal for over 25 ambit. Z force requirement is the peak of the transistor, Z maximum current of the transistor, which means we need I2 of our transistor. Ok, remember that this transistor during the on period during embittered changes from I1 to I2. Or you won. And he has the maximum value 2m. So it's changes or recharging Z during the on period that charging is the inductance from I1 to I2. So we need the peak value of transistor, which is i2. So what is the value of z transistor here? I2, i2, as you remember that equal to z delta i over two plus i average delta i over two plus z average current of Z, of Z inductance. This value because this one and this one during on period our series. So I m, which is the average of the inductance, is same as average of the supply. And you'll notice that z relation between i sub lawyer and I would average is k over one minus k. This one representing this apply current or the inductance current average. So we broke from Z lossless converter that IS equal i average a k over one minus k delta i over two, which is Rebel, which you obtained. We will have 1.65 as an amber for Z maximum current. Velocity requirement is Z critical inductance and a critical capacitance. Critical inductance. By substituting in our law and critical capacitance, we can get the value of Z minimum capacitance and inductance and Z minimum capacitance. So this was our example. Oz boast owns the buck-boost converter. If you have any questions, don't hesitate to ask in Zack witches part. And if you don't understand something, you just ask and I will I will be there for you. Okay? So this was our course for DC shoppers and hope you benefit from this course and see you in another course. 123. Definition of Inverter: In this lecture we are going to discuss is the definition of inverter. So what is an inverter? And inverter is simply and power electronic equipment, not otherwise. And we will understand why it's called an equipment, not advice. This inverter, this power electronic equipment is used to convert a z DCM boat supply into NAC Albert supply. So we simply, we have here our V minus V DC or direct current, or one directional current in both as an input. And we have a variable output AC current by using z power electronic equipment. The gold is the inverter. This inverter is also known as a DC to AC converters because it, it changes z DC or direct current into alternating current. Z Albert off as the AC or the abbot of the inverter is N c, which is symmetrical with add desired magnitude and frequency. So the output voltage, or ISI voltage, we control it's a value or magnitude, and we control its own frequency by using an inverter. So as the inverter converts DC to AC or the direct current to alternating current. And we have also a variable magnitude and variable frequency. We can control z magnitude or z value of the voltage Z maximum value. And we can also control the frequency or how many cycles per second. Here is an example of an inverter, 1900 watt inverter as power. It converts that input 12-volt DC into a 100 volt a, c. So c is, this is a well-known inverter and you can find it even on Amazon. Okay, you will find here is that we have here Z album. We'll see Albert AC 110 volt, 220 volt. And you'll see here is that we have here our output here, and our output here. You can connect Z as the output here we will see that we have Z line as the phase and the neutral and the Z airs phase neutral and the Earth has three terminals. And you will see your fears neutral and the earth. Okay? And you'll see on the other side, we have here a Boston undermine us which are presenting is the DC. So he bought here our input DC voltage is 12 volt. Okay? This a mod z is positive is connected to Z red bar, and z negative is connected to z negative here, or Z black one. Okay? So z positive undermine US representing a Z 12 volt of z DC supply. We connect to z positive here and negative here. And we take z would EC, from here or from here, phase, neutral and earth. Okay? This is a very simple power invert. Ok, so in this course, so we are going to learn how well does an inverter work and how does the circuit looks like? 124. Importance and Applications of Inverters: Before we go into Z circuits, so we need to understand the importance of the inverter. Where do we use inverters? We use inverters in photovoltaic or PV cells or solar system in order to convert a CDC out into AC. So as you remember that in solar systems or as BV cells, Zai converting the sunlight, Z, converting the sunlight here you'll see that we have here a solar panel or PV cell. This BV module converted Z sunlight into electricity. The type of the electricity is DC electricity. Now we will have here and an equipment called disease solar regulator. As this one regulates a z charging the batteries. Okay, so we take z output dc power fit into Z solar regulator which regulates and z charging of z batteries, Zhi De Si batteries. Xunzi DC battery is a DC supply here, is supply to an inverter to convert his EDC into AC. Now after this, we can use the AC power to feed our equipment. So z inverter here is used in our solar system or PV system by converting the DC battery power into an easy all alternating current. Now let's see another example. We can use it in UBS or uninterruptible power supply. So let's see now, how does UBS ward? Ubs uses a set of batteries along Go is an inverter to supply power winds ME. Empower is no longer available. What does it mean? It means that if we have here at critical load are very important to load here like computers, computers which he cannot sustain or will be terminated off our electrical power. Okay, so we need onesy manpower is cut off. We need to still supplying exists computers as to prevent as their shot down. So we'll see here is that we have here a switch which turns off and on, on and off fuzzy men supply. So that means supply goes from here and goes through siloed. Ok. So z AC supply goes through Z load to supply it. This is Z men supply line. Okay? And at the same time, we use the power coming from here into rectifier, rectifier, converters, Z AC into DC. Remember that from our first course, we discussed dizzy rectifiers, Z rectifiers, converters in input, AC power into IGC power, Zynga, DC power is used for charging the batteries. Okay? This is the normal or Z man condition where we have our supply, supplying is allowed and the charging Z batteries. Now assuming that Ximen supply have a problem or turn it off suddenly, okay, this switch is turned off. Okay. Now's the batteries. Stuart is discharging or supplying power to hell by using an inverter. Z inverter converts the DC power from the battery into AC, which is used for Z loot. So again, during our main condition, we have our supply or AC supply, supplying our load and charging the batteries using rectifier or AC to DC converter. Now, during the fault the conditions, we use the batteries or DC supply to convert the Z DC to AC inverter, zen supplying our loop. Now, another application for our inverter, you will see is that the inverter can produce pure sinusoidal waves. Okay? You will see here is that we have a pure sine wave. This sine wave is the same, which is taken from Z supply. Okay? If you look at your own home or at your own office or anywhere, you will find that z power or Z voltage from Z socket is a pure sine wave, ordinarily pure sine wave, not absolute but nearly pure because we are talking about a commercial sinewave not appealed. Okay, so we'll assume it is a pure. So this wave, which is the blue one, is the sinewave, which is taken from our supply. Now you'll see that we have another type of waves which he can be produced, the FISA inverters, which is a square wave, which is market, buys a green line. And we have another one which is a modified square wave or ques high square wave. It can be said as this one, this red one is called the acquiesce high or as modified square wave. Okay? So it sees that was a difference between a square and a modified your swine zeta square from 0 to pi is always existing as abosDF. And from pi to two pi existing as a negative source. There is no period where the square is here. But zig zag, read one, you will find here this region is 0, this region is 0, and this one is region, or this region is 0 and this one is controlled also buys a inverter or our inverting system. Ok. So we have C vibes here, sine square or modify, a Zen modified or Akwesasne chi square wave. So you will find that here. For a law and a medium power application, we can use as square wave or a square psi square wave voltage. Okay? So during law and the medium power applications which have low power or a medium power, we can use this square wave, which is coming from any type of an inverter. Or we can use the modified or Zack Wes high square wave. Now for high power applications, we need a load store, the sinusoidal wave, which means that we need at where like this or nearly pure sinusoidal, nearly pure. And we'll see how to get this square and how to get the xhat low distorted sinusoidal wave Zillow distorted as your nor in this course is done by using pulse width modulation. This 5p squared is can be obtained by a normal inverter. And you will see here is that we can reduce the harmonics by using pulse width modulation techniques, which you are going to discuss in z scores. Now remember that in z as a square wave, if we analyze this wave using Fourier series, you will find a 40 year or four year. Sometimes it people say four-year and sometimes CA, pour 40 air. But Fortier can be analyzed this square wave into a finite number of sine waves. Okay? So is this one has only one component, which is the fundamental. Okay? So we use pulse-width modulation to reduce the amount of Z harmonics or higher order harmonics inside the square wave. And we will understand this and also in our course. Now here are some types of waves there too. You can have in your own mind or memorise it for yourself. You'll find DNA and this is a sine wave or a pure sine wave. This one is a square. This one is called a ram function. This one is called the ASU 2s. And finally, triangular, which we are going to use in the course, is a square and sign. Remember this serine widths, okay? 125. Single Phase Half Bridge R Load: Now let's discuss a Z types of inverters. We have two types of inverters and we have a single-phase inverters and three-phase inverters. Now in this lecture we are going to discuss is E single-phase how rich or load inverters. So for the single phase, single fans means is that our output here, which is a resistive load, okay? Having as single fees or one phase, okay? So we have a single phase here and our load, a resistive load here, plus or minus this is z v Albert. And we have a half bridge and we'll understand the why it is called how pitch. So we'll find that we have here VS over to VS over two add to dc supplies. And this supply is VS overdue and the hub of the supply here. And novels supply heat. Okay? And we have here two diodes, Jocelyn eclecticism, Schweitzer, We will find that we use this diets in case of we have R, L load. Now we have here Q1 and Q2. Q1 and Q2 are transistors used the for switching on and off. This transistors or power electronic switches can be MOSFET or IGBT in order to switch fast. Okay? So we have here VS over to NBS over 2.5 of the supply here, and hover over supply here. Here we have our Albert resistive load. We have two diets. This diodes used in case of we have resistive and inductive load. At the same time. Q1 and Q2 transistors used for switching on and off. Okay? So we use transistors in instead of normal or mechanical switches, because as you knows that, that as the sine wave has a frequency of 50 hertz or so, we cannot use normally switch or a mechanical switch to switch 50 times in one psych. Okay? That's why we use a transistor to switch on and off very fast. Okay. Now let's see how it does resist circuit of Z, how Bridge walk. But first, we need to understand that the transistor, how does our transistor walk? Okay, so we'll see here is that this is our transistor will find it has two terminals here and z gate. So let's get back. You will find two terminals here and z gate signal. Okay? Remember z gate signal. If we have here I have high input or a high gate signal, then this output will be equal to 0, or the transistor becomes assault sect. Or if z m would equal to 0. Zig zag, good, does not have any signal, then this transistor will be an open set. Okay? So simply, again, if we provide z gate of D01 with a high signal or high voltage, then Zao transistor Q one becomes assault set. If we don't provide any signal, one is in the transistor will be an open circuit. Now let us see how does this transistor or this circuit walks. So adversity during the period from 0 to T over two, okay? From 0 to T over two, you will find that we provide signal to Q1 and we don't provide any signal Tokyo to. So when we provide a signal to one, it becomes assault circuit. And we says that Q1 is on and the Kyoto does not have any signal, so it is off. So during the first half cycle, Q1 is on and the Kyoto is off. So let's see our equivalent circuit. So this part is a short circuit and this one is an open circuit. So VS over to VS over to the resistive load. X21 is a short-circuit and Kyoto is an open circuit. Now we will find that the V output from 0 to T over two by using simple query, AN equal VS over two. So here we have our v, v s over two from 0 to T over two. Okay? Very simple. And Z current absorbed by solute assembly equal to V supply or Z voltage over resistance. So it will be VS over R, because V s over two divided bys R resistance, okay? Vs over two is of supply and dizzy, total resistance is R. Okay? This is our current, or it is a same as the current of Q1 or i1. Ok, now, from t over two to t, or from the hub of the period two Z total period. Now's the reverse happens. Q1 is off, Q2 is on soil, find Q1 is off. Therefore it is an open circuit. Kyoto is on a short-circuit. And by applying KVL here, you will find the assemblies at V-out is equal to negative VS over two. So here you will find from t naught over two to t, you will find that V out is equal to negative VS over two. So negative VS over two from T naught to T naught. Okay? And z current or current or current off Q2 is equal to negative VS over to our z voltage over resistance. Okay? So I'll finds that again. K11 Q1 is on your funds as Abby, our what is called the VS over two. And here V negative VS over two from T naught to T naught. The current here is VS over two, or the supply offers a resistance. Here. I2 is again V supply over 2R. Okay? This current is the output current is negative current, okay? Because our original current was in this direction, in the same direction of one. But q i2 is I reverse to I naught. So i2 is equal to VS over to R i2. Remember that i2 is in this direction from up to down, the negative or equal to negative i naught. Okay, That's why you'll find here Apple step value. Now, remember that this is a very important information that Q1 and Q2 are controlled. We control Q1 and Q2 will remember that we provide here a gate signal and all provide here I get signal. So Q_1 becomes a short-circuit during having its own good signal. During from 0 to T naught over two. This one is assault circuit or having a good signal from t naught over two to t naught. Now remember that we cannot have most of Q1 and Q2 switch it on at the same time because it will cause a short circuit on the sublime. So imagine that we have here one having its own signal. So it will be a short circuit and the Kyoto at the same time having its own signal. So it will be a short circuit. So by applying a largely KVL here, you will find that V supply is over nearly 0 resistance or the resistance of the wire. So the current will be very large and it will people on it. Or simply we can say we have a short-circuit on the supply. So again, if Q1 and Q2 are switch it at the same time, Zen Zell will be assault circuit on our supply. Very simple from circuits. Now how does something like this will happen? You will see is that at t naught over two z, Q1 transforms from short-circuit to open circuit. And Kyoto transforms from an open circuit or short circuit. So there is a moment where z are operating at the same time. So what does he do in this case is a make a small delay time. So Z way. So for example, Z turn off Q1 and after every small time delay, time milliseconds or microseconds, Zai switch on Q two. Okay, that's all what z2. Now let's analyze our sector. We have here our V out is a square wave, okay? And it's called half bridge because you only have half of the supply VS over to. Now, remember that the root mean square for a function is equal to z function as root mean square is equal to the square root one over z period integration from 0 to Z total period for the function square. So this is a symbol sink which we all must do know about Z root mean square or RMS of any function. Now, an order to get the root mean square out of our air. Voltage for this supply. Okay? It will be the integration of square root of z, integration from 0 to T naught, okay? Square root of one over the period from 0 to T naught over z function or the voltage square. But remember that that's the Boston here. Vs over two is equal to z negative time. So we can integrate from 0 to T naught over 20 to T naught over 21 over T naught, but multiply it by two and the function square, remember that the function is a VS over two. And having it all square will give us VS square over four. So the square root of this function will give us v s over two. Or you can find here is that if we square this function, it will be here. And you'll find we have IDC VS over T2. So we can say finally is that the root mean square of z V output is VS over two. Now, there is an important thing you have to know the difference between a sine wave and a square wave. And I will tell you why. You will find that the sine wave is what comes naturally from the generator, and that is how the supply voltages look slow. However, the square wave is considered as ASUM mission or the output of some types of inverters consisting of lots of harmonics. So we have here z sine wave have only one component, which is the fundamental V1. For example, we are saying this sine wave is at a frequency 50 hertz. But the square wave, by using ZAP Fortier, can be analyzed into lots of harmonics. And you will see that we can have v1, v3, and v5 is, is this one is at 50 hertz, for example, Z as sort of the harmonic as add stream multiplied by f, z if f is a harmonic fuzzy 7's harmonic and so on. This type is the type of wave which is used in high power applications does not cause noise. Of course. This square wave or quiz Hai Zi modified, can be used in low application which does not require pure sine wave. Z. Problem of z harmonics is that z cause, for example, in motors, z equals a rotating magnetic field, which will produce a torque opposite to Z, original torque of z mode. So it causes noise incisor motto, that's a YZ harmonics is not a good thing. Another thing about harmonics, monica causes overloading in cables in power lines increases z i square r losses or z power losses is a harmonics is not angled saying are not a good thing in our power system. So we use different types of filters in order to remove the harmonics. Now remember Z instantaneous output voltage can be expressed in z fourier series S, z square wave, XAMPP previous square wave can be analyzed as V output is equal to the sum of 135 to infinity do VS over n pi sine omega t. And remember that at n equal to or z, even harmonics are equal to 0. So I will find that we only have 135 and so on to VS over analyze this Fourier series and represent is a square wave of Z L. Now, if we would like to find the z root-mean-squared of onlys fundamental. The fundamental which is used or consumed by our supply, by our loot. At n equal one, or add Z, normal frequency on power frequency 50 hertz or 60 hertz and so on. So at n equal one gives us Z fundamental. So add Z fundamental V out is equal to two VS over pi. Okay? But this value is the maximum. In order to transform it into root mean square, we divide by root two. So we have to VS over pi representing Z maximum of the sine wave. Dividing it by root two gives us the root mean square value. So V1, or the fundamental component, is only 0.45 of savvy supply. Okay? And remember these values because we are going to convert between these values and Z bridge rectifier. Okay, now, the most important thing is this lecture is that we have half bridge, which means we are taking just half of the supply VS over two. And we will find that z function or Z square output can be represented by a Fourier series to VS over Enterprise on n omega t. And remember this function because we can use it in defining something which is called the distortion factor, harmonic factor, something which we are going to call this the performance parameter we are going to discuss in the next lectures. Okay? 126. Single Phase Half Bridge RL Load: Now let's discuss a z single-phase, how a bridge or a load shows a previous video we discussed disease, resistive load. Now we are going to discuss a Z or involute resistance and an inductance. So far an inductive load, you have to remember that the current cannot change instantaneously. So what happens here? So first, we have at the beginning, we have Q1. So starting from here at 0, where Z resistance and inductance, or the current year is equal to 0. Ok? So at the beginning or we give gate signal to Q1 0 for K1 conducts and it becomes assault circuit. So V output or the current through Z resistive load will increase from 0 until Q1 stops conducting. Okay, so this diagram or this axis represent the Z current or the load current with respect to time. Nazi voltage. The voltage is V Albert, Of course, when Q1 conductors, therefore the output will be equal to VS over two. But now in Z RL we are discussing visitor. So current starts from 0 until Q1 stops conducting. So at this moment, Q1, this inductance is a charged by a specific amount of charges. Okay? So the inductance contain magnetic field. Okay? Now's the inductance. Want a store discharges this amount of torque, okay? So adds a beginning Z current goes from VS over to throw Q1, and then in this direction from right to left. Now, after Q1 is off, the current will discharge in the same direction. So the current will go in this direction. Okay? So I naught has two ways, is or to go up or go down. So if it goes up, D1 is off, and Q1 is off. Saw as the only way here its score downward through the supply voltage and through D2. Because also Q2 is off. So the current will go through this loop. So during busy period where Q1 and Q2 is off, after Q1 is conducting, D2 is on and the current is discharging, okay? When Q2 is on or giving a signal to Q2, Xunzi charging in the reverse direction and z current or the charges will increase in the opposite direction until again z negative maximum value Xin Zhi Q2 is off, D1 is off. And the direction of the current in this direction from left. Right, reverse it to I naught, okay, from left to right. So it will go through D1 Zen cirrhosis loop, okay, throws this law. So D1 and discharging its negative value until 0. So again, we have here and the cycle of coarser bit itself. Now we have four parts. We have when q one is conducting, when, when T1 is conducting, Xin, D2 conducts zen q2 or dq2 conductors, and then D1 conducts. Ok. So now let us see our equations in something as output voltage will Baeza seem to VS over n pi sine omega t, o and N equal to 40. Either all of them are equal to 02468, all the even functions, or is even in. Now for Z current, remember that the current is equal to z voltage overseas at, okay? So we have here is a voltage 2V S over any boy, or for Zed hours, Nt is the square root r-squared plus n omega square y, because the inductance a changes with z order or harmonic order. Ok? So we have here XL depending on n, OK, because he Excel is equal to omega L ands omega itself or the frequency it changes. So at each end we have a different frequency. Therefore, we will have a different inductance. So do VS over N The by divided by total impedance zed r square plus n omega l square y. Because z omega or as a frequency dependence on n. Okay? Remember that at in equal one, we have omega over two pi f. At n equals three, we have, sorry, F or Z triple of frequency. So that triple of the inductance and so on. Sine n omega t minus theta n. Why? Because as you remember that for an inductive load or, or elute, Z current will be lagging from z at a voltage by an angle called deceit. Okay? As this Sita also depends on z harmonic factor n. And Sita n will be equal to ten minus one n omega L over R. C then is equal to ten minus one n omega L over R. So what is important for us is the fundamental component. So if i node one is Rotman square fundamental, okay, at N equal one and root-mean-squared. So let's substitute at n equal one. We will have to VS over pi root R squared plus omega L square to VS over by root R squared plus omega L squared. Okay? Add in equal one. Ok? Now for Z current is this value is called the Z maximum value because as you remember that i max sine, i max sine, any function is the maximum value sign. So in order to get Z root-mean-squared or the effective value, then we will divide by root two. Remember that the differences between Z maximum value and the root mean square value in a pure sine wave is that we take z function or Z maximum value and divided it by root two. So this part which represent a z fundamental component of the current is symbol is this part at n equal one divided by root two to get 0 mean square value. Now Z fundamental hour-by-hour, which means that Z benefit Bower or Z important power, or z power at the fundamental component at n equal one, which is a useful Bart Nazi harmonics, is equal to V1 Z voltage as a fundamental multiplied by Z load current as a fundamental multiplied by cosine theta on, ok, v psi of cosines here, okay, as you would remember that z power VMM or the blood by Z current multiplied by cosine theta between them. Or we can say since we have a resistive load, therefore Z effective value will be equal to I square R. I square is the fundamental component which will discuss the year all square multiplied by our simple as slices. 127. Performance Parameters of an Inverter: Now let's discuss a z brief formance parameters. We have some parameters which should represent a z performance of our inverter and help efficient is our inverter. Z quality of an inverter is evaluated by some parameters. Z harmonic factor of n harmonic or Hf, x0, Total Harmonic Distortion, THD as distortion factor df. And finally is a lowest order harmonic l o h. So this for, but I might also represented as equality or the efficiency of an inverter. Let's discuss each of these factors. Number one, z harmonic vector. Z harmonic vector is simply a ratio between Z contribution of an harmonic number n over z fundamental component. So HF representing simply z magnitude of z harmonic factor number n with respect to 2Z fundamental. Okay? It measures z individual harmonic contribution. So v1 is the root mean square value of the fundamental component. Vn is the root mean square value of Z, n harmonic component. So this factor HF simply represents how much Z value of Z harmonic with respect to two Z fundamental. Now let us see another factor, z Total Harmonic Distortion. This one representing Z total harmonics, z all zeroes mean square value of Z total harmonics with respect to z fundamental. So we see here how this value or this harmonics value compare the two Z fundamental. It measures the clauses needs in z shape between wave form and it's a fundamental component. We will see is that the harmonic factor here, we had only one component with respect to those fundamental. But Z total harmonic distortion representing all of the harmonics with respect to 2Z fundamental. And of course, all of these values are in a root mean square value. Now, our surgery term is that distortion factor, Xa distortion factor is simply we have here for Z total factors or Z total harmonics is VN over n squares or some mission, of course, of vn over n squared, all square under square root. Okay? We simply, we dogs a root of all of the harmonics, except is at, each of these harmonics is divided by n squared. And all of this is divided by V1 or the fundamental component. Now if you'd like distortion factor for only one component, we take the n over n square over v1, vn over n square V1. Ok, so what does add social and vector means? It measures the effectiveness in reducing unwanted harmonic corridor, which shows us CSI effect in reducing exists harmonic. For example, if we reduce disease certain harmonic it will shows us how effect or how it is effective to reduce this harmonic. Now, our loss term is the lowest order harmonic, this harmonic component whose frequency is the closest setosa fundamental. For example, we have Z fundamental, Z served, Z fifths, and so on. So the closest photocell fundamental is SERT. But the important thing is that z value of it or z magnitude should be greater than 3% of z fundamental components. So what does it mean? It means that that Z harmonic vector should be greater density percent, as you would notice here, is that Z harmonic factor is the ratio between VN over V1, or it's always z value of Z harmonic with respect to those fundamental. So if this value greater sensory present and z clauses the frequency to the fundamental Zen it's considered as Z lowest order harmonic. Now, in the next video, let's have an example on Z, single-phase how Bridge and of course on the performance parameters. 128. Example on Single Phase Half Bridge: Now let's have an example on Z single phase half bridge inverter. So we have here a single phase half bridge inverter as he having a resistive load of R equal to 0.4 ohm. And the DC input voltage is 48 volt. Determines or data mine x0 root mean square value of the output voltage V1, Z, output power b naught z average and peak current of each transistor. Zb, reverse blocking voltage VBR of each transistor, Z total harmonic distortion, that social factor and the harmonic factor and store some factor of Z lowest order harmonic. You'll see zeros are a lot of requirement here, but you will find that it is easier Zan your sync. All of the problems about single phase bridge are like each other. Okay? Now, let's see. Given R equal to 0.4 ohm, source resistance to wait for all z V S is 48 volt. So here we have 48 volt R equal to 0.4 ohm. Now if we redraw the circuit and the waveform, as we discussed before, you will see it will be like this. And I will show you why I keep z's in this light. Now is the first requirement is that we need to find z root mean square value of the output voltage. So we need to find the v1. As you remember that V1 is equal to point the 4-5 OCV supply. Or we said before it is too VS over pi or root pi. Ok? So V1 or is a fundamental component of zeros, which we discussed before, is equal to 0.45 V supply. So it will be two into 1.6. Now the second requirement is that Z output power p naught. So z power output, as you know that z power is equal to IR would square multiplied by r or v square over r. So z v square is v s over two. V out is V s over two as of, as of course, Ottoman square value root mean square of z is equal to v s over two, okay? Which is 24 volt. So the output power is equal to v squared over r. We took Z output voltage over R. So V output is 24 over r, 2.4, which will give us 241 hour surgery requirement is the average and DB current of each transistor. So we'll see is that here we have, for example, I1, fuzzy current of the transistor Q1, Z maximum is a VS over to our very simple ZB current goes a transistor is 24 over 2.424, which is VS over two, okay? Divided by the resistance, which is VS over to R, which is ten ampere. And now for z average, as you know that here, the current is VS over two r. Okay? So this current is operating for half of the cycle. Therefore, Z. Transistor conduct Sura, 50% duty cycle. Therefore, the average will be 0.5 multiplied by this current. Or you can get it by exhaust integrating this function from 0 to T naught over two divided by t. Or is the total period one over t integration from 0 to T naught for the function. Okay? Simple mathematics, all simple integration, same as we did in rectifiers, Of course. Okay. Now our force requirement is that we need to find that peak reverse blocking voltage. So as you remember that in our previous course of rectifiers, you will find that the z b value or Z breaking a value, it means that we would like to find z voltage across, for example, Q1 when Q2 is conducting or Q1 is reverse biased. So when T1 is reverse biased and the Q2 is conducting, you will find that V output is equal to VS over two and the year VS over two, or negative VS over two, N over two. By applying KVL, you will find that the voltage across key u1 in this logic EBL, when q one is reverse and the Q2 is conducting, it will give us V supply. Peak reverse voltage is tomato blah By 24 or two metabolite by VS over two, which is 48 volt. Where did I get this value as simply by doing a KVL here when v-out or Q2 is conducting. So V out is equal to VS over2 and DVS over two. Okay? Now our number e is a Total Harmonic Distortion. So an order to Giza, total harmony, social nuts, right, Zeno first, as you remember, total Hermann social is equal to route all of the harmonics divided bys a fundamental. Okay? The fundamental is equal to 0.45 V supply, okay? Which we will discuss the before here in number one. Now for Z harmonic itself equal to roots on machine all harmonics. So this function, and instead of zoster substituting with 3579 and so on DNA TO infinity. We can Justices at z harmonics is equal as a root mean square is equal to z v Albert square Z total value of the output voltage minus z. Fundamentally, Suo Zhi Dao was a total Albert minus the fundamental will give us z harmonics. So V out square is V sub ly square minus v1, which is 0.45. Whistleblowing will give us this value. We can take this and divide it, buys a fundamental, we get 48%. So we'll find the year we have a larger distortion of nearly 48%. Now, let's see z and extra requirements, that distortion factor and the harmonic factor. And so some vector of zeros to all the harmonic. So Galaxy against the distortion factor V output is equal to the summation of these value. Okay? This function which we will discuss the before. Now, in order to find z distortion factor, we know that the distortion factor is equal to one of the fundamental, the fundamental we already got it. Or Wentworth 5V supply. Here we have some measure of 23, which is dU is equal to 0, of course 357 and so on. V n over n square or square to the power half. Okay? So we can substitute with this value by Vn, Which is what? V3, for example, over three square V3 over city square, V5 over five square V seven over seven squared, and so on. Okay? V3 is equal to two V S over root two pi multiplied by N equals 35 is equal to two V S over five pi multiplied by root 272 VS seven by, multiplied by root two. So we can substitute here and we will finally get this value. Ok? You will want to know is this summation we can do something else. What is it? We can exhaust symbol phi or exhausted divide a gets this value and this value and this value and number nine, zoster likes it only to mine. After substituting under a square root. We can get close value to this. Then we will divide by v1 and the it's 3.8 as a distortion factor. Now let see Z final requirement harmonic factor and social factor of Z lowest order harmonic. So z harmonics that we have here is z assert waves and seven. Susie closest one is three. So let's see 3 first. So the harmonic factor of number three is equal to V3 over v1. And as you remember that v3 to v1 is one of our.srt. Okay? You can get this value from the function heat to VS over n pi and to VS over three boys. So z division of these two functions will give us one over three. So to give us 33%, now is a distortion factor is simply 0. Third, V3 over three square multiplied by V1, which is one over 27, which is 3.7. So Z harmonic factor OZ served is, is searched 3%, which is greater Zen Of course, we present as remember that the lowest order harmonic shouldn't be greater than 3, 3% year is representing the Z harmonic vector. And the harmonic factor of Z served is surplus represent greater sensory present. That's all. This is a very simple example on Z, single phase inverter z. How Bridge, of course. 129. Single Phase Bridge Inverter R Load: Now let's discuss is the single phase bridge resistive load. So again, we have here a single phase. So we have here our load plus or minus V Albert or plus minus V output from here, positive minus prostate VCA and z negative is B. Ok? So this load is single phase load bridge because this time we have Z total voltage here is V supply, okay? As you remember that previously this point of zeros was connected to Z, neutral of z two supplies. But here you will find here VS over to see it as it was VS over two, which means the total voltage is V supply. Now in this time, since we are talking about a full bridge or, or bridge, we will have four transistor, Q1, Q2, Q3, Q4, d1, d2, d3, and d4. So we have four transistors and 4-byte. Previously in the half bridge, or we had two transistors, two bytes. But in z for a bridge or bridge, or we have four diodes for transistor. So now let us see how does this circuit works. So starting from 0 to T over two, you will see that we are going to give apples to Q1 and Q2. So Q1 and Q2 will be assault circuit. Q3 and Q4 are off. So Q1 short-circuit, our resistive you'd Kyoto is a short-circuit. Therefore z v output across by applying KVL will be equal to V sub block. So during off cycle Q1 and Q2 is operating or are operating. And during design negative cyclin, Q3 and Q4 are operating the same as a rectifier or a controlled circuit. As you remember, is that for an AC supply during this hour boast of cycle zeros here, rectifier and another hero rectifier and doting isn't negative cycles. The azar torque departs. Same here with Z bridge, B2 is an inverter. Okay, so during busy from 0 to t over Tokyo one I dequeue to operate. Therefore, this one short-circuit, this one short-circuit, therefore V out will be equal to V supply for z. Second cycle t over two to t, or the second half of the period Q3 and Q4 are operating. So Q3 and salt circuit, PO4 short-circuit and between them is our load. So V Alberto will be two negative V supply. So during busy boast of psych and V supply during the second half is negative V supply. So our, our, what would be simple like this. The V supply from 0 to T naught over two, and from t naught over two to t naught is equal to negative VS square wave with a maximum value v s. So the first thing you will not see here is that ends up bridge. Our maximum output is V supply. But Lindsay-Hogg bridge, we had VS over two. Now the second thing you'll notice is that here we have v AB Z load voltage is VAB. Now, Let's see our laws. Z root-mean-squared output voltage can be found again with z integration. Two over t naught integration from 0 to T naught over two. For z function here we have a V supply, but previously we had recently over two. So we'll find here is that V out as a root mean square is now recently in South Bridge was V supply over two z and instantaneous output voltage you'll find here it is simply, is the same equation to VS over n pi sine omega t. But here it is multiplied by two. Okay? We reversely had VS over two. Now we have VS, so finds that the equations is simply multiplied by two. So we have here for VS. And instead of to VS for VS in Z as Z bridge inverter. But foreseen half bridge we had to VS over n by. So now for Z fundamental component, as you remember that we will just substitute here by one. So we'll have war for a VS over pi and the multiplied by root two. Okay? So this will give us finally point time VS. Remember that in z half bridge we had 0.04 or five years. So you will not see those things here is that in the single phase bridge, that z fundamental component is multiplied by two. Z V output is multiplied by two. Okay? So we have doubled the Albert W z four-year or z v square. Okay? 130. Single Phase Bridge Inverter RL Load: Now in this video we are going to discuss is the single phase bridge inverter or a lot. But I just want you to correct this mistake you'll find here is that I said here rectifier, but it was inverter, of course, who are talking about inverters in this course. Not rectifiers as that was assembled mistake. So now in this lecture we are going to discuss single phase bridge are a little bit. So we have this circuit again but our elute. So now what is happening here? We will assume that we are starting here from 0, where Z inductance is not conducting Q1 and Q2 is on. And therefore the current will increase until Q1 and Q2 is off. Then adds his value, we will find that the current will continue in the same direction. So it will go through D3 and D4, okay? Then it will be, reverse it by using Q3 and Q4, Xin, again D1 and D2. So I want you to see each step right now. Is the first step is that we have Q1 and Q2 are own, okay? For Z first apart here, this point k. So z inductance or 0 load will start to conduct Sarkozy v i would also Rosie supply voltage. So V supply will start supplying through Q1 and Q2 as this lewd, okay? So V out will be equal to V supply and Zika will start to charge. Current, starts to charge. Now, Zen for z second Berio at t equal t over two here, as you have observed period here, you will find that Q1 and Q2 are off. Q3 and Q4 are also of. So what happens here? You will find here Q1 and Q2, Q3 and Q4 off Waze current direction was in xyz direction, okay? From this apply. So rosy wrote and exist. So z guarantees from here to here, from left to right. So the current, since it is an inductance, it wants to store continuum Z. Current does not change instantaneously in celebrations of an inductance. So what happens Z inductance here as it moves from left to right. So it will go through these three. Q3 or Kyoto or D2. D2 will be reverse biased, so it will not bass solo here. Kyoto is off, Q three is off. So it does not have any way except going through D3. Now going through D3, I would like to go like this. So Rosa supply, D1 is off, Q1 is off. So I will go through their supply. Now at this point, I want to go against Rosie wrote PO4 is off. D2 will not go through zs is point of zeros. So D3 and D4 are on z naught current flow through D3 and D4. The literacy 0, find here something. The current year is assumed direction of this current. This current is charging, this current is discharging. Okay? Now, for the sort of the border Q3 and Q4, our own soul finds that the current goes in Z direction of clockwise from here, from B to a. So now I would like to Mexican onto more from B to a. So during the period where Q3 and Dequeue for our off at t equal t capital or z whole period. Like here. This is a first, as this is half of the period, this is three over four of the video. This is a total period. So at this point, z, like this point, Q3 and Q4 will be off. So what happened? Q3 and Q4 are off. So the direction of the current, as we remember, is in the direction of clockwise Q3 and Q4. So from B to a, so from b, I will go through a. Now I have two directions. D1, d4 is of, Q4 is off, Q1 is off. So I will go through D1, rosa, right? Dc-3 and q3 will be reverse in this direction through a resupply. So zeroed D1 throws us apply. Default will not give us z Lyrae, same direction of current, okay, so direction of cornets from B to a. So Rosa supply, so z negative Zen, Sorrow Be, So in order to get back to B, so we'll use d two, okay? So D1 and D2 will conduct during this period. So at the beginning, Q1, Q2, Xin, we see what is the direction of current from a to B. So Z current will discharge from a to b again. So from a to B through these three throws us apply through D4 back to a in z, Q3 and Q4 off from B to a, D1 supply d2 to be again. So we'll just have to understand, whereas the current is going or is a direction of the current. And you will understand why d1 and d2 is here operating and D3 and D4 are operating in the previous one. Now holds the equation Z load current, as you remember in the previous one or z hub bridge to VS over root R squared is the same equation. But here we will have for VS, since Z out is VS, VS over to see the end is the same, then minus1 n omega L over R. Now, let's have unexamined in Z. Next to video about z, full bridge inverter. 131. Example on Single Phase Bridge Inverter: Now let's have an example on Z single phase bridge inverter. Now we have a resistance of 2.4 ohms, and the input voltage is VS equal 48 volt. So it is the same as the previous example, but the difference here is that we have full bridge or a bridge inverter instead of half bridge. Now are theorists requirement is R square value of the output voltage V1. As you will remember that the root mean square is equal to 0.9 VS 0.09 VS. So 0.9 multiplied by the supply voltage will give us this value. Now, the second requirement is the output power as well. Remember that z power output is equal to v squared over r. And there we said here in zee bridge rectifiers that V-out is equal to V supply as ultimately square of course. So z v output or the power output will be equal to V out square, which is v square over the resistance R. So we'll see that non-Han 60 watt. If you compare this value with the hub bridge all seasons, this value is equal to four times z previous value. So in Z bridge rectifiers, Albert Bower is maximize the boy four times. And z fundamental component is z w. Now our requirement is the average and the beak current of each transistor. So z average here, as a big current is equal to VS over R, VS over R, the voltage over resistance. But remember that in the previous one where we had VS over to our VS over 2R. But here we have VS, since it's a rheometer, the Quanto VS over R, which will give us 20 MB. You will see that the current here of the transistor is z, w. Previously it was ten and bear here, it was 20 MB. Now for the average is 0.5, multiply by Z maximum value or civic value, which will give us ten and beer because it should transistor warp is only for half of the site. Now for our served requirements, A-B-C, reverse blocking voltage. Okay? So let's assume that we are in the second case where Q3 and Q4 our own. If Q3 and Q4 are on, and therefore gives you salt circuit Q4 short-circuit Q1 and Q2 are reverse biased. So u1 is a reverse it right now. So we'd like to find z reverse voltage. This Q4 is short-circuit. So by applying KVL here, you will find that z v reverse or the voltage across Q1 is equal to V supply. So let's see it. You will find here that's a big voltage is equal to V supply. This value is the same as in z half bridge. Now let's see our requirement. We need to find the Total Harmonic Distortion. So again, as you remember as, as a total harmonic distortion is simply equal to Z total harmonics as origin square over z fundamental component. So Z total calm harmonics, which is from sitting in 57 and so on, is equal to v square minus z fundamental. Ok? So this will give us 0.4, 3.5.2 V supply over the fundamental component. We will finally get total harmonic 48%. So we'll find here is that the total harmonic distortion is as same as the previous example. Ok, of the half bridge. Our next example or next requirement is that distortion factor. So the distortion factor will be equal to, again, as we remember, is that we have here for VS over n a by n omega t. Ok. And do we know that the distortion factor is awesome mission of z, each harmonic over n square, or a square under 0. So we will do is the same mathematics as before, but the difference is here we have four AVS and instead of OVS, you'll find again that the distortion factor is the same as before. So that's torsional effect. Z total harmonic distortion in case of z half bridge or full bridge is as him. Now lets us see z value of Z harmonic factor and store some factor for z lowest order harmonic. You knows at z lost order harmonic here again is the third harmonic because the harmonic factory is equal one over three, so it's 3%. And the social factor is one over 27 is 3.7. This is values are the same as before. So the harmonic factor, the distortion factor, Z total, social orders him. The difference between z half bridge and the full bridge is that Z bar is increased by four times here. And z fundamental component is doubled. So that's a benefit of a bridge rectifier. Okay? So I hope you benefit from this example and the previous one so that you can compare between them. Now, in the next video, we are going to store discuss a very important application of the inverter is a three phase inverter. And you have to come to concentrate on this lecture. 132. Three Phase Inverters and Obtaining The Line Voltages: Now let's discuss Z three-phase inverter. So we have here our three-phase inverter. In this way, we have our output is as three phase. So we have here, our combination here, S1 or switch number one, switch on for 23456. Each of these switches consisting of one transistor and one diode. Ok. So this part can be represented by this circuit. We have here our switch one, Q1 and D1. This toolbars or sharing the same node, ok? And the output from here is phase number 84 stringing, you will find the Esri and S6 and be having the same node. So we have here as necessary or switch number stream. So it's the number six and B having the same node. Four, switch number five and number two, and C having the same note. So we'll see number five, number two, and see here. So our input was a DC supply VS. And our Albert phase a, phase B and facie. Okay? This three phase is usually connected to motor. So we use the three-phase inverter in order to control the voltage and the frequency of motor by using something which is called Z variable speed drive. Okay? This all be understood in another time, but for now you have to understand that three-phase inverter converts z dc into AS3 phase out. Ok, so now let us see how does this circuit walks. So we have here our conduction method, or Z first method here in order to produce as three-phase our what is called Z a 180 degree conduction. What does a 180 degree conduction means? It means that each of these transistor is conducting for a, for a 180 degrees. So again, transistor Q1, Q3, Q5 due to q six q for each of these transistors having air conduction period of a 180 degree. And for Q1, we have a good signal called G1 or gate signal number one, ok? For q three, we have z gate signal g string. For Q5, we have z gate signal gene 54, number four, g4, gx, g two, and so on. Ok, so each of this transistor having its own gate signal for a 180 degree in order to make this transistor conductor for a 180 degree. So we'll see here is that we have here G1 from 0 to pi. A 180 degree z2 is conducting from boy over C3 by RC after 60 degree until pi plus pi over three. Okay? So you will see that G1, G2 after it by 60 degree, resists three after g2 by 60 degree. Again, z4 by after g3 by 60 degree, G5 after z4 by 60 degree, and so on, G6 also after 60 degree. So you will see that G1 from 0 to pi for a 180 degree conduction. G2 is also a 180 degree degree, but shifted boiling biodiversity, okay, or shifted by 60 degree. So you will see that here, each transistor is conducting for 180 degree. Three transistor remain on at any instant of time and all see how. For Wednesday transistor Q1 is switched on. For example, here, if Q1 is switched on, you will find that this transistor becomes assault circuit. So a will be connected to positive of the supply. Okay? So when Q1 is sweatshirt on terminal a is connected to the positive terminal of the DC input voltage. If Q4 is switched on, that CEQ for Q4, which is this one. If this one is switched on or it becomes a short-circuit. Zen Z AA or phase a is connected to the negative of z supply, negative terminal of the DC input voltage. Okay? Now lets us see held us this circuit walks. So we have here Z1, each of this gate signal on doctors for a 180 degree. Now let us see from this graph or from this signals. So we can get Z line to line voltage. What I mean by a line to line voltage, I mean VAB, Vbc and VCA. So we can get VAB, Vbc and the VCE from this gate signals. Okay? So this gate signals, again conducting for 180 degree. Each of these signals is shifted by 60 degree. And we can get Z line voltage, VAB, Vbc, and VCE. Okay? Now how to get VB? So VAB, as you'll remember from circuits, VAB means that we need to get Z terminal voltage VA. And the terminal voltage VB, and subtract them from each other. So VAB is simply equal to VA minus VB. Okay, so Va. Va is both step when Z1 is conducting or T1 is conducting, right? So we need VAB is equal to Va minus Vb. Va is z boast a voltage of E, which is happening when D1 is conducting. Okay? Now, for B, when does B is equal to a value when Q3 is conducting, U3 means that B is connected to the positive terminal of the supply. So a VAB will finally be equal to Z1 minus g3 multiplied bys IV supply, okay? Which means that G1 VS is simply VA, order. Suppose the value of a g3 multiplied by V supply means z boasted value of VB. Okay? So VAB assembly Z1 minus G3. So let us see now Z1 here minus G3 game. So Z1 here from 0 to two pi over three will be equal to a positive value here until two pi over three. So we are talking about g1 minus z3. So from here on till two pi over 32 by a three is a second dotted line is equal to a positive value from TBI varsity until by. You will find that here containing a signal and year having also a signal. So z V output will be equal to 0 from two y over three to pi. Remember that we are for VAB, we are subtracting g1 minus G3. Okay? Why? Because g1 is the positive terminal of a and G3 is the positive terminal of b. Ok. Now for let's continue what we have here from pi and until two pi over three plus pi, okay? From here to here, we will have 0 Z1 and the Boston z3. So 0 minus this value will give us a negative part, and so on. Okay, from here, from pi plus two pi over three to two pi, you will find that we don't have here a signal, and here we don't have any signal. So the output is equal to 0, okay? And then the cycle repeats itself. So you'll find that the Albert here, VAB is modified square or a square wave. Okay? Now for Vbc, Vbc symbol equal to VB minus VC, VB, whereas the positive terminal of b is q three. Z positive terminal of C is Q5. So we also cracked. V or q as E3 minus G5. So lets just see G3 minus G5. So z3 is equal to 0 here. And here we have a signal. So 0 minus a signal will give us a negative value here. From here to here we will have a 0 from two pi over three. Let's see two pi over three here until and here, pi plus pi over three, you will find that g five equal to 0 here. And here we have a signal, so it will be a positive value, and so on. Now for VCA, VCA means that Vc minus v K. So v c z positive of C, which is signal number five minus f of a, which is Q1. Okay? So z boasted of seeing which is five minus 15 minus g1, z5 minus y1. So we'll see here in this part of content, 60 degree will be equal to 0. From here until by, let us see, from here on till pi g five minus Z1, which will give us a negative value again. So finally, from this we reduced from ADC supply as three phase voltage, okay, VA and VB and VC, and we got Z, VB, VC, VCE, z line to line voltages. Okay? And we need to find z phase voltages and currents and so on. Ok. So you will notice that here, for example, this one is a square wave Vbc. You will find that it will be shifted from Xen, original one by a 120 degree. This part. Or find it starting from 0 until two pi over three, or a 120 degree Xin extra signal Vbc, starting after a 120 degree VCA, starting from after this one is started by a 100 also 20 degree, 20 degree. Okay? So as you remember that a three phase voltage is shifted by a 120 degree. So VAB a shifted from Vbc by a 120 degree, VCA shifted from Vbc by a 100 and degree and so on. And you will find that we can draw VAB and other Vbc and VCE will be shifted by a 120 degree roster like this, we can find one and Jeff disease signal by a 120 degree. Now let's discuss Zhi De France between x0 and delta connection. You will see that we have a star connection or Y connection and delta connection. I just want to remind you is that for a y connected load, for this time, you will find that the line to line voltage must be obtained first in order to find the phase voltage. So in the previous week, got VB, VB VC, VCE, which is the line to line voltage. Okay? Then z phase voltage will be this value divided by root three. Now four z delta connection. Remember that z phase, phase voltage is equal to line to line voltage. So V AB is the same as V a or z voltage across Z. Once is the phase currents are known, Z-line columns can be obtained here you will find that z phase current is equal to Z line current. For z delta connection, you have to know that we can find the phase currents. Then we can obtain the line currents are not equal to h OSC. Okay? So here in z, y, z, line current and FiscalNote are equal, but the phase voltage and the line to line voltage are not equal. Here is the phase and the line are equal. But a Z line current and current is not important. 133. Three Phase Inverters and Obtaining The Phase Voltages: In this video, we are going to discuss how to get the phase voltages in Z's three phase inverter. So in previous video, we discussed how this inverter walkers and we go to z line currents have VAB or Z line voltages, VBE, VBE, seem, and VCE. So now let's see how we can get z phase voltage. So first you will find that we have as three, where we have a three or six modes of operation in cycle or less three modes of operation in each half cycle. So we have a six and mode of operation in one complete cycle or a two pi. And each mode of operation is 60 degree. Or for each half of the cycle, we have only three modes of operation for our y connected load. So now let us see. You will find that here from 0 to pi over three or 60 degree, you will find that we have Z1 is conducting, the G5 is conducting, G6 is conducting. So we have a surrogate transistors are conducting from biodiversity to two pi or Syria or another 60 degree. We have z1, z2, and g six. Now from two pi over three to pi, you will find G1, G2, and G3. So the first thing you will notice that each mode of operation, for example, Z, 60 degree here, or 60 degree here, or 60 degree here, you will find serine transistors are operating at the same time that what we said in z naught, which we said that we will explain it later. Okay, so we'll find for each second-degree, We have a sorry transistor always on. Okay? And you'll find here one, do 33 modes of operation in half cycle. Or for one complete cycle, you will find six modes of operation. So in order to get the VCE voltage, we will study each of these modes. So from 0 to pi over three, which is our first moment, okay? Omega T, or our angle from 0 to biodiversity or from 0 to 60 degree. So find G1 is on, G5 is on, G6 is on. Okay, so let's see if Z1 is on. Z1 is on. So air connected to 0 positive terminal of the supply, okay? Z five is on. So G5, This one is on. So C is also connected to the abosDF determinant of supply. Now, for 66 means that B is connected to the negative terminal of the supply. So in order to draw the equivalent circuit during the period from 0 to pi over three, we will say is that a connected to positive of the supply, C connected to z positive and be connected to the negative. So now let us see. We have here our supply, V, supply a and C connected to the positive terminal, as we said here. Because g1 is on and G5 is on. So a connected to positive, C connected Tomas, okay? A and C connector 2Z Boston. And B is connected to the negative terminal, as was heads at GE sex is on, which means that B is connected to negative terminal of the supply. Ok? Now we will find that a and B and C, each of these phases have, are a little bit, or we will assume I resist a pure resistive load. Okay? And all of them are connected in 1 called the neutral, since we are talking about a star connection. So you will find here our neutral point and resistance, resistance and honest resistance. So AC connected TO boston, be negative, sorry, resistance afford each of this phase and all of them are collected in one neutral point. Now, we would like to find Va, Vc and VB remembers that VA is the voltage between a and a neutron, VC between C and a neutron, a B between B and a neutron. So we have here our circuit, we have V supply and resistance. So we can find the equivalent resistance of our circuit, which will be 3R over two. R and R parallel to each other will give us r over two. And another series resistance R will give us 3R over two. Z supply current here will be V supply over the equivalent resistance symbol from Ohms law. So it will give us two VS over 3r. Okay? Now we would like to find a z, a voltage of V n here. So vn is simply equal to z current here multiplied by the resistance. So the current here, since we have two equal resistance, therefore, the current here will be one over two or u one over two. So v a n will be equal to V CAN sensors the R parallel I1 over two multiplied by the resistance. Okay? Are you one divided by two will give us VS over three. R multiplied by R will give us V supply over three. So we got here VN called VCM VS over three, where we got to the current year and the matplot advisor resistance. Now for Z voltage VBN will be equal to negative y because the current year is opposite 2Z phase. You will find here is that v AN assembly equal 2i current from left to right multiplied by resistance. Vbn is equal to Z current from left to right multiplied by resistance. But I1 is in opposite direction from right. Left here, okay? So it will be equal to negative I1 multiplied by the resistance. So it will give us negative two VS over R. So this is the phase voltage, and this is the phase voltage for each of these phases at 0 to 60 degree. So this was the first mode. Second mode, which is from 60 degree to a 120 degree. Now let us see it. From here to here, you will find G1, G2, G6 is on. So G1 is on, G1 is on. So a connected to the positive, G2 is on z two, so C connected tos inactive. Now number six be connected to a negative. So we have here a, B, and C connected to the negative, is connected to the positive. So now let's see the equivalent circuit. We have Boston negative, okay? We have a connected to the positive terminal, B and C connected to z negative terminal. So and the city resistance, as we said before. Now we can get the equivalent resistance against the order of R2. And we can get the total current. And if we can get vn here, because a connected to the Boston. So Z current from left to right is the same direction of i2. So vn is Avastin value, VBN and VC. And here it will be a negative value. Okay? Again, by using Ohm's law, we can get this currents and devalues is now from z, for z mod number three, from Doughboy varsity two pi or from a 120 degree to 180. You will find that here. We have z1, z2 on g3, on Z11, a, Boston, G21, see negative G3 on be positive. So a and b are positive, c is negative. A and B are Boston if c is negative. So by using, again, give you N, We can get Vn, VBN equal via sorcery and V c will be equal to z negative Z current multiplied by resistance. Okay, very simple. Now if we draw the equivalent voltage wards, we draw the voltage with respect to the omega t or suspected to time. We'll find years at Vn from 0 to pi over 360 degree as well. Remember VS over string here to VS over three here, voiceover S3. For VBN, it wasn't two negative and one Boston for this end positive, negative, negative. Okay, this is the first second modes or the moon. First, second son's first, second.srt. Okay. Now, instead of zoster finding mode number 456, we can draw this shape, but with a negative value. Okay, remember that z voltage or AC voltage is symmetric around the z x axis. Ok? This one will be reversed again. This one will be reversed again. So we found z phase voltage using a 180 degree conduction method. Now, for Z line to line voltage, VAB can be expressed as and the Vbc and VC can be conceived breast as they're following Z n equal 1354 VS. And by citing the Bible three sine n omega t plus pi over six. Okay? Need to memorize this equation. And VC, VCE will be the same, but zoster shifting the boy a 120 degree from each other. So this is a 60 degree minus a 120 degree, okay? And will give us pi over six, which is as biodiversity. And certainly the green minus a 120 degree will give us minus 90. And here, minus 90 minus 100 degree will give us minus 210. Okay? So as this value shifted by this one by a 120 degrees, shifted by this one by a 120 degree. Now you will notice that sign in nearby stream at N equals three, okay? Sign city pi over three, which will give us sine pi. And as you know that signed by is equal to 0 at six, it will give us sine two pi. So this will give us 0. At line. It will give us, of course is three pi, which is also 0. So what does it mean? It means that z treble harmonics inequal city mine 15 are 0. Why? Because sine in the bio varsity would be equal to 0 at n equal city will give us 0, mine 0150 and so on. And of course, six will be equal to 0 because it is an even harmonics. Okay? Now, let's define Z line to line or minuss square. Ok? So let us see the function line to line. Okay? We have to get back to this one, okay? From 0 until Toby over three, okay? So the integration of the function is from 0 to two pi over three. And we have another here from pi to two pi over three plus point. Okay? So we can integrate this function and multiplied it by two. We are getting here is our root mean square Nazi average. So let's get back to the root mean square value. From 0 to two pi over three for XVIII line to line, which is VAB, vbc, Vcd, any of them for the function square, which is VS square. And two, because we said that we have a wall step and the negative cycles over two pi the whole period. All of this under zi square root will give us finally root two over three V supply, okay? Which has this value. Now for Z root mean square of z component, it will be equal to. Now let us see z line to line here for VS over n pi sine in the biodiversity. So add any value or any N 0 square will be Z maximum value divided by root two. So it will be for VS sine in a bio sorcery and thereby divided by R2 in order to get or transform it from maximum 2s or root mean square value. Now, at n equal one, we can get the fundamental component which will give us this value. You don't have to memorize this or this, or this one. Just remember that root over, sorry, V supply is z, v line. And just remember that here. This function representing VAB, okay? And you can get vbc Vcd bisecting this function. Now if we'd like to find is the root mean square value of z phase value. We said that as we remember that from Z voltage or z phase and the line, remember that z phase voltage, line voltage divided by root three. So Z line voltage, as you remember, R2D2 over string. We'll divide this by root three will give us Ruto VS over three, which would give us this value. So by remembering, getting or getting z-value offline, we divided by root three and they get z phase. Now Z line current I0 for our elute is given by, as you remember, for VS over enter by sine in the biodiversity sine n omega t minus z lagging ganglion, which will give us ten minus one and omega L over R. Now for Z zed here is root R-squared plus n omega l omega square. As we said that here we have n because z inductance depends on the frequency. Here we have root three because we are talking about Z line voltage. As you remember that z IE in z star connection or the phase current is equal to Z line current. So z phase current is equal to z phase voltage divided by root divided by z. So z phase voltage is simply Z line divided by root serine, as we said here, V line over root three. So this root city is used to convert the Z line voltage, phase voltage. Okay? As you remember that in star connection, IE or Z line current, Orpheus current at the same time is equal to z phase voltage divided by present. Ok? So that's why we divided here by root three, because we have seen line to line voltage here. And what it by root three to transform into phase voltage. Now in the next video, we are going to have an example on Xin three-phase inverters. 134. Example on Three Phase Inverters: Now let's have an example on z3 phase inverter. So we have here as three-phase inverter with Y connection or r star connection, having lewd resistive load of their own and an inductance of 23 milli Henry. So we have our Lord as our elude. Z inverter frequency is 60 hertz and z d c m with voltage is V S or Z DC voltage equal to 120 volt. Find Z following requirements. So the first requirement here is that we need to find z in something. Yes, line voltage VAB as a function of time or omega t and the line current as a function of time or omega t. So the first thing you have to know that sensor we have here Y connection. Remember that z base current is equal to Z line current and the phase current is equal to V phase over Zed. Now, let's define the first v a v as a function of time. So VAB as a function of time will be equal to Z following how we got this function or this information assembly. So remember that VAB is symbolic. What 2a summation of n equal 135, and so on for VS over N a by signing a bio for three sine n omega t plus pi over six. So what we did is that we took this information or this equation. We have V sub one to a 120 volt and we have n equal 135. But remember that's at treble is equal to 0. So we have one, we have five, we have seven, we have 11, and so on. Omega t plus pi over six, this is a constant, N will be equal to 15711 and so on for VS, all of this is known. So that is what we did simply we just substitute in this equation now for Z current for our, a lot. As you remember, IA is equal to four VS over root city in a by root R squared plus n omega l or square sine n over three sine omega t minus theta n. So what we did that we have V supply, which is 220 volt r square r is given insights or problem, or omega is two pi f And we know that the frequency is 60 hertz. Z inductances also given Z, n is the variable, n is variable here and seed the n or z angle will be tan minus one. And omega L over R. Omega L is known, r is known, n is variable and it changing 15711 and so on. So let us see our problem. You will see that here b, a b by substituting equation, we got this equation. And for z, is it dilute or z Zed, which we needed for Z current root R squared plus n omega n square and z angle tan minus one and omega L over R. Now we have omega L is given two pi frequency, which is around 77 or z omega itself. And z inductance 23 millihenry 0 resistance equal ten Ohm Z. Resupply 220 volt zoster by solving this by your hand. And you will find that this equation is very easy to obtain. You will find here the misrepresenting Ghazi fundamental component at omega T plus 30 degree. And you'll find here five, which means you are talking about the fibs, the harmonic. Here we have seventh harmonic, Zen Z 11. Okay? Remember that we have here one is that even harmonics are equal to 0, z sort of the harmonic is equal to zeros. And cities at treble harmonic 57, mine is also a trivial harmonics or it will be equal to 0 then is uneven harmonic, 11 is known. And Z current by subsisting also in the equation, we can find this equation of Zika. Okay? So zoster by substituting, you will find this values. Now for Xen. Second requirement here is that we need to find z root-mean-squared line voltage V L. So as you remember that the line voltage equal to root two over three multiplied by V S, which is this value, 0.8165 V supply. So let's get back to it. So you can remember. You will see that V line is equal to root two over three VS, which is 0.8165 years. Okay? Now, as second requirement is the phase voltage. So as you remember that z phase voltage is equal to this value or equal to z line voltage divided by root city. Okay? So you will see here again, if you don't remember that v phase is equal to v line over root three, which means is this value of VS. Okay? Now, our served requirement or force requirements, root-mean-squared line voltage adds the fundamental frequency. So we need to find v line at x0 fundamental. Okay, so you'll see here is that we have this one. This is the line voltage as a function of time. This one representing the fundamental component, this one representing z faves, this one representing the Z sevens, and this one representing Z 11 or 11's or as 11 turn or 11 harmonic. Okay? Now we will see that this one is a fundamental and this one is the maximum value. And this one is the fundamental and Z maximum. And we need to find z root-mean-squared line adds the fundamental. So we can take this value and divide it by root three. Okay? So by taking this value divided by root three, it will give us this value. Or Z line voltage V one can be obtained from this equation. Here we have v L1 is as this equation, this one at n equal one, which is for VS of our boy soin biodiversity at inequality one of course. And after taking it, we divided by root two and order to convert it to root mean square, which is this value. Okay? So by knowing this equation, you can find anything you would like, okay? From this equation you can find the fundamental Z verbs, any load like okay, you don't need to memorize this values. So by knowing is at line is root two over three VS and Z vales is lying divided by root three. And the z v root mean square of any voltage. We can get it from this equation, getting Z maximum and dividing it by root two. Now let us see z. Next requirement. Now would like Zao root-mean-squared phase voltage. So we got here z line and do we need here to z phase at all sorts of fundamental. So we know that the relation between the phase and the line is that z phase voltage equal to Z line voltage divided by root three. So let's see. Phase voltage is equal to Z line divided by root three. Okay? Now let's get back to Xin extra requirement Z Total Harmonic Distortion. So the first thing we are going to do is that we write z equation of Z total harmony social, and here is a distortion factor. Ok, so let's write the equation. You will find here is that the total harmonic distortion is equal to z summation of all harmonics divided by Z fundamental component. So z fundamental here, as we said before, fundamental here is 0.7797 VS. OK. And you remember that this value is root mean square value. Now for Z harmonics, you will find some mission 5711, because Z one is a fundamental, three is 02, is 04 is 0, and so on. Ok. So this value is simply equal to z supply voltage or Z total line voltage of the output minus v1 square. Okay? So v l or minus V fundamental VL is total load voltage, or V0 line of the output. Okay? It will be equal to, from this equation, as you will remember, that v l is equal to root two over three VS. This is Z total line output voltage. Okay? So we can take this value minus the fundamental will give us this value and we can get the Total Harmonic Distortion. So remember that this one is obtained from z square wave of z line to line voltage. Okay? This representing z root mean square of z, line voltage of Z out. Now, by subtracting this from the fundamental and dividing it by z fundamental, we can get the Total Harmonic Distortion. This problems are very, very symbol exhaust. You need to solve it by your hand. Ok. Now let's define Z distortion factor. Remember that distortion factor is simply equal to one over V1 z, some measure of two, three and so on. All of the harmonics a V n over n square, a square on the zeros. Okay? Z, same equation which we used in z as single phase half bridge, gay or single phase bridge. Now vn. For example, V five over five square V seven over seven square v 11 over 11 squared. So z voltage here. Remember that we already obtained it here. Ok? This one is V1, V five, V seven, V 11. So we talk xyz values, okay? And divide it by rho to remember is that again, these values are root mean square values. So we talk z values in the equation and divided by root two. Okay? Now we can get the total distortion, distortion factor 0.8, 5-7 percent. Okay? Now, z in extra requirement is that we need to find z harmonic factor and distortion factor of Z lowest order harmonic. Remembers at our harmonics is from after one of course two, which is equal to 0, city equal to 0 here since the z treble harmonics are equal to 0, for z equal to 0, but five is existing. Okay, so to start with z order harmonic five and defined its magnitude. Okay? So the lowest order harmonic is z fails because three is equal to 0. Since as the treble harmonic is equal to 0, its value as a harmonic factor will be equal to V, five over V1, okay? Or Z favors a harmonic of the fundamental, which is 20%. The in-store some factor will be v five over n squared over the fundamental, which will give us one over a 125, which is 0.8%. Ok, so that was easy requirement. Now let's define our next requirement, or Z lost one, z power out. Okay? Remember that z output power is equal to v square over r, or i square multiplied by r. So we know that of course is z a v is phase value. So we can see is that our power here is equal to four, a y connection. Our power is equal to three V phase, I phase. Or we can say city V squared over R. Or we can say three i squared multiplied by R z law of z power, of Z load power or the act of bar. Okay? So three V phase I, phase VIV is, is already obtained z phase value, you will find here we already obtained it. V phase of the output is a 103.7. Okay? 103.7 or 0.9 whatever. But it is 0.7 for Z line current for the phase current is equal to Z line current divided by line current is equal to phase currents. And so we are talking about Z Y connection. Why connected loads Z line current is the same as the phase current. Okay? So Z line current will be equal to V phase over z resistance. The current is equal to V over R. So the voltage here is V phase a 103.103.7 divided by the resistance R. So we can get Z current 10.37 and bear. So our bout Albert is three V phase, I phase. And it will be equal to like this. Ok? Now, as the important thing is that we have to know is that power can be obtained also boy, stream, v squared over r. We can say, sorry, v squared over r. Z voltage square root is a 106.09 square divided by the resistance, then on, okay? All are the same. So in this example, you may find it a little bit difficult about justify solving by your hand and you will find everything is your Zan using Ok. So in Xenakis video, we are going to discuss is the pulse width modulation and how we can use it to produce as three-phase out, ok, nearly or close to assign with. 135. Single Pulse Width Modulation: In this video we are going to discuss Z balsa weeds modulation or BWM. So z pulse-width modulation is a technique used it to generate a DAC from DC input voltage. So let's see how does pulse-width modulation walks. So first, we have three types of pulse width modulation techniques. We have Z single pulse width modulation, z multiplied pulse width modulation, enzymes sinusoidal pulse width modulation. So let's discuss first the single pulse width modulation. How does a single pulse width modulation walks? So in this method, you will find that we have only one pulse per half cycle. That's why it's called single bonds, because we have only one pulse generated. Bear half cycle, okay? And z width of this pulse is varied to control the inverter Albert voltage. So let's see. We have here two signals. We have z for a signal called Dizzy carrier signal. And we have another signal called z reference signal. Z carrier signal is the form of a triangular wave. Form. 0 reference signal is the form of a square wave. Okay? This carrier signal which is ask triangular shape, this wave has a maximum amplitude of ISI or a, or the amplitude of the carrier signal 0, France signal having a maximum value called the AR or z reference amplitude. Okay? Now held us AC signal is generated or how does, how do we generate this output signal? Simply by converting the carrier signal with the reference signal. Well, when z reference signal greater than Z carrier signal 0 will be up faults. Okay? So we have here from starting from 0, z carrier greater than 0, France. So z i would signal will be equal to 0 here until this point, z carrier equal to z reference. Okay? So we have here our balls is a starting, okay? And from here to here you will find that z reference signal is greater than Z carrier signal. So when z reference signal and greater Zan Zi carrier signal, Zeno will have a waltz. Okay? Now, the second one here when z carrier signal also as the reference signal, greater Xunzi carrier signal. And then we also will have an output in the negative direction. So in this direction you will find here is this is the carrier and this one is the reference signal. So z reference signal here having an negative value more than this. A career signal. So during this period two, we will have another pulse. Okay? This one representing z gate of z positive transistor or q one. This one representing z gate for z negative one or q four. Okay? So as this one, but reduces z positive value or Boston Albert voltage, this ball's produces the negative output voltage. Okay? So when z carrier, during positive half cycle, when z reference greater than the carrier, we will have our pulse. When z carrier having an negative value, more Xunzi, when z reference having a negative value more than Z carrier signal, then we will have another pulse foresee negative part. Now our Albert will be this signal when it is positive, it will be positive when it is on Z. Z four also transistor number for all z negative transistor, Xin will have a negative volts. Ok? So this represent as our output voltage, this value as V supply and this one is negative resupply as this method use the torque control z gate signal generation. Ok. Now we will find this pulse having a width of delta. Okay? This point, which is the Medal of z carrier signal here, you will find this is pi, this is pi over two. So this point is also pi over two. And this signal is delta, okay? And this signal is also symmetric around pi over two. So this part is delta over two, and this part is also delta over two. But this value as an angle, it will be pi over two plus delta over two. This one will be pi over two minus delta over two, because z by over two minus two is less than pi over two. And this one is greater than pi over two. Okay? So is this point pi over two minus delta over two, y over two plus delta over two. Okay? And for z negative part will be the same but three pi over two. Since the z, the France between here and here is pi, okay? Or a 180 degree. Now this part is also that overtone and this one is delta over two. So again, single-pass wouldn't be duration pulse width modulation means that we have only one pulse per half cycle. One pulse during this half cycle, which is from 0 to pi. And Azara lifecycle from pi to two pi, which is also another one pulse. Now again, z gating signal through throws us from the stores are generated by comparing rectangular reference signal, which is a square wave of amplitude AIR. Okay, reference amplitude with add triangle carrier wave of amplitude abc. So we compare this tool way of z carrier, which is triangle and disease, rectangular or a square wave of reference signal. Now is the frequency of z reference signal. Neutral mines or determines z fundamental frequency of the output voltage. Okay? So z frequency here, you will find that this is z1 complete cycle, okay? And you'll find you owning Z1 complete cycle. We also have the same frequency, one complete cycle. If this reference signal frequency increased XeF4, what will happen? Our Albert will also have a higher frequency. So z reference signal control frequency, control the output frequency by varying the Z reference from 0 to a carrier or the amplitude of the carrier signal z balls. So into delta can be varied from 0 to a 180 degree. Now, let us see at a reference equal to 0, okay? Era France will be not greater Zen a carrier. Okay? So always there will be no output signal. But if we have a reference equal to a carrier, then we will have a signal from 0 to pi and from pi to two pi. Okay? This is known as XM. We do lotion indexes and modulation index, which is the ratio between AR or 0 France amplitude to Z carrier amplitude. Okay? Now Z outward root mean square voltage in z, single pulse modulation is equal to V. Output is equal to one over t. Integration from pi over two minus delta over two boil of R2 plus delta over 2m, okay, which is same as y plus delta over two came for Z Albert which is resupply or square. All of this honours or square root multiplied by two. Again. So we have here z integration for the arg min square integration from pi over two minus lambda over two to pi over two plus delta over two. For z function which is VS square. And the multiplied it by two because we have a signal cycle and another signal in the negative direction. So this will give us finally V supply multiplied by root delta overboard. Okay? So by controlling z delta, we can control or Z pulse widths. We can control Z root mean square of z v out. Now, z function can be determined by as he Fourier series V out as a function of time or omega team. Some mission from 135. For VS over n pi sine delta V2 sine n omega t. So this represent a z square wave varying with z delta. Okay? Now, if we draws a relation between the modulation index and store some vector and z ratio between V output or Z harmonic with respect to two VS. OK. So now let's assume we have here at modulation index one, here at modulation index 0 and Z values between it. Okay? Now you will find here is that the fundamental component V01 is maximum at modulation index equal one. This is true because we have at modulation index one xin, our V output will be equal to V S or V output as a root mean square root will be maximum. So here you will find V1 will be maximum xin as the modulation index decreases, V1 decreases. Now let's see is that store some factor. You will find here's a distortion factor is varying. We will find here as a percentage modulation index equal one, it is nearly 3.5 and it decreases. Xin increases again. So all finds that x_hat and would you relational balls. So when someone duration, it changes zed distortion factor. Now here you will find for each of z component V seven, V5, and v3, which is the harmonics of our wave in our, in, using Azim, well so-and-so modulation. You will find for examples that v3 here, having a high value at one, then it decreases and then it increases again. So we'll find that by changing disease balls so wins itself. We can control the amplitude of the harmonics. Okay? So add for example, at modulation England index at 0.75, for example, we will have certain harmonic nearly equal to 0. Okay? Because we here you will find that v3 is equal to 0, same as V7. We can control and make it 0. V5, Susie Bolshevism undulation helped us in many things. Number one, we can control Z albatross. We can change it and distortion factor, we can change the Z fundamental amplitude. We can also change zinging harmonics or decrease as the value of the harmonics. And the problem is with small modulation. Pulse-width modulation, or modulation index, is that as the fundamental component decreases, as the, as the modulation index decrease. Okay? Now in the next video we will discuss Zim, multiple pulse-width modulation. 136. Multiple Pulse Width Modulation: Now in this video we are going to discuss a z multiple A-ball Swedes modulation. So in the first one we had one pulse or single passwords omen duration or two means. One pulse bear half psych. Nows on multiple levels will have a multiple number of pulses per half cycle. Okay? So what we'll find here is that instead of having only one pulse, we would have our multiple. Now, how does this works? You will find that we have here is the frequency of the carrier signal or Z carrier signal here will have, will not have the same frequency like z reference signal. Okay? What does it mean? Let's take it back to a Z single pulse widths and see what is the difference. Ok, you will find here is that for Zhi He single pulse, you will find that we have here is this one. And this one represents one complete cycle or one triangular wave. Ok, you will find here z hub here, and another hub here. Okay? So this half and this half representing one pulse of Zakaria. This half and this half also representing one pulse, which means that we have here one triangle in z positive cycle or negative cycle, and the one triangle in z positive cycle. So again, we have here half of Z triangle, Howard Zehr triangle has the triangle and the triangle, which means one triangle in the Boston, one triangle in z negative same as z reference. So we have one Square in Boston and one node, one rectangle into Boston, and water rectangle in z negative. Okay? So that means that we have frequency of the carrier signal is the same as the frequency of 0 from signal. Okay? So now let us see in z multiple pulse-width modulation, what is happening. You will find here is that we have here one triangle, hub of triangle here, one triangle, one triangle, another one, another one, another 1.5 triangle, 1.55111, okay? But so we have a total number of triangles here is ten triangles, okay? So x1, x2, so z carrier signal repeats itself ten times in one complete cycle. But z reference signal, you will find here one rectangle in z positive cycle and the one in z negative cycle. But z carrier having five into positive cycle and five in z negative cycle. So I will find that if we compare z reference with Zim carrier signal, when z reference greater Xunzi carrier, we will have an alma pulse here. So this was one. Here we have z reference is greater than the carrier. So we have on Assad balsa and of course is the width is delta. Another pulse here, another pulse here you will find that the total number of pulses is equal to Z, total number of triangles. Okay? We have ten triangles. So we'll find here 55, which means ten pulses. So it's a single pulse modulation. We had only one pulse, but here we have five balls is bare half psych. Ok. Now we will assume here something which is used for Fourier series that z, this wave form or this pulse is add alpha from 0 axis and z negative one here is at alpha plus pi. Okay? Now, you will find that the frequency of the carrier wave, which FSC determine mines or determine Z number of pulses per half cycle. Okay? Beam is number of pulses per half cycle. B is equal to x cosine over T over two f naught. Ok? So FVC and representing is a frequency of the carrier, f naught, representing the frequency of the square wave or the output voltage. Okay? So the frequency of the carrier here, you will find the triangle repeats itself ten times. Okay? So if we divide z triangle ten times over to, okay? This means that we will have five pulses bear half cycling. Okay? So z square repeated itself one time in one complete cycle, but this one repeats itself ten times. Ok? So FSC over ethanol is called MF, or it's called disease frequency modulation ratio. So f naught is the frequency of the reference signal or the output frequency. And the MMF, which is FVC over f naught, is a frequency modulation ratio. Now is a modulation index is also as before, AIR over AC. Now Z out root mean square voltage can be found from, okay, is the integration of z function in a sin of pi over two, it will be by over b, which is number of pulses bear half cycle minus delta. And here plus delta and z, our touch is VS square. But here since we have 2pi, which means a, b is the number of pulses per half cycle to be representing Z total number of pulses. So this gives us VS root b delta over by. So in this sinking pulse modulation, we had b, which is a number of possible OFF cycle. We had only one balls. So Zillow in single pulse modulation was the output equal VS root delta over pi, but in cents. So we have here several number of pulses per half cycle. We add a term equality beam. Which is number of pulses per half cycle. Delta is equal to m or z modulation index over f c. Now, let us see. You will find here is that we have 12345, which is five, which is called B or Xena pr of pulses per half cycle. So as the total number is total tubing, okay, total number of pulses. Now, this one having a reason or Z widths of it is called delta, same as before. Okay? Now from this, we obtained several number of pulses and all understand how does this effect z as r square for your series and everything. Now let us see you Fourier series we have here for Fourier series, our voltage is equal to some mission one, sorry Vive, so on. Bn sin n omega t, where Pn is the determined the pie considering a pair of pulses such that z Boston pulse of duration delta_t starts at omega t equal alpha and z negative one starts at omega T equal prime plus alpha. So pn is simply considering z pulses from alpha, from this pulse on till alpha plus pi two, this box, okay, bn, representing this process. So let us see now, pn is equal to this long formula or formula, okay? For this formula, you will find pn is equal to this part. Do we take this part inside sine n omega t, we will get V out as a Fourier series. Okay? This is not important for us as the most important thing is to understand that z V output is equal to OVS root b delta over pi. How does Z Albert as fundamental? And two, comparing the two Z single pulse modulation, how it is changing. So compared to the single pulse or zoomed duration, you will find that switching losses increases. Why? Because our, our transistor is operating on and off, on and off, on and off. Each pulse is on and off. Okay? So z switching losses in z power electronic devices is increasing. Also z. Another disadvantage is that the amplitude of some high frequency harmonics order increases. Z amplitude of low-frequency harmonic order decreases. This is a good thing. Why? Because the lower frequency harmonic means that z frequency, our harmonic say close to one. It means that we are talking about the harmonic order number three, number five, number seven, and so on. So this amplitude is decreasing, which is I would sing, means which having a z highest effect in the harmonics, ZI frequency harmonic is not really important because it is already a low value. Now we feel cumbersome single pulse modulation with z multiple Abbas modulation from z modulation index with z variation of z harmonics. So let's see is the fundamental. You will see that ad modulation index equal one. Here, modulation index number one, z fundamental compound is nearly equal to each other. Ok? Now for Z single pulse, you will find that the distortion factor is varying from nearly 9% to 3%. Now, here you will find it is varying from 4.5. All sorts of distortion factor is decreasing at z modulation or low modulation index. Now for Z frequency here, or Z harmonics or the harmonic, we said that low order harmonics decreased in Zomato. Now it is varying from 0.3 to 0.2. Now let us see here, you will find that it is varying from point to stream to again 0.3. Okay? So at low modulation index or find it is lower Zan here. So v3 year as lower than here. Now for Z sevens, you will find that it is already low from nearly 0.1 line, for example, until itself. So let's see here it is having one or 0.10.152.25. Okay, it's just varying goals, So here, but you will find that something here, V7 is considering as a higher order harmonic. So it is increased, increased as austin pi, a small value, v5. It is nearly also the same here. Ok, there's also a small difference, but you will find that this one is nearly constant. Z values of v phi V7 is nearly constant across as either France modulation index. But you will see is that here V7, V5, v3, all are playing. They are don't have a constant value. And benefit here in Zomato blue balls, so finds a distortion factor decreased. It was mine percent. Now it's 4.5% until less than four here until great Zen for plasmon value. So that is was the difference between the single pulse and the multiple pulse width modulation. 137. Example on Multiple Pulse Width Modulation: Now let's have an example on zing alterable width modulation, all this England pulse-width modulation and this are same. If you understand the multiple you can solve for z sink. Okay? Now we have a single phase full bridge inverter. This inverter is working with pulse width modulation technique. Okay? So we have here at Control Z power in a resistive load, ZL input, dc voltage VS equals 220 volt. I uniform a pulse-width modulation with five pulses per half cycle. So b is equal to $0.05. So we have five pulses bear half cycle. For Z require to control Z widths of each pulse is 30 degree or pi over six. Now, z for a set of Guam is determine Z root mean square voltage on Z. Rude. So let's affine 0-20 square. As we remember that the root mean square is simply called V supply route B delta over pi. Let's just see what root mean square is equal to VS supply to a 120 volt route B, which is five pulses multiplied by delta, which is a certain degree of R pi. So certainly over a 180, which is the same as pi over six over pi, okay, is the same. So the output as a root mean square will give us 208.8. Okay? Now z, second requirement is if the DC supply increases by 10%, so our supply is now 1.21 multiplied by 220 determines z also needs to maintain the same load power. Ok? So we changed our V supply and we wanted to make Z outward powers as him. So let us see. First. Remember that the output power is equal to v squared over r. So in order to keep z output constant or seem Bauer constant that we have to keep VS root b delta over pi constant. So VS, it change it. Okay, VS became 1.1 multiplied by 220, which is 242 root. B. Number of pulses is the same. Five pulses. Z widths changing, okay? So delta is changed, auditor's are acquired over pi. This will give us 200.8 volt, okay? Because in order to keep blood power constant, we need to keep the Albert Cornerstone. So VR with here constant VS 1.1. VS root five delta h, So delta decreased, it became 24.75. Our third requirement is Z. Maximum possible pulse width is 35 degree. So delta became 35 degree, okay? Determine Z minimum allowable limit of the DCM mode source. So now our requirement is V supply. So we will keep the 100. It will be equal to this supply which is now unknown. Wrote five delta is certified degree or one pi or a 180 degree. Now, our supply must decrease to 0.643. So we have here at 200.8. Now also apply to a 100 and t 0.64, which is the minimum allowable input voltage. Why? Because our supply was originally 220 volt now which is at a 103, which is lower than the previous value, because our delta increase the from inserting those 35 degree. Okay, now winds and extra video, we are going to discuss z, sinusoidal pulse width modulation. 138. Sinusoidal Pulse Width Modulation: Now let's discuss Z Law. So type which is the sinusoidal pulse width modulation. Okay? Now in this course who are going to discuss a pipe called Zen uni directional. So first you will find here we have a reference signal and carrier signal. Before we had the carrier signal as a triangular waveform. Okay? The referencing none was a rectangle, its waveform. Now our enzymes or the other, our BWM, we have a reference signal of sinewave. Okay? We have here two sine waves. We have our positive sine wave and negative sign wave. Okay? So we compare z positive and z negative with z balls itself. Ok? Remember that G1 and G4 or Z transistor Q1 and Dequeue for, remembers that, that one was Z transistor, which you connect this node number a or phase a to the positive of the supply. G4, connect this a to z negative of the supply. Ok? Now, remember that also that a o here and will be g1 minus z four if we're talking about his number eight only. Now, let us see how does this pulse-width modulation technique works. So first, let's compare the boast of cycle, though in order to get z, z0, y1 and compares the negative cycle to get G4. So comparing Z boast of cycle, you'd see here is that when 0 from sigma greater than Zachary or signal, we will have an output equal one. So here, carrier signal, or the reference signal is Xunzi carrier. So the output will be equal to 0. The starting from here to this point, you will see that reference signal is greater than zinc carrier signal. So we will have an output. Okay, have adults. Now for here, you will find here from this point to this point on other pulse because z reference signal, greater Xunzi carrier signal. And same was this pulse. But you will notice that this pulse does not have the same width of this pulse, that does not have the same width of this pulse. Ok? Now, for this part, you will find that this part, which is this one which is the negative. Ok, we are talking about the Boston sine wave. So this part is less negative Zan Zi carrier signal. Ok, we are talking about 0 positive value. Okay? So this one is more Boston Zhuangzi carrier signal. So we'll have another pulse. At this point, you will find z reference signal is more Boston Zen Z carrier signal. So we'll have another pulse and so on. Now for z4 lets you compare z negative with it. Now you will see that this point is more positive then the carrier signal. So we'll have about at this point z reference signal having more positive values, n z carrier signal. So we'll have another positive pulse here also opposed the pulse. Now for this part, again, you will see that this part is greater than this part. Z reference signal here of z negative sine wave is more boasted Zamzee carrier signal. So we'll have another pulse here. You will find from this point to this point, z negative sine wave is more positive Zan Zi carrier signal. So we'll have another pulse and so on. Now in order to find the V output, it will be z1 minus U4. So g1 minus U4 will give us hear a pulse, another pulse and betweens MCO. Here we will have also two pulses and between 0 here will be the same. This part will be negative, ok, because this part or this pulse is greater than this pulses. So I'll find the ER negative values. So you will notice that finally, each of this pulse having a different width, ok, this is number one. Number two, you will find is that easy medal. At this point in the middle of Z out, you will find it has the largest pulse or largest wins. This one having lower widths. Lower wins. Zen you'll find here lower and lower. Same here you will find the maximum is between z in z medal and going to the right or the left, Z pulses starts to decrease. So why do we do this in order to increase Z, Albert and produced a more fundamental component, or z will be close to the fundamental component. Okay? Because here, which is a high value, representing a high value and sign, and you'll find here are the sign decreases here and decreases here. Here's the Waltz's decreases. The bosses also decrease. Here you will find the maximum in z negative. Then smaller value on the right and smaller vary on z left. A smaller value, smaller value. Okay? So now we will able to produce Xen, sine wave over shape gloss to sinewave not appeal, but close to it, lower harmonics. Now z getting signals are generated by comparing a sinusoidal reference signal of a frequency FR with a triangular carrier wave of effacing. Now CBK amplitude. Our controls, of course, I'm intuition index and, and turn and controls the output Whitney square, the same as we did before. Because we said before that as ER increases, then the output root-mean-squared will increase. Z winds of each pulse is valid in proportion to the amplitude of sinewave evaluate as central office evolves, as we said before here, is that at this point that we have in Z maximum value of z sine. So here we will have the largest pulse. Here. We have a low value here, low value. So this one is the smallest pulse, and this one is a smallest pulse. Now it was, you will find is that the distortion factor ends. The lower order harmonics are reduce, significant, okay? Very reduced. Z distortion factor is reduced greatly compared to Z multiple a pulse width modulation. Okay? Zao modulation index is given by a R over AC, as we said before. Now z equation of the output root mean square V out is equal to V supply. And in bracket you will find that some measure of all Sebald says, remember that z pulses are not equal to each other. So each balls have its own width, and so we will sum all of them together. Now, z fourier series will have this long equation again as z multiple adults who is a modulation. Now we will find that that single-pole also sinusoidal pulse width modulation eliminates all harmonic orders less than or equal to p minus one. So you will find that, for example, if we have five pulses, therefore it will be two multiplied by five minus one. Therefore, it means that the lowest order harmonic is nine. So it means that z served Z verbs, Z7 star or eliminated. So it works as a filter of harmonics. Okay, so x0, x1, and so their boss was immigration reduces the harmonic secretly. You will find that the equation which represent a z reduction in harmonics is sinusoidal balsa wood emulation or bushes the harmonics of the outward into a high frequency range or finds it here, z squared z phi of z 7s are removed. So I'll find z higher-order harmonics is moved a to Z. Right? Now we will find that the equation is Fn equal j MF plus or minus k and multiplied by fc. Fc is known is if carrier frequency k is certain constant, it will be 135. J is another variable, or this one is our variable. This one is not a constant. Variable 1-2-3, MF is undulation frequency or frequency index of frequency modulation index, which is FC over ethanol. Ok, so we can replace m f by two ping. Now, if I would like to find z first harmonic, okay, I would like to know z first harmonic, the fundamental. Okay? So we'll assume we have five pulses. So 2pi, which is ten, okay? And j is 0 severely one, K1. So it will be 2pi, which is ten multiplied by one, which is ten plus or minus one. So it can be z, first order harmonic after all her all first harmonic is Zn 11s and Z lines, which is here z lost order harmonic. So again, where did we get this? We got this from this equation by substituting by soloist variable z equal one and k equal one. So it will be two p plus or minus one. But we would like to find z lowest. So we bought here and negative sign. So it will be two p minus one modulation index. And we can find that z beak. I'm talking here about ZB, fundamental output voltage. Remember that z big value here, for example, v1 is less than 0.8, okay? Z, this is a fundamental component at modulation index equal one. But remember that z V1 at this point is considered as a root mean square. But here we are talking about what CBG value. So we take this value and multiply it by u root two. Again, v1 here is 0 to mean square value. So for CBT, we take this value or V1 multiplied by root two. Okay? So this one will nearly be equal to Z. Fundamental component will be equal to m and VS. Okay? At M equal one, Vm1 equal V supply. Remember that this value is the maximum of Z fundamental component. Here. This is the maximum and this is the maximum of 0 minuss square, okay? Z root mean square value of this one is the maximum value, not Z maximum Friedman squares. This one is a maximum and this one is a root mean square. You will find user distortion factor is greater than 0.8 by small percentage and going Li Zan 0.4. but in z balls, multimodal pulse-width modulation you will, as you remember, it is nearly from 4.5% to 4%. Okay? So our funds that store some factor here is reduced significantly. Now, as there is another relation between Vm1 over V S or Z, maximum, obviously fundamental over VS. With respect to Tools, I'm modulation index. You will find here we have a linear period and nonlinear region, okay? Z, linear is from 0 to one, as here, from 0 to one, you will find at one modulation index called One, we have Z fundamental component equal to V supply. But in z nonlinear reason for five pulses here at modulation index, obviously, we will have an alphabet of four over pi again. So as this region is called over modulation, okay? This reason where we have a square pulse. Now let us see. Remember that the square pulse v out equals the summation for VS over n pi sine omega t. And this is z is a measure of the single pulse or Z fundamental component in case of, say, in single phase bridge, okay, single phase bridge and odd z balls. Okay? Now z, therefore you will find that for, for getting Z maximum, obviously fundamental V maximums of fundamental is equal to four VS over pi. For VS over pi, which is 1.27. And VS. In order to produce a square wave, M must have increased beyond one. We'll see is that here, in order to get four over pi, which is required here, four over pi, we will need to have m greater than one at city. So this is called the Z over modulation when you are operating on m greater than one, z value of m, where Vm1 for VS over pi is dependent on b, which equals 34 equals seven. So this one is at B equal seven or non profiles is perhaps I can equal seven. Z over modulation leads to a square wave operation. Okay? Since our fundamental will be for VS over pi, or this equation which is for VS over n pi sine n omega t. So as a square wave means a more harmonics degenerate compared to two Z linear region. So over modulation is not suitable for abrogation requiring lords torsion or low harmonics. So that was our examine or our, our explanation on z sinusoidal ball sewage Xun duration. So you have to know Zan balls. Sinusoidal pulse width modulation reduces the harmonics compared to Z multiple pulse-width modulation. And z balls. So in some undulation or dishonest. So they are also in simulation produces where Glassdoor, zinc or original or Z fundamental wave, as you see here, at M equal one, you will find that z value is equal to or nearly equal to V supply. Okay? At M equal one, z fundamental or Z maximum is the sum of the momentum is equal to V supply.