 In the last class, I finished discussing about DC to DC converters. I briefly told you about the soft switching inverter, soft switching DC to DC converters as well. Soft switching generally is used for to reduce the switching stress, switching stress on the device. So, if I use soft switching technique, it is possible to increase the switching frequency. Therefore, the size of filter components comes down. Since I am doing soft switching, there are only conduction losses now. Your switching losses are practically eliminated by using a separate L and C, what is known as the resonant elements. So, there is no EMI has been taken care. Since there are losses of reduce, my entire cooling requirement also has come down, but then the complexity has increased. So far, we have discussed about DC to DC converters depending upon the power level and depending upon the voltage level. As of now, we are in a position to decide on a suitable circuit topology. Basic circuit topologies we have discussed. Let us see sometime later whether we need to do some modification in the basic circuit topologies. Now, we will see given an input given a DC voltage and if I want to convert it to AC, what are the possible ways of doing this? Well, initially we will do the basic inverters and maybe when you are doing the overall system integration, the modification that are generally made depending upon the power level, we will see. I said basically there are two types current source and voltage source. I also told that the literature says that the weakest link in voltage source inverter is the electrolytic capacitor. So, in order to improve the total lifespan of the inverter, we may have to find a solution to this electrolytic capacitor. Can we replace this by some other more reliable storage element? A basic, still an open ended question. Second is a current source has certain advantages here. Here the voltage is practically at this in the voltage source inverter, current can reverse, current can reverse instantaneously like a capacitor. So, basically I need, but then voltage at this point cannot change. Similarly, here current cannot change instantaneously, whereas voltage across the inductor can change instantaneously. So, I need to, I cannot use the same type of devices in current source and voltage source. There are two different types of devices. Here it is a bider, the switch should be able to carry current in both direction, but it need not block the voltage in both direction, whereas here it need not carry the current in both directions, but then it should be able to block the voltage in both directions. We started discussing about half bridge voltage source inverter, basic building block. I need total volt, I mean the total input voltage is V dc, but then at the output I have a square wave whose magnitude is V dc by 2 and minus V dc by 2 is a square wave. It has its own fundamental component, given by 4 by pi into V dc by 2. Then I also found that though I am switching S1 and S2 are complementary, I need to turn off S1 and there has to be a finite time for which neither S1 and S2 are gated, but then even if I get S2, a finite time current may not flow through it because of the load power factor, for sometime current flows through d2. Now if I replace the load by an inductor, an ideal inductor, how does the current waveform look like? I said I replace this load by an ideal inductor when S1 is on, voltage applied to the load is V dc by 2 and it remains constant for t by 2 seconds. See I will repeat, I have connected an ideal inductor across AB that is the load. I close S1 for t by 2 seconds, voltage applied to the load is dc and it is equal to V dc by 2, a constant DC voltage is applied to an inductor therefore, current increases linearly, provided it does not saturate. Of course, we should not allow the inductor to saturate. After t by 2 seconds S1 is turned off and S2 is turned on, voltage applied to the load is minus V dc by 2 remains constant for t by 2 seconds. Therefore, current decreases linearly and at steady state, current waveform looks something like this, it is a triangular wave. So, this is t by 2, so this will be t by 4, for t by 4 duration seconds d1 conducts and after that S1 conducts for t 1 by 4 seconds and so on. So, if I see the compare the voltage and current waveform, see this is the voltage applied to the load and this is the current flowing, voltage applied is positive, current is negative. Therefore, the product that is nothing but the power is negative. So, source is absorbing power, in other words power is flowing back to the DC link. Similarly, here voltage positive, current positive, so product is positive, load is load, source is supplying power to the load, in this zone current positive voltage negative, so therefore, power is negative, source is absorbing power and the remaining last t by 2 seconds voltage negative, current negative, therefore power is positive, source is supplying power. So, the average power consumed by the load is 0, it makes sense because load is an ideal inductor. So, average power consumed by the inductor is 0. I have one, as of now let us neglect the devices, device losses as well, the so called the switching and the conduction losses. I will assume that the inverter is an ideal one. So, an ideal voltage source inverter feeding an ideal inductor, power, average power required or average power, average power input is 0. So, if the average power input is 0, so we can replace this battery by, battery by a capacitor, charge the capacitor once and forget about it, it will continue to work for ever. The sense I am neglecting the losses, there are no losses, loss system is lossless for sometime capacitor is supplying the energy and after sometime the remaining half cycle in the entire one capacitor receives power. So, capacitor voltage will change, when it is charging capacitor voltage will increase and when it is discharging capacitor voltage will fall. So, there is a going to be ripple in the DC link voltage very enough, but then I do not need to an active source. If I consider a non ideal system, in the sense this inductor has a finite resistance which is very small, devices have finite losses, again these losses are very small compared to the k v rating. So, the net losses they are finite. So, if I have a capacitor here in order to maintain the voltage constant, I need to have some input active power input to the DC link. So, this is a very important concept that is required when we connecting this inverter to the grid. If it is an, if the inverter is feeding or supplying power to an ideal inductor and if I consider ideal devices, so average power required is 0. If I take the non idealities into account, active power that is required to maintain or to sustain this DC link is very small compared to the k v a r supplied by the source. Why k v r? It is supplied by the source or supplying reactive power or supplying an inductor, v a is watts plus war, j war. So, you require a delta p to sustain or to address this or to account for the losses, you require delta p to account for losses while supplying q, supplying q. So, this power active power may be equal to delta p has to come either from the AC grid or I need to have an active source to supply this power. So, if I do not have the grid here and if I connect a load and connect an ideal inductor, then I need to have some sort of a either, I need to have either a battery or may be power input from as part of the power input from solar is being utilized to address this losses. Remember an inverter can supply any q, any q in the sense as long as it does not exceed its rating. To do that, it requires delta p, a very important result I will just write here inverter requires delta p to supply q. So, in a grid feeding inverter as up now we do not know how to connect it to the grid. Let us assume that we know how to connect the inverter to the grid. Having connected the inverter to the grid, there is no solar insulation in the sense in the night grid requires reactive power support. At that time inverter can supply q, to supply this q it requires a very small amount of power. So, this power can be drawn from the grid. So, at night there is power, so if there is no input from the sun grid requires reactive power support at that time inverter your existing setup or grid feeding inverter can supply q which can be equal to its kv rating of the inverter. But to do that it draws a very small amount of power from the grid that is equal to delta p. The control loop, the control structure we will see sometime either on Tuesday, the second half or on Wednesday. What are the limitations of off bridge? I said input voltage is total is V dc, but then output is V dc by 2. And may be only one device is conducting at a time, so may be losses are less. Now, see here off bridge inverter I have actually V dc divided into two parts, but then I get V dc by 2. So, has a fundamental component which is sinusoidal which is equal to 4 by pi into V dc by 2. Now, if you want a 230 volts RMS 4 by pi into V dc by 2 is 300. So, calculate what is V dc is required. Now, instead of using that can I use full bridge inverter. So, what happens at full bridge inverter? There are four devices, two are conducting at a time s 1 and s 2 on. Voltage applied to the load is V dc, turn off s 1 and s 2 turn on s 3 s 4. Voltage applied to the load is minus V dc, but then two devices are conducting at a time. It could be an issue, see two devices are conducting at a time. If the input voltage is say around 400 volts, device drop is of the order of 2 to 2.5 volts, one device conducting a two device conducting it may not make any difference. See I will repeat, the input DC link is high say is of the order of 400 volts. How did I get this magic figure 400 volts? I will tell you later. Right now we assume that it is 400 volts. So, if it is around 400 volts, device drop could be of the order of 2 to 2.5, one device conducting or two device conducting makes it may not make any difference. Now, instead if I have a 12 volts solar panel or set 24 volt solar panel, V dc by 2 is going to be 12, 12 minus 2 is going to be 10 or means or not 12 minus 4 is going to be 8 or if I have a 24 volts panel, 24 minus 4 is 20 is 1 over 6, really significant, really significant. So, it depends on the input voltage. Another limitation is output voltage is a square wave. So, if I take the 4 year, if I write the 4 year series, the harmonic spectrum is going to be fundamental. If T is 20 millisecond, I have a 50 yard supply, then I have third, fifth, seventh, ninth, all odd harmonics are present, all odd harmonics are present. So, if you require, if you want a sinusoidal supply and the harmonic content should be as low as possible, I need to do something else. I just cannot allow the device to conduct for 180 degrees. See, each device is conducting for 180 degrees. Having turned on, it is maintained in that state for 180 degrees. I will repeat, having turned on S 1 and S 2, they remain on for 180 degrees. So, limitation here is it has all odd harmonics, three phase inverter, single phase we discussed. Now, let us see a three phase inverter. We require three legs, three legs of an inverter. Where is that? Here is it, three legs, there are three legs of the inverter is an IPM. I require three such modules, three such modules and I do not know how many such devices are required because these are four amperes, this is 75 amperes. I do not know what could be the number of devices, what could be the size, whether will this such a small heat sink will do. I do not know, we need to find out. I have three legs are connected in this fashion. As of now, I have taken center point as a reference. It is not required for analysis I have taken. The rule here is at any given time, one switch is on. At any given time, one switch is on in each leg. So, that is the principle of operation in VSI. At any given time, one leg, one device is on. So, each leg, one device is on. So, therefore, there are three devices on. I can have any combination, no issues. By the way, here S 1 implies a control switch. In addition, there is a diode across it because this switch should be able to carry current in both direction, but then voltage in only one direction. It need not block the voltage into this. So, I need to have three phase supply. I will turn on S 1 after 120 degrees. I will turn on S 3 and after 120 degree S 5. And when S 1 is on, V A 0 is V A 0 is V dc by 2. When S 3 is on, it is V B 0 is V dc by 2. So, when S 5 is on, V C 0 is V dc by 2. Similarly, the complementary switch is on, voltage is minus V dc by 2. Here are the waveforms. V A 0 is a square of V B 0, each of magnitude V dc by 2 and V C 0. Though V C 0, S, see here, do not get confused here. S 5 is turned on after 240 degrees because after 120 S 3 and after 120 S 5 and that will remain on for 180. So, compared to the positive 0 crossing of S 1, this turns on after 240 and it is allowed to remain there for 180. So, 240 plus 180 is how much? 240 plus 180 that completes one cycle and 60 degrees still remaining. So, this is 60 degrees where S 5 is turns off and S 2 is turned on. So, this is one complete cycle, each of 60 degrees pi by 3, 2 pi by 3 and pi. So, 0 to pi by 3, see here, attaining given term 3 devices are on. 1, 6 is on in the lower half and S 5 upper device in C phase. What is V A 0? See this voltage, when the switches are on, see switches are on in a voltage source inverter, the so called pole voltages are known. Pole voltage means voltage of A or B or C with respect to either the midpoint of the DC link or with respect to the negative DC link, it is known. I told, I will let me repeat, if the devices are, if I know the conducting state of the device, if I know the conducting state of the device, I know the pole voltages. Pole voltage is nothing but voltage of A or B or C with respect to the negative of the DC link or center point of the DC link. So, had plotted the pole voltages. Now, how do I determine the line voltage? Line voltage is nothing but V A B. V A B is V A 0 minus V B 0, V A 0 minus V B 0, V A 0 is V DC by 2, V B 0 is minus V DC by 2. So, V A 0 minus V B 0 is V DC. So, for 2 pi by 3 radians, V A 0 is a square wave of magnitude V DC. From 2 pi by 3 to pi, V A 0 is also positive, V B 0 is also positive. So, V A 0 minus V B 0 is 0. So, there is a 0 voltage period in line to line voltage wave form and similarly, in the negative half. So, line to line voltage wave form, I have pulse width of 120 degrees, the magnitude is V DC and 0 for 60 degrees. Now, if I know the voltage wave form, depending upon the type of load, I cannot draw the current wave form, no issues. Now, I have a 3 phase inverter, the load could be either delta connected or star connected. Now, star connected, if it is a star connected load, I need to know the phase voltage. If it is a delta connected load, I cannot determine the phase current if I know the phase voltage. So, I will repeat, if the load is delta connected, if I know the voltage wave form, line to line voltage wave form and if I know the type of load, it is possible to determine or it is possible to draw the load current wave form. I mean the phase current wave form, phase current not the line current, if I know 2 phase currents, I need to add them, wave forms I need to add to get the line wave form. Please do not say that root, phase current into root 3 is line current, no it is not valid. So, that is how it looks, constant voltage is being applied to VAB, depending upon the type of load, current wave form will change. If it is a RL load, voltage applied is constant DC. So, current will exponentially increase at steady state. If it starts from at t is equal to 0, if I energize, current has to start from 0. At steady state, it has a negative value, exponential increase, voltage applied to the load is 0. If the voltage applied to the load is 0 and if the load is RL, current will decay because applied voltage is 0. If the load is purely inductor, this will be linear in this zone, current will remain constant because voltage applied to the inductor is 0. So, current will remain constant and here current will decay. So, this is the current wave form IAB. I can draw similarly IAC, IA I can determine by adding these two wave forms. I have to add these two wave forms to get IA, this is how it looks. I will not discuss much time on this. The observations, there are six steps in each cycle. The number of steps in each cycle are six, hence the name six step inverter. V s 0, V b 0, V c 0 are pole voltages. Predominant harmonic, what is the predominant harmonic in the pole voltages? So, it is a square wave, definitely all order harmonic 3, 5, 7, 9, 11, everything. But then, line to line voltage is a square wave of 120 degrees duration. It can be proved that harmonic only 6 n plus r minus 1 harmonics will be present, 5, 7, 11, 13, so on. One way is to write the Fourier series or write the Fourier series for V s 0. We all know that this is 1, 3, 5, 7, 9, 11, everything is there. Here also 1, 3, 5, 7, 11 harmonics. Now, V AB is V s 0 minus V b 0 and then you get the harmonic spectrum. Third harmonic and third harmonic get cancelled. It can be proved. You just prove it. So, in the line to line voltage wave form, we cannot have third harmonic. It is a very popular or very important result that we have studied. So, predominant harmonic is 5th, 7th, 11th, 13th, 6 n 1 plus r minus 1. So, this is the V s 0. This is the V b 0 harmonic spectrum. This we can get by writing V b 0 and etcetera. So, the r m s value of the line to line voltage is square root of 2 by 3 into V d c. What is the r m s value of this? We will get as 2 by 3 into V d c of this wave form. And the fundamental component here is root 6 by pi into V d c, 0.78 V d c. Fundamental is 0.78 V d c, r m s value of the line to line, r m s value of the line to line voltage is 816 V d c. For star connected load, I need to determine the phase voltage. There are 2 ways to do it. One is depending upon the switching combination, draw the equivalent circuit and determine what is V a n, what is V b n, V c n. I mean, it is something like this. Say, SCA attaining one time 3 devices are on, SCA is on, this is on, lower switch of SB and lower switch of SC is on. I will take this example. See, attaining one time, one of them is on. I will take SCA is on, lower is lower of SB, lower of SC is on. So, when SCA is on point, what could be the equivalent circuit? What could be the equivalent circuit? When SCA is on, A gets connected to the positive DC bus. When the lower switch is on, B gets connected to the negative DC bus. Similarly, I have assumed that lower switch is on, C gets connected to the negative DC bus. So, how does the equivalent circuit look like? It looks something like this. I am not going to draw, explain to you each and every step. One is on. So, this is A, this is B, this is C, finally connected to positive, negative, this is V d c. Assuming the load to be balanced, assuming the load to be balanced. So, what will happen? I have the magnitude of and the phase angle of the load impedance is the same. So, I can prove that V a n is 2 by 3 V d c. V b n is equal to, V c n is equal to minus one third V d c. If you say, assume that this is R, R, R. So, combine this, this is going to be R by 2, this is R. So, voltage at this point comes to be one third, this comes to be two third. So, can I make a generalized rule? I am not going to derive for other combinations. See, I have three switches. They can take two values. Basically, there are eight possibilities, they are all three of them are on and all three of them are off. There are two possibilities. Under that condition, when all three of them are on, A gets B and C, they get connected to the positive d c bus. So, all three line terminals are at the same potential. So, therefore, source does not supply any power to the load. It is also known as load is freewheeling through the positive d c bus. Similarly, when the lower switches are on, these three points are connected to the negative d c bus. Power supplied by the source is 0, load freewheels through the negative d c bus. So, when A, B and C are at the same point, B A B line to line voltage, B A B is 0, B B C is 0, B C A is also 0. All three line to line voltages are 0, A B B C C A and therefore, you can prove that B A N is also 0, V B N is also 0, B C N is also 0. So, this is also known as, these two vectors also known as 0 voltage vectors, 0 voltage vectors. Why? Magnitude of the voltage applied is 0. All three terminals are connected, either they are connected to the positive d c bus or they are connected to the negative d c bus. The remaining six, what are they? One is on on top, two are on in the bottom or two are on in the top, one in the bottom. So, there are six such possibilities, six such possibilities. So, one such possibility is, I said I assumed A is, upper device A is on and lower switches of B and C are on. What is the magnitude of the voltage that is obtained? If one device is on, we found that this voltage is 2 by 3, 2 by 3 and when two devices are on, we found that this magnitude is one-third, one-third. Remember, if one device is on, magnitude is 2 by 3 and if two devices are on, magnitude is one-third. What is the sign? They could be either positive or negative. If only one upper switch is on, the corresponding phase voltage is positive. Similarly, if one lower device is on, the magnitude is minus, sorry the magnitude is 2 by 3 and sign is negative. Instead, two upper devices are on, magnitude is one-third, sign is positive. Here, if the two devices are on, magnitude is one-third, sign is negative. That is all. V B N and V C N are minus one-third V D C. Why these two devices are on? V A N is one-third, sorry V A N is two-third V D C. So, there are six active states and two null states. So, I can draw V A N, V B N, V B N, V A N. And V C N, how does the waveform look like? How does the waveform look like? See, I will go back, can draw from here also. In this zone, see here, V A 0 is positive, V C 0 is positive, but V B 0 is negative. So, therefore, V A N is plus one-third, plus one-third. Why upper two devices are on and V A is also positive, so plus one-third, whereas V B N is minus two-third. Now, what happens from pi by 3 to 2 pi by 3? Only once which S 1 is on, B in the second leg S 6 is on, that is the lower device and in the third leg S 2, which is again a lower device. There are two lower devices in the lower half are on, so therefore, V B N and V C N are minus one-third each and only one upper switch is on. So, therefore, V A N is two-third V D C. See in this zone, A and B, there are two devices are on. So, V A N is plus one-third V D C. You can draw the equivalent circuit and derive. So, the phase voltage waveform here is one-third V D C between 0 to pi by 3, pi by 3 to 2 pi by 3 is 2 by 3 V D C and 2 pi by 3 to pi it is again one-third V D C. So, again there are six steps, hence the name six step inverter. See the entire analysis is there. You can go back and read it is simple, one is zero vector vectors. This is equivalent circuit approach. This is how the waveform look like, one-third V D C, two-third V D C, one-third V D C hence the name six steps. The harmonic spectrum, you can prove that, I will just show you. The harmonic spectrum must be, again the phase voltage has 6 N plus R minus 1, harmonic can be proved. So, as of now, we learnt how to convert D C to a higher voltage D C, any voltage magnitude that I can get now. If I want to have a very high voltage, I need to use a transformer in the sense and depending upon the power rating, I can choose either a fly back forward or a push pull and there are other topologies also exist. Now, given a AC voltage, I can convert to AC. The output voltage waveform is a square wave. If it is a, the line to line voltage has 120 degree duration, whose magnitude is V D C or 60 degrees it is zero. Phase voltage waveform is one-third V D C and two-third V D C. The harmonic spectrum is approximately the same, 6 N plus R minus 1, harmonics are there. So, we are using it as a UPS here to feed, assuming that, now there are, you need to make two assumptions. What are they? I have solar panel over that is coming here. This is D C to D C. This is D C to D C. This is D C to AC. Now, I can, sorry in the sense depending if I want to have a 50 hertz supply, 50 hertz supply, the predominant harmonic is 6 N plus R minus 1 harmonics, nothing but 250 and 350, 550 and that is 11 and 13, 650 hertz. This is the predominant harmonics, assuming that you want a 50 hertz supply. If you want to have a variable frequency supply, then therefore, as the frequency of the fundamental changes, frequency of the predominant harmonic also changes. As of now, I will assume that it is a 50 hertz supply. Predominant harmonic is 250 and 350. I do not think we can directly connect this inverter to the grid, because harmonic magnitude of this fifth and harmonic is substantial. So, the total harmonic distortion T S D of the voltage waveform is very high. So, your, the entire black box may not be able to satisfy the given norms of I triple E and I E C. So, you may not be able to connect it to the grid. So, if I have to connect it to the grid, definitely I need to reduce the harmonic content. How do I reduce this harmonic content? That I will discuss in my next lecture for another 10 minutes or so, I will take up the questions. I will encourage you to come on chat mode, because it reduces the time. In this, I can directly read the question. I will repeat it instead of going into each and every center and repeating the question. Please feel free. War angle, over to you. Sir, the one leg of a 3 phase inverter. Both the switch are complementary. That means, if the upper switch is on, then the lower switch should be off. If the lower switch is off, then upper switch will be on. The module that you have given, where the three legs which we have shown, it is always that condition or by any means we can switch off both the switch, the upper switch as well as the lower switch, so that you can do any other modifications. So, I hope you follow my questions, sir. Now, I do follow in a voltage source inverter, each leg, the devices in each leg are complementary. If S 1 is on, S 4 is off or vice versa, they are complementary. They are complementary. Say that again? No, both the switches can be made off. Both the switches can be made off. No problem. You can turn it off. What is going to happen? See, if both the switches are off and if there is some current flowing this side, yes, it can flow. Are you with me? Some current can flow provided there is some current coming from the load side. In fact, it happens here, is not it here? See here, you turned off the switch here, S 1 and you turned on S 2. Assume that you do not turn on S 2. What will happen? Whatever the current that was flowing through the inductor, this was the direction of current, it will start flowing through D 2 and when this current becomes 0, conduction stops. Are you with me? So, if there is some current coming from the load side, yes, it can flow through the diode. If there is no current coming, that is it. Is that clear to you? Yes, actually why I was asking means suppose if I want to go for other topologies, higher topologies, higher levels. Higher levels also the principle is the same. In a 3 level inverter, higher level also has in a 3 level inverter, 2 devices are on. There is a rule to be followed, rule to be followed. Here, one device is on at any given time. If the both the devices are off, there is no power flow. It is only when if there is some current coming from the load side, it can come and charge the DC link, that is all. In fact, that is being used sometimes. Suppose it is like this, I have, say for example, I have a voltage, a 3 phase voltage source inverter. This is line A, this is B and C and I connect a capacitor here. I will not give any pulses to the active switches connected to AC supply, connected to AC supply. I have an uncontrolled bridge charging the capacitor. I have an uncontrolled bridge charging the capacitor. So, we will see sometime later. This is being used initially to charge the DC link capacitor in a voltage source inverter, feeding power to the grid. Is that okay? If I do not switch on, I have an uncontrolled bridge, 3 phase uncontrolled bridge. So, there is a question in the sense inverter, the question I will read out. Sir, you said we can apply inverter for supplying reactive power, whether it depends upon our load or on the inverter can direct to supply Q. I have a 3 phase source supplying power to the load. As of now, I will, load is, I will just assume that, as of now I will assume that it is a linear load. Power factor is less, not equal to 1, linear. After sometime we will see it can have a non-linear load as well. So, if, when I am saying linear load, current drawn is sinusoidal. If the input is sinusoidal, current drawn will be sinusoidal, but then power factor is not equal to 1. Under this condition, your voltage source inverter, how you do not ask me right now, can supply this Q. Here is, if the load is, sorry, SL is power load plus or minus j QL, as of now I will assume that inductive load plus QL. This inverter can supply QL, this inverter can supply QL and thus QL should be less than or equal to, I will repeat this QL should be less than or equal to S inverter, K V rating of the inverter, K V rating of the inverter. So, Q supplied by the inverter is, could be as high as the K V rating of the inverter. To do that, a small amount of delta P is required, to do that a small amount of delta P is required. The delta P is to account for losses in the inductor and the losses taking place in this inverter, because whenever I switch the device, there are losses. The circuit theory base, for this I will explain to you. Right now, I proved that inverter can supply any Q, Q could be as high as its inverter, its K V rating. To do that, it requires delta P. Suppose I do not have a load here, I have just three phase transmission line and if it requires Q, yes this inverter can supply Q, that Q is equal to or less than or equal to inverter, it is inverter, it is V rating and so and this inverter will draw a small amount of delta P from the source. So, question is you do not need to have a load. Now, load could be grid or load could be some whatever. The question is what are the advantages of three phase inverter over single phase inverter? Now, how do I answer? What is the advantage of single phase supply over a three phase supply? That is all. If you want a three phase inverter, three phase supply, you need to have a three phase inverter and if you want to have a single phase supply, you need to have a single phase inverter, that is as simple as that. So, when do I go for a three phase? For low power application, I use single phase supply. For as the power rating increases, I go in for a three phase supply, go to the market and find out, can you get, can you get say more than three HP pump, irrigation pump, a single phase three phase, a single phase three HP irrigation pump, you may not get. So, you have to go in for a three phase supply. So, if you want a three, if you have a three phase load, I need to have a three phase supply. What are the reason that makes electrolytic capacitor unreliable? Come on, that is a device physics or whatever the capacitor construction or design aspect, you go and read the construction of a capacitor, why they fail? What is the reason that makes an electrolytic capacitor unreliable? How to generate a sine wave from the output from the inverter? Do not worry, I will discuss right now. How to generate a sine wave from the inverter? I will discuss right now. Sir, out of two modes namely 120 degree and 180 degree mode inverter, which is good for star and delta connected load? Let us not talk about 120 and 180 degree, this is just a basic block. If you have to sell this, sell your inverter in the international market or in the market, neither your 120 degree will work, neither 120 nor 180 degree will work, you have to do something else. In the next class, we will discuss after this. Please, 120 and 180, they are simple to understand. Since they are simple to understand, I did this and I told you the limitations. Now, we will discuss what are the limitations. I told you the harmonic content. So, if it is a 50 hertz supply, I have a 250 hertz and 350 hertz harmonic. So, you may not be able to connect your load. If you connect your load, there will be harmonic torques that will be generated, which I am not going to discuss. You may not be able to feed it to the grid either. So, you have to do something else. What is that? We will see in my next lecture. What is the advantage of using split power supply in 3 phase inverter? Who is using the split power supply? I did tell you that you do not need to use center point of the DC link as a reference. In fact, it is not there. It is not available. I can use one single capacitor. It is just for convenience sake that I have used a center point. Of course, by the way, I do not know in what context you are asking this question. I can use this if I have a split capacitor topology and if I connect the center point to the neutral of the source, there are certain advantages. Please, KJSM, I do not know in what context are you using. In what context you ask this question? If your question is based on this slide, let me tell you one thing you do not require this point at all, do not require this connection. In fact, in a conventional 3 phase voltage source inverter, this point is not available. But then there are certain topologies. I can use this point and if I connect it to the neutral, there are certain advantages. What are the advantages? I will discuss it. I hope you meant that. You can confirm through chat. Now, 36, I will take two more questions. KJSM asks, sir, I was asking about the split power supply. Where is the split power supply? We asked the question about split power supply. Where is the split power supply? I know that you already answered. No, no, I have not answered in the sense. Wait a minute. I have only one part here I have answered. Assuming that the question is based on that photograph. But then there are circuit topologies wherein we connect the center point of the DC link to the neutral. When you do that, there are lot of other advantages. What are they we will see? Now, which question is that? Did you ask the first part or second part? Now, what? Yes, first part, first part you do not require. What are we doing still in DC? How much efficient to apply a multilevel inverter rating? How much efficient to apply a multilevel inverter for obtaining a sine wave inverter? Now, why multilevel inverter? We are still in 2 level. We will see multilevel sometime later. Multilevel generally used for high power application. What are the advantages we will see? First we will try to understand 2 level of p-dolume technique and we will see 3 level or multilevel later. For DC to DC conversion, how to select whether to go by buck, boost, buck, boost, fly back? Tell us formula. Sir, I am not going to tell you the formula for DC to DC conversion. How to select whether to go for buck, boost, buck, boost, push, pull, fly? Kindly tell us formula related to converter selection. No, sir, depending upon your power level, depending upon the input output voltages, you have to select one of them or you have to modify them. I do not want to tell you and nor I will tell you which one you should use. You choose fine, fine. You take one topology and if it is wrong, what price will you pay? Fine, you will learn. You will gain some experience. That is more important. So, I am not going to tell you which one you should use. It depends on the power level as well as the input and output voltage ratios. Sir, suppose the inverter is connected to the grid, then on what basis we can select the coupling inductor value? Wait a minute. Let us connect the, first we will connect the inverter to the grid and we will see later. We have not connected the inverter to the grid so far. What could be the value of the T H D in ideal sine wave inverter? Hey, Bhagawan, what should be the value of T H D in an ideal sine wave? What do you mean by T H D? T H D is total harmonic distortion and you say in an ideal inverter, ideal sine wave. I think there is a inner contradiction, I guess. Total harmonic distortion and ideal sine. Anyway, wait a minute. This class, what is A S, A S, E B? The value of inductor, value of capacitor, we will see. Let us connect the inverter first to the grid. Then we will find out what should be the value of L.