 Good morning. I will start my today's lecture. Before doing that, I will just briefly recapitulate the points that we discussed yesterday. We did a single phase inverter, then we started discussing the three phase inverter. We discussed voltage source inverter, wherein the input voltage is the inverter is held constant by a capacitor. In a voltage source inverter, in each leg, one device is conducting at a time. At any given time, one device in each leg is conducting. Having turned on the device, if I allow the device to remain in that state for 180 degrees, line-to-line voltage waveform has a pulse of 120 degree direction. The magnitude is the supply voltage V dc and for the remaining 60 degrees, the output voltage is 0 in the positive half and the negative half, it is just the opposite. So, if I take the Fourier series of this waveform, the harmonic spectrum is going to be 6 and plus or minus 1. If I want just a 50 hertz supply, the predominant harmonic is going to be 250 and 350 hertz. 11th is going to be 550 and 30th is going to be 650 hertz. When it comes to a star connected load, the phase voltage has a 6 steps waveform. So, hence the name 6 step inverter. In order to reduce the total harmonic distortion, what we concluded was instead of having one pulse of 120 degree direction in the line-to-line voltage or having turned on the device allowing it for 180 degree, we have a poor harmonic spectrum. Instead of having one pulse, I have a large number of pulses. So, if I do that and if the load is inductive, my current waveform is going to be approximately a sinusoid. Let me tell you one thing. Voltage is made up of large number of pulses, but the load is inductive. I can have a current power which is approximately a sinusoid, but then if you want a purely sinusoidal supply, a voltage source as well, voltage waveform should be sinusoidal, then I need to use an LC filter at the output of the inverter. I will repeat. If you want a voltage waveform also should be sinusoidal, then we need to use a separate LC filter at the output of the inverter. I told you there should be a large number of pulses in one's in half a cycle. How do I or when do I turn on and off the device? So, for that I need to use a PWM strategy. The most common PWM strategy being a sinusoidal PWM technique wherein there are three sinusoids which are displaced in time by 120 degrees. The frequency is equal to the 50 hertz provided if I want a 50 hertz supply is compared with a high frequency triangular waveform. Both magnitude and frequency of the triangular waveform is held constant. The frequency of the triangular waveform determines the switching frequency and therefore, the frequency of the predominant harmonic. So, this frequency is fixed based on the power level. At low power levels it could be high and as the power level increases, we need to reduce the switching frequency. The magnitude of the output voltage depends on the magnitude of the sinusoid or the modulating wave. If I want to have a constant magnitude, I need to keep the magnitude of the sinusoid also should be held constant. This is sinusoidal PWM technique. Second one which is very popular is the space vector PWM technique. What is a space vector? As a three phase sinusoid can be represented by one vector what is known as a space vector. In a two dimensional space if I plot, it has a magnitude as well as under angle theta. At any given time, if I know the magnitude and theta, I can get d axis and a q axis component and if I know the d axis and a q axis component, it is possible to determine the instantaneous value of V a V b V c. And we found that in a three phase voltage source inverter, as a three phase voltage source inverter, there are three legs. At any given time three devices are on. So, there are eight possible conducting states, two of which are zero voltage vectors in which all the three upper devices either they are on or off. So, in both the cases, the magnitude of the space vector is zero and the angle is also zero. In other words, we are at the origin. The remaining six vectors are known as the active vectors and we found that they occupy the six vertices of an x again. See here, V 1 vector is obtained by turning on phase a alone and the lower devices of phase b and c are on. See, right now I am monitoring only the upper devices conducting state of the upper devices. The upper device of phase a is on here in addition to phase a, phase b also turned on. And if I see from the difference between these two vectors is the conducting state of only one device has been changed. Conducting state of only one device is changed. See, from V 2 to V 3, I can go by changing the conducting state of one device. Phase a turns off, nothing happens to phase b and phase c. And from V 3 to V 4, phase c is turns on, nothing happens to b and a. V 4 to V 5, b turns off, nothing happens to c and a. And 5 to 6, a turns on, nothing happened to c and b. So, from one vertex and other vertices can be travelled just by change the conducting state of one device. So, if I do that, in one cycle is completed in 6 steps. In other words, it is the 6 step operation itself. The harmonic spectrum is going to be same as that of the conventional square wave inverter. I said we have to decrease the total harmonic distortion. Therefore, we need to have large number of pulses. Somehow, I have to move in a very smaller steps. In other words, I need to discretize the given sinusoid or I have to digitize this given sinusoid. What I will do is, there is a sample and hold circuit. Every time it samples the sinusoid and that values held constant for a sampling time T s. Somehow, I have to realize this, the variable magnitudes using the space vector PWM type. Now, the important result I formed you was, if the 3 voltages are sinusoid, then V s can be represented m into e to the power j omega t, where m is the modulation index, omega is the frequency of the output voltage. And locus of V s is circle. Now, how do I digitize or how do I realize this? Here, the magnifying V by f or if I have constant frequency 50 at supply, I need to keep V constant. The philosophy is the space vector lies in between any two active vectors. Then these two active vectors and 0 vectors are used to synthesize V s. If the vector lies in sector 1, I need to use only V 1 and V 2 and 0 voltage vectors to realize V s. How do I do that? Assume that it lies in V s. I told you and V s is assumed to remains stationary for T s. T s is nothing but the sampling time. When this V s is flowing in a coil, it produces some flux. The same change in flux I should realize using V 1 and V 2. So, what do I need to do or what do we need to ensure? There should be a volt second balance. At any given time, if V s is making an angle theta, magnitude is V s. So, volt second of that is V s into T s. T s is the sampling time. That should be equal to V 1 into T 1, where T 1 is the time for which V 1 is applied, V 2 is the time for which or T 2 is the time for which V 2 is applied. These are the two active vectors and the remaining time I will apply the 0 vector. Since the voltage applied is 0, there is no change in the flux. So, if I satisfy this condition, I have realized V s. How do I ensure that this condition is satisfied? What I will do is I will resolve V s along x axis and y axis. That is V s into T s into cos theta is the x axis component of the space vector. V s into T s into sin theta is the y axis component. So, right hand side of the x axis component is V dc into T 1, where V dc is the magnitude of V 1. V 1 is along x axis. It does not have a y axis component and it is on for T 1. V 2 vector makes an angle 60 degrees. So, the x axis component is V dc into cos 60 into T 2. So, this is the volt second of x axis component of the right hand side. Left hand side is V s into T s or T c does not matter. T s into cos theta. I liquid it. Similarly, I will do for y axis as well. I have two equations and two variables. So, T 2 is given by time for which the V 2, time for which V 2 is on is given by this equation. T 2 is equal to T c is the sampling time. It is known. A is the modulation index. Sin 60 is the value is known. Only the variable is sin theta. And this theta is the angle which the space vector makes with x axis. Since this vector is rotating and if I know the time, I can determine angle theta. If I know the frequency of rotation, that is 50 hertz here and at any given and I know the time as well. So, I can determine sin theta. So, if I once I know T 2, I can calculate one as well. So, T 1 is given by this expression T c into A into sin 60 minus theta by 60. And the time for which 0 vector is applied is given by this equation. Do not get confused. I have been saying T s, but here it is T c. T c is nothing but T s by 2. Why it is T s by 2? I will tell you. So, these are the following observations. T s is the 0 voltage vector. T c is the sampling time by 2. T 1 is the time for which B 1 is applied and T 2 is the time for which B 2 is applied. And T z could be either 0 0 0 or 1 1 1. Does not matter. So, the observations are irrespective of the ratio in which T z is divided between the 0 sets. I can apply 0 0 0 or I can apply 1 1 1. You can divide as per your convenience. It does not matter. It does not affect your whole second balance. Active vector V 1 need not be applied over a continuous period for T 1 seconds. It need not apply, but then total time should be T 1 and total time for which B 2 is applied should be T 2. It need not be applied continuously. Let us not get into this final aspect. See here. The implementation period is here. Initially, I apply 0 0 0. The philosophy is attaining given time only once which is should be turned on or off. Those switches are off. In other words, I am at the origin. I will turn on phase A. Nothing happens to phase B and phase C. So, I am along the x axis. V 1, voltage vector V 1. After some time, T 2 turns on. After some time, phase B is turned on. I will apply 1 1 0 and at the end of T 2, I will turn on C as well. Where do I go? I go to the origin. So, you see here 0 0 0, 0 0 1, 0 1 1, 1 1 1. This is half of the sampling time and again I will go back the same manner 1 1 1, 0 1 1, 0 0 1, 0 0 0. So, if I see this figure here hexagon, what am I doing? What are we doing here? 0 0 0, 0 0 1, 0 1 1. I come back to the origin 1 1 1, go back in the same direction, something like this. That is what I have shown. You do not need to do this. What you can do is 0 0 0, 0 0 1, 0 1 1. Come back 0 0 1 and 0 0 0. You do not have to turn on C phase and go to the origin, not required because if there is no change in the volt second or it does not affect over that volt second balance condition. It is very simple to, why it is so popular is very simple to implement in real time. I need to calculate only T 2 and T 1 and only one variable is not known there or I need to have the truth table of sin theta. I will digitize 60 degrees, I will divide into some n parts. So, n by 60 I will make a table and depending upon the switching frequency. Rest all are known. T C is known, T sin 60 is known, A is the modulation index. If I want a constant output voltage, A also remains constant. Now, T 2 is determined by sin theta itself. Theta is the position of the space vector and it is known. For each sampling time, yes, for each sampling time space vector rotates by an angle delta theta. If T is equal to 0 space vector is along x axis, after T C it moves by an angle delta theta. So, I need to know only sin theta. I can have a truth table and I will save it in the memory. Similarly, for T 1 I need to know only 60 minus theta. So, very popular and very simple to implement. I do not want to discuss more details of the sinusoid, sorry, most details of the space vector PWM technique. There are finer issues. What is the magnitude of the voltage? What is the maximum value of V s that I can get? For which the output voltages are sinusoid? I said what is the maximum value of V s for which the output voltages are sinusoid? How do I determine? I know the condition. If the output voltages are sinusoid, then locus of V s is circle. So, that is nothing but the radius of the encircle being circle. What is that value I can determine? Because I know what is this, the radius of the circle is given by this value. So, I can determine or it is possible to determine the maximum value of V s for which the output voltages are sinusoid. More about it, you can read the literature. It is well reported. So, you can go to the slides also. They are very simple to understand. Here I have told you what should be the number of samples per sector? What should be the number of? I told you if there are n samples. So, I need to have 60 divided by n, the size of the truth table. Now, what is that n? I have explained here. All the sectors should be equally used in entire cycle producing a symmetric line voltage and samples of positions symmetrically about each of the sector with one sample at the center. If n is equal to 3, the samples of positions are 10, 30 and 50. So, these are rules should be followed for the three-phase limit. So, if I know n is equal to 3, I need to have sin alpha, sin 10, sin 60 minus 10, sin 30 and sin 50 and sin 10. That is all I need to have. That is about it. So far, we discussed DC to DC conversion and DC to AC conversion. Now, before connecting this AC power to the grid, let us see what should be the most elegant method of converting this low voltage or low voltage output from the solar to 230 volt supply. We know now, if I know the input voltage, I can get any output DC voltage that I can get by using the various, one of the various power electronic interfaces that we have discussed and given an input voltage, I can convert it to AC. Now, what sort of a configuration to use? Say input is, I have only a 24 volt solar panel, I want a 230 volts AC. Now, to get a 230 volts AC RMS value, what should be the DC link voltage that you need to calculate from this relationship? It depends on the PWM technique. See here, everything is given here. So, if I know, I told you we need a 230 volts RMS. So, therefore, peak comes around 300. To get a peak of 300, what should be the DC link voltage? Now, how do I get this a high voltage DC that I can get? If I know the output voltage of the solar panel and if I know the required DC link voltage, if the ratio is too high, depending upon the power, I can calculate, I can use either fly back, forward, push pull or full bridge or if you do not want to use this, use or if you do not want to increase a DC link voltage or if you do not want to boost the DC link voltage, what you can do is a low voltage DC, you can invert it to a low voltage AC, then use a 50 hertz transformer. So, I will show you the circuit diagram. Solar panel, low voltage DC, whatever I can get a low voltage DC here, invert it. I have a low voltage AC, step it up and then get a 230 volts, 50 hertz AC. Everything is being done at L V low voltage. It should be extremely careful while calculating the turns ratio because 1 volt, see if I have a 12 volts or a 24 volts, depending upon the power circuit configuration here 1 or 2 devices may be conducting. You need to take into account the device drop here plus the winding resistance here. It is not V 1 by V 2 is equal to N 1 by N 2. It is E 1 by E 2 is equal to N 1 by N 2. What is E? The voltage across the magnetizing inductance in the equivalent circuit. V 1 by V 2 is equal to N 1 by N 2. It is true only for an ideal transformer. It is not a non-ideal transformer and unfortunately we are using a non-ideal transformer. So, you need to take into device drops, but then entire electronics is working at low voltage. So, life is relatively easier. You can handle low voltage, but then you are using a 50 hertz magnetics, 50 hertz transformer, 50 hertz transformer. So, the moment I say 50 hertz transformer size increases as the frequency of operation decreases, size increases. How do I reduce the overall size? I need to increase the frequency of operation. Here I have to use 50 hertz because the fundamental frequency is 50 hertz. Now, can I reduce the size? To reduce the size one thing is I need to eliminate this 50 hertz transformer. So, what will I do? I will have a high voltage DC. From this low voltage, how do I get a high voltage DC? I may have to use a again a transformer there, but then it is a high frequency transformer. See, we convert this a low voltage DC to high voltage high frequency AC, high frequency AC using either using a push pull or a full bridge whatever configuration high frequency AC, very high frequency. I will rectify it again. I will rectify it and convert it to high voltage DC. By the way high voltage means the voltage the value of voltage that is required to get 230 volts RMS. Please when I say high voltage DC it implies the required DC link voltage to get 230 volts RMS. Now, this high voltage DC I will directly invert it. I will directly invert it and if you want a purely sinusoidal supply then I may have to use an LC filter. I may have to use an LC filter and if you want an isolation of course, you can use again a 50 hertz one is to one transformer. So, the entire configuration will change depending upon your requirement depending upon the application one or two typical examples I will we will discuss. So, far we discussed DC DC conversion block DC to AC conversion block and now irrespective to our independent of this voltage I have a here either a single phase 230 volts 50 hertz supply or 415 volts 50 hertz AC. Now, if you want to connect it to the grid what do I do? Now, the so called the hot topic there was so many questions being asked how do I connect it to the grid? How do I connect it to the grid? Now, we will discuss that topic. I have a source AC source a single phase AC source this is nothing but the grid and this is nothing but a voltage source inverter a voltage source inverter a single phase voltage source inverter. I am connecting this voltage source inverter to the source through an inductor what should be the value of an the question that was asked yesterday we will see. I told you it is a voltage source inverter basically is a volt two quadrant converter two quadrant converter current could be positive or negative but voltage will remain the same. Now, the power coming is coming from the DC side from the solar side this is the direction of power and it has to be fed back to the source how do I connect it? Now, let us not worry about how to connect it we will see assume that it has been connected assume that it has been connected how do I transfer the power? How do I transfer the power? Because this inverter does not know how much power is coming from the solar cell? Here as of now we are assuming that there is an MPPT maximum power tracker depending upon the sun insulation it extract the maximum power from the solar cell and it dumps to the capacitor that could be either at the input of the inverter or it could be the output of a DC to DC converter. So, only the MPPT knows the amount of power that is being drawn from the solar cell inverter does not know that power has to be fed back to the grid how do we control that power? Now, before doing that let us do go back to circuit theory by the way what sort of a waveform will I get here? What sort of a waveform will I get here? Voltage at this point is a pulse width modulated waveform that is what we have shown that is what I had shown you yesterday. This is the output voltage waveform of the inverter voltage source inverter there are large number of pulses. So, if the load is highly inductive I get a sinusoidal current. So, it implies that this square wave large number of pulses they have a fundamental component whose frequency here it is 50 hertz. So, if I use a LC filter I will get a purely sinusoidal voltage. So, I am assuming that there is an LC filter or I will neglect the higher order harmonics and I will draw the only the fundamental equivalent circuit. I will repeat I will draw only the fundamental equivalent circuit. I am assuming that there is a LC filter after we will see if there is no LC filter what happens do not worry. So, voltage at this point is a sinusoid. So, in other words I have a AC source here. So, this is the equivalent circuit the can I so this is how the equivalent circuit looks like I have a grid voltage V 1 voltage source V 1 V A B is the output voltage of the inverter output voltage of the inverter. This is nothing but again a sinusoid it is straight they are connected through an inductor as of now I am assuming that this inverter is connected how to connect it we will see the so called the synchronization. So, the equation for power transfer is nothing but two sinusoidal voltage source are connected together tied together by an angle by an inductor L power transferred from source V A B to V 1 is given by V I V A B omega L divided by sin theta V 1 V 2 divided by x into sin delta a very popular expression we have studied in either in the second year or third year V Tech power transferred when the two sinusoidal sources are connected together through an inductor L power transferred is given by V 1 V 2 divided by x into sin delta. Delta is the angle between V 1 and V A B so by controlling delta I can transfer the power or I can control the power fed back to the load how to keep that or how to choose the delta how to determine this delta what could be the control strategy the question number one that because inverter does not know the amount of power drawn from the solar cells inverter does not know only the MPPT knows how do I choose the correct value of delta here I am keeping see here V I is the input voltage of this the grid V A B is the fundamental V A B is the magnitude of or the RMS value of the output voltage V A B V 1 is the RMS value of the supply voltage V A B is the RMS value of the output voltage of the inverter omega L is this reactor how do I choose delta we will see see here is the two wave forms this is V I this is output voltage of inverter the fundamental equivalent circuit if there is no harmonics if there are sorry if there are there is no LC filter what is the equivalent circuit I need to draw the superposition I need to use the superposition theorem I am assuming that I am assuming that input voltage is pure sinusoid and there are harmonics here what is the equivalent circuit equivalent circuit is this is I need to consider the voltage of one frequency at a time this is the equivalent circuit at fundamental frequency at any frequency this is 0 here why input is a pure sinusoid here is the effective inductance now it is omega 1 into L where omega is the frequency of the harmonic that is present and here is the magnitude of the harmonic of the RMS value and I can determine the current that is being generated by this harmonic voltage I will repeat the harmonic equivalent circuit is going to be something like this omega n omega into L where n is the number of the harmonic that is present and here is the RMS value of the harmonic voltage that is present I can determine the current. So, I need to do for each and every frequency that is present as the frequency increases this inductor will practically will appear as if it is an open circuit. So, high frequency components get filtered out high frequency voltage components will get filtered out. So, if I know the various harmonic components I can determine the total harmonic distortion. So, if I know if depending upon the power level and depending upon the grid voltage level the with the various standards will tell me the total harmonic distortion and accordingly I will choose omega L and the switching frequency by the way switching frequency I will not be able to change because it depends on the power level. Generally, we do not change the frequency frequency we will keep it constant, but then I need to T h t has to be the total harmonic distortion that is present in the current waveform that is fed to the grid that should be satisfied by or this is given by the standards various standards. Therefore, I can calculate this inductor that is one of the ways that another criteria as well higher the value of inductor slower is going to be the response because if I there is a change and how do how does this circuit behave you will see once one of the ways to determine is I need to control T h d switching frequency is known. So, I will know the harmonic spectrum once I know the harmonic spectrum I can determine the current and therefore, yesterday there was a question how do I control the reactive power as well or I told you that voltage source inverter can supply any q it requires only delta p it has to either the delta p can be used from the d c side provided there is an input or in the night if there is no input power active power input this delta p can come from the from the grid and this inverter can supply q the reactive power what should be the. Now, if delta is equal to 0 consider an ideal case delta is equal to 0 and v i is not equal to v a b v i is not equal to v a b delta is 0 what happens if delta is 0 v i is in phase with v a b magnitudes are not the same. So, definitely there is going to be a difference potential across the inductor there will be a current flow what is that of a current now since delta is 0 active power transferred is 0 if the active power transferred is 0 there is no in phase component, but then there is a current flow why these two points are at different potential I told you v 1 is not equal to v a b and delta is equal to 0 if delta is 0 active power transferred is 0 if the active power transfer is 0 in phase component is 0, but then there is a current flow. So, it has to be reactive current it has to be reactive current. So, either it could be leading or lagging either it could be leading or lagging that depends on the magnitude of the inverter output voltage. I am assuming that grid voltage will remain constant by changing this magnitude I can make this inverter to supply either a leading current or a lagging current in other words this inverter can appear behave like a capacitor or like an inductor that I can do just by controlling the output voltage of the inverter. This expression says I will make delta is equal to 0 there is no active power transfer by controlling v a b if the magnitude of v a b is higher then there are two possibilities why two there are three if v i is equal to v a b and they are in phase there is no current. Now, v i could be higher than v a b or v i could be lower than v a b. So, these two cases corresponds to leading and lagging current that can be determined from this equivalent circuit see magnitude of source voltage is less than the magnet is less than the output voltage of the inverter the grid voltage rms value of the grid voltage is less than the rms value of the inverter at that time we found that I s the source current leads the source voltage. So, this is nothing but see it is a capacity circuit current leads the voltage if it is less current lacks the voltage this is nothing but an inductor. So, therefore, a voltage source inverter can act like an inductor or like a capacitor to do that it has to draw a very small amount of power that power which accounts for the losses here and the losses in the inverter. So, in non ideal case this delta is not equal to 0 this delta is again determined by the system losses. So, for both the cases we will see what are the both the cases one is when there is an active power transfer how to choose the value of delta and when there is no active power transfer it has there is only a reactive power delta value of delta approaches 0 it is very small just to account for losses how to determine this value of delta we will discuss this is the equivalent circuits yeah this is about it. So, here is an schematic I have a source load this is a voltage source inverter tied to the grid through an inductor as of now we have not discussed how to tie this inverter to the grid have patience we will see there could be a power coming from the DC side from the DC side that power has to be fed to the grid assume that there is a spare capacity because sun insulation has come down and grid requires grid requires reactive power in addition to this active power this grid can supply q how to supply q we will see if it can supply active power it can supply reactive power it can also supply harmonic current as well now how to how it can supply the harmonic current we will see it is not a issue here see here this is equivalent circuit load is non-linear in nature assume the load is non-linear in nature the moment I say load is non-linear in nature current drawn by the load is non sinusoidal when the current drawn by the load is non sinusoidal using the Fourier series I can get the various harmonic components input voltage is sinusoid I want that the harmonic source the harmonic current supplied by the source should be 0 what we should do what we need to do is this inverter should generate the required harmonic current assume that the load is drawing the fifth harmonic current and the seventh harmonic current depending upon the impedance of this inductor my inverter has to generate a fifth harmonic voltage and a seventh harmonic voltage how to generate it is possible so in principle we can control this inverter such that it can supply active power it can supply reactive power it can supply q as well it can supply the harmonic current as well I am assuming that I have used a faster devices power handling by this inverter it is relatively low therefore it can compensate the harmonics that is about it I will take a break I will take few questions now we are ready Nagpur go ahead for your question sir my question is on SV PWM question on question on in case of state space vector PWM in space vector PWM while implementing the vector we put 0 vector at the starting as well as at the end of TS okay go ahead go ahead isn't it so is it necessary to bother about the dead band I told you 0 vector need you need not put 0 vector in the beginning as well as at the at the end this is I have just shown you I do not you do not need to have 1 1 1 you can have 0 0 0 0 1 1 0 1 1 and again come back straight away 0 0 0 0 0 0 1 0 1 1 and go back you do not need to have 1 1 1 you do not need to have 1 1 1 so at the end of TS oh yeah okay fine end of TS I am at the origin fine we do have 0 0 0 0 yeah okay all the is it necessary to implement the dead band oh you mean dead time oh you mean dead time yeah yeah yeah voltage is inverter 0 0 0 1 1 1 what does it mean either three devices are on or three devices are off I am measuring the load to be highly inductive load is load all the three points are connected to the positive DC bus and load is freewheeling through either to the positive DC bus or to the negative DC bus devices are carrying current there is a current flow depending upon the direction of the load all three points are connected to the same point load is a current source highly inductive there is some current flowing there is some current flowing in the load now after 0 0 what sort of where are you going in the same what is the conducting state it could so happen that you may be going to the second second sector and there has to be a voltage there has to be a dead time between these two there has to be a dead time between these two here 0 0 0 fine but then there was there is current flowing through the load devices are carrying current either the current is flowing through the device or it may be flowing through a diode I do not know it depends on the load we have assumed that load is highly inductive on there is current flowing through the load just because you have applied 0 0 0 or 1 1 1 it does not mean that there is no current flowing in the load is that ok and if you are switching the complementary device there has to be a dead time in a voltage source inverter see for example, if you are going from 0 0 0 if you are going 0 0 1 you are turning you are turning on this switch initially there was some current flowing and when I turn off this current has to flow through this switch either to the switch or diode I do not know it depends on the load power factor it depends on the load power factor immediately when you turn on the switch I do not know whether this switch will start carrying current I do not know that is a load current load current but then what will happen if there is a voltage source here there is another path as well oh my goodness the question is question is you said that high power level we need to reduce the switching frequency on what basis this is done now question my question to you is why we should reduce the switching frequency depending upon the power level what happens if I if I keep the switching frequency same and gone increasing the power what will happen as the power level increases current flowing through the device also will increase and every time I turn on and off there is going to be the switching losses turn on losses turn off losses and at any given time the operating point should lie within the safe operating area and I told you that there is what is known as T j max junction temperature maximum junction temperature has to be controlled so that is the reason we need to reduce the switching frequency so as the switching as the power level increases current increases voltage also may increase voltage also may increase because kilowatt rating I may have 450 volt input as the power rating increases I may go in for 1.1 kv 3.3 kv or 6.6 kv so every time you switch the device there is going to be on turn on losses and every time I switch off there is going to be a turn off losses I need to control the junction temperature and this may be the third or fourth time I have been saying this same question is being asked in power circuit why third harmonic is not applicable what do you mean by in power circuit why third harmonic is not applicable I never said this could you please frame this question properly in power transfer circuit why third harmonic is not applicable in a 3 phase 3 via system can line voltages have third harmonic in a 3 phase 3 via system can I can the line voltage have third harmonic this is being thought in may be in the third year BTEC what is advanced sinusoidal PWM for single phase inverter I do not know what is advanced sinusoidal PWM for a single phase inverter I do not know be more specific what is advanced yesterday there was a question how do I how how do I have PWM what is a PWM technique for a single phase inverter single phase inverter has take one sinusoid of required frequency and magnitude if you want to have a 50 hertz supply the frequency of this sinusoid should be 50 hertz you want a constant magnitude therefore you keep this magnitude constant compare with the high frequency triangular the intersection point determines the switching instant for the inverter if you use half bridge only turn on you there are only two devices and if there are if it is a full bridge you turn on s1 and s2 together s1 s2 s3 s4 if you have half bridge you turn on only s1 at a time when s1 turns off turn on s4 and if you use full phase full bridge s1 and s2 are turned on and off together s3 and s4 are turned on and off together that is all this is for a single phase single phase inverter now what is advanced for single phase I do not know in power factor improvement using a VSI which factor we need to vary for smooth control of power factor improvement let us see in power factor improvement which factor we need to vary for a smooth power factor come on that you need to answer power transferred is V1 V2 divided by x into sin delta this is the expression for power what is the expression for q forget about if you do not know the expression for q it is ok now you tell me I am assuming I am taking an ideal case delta is equal to 0 V1 and V2 are in phase V1 is 1 voltage source this is V2 this is x so this so this is V1 V2 this will be your I current flowing now you want to have a smooth control of this magnitude of current how do I do that that I can do by smoothly changing the difference of V1 minus V2 difference of V1 and V2 V1 minus V2 I need to change smoothly so if V1 remains constant I need to change V2 how do I do that assuming that V2 is the output voltage of the inverter that is V2 is given by m into VDC that is what I told you this is the output voltage of the inverter now I can change VDC or I can change m I will keep VDC constant I will change the modulation index m so if I change the modulation index m in a smooth manner out V2 will change accordingly since V1 is V1 is assumed to remain constant I will change smoothly that is what you want and for non-ideal case delta is finite and this now you derive an expression and you find out or the controller will do all this you have to change modulation index sir this next question is sir how VSI can supply both lagging and leading reactive current not at a time it cannot supply together it either it can either it can supply either it can supply leading or lagging even I do not know how it can supply but from this vector diagram you can infer isn't it this is the equivalent circuit both are in phase so active power transfer is 0 but the magnitude is not the same if the magnitude is not same there is going to be a current flow but it is not active current in phase current so that has to be quadrature current so either it could be plus 90 or it could be minus 90 so this vector diagram tells me in this case when output voltage of the inverter is higher than the grid voltage current is leading if the output voltage of the inverter is less than the grid voltage current is lagging so I need to just control the magnitude of the grid voltage with the rest the magnitude of the invert inverter output voltage with respect to the grid now how do I do that BAB is proportional to m into VDC if VDC remains constant change m that is all yeah any other question yeah I will it is 10 11 now we will start at 10 15 a 4 minute break I will take