 lecture before doing that let me quickly recapitulate the things that we did in the last class. We discussed the fly back converter wherein I am using though I am using a transformer only one coil is carrying current at a time. The transfer function to the fly back converter is given by V naught is equal to N 2 by N 1 into V in or V dc d divided by 1 minus d. One point to be remembered is that provided flux is continuous, flux is continuous this is a transfer function, but I also told you that generally fly back converter is operated in discontinuous maximum value is around 0.5 or so. So, at that time what could be the output voltage 1 has to derive 1 has to derive please remember this function this transfer function is valid only if only if flux in the core is continuous. See how did it derive this equation we used volt second per turn balance volt second per turn balance. So, I said primary voltage applied to the primary is V dc for d t seconds and number of turns is N 1 and we equated this equal to volt second volt second per turn of the secondary I said V out 1 minus d into t divided by N 2. See I am assuming that I am assuming that voltage applied to the secondary is V 2 for 1 minus d into t seconds that can happen only when diode is conducting in the secondary see in this circuit dot current enters the dot current should enter the dot now this is how it is. So, switch is on for d into t seconds and switch is turned on only after t seconds is like you know this is what I am trying to tell you d into t and this is t second time is being turned on. So, one from d t to t voltage applied to the secondary is V naught provided this diode is on. If the current becomes 0 in the secondary before t then this transfer function is not valid this function transfer function is not valid remember I will repeat voltage applied to the secondary is V naught only when diode d 2 is conducting and in case diode turns of before t then volt second per turn for the secondary is not V naught into 1 minus d into t divided by N 2. So, you have to find out where the diode turns off. So, if it turns off somewhere here then volts second per turn is beta into t minus d into t remember. So, this is this trans equation I have derived neglecting the winding resistance device drop. So, if this device drop is comparable or comparable to the input voltage say I have 12 volts solar panel and device drop is 2 then the 2 is comparable to 12. So, we have to take care of this 2 volts similarly here in the secondary. So, all the idea non idealities we need to take into account while determining the turns ratio. So, if time permits we will take one design problem all those issues will be discussed. Similarly, for a forward converter as well forward converter as well. So, in both forward converter as well as in fly back converter I am using a transformer, but then if I see the BH characteristics we are operating in the first quadrant itself. So, in other words the magnetics are not utilized fully can I use bidirectional core excitation something similar to AC. So, if I use bidirectional excitation is something similar to AC therefore, voltage induced in the secondary will have both positive half as well as negative half finally, we require a DC. So, what we need to do the voltage induced in the secondary we need to rectify it and filter it to get a DC. Please try to understand the moment I am using bidirectional excitation voltage induced in the secondary is going to be AC because it will have positive half as well as a negative half I want finally I want a DC I have to rectify it and filter it. So, why are we doing this first we are converting this DC to AC and again again to DC. I did say that I am using a bidirectional excitation, but one thing that I did not tell you is the switching frequency is very high or the frequency of the voltage induced in the secondary is very high. So, as the switching frequency increases. So, frequency of the voltage induced in the secondary also increases. So, there are two advantages one is size of the magnetics the inductor or the transformer reduces as the frequency increases number of turns come down. Second I have to rectify this AC to DC as the frequency increases the ripple frequency also increases see what I mean is if I take say full wave rectifier single full wave rectifier it looks something like this. What is this? This is T by 2 and this is T by 2. So, if I do if the frequency is very high. So, this the frequency of this pulses also increases. So, my filtering requirement comes down. So, if I have a three phase six pulse conversion what will happen? I will have something like this. So, my filtering requirement in this case is much less compared to filtering requirement here. So, what I will do is I will convert this DC to high frequency AC magnitude is determined by the application. So, number of turns will be magnitude will be determined by the application frequency is very high I will convert it to DC then I will use a very small filter to filter out the ripple and get a constant DC that is the philosophy. So, first one in this class is push pull converter push pull converter the circuit configuration is here. See the secondary sorry see the transformer connection I have a center top and at the secondary you just see it is nothing but a center top full wave rectifier is with this circuit would have seen in while fabricating a regulated power supply in the lab something like this. It looks something like this what we did to get a 5 volt DC in the lab what we did a diode and diode capacitor this what we did and we connected to AC. So, the secondary here what this does it rectifies this is a filter something similar one here there is no difference at all see the secondary there is a filter where did we see this stage this filtering stage in buck converter. See in the primary side I have two switches and see the connections a center top primary center top secondary even in T 2 or two switches V DC is the supply voltage here it could be voltage from the solar panel through a capacitor filter or whatever. See the dot conventions let us see how it works now from 0 to DT by 2 the first half T 1 is on when I close T 1 this is the path positive the direction of current. Current enters the dot let us see what happens in the secondary current enters the dot current can leave the dot. So, yes current can leave the dot D 1 can conduct can D 2 conduct no. So, when T 1 is turned on current enters the dot current can leave the dots therefore, D 1 conducts. So, what is the path D 1 secondary the filtering stage back to. So, what is the voltage applied to the N 1 turns it is V DC. So, what the corresponding there will be voltage induced in the secondary with the dot as positive and that will that will force the current in the secondary after sometime open T 1 here voltage across N 1 is equal to V 1 1 that is nothing but V DC with the dot as positive. Now, this dot is positive and this dot also will be positive. So, what is the voltage rating of T 2 you just stress the circuit V DC positive of the V DC is connected to 1 terminal T 2 negative here to the positive of the voltage source which is induced a positive voltage source which is induced in N 1 2 with the dot as positive I am calling that voltage as V 1 2 if N 1 1 is equal to N 1 2. In fact, they are the same the voltage induced here will be V DC itself. So, this is the voltage. So, similarly what is the voltage induced in D 2 when D 1 is conducting same concept we need to see this is positive this is positive voltage induced in N 2 1 is proportional to the number of turns in the supply voltage similarly N 2 2 stress the circuit positive positive D 1 is on the entire voltage D 2 has to block. So, voltage across D 2 is 2 V DC N 2 by N 1 voltage across D 2 is 2 V DC. So, voltage rating of the switches is 2 V DC after sometime open T 1. Now, what happens in the secondary the moment D 1 is on D 1 is positive after sometime open T 1 what will happen. So, this is the duration from 0 to T is equal to 0 to T T by 2 T 1 is on T 2 is off from D T by 2 to T by 2 see here D T by 2 to T by 2 I am saying T 1 and T 2 I am saying T 1 and T 2 are off I am not saying anything about the secondary from D T by 2 to T by 2 both T 1 and T 2 are off what happens in the secondary we will see from T by 2 to 1 plus D into T into T by 2 T 2 is on T 1 is off and from this period to T both T 1 and T 2 are off nothing has been mentioned about what happens in the secondary N 1 1 is equal to N 1 2 and N 2 1 is equal to N 2 2. So, when I close T 1 current enter the dot current can leave the dot in the secondary. So, therefore primary current I 1 has 2 component 1 is the magnetizing current plus the equivalent secondary current. Now, when I open T 1 what happens 1 there should be a path for for the magnetizing flux or the magnetizing current and there should be a path for the inductor current as well I L while discussing forward or fly back what we found that when the if the current enters the dot the correct direction the polarity for the magnetizing flux to be continuous is current should enter the dot in some other winding I will repeat the correct direction is for the magnetizing current if current enters the dot current should enter dot dot dot even in the secondary or even whichever winding same thing is in forward as well this what we did V D C current current enters the dot current leaves the dot. So, these 2 windings carry current simultaneously. So, the the source current or the primary winding current has 2 components magnetizing current plus the equivalent secondary current. So, when I open S if current enters the dot current can enter the dot here. So, this is the correct direction or direction for the magnetizing current current should enter the dot something similar same principle is applicable everywhere even here if current enters the dot current should enter the dot even in some other winding by the way T 2 is not closed. So, what happens in the secondary you just see see in the second half. So, current can enter the dot if current enters the dot D 2 it can flow through D 2, but then here if the moment D 2 starts conducting D 1 also starts conducting D 1 also starts conducting see the what will happen to the flux linkages in this 2 windings you need to check the moment D 2 starts conducting D 1 also will start conducting D 1 also starts conducting. So, this is the equivalent circuit I 2 enters the dot I D 1 can leave the dot if D 2 conduct D 1 also can conduct see I 2 enters the dot these 2 are mutually coupled can if I 2 whatever I D 2 can enter the dot some current can leave the dot I D 1 can leave the dot yes D 1 can leave the dot see it is bit tricky here just try to understand in this circuit nothing no current can flow in the primary, because both the switches are off when I close the switch no current can flow through D 2 because dot is positive dot is positive dot is positive voltage twice the supply voltage appears twice the supply voltage into the number of turns will appear across D 2. So, D 2 has to block that voltage when I open the switch correct direction for the magnetizing current is since this enter the dot in the primary it should enter the dot enter the dot in any winding which is mutually coupled yes it can enter the dot here, but then if it enters the dot here it can leave the dot can leave the dot here because of D 2 1. So, therefore, both D 1 and D 2 are on and this is the equivalent circuit both D 1 and D 2 are on now see this is what this is the equivalent circuit N 2 2 into D 5 by D t is the is the voltage here similarly N 2 1 D 5 by D t is the voltage here R is the winding resistance of this 2 coil this is the equivalent. Now, can we write the voltage equations this are the 2 volt N 2 1 D 5 by D t minus I D 1 I D 1 into R 1 is equal to V o 1 this is the voltage V o 1 and that voltage is also equal to N 2 2 D 5 by D t plus I D 2 into R is equal to minus V o 1 voltages are same and if I voltages are same therefore, if I equate it equate it I get this equation if I D 2 minus I D 1 of course, I D 2 should be greater than I D 1 why entire the magnetizing current current can flow only through this. So, I D 2 should be greater than I D 1. So, this is the voltage average voltage across L equal to 0 V o 1 is equal to 0. So, voltage induced this how it looks T o T 1 on T 1 on for D into T D T D into T by 2 in this period both are off. So, if the flux is continuous both T 1 and T both D 1 and D 2 are on here both the diodes are on both the diodes are on that is the reason see here I D I L 1 is the inductor current when T 1 is on I L is the current that is flowing through D 1 same current. The moment you turn of T 1 T 1 current become 0 current flowing through D 1 also falls by falls to approximately 50 percent. And correspondingly current through D 2 also rises this how at T by 2 to. So, what happens after T by 2 I will turn on T 2. So, when I turn on T 2 what happens in the primary this is the path current leaves the dot current leaves the dot if current can leave the dot current should enter the dot not possible here this is the path. See please do not get intimidated only principle that you need to understand is you need to know is mutually coupled circuit. Yes it is not very straight forward, but then if you can understand this dot convention the it is becomes relatively simpler. So, if current enters the dot current should leave the dot if there is a path if there is no path when I open the switch if current enter the dot current should enter the dot in some other winding that is all only those concepts one need to remember. When I close the switch current leaves the dot N 2 2 when I open T 2 I am not close T 1 we have to wait for some time I said in this period none of the switches T 1 and T 2 are off 1 plus D into T by 2 2 to T T 1 and T 2 are off something similar to D T by 2 to T by 2 both T 1 and T 2 are off the transfer function is you can derive it as 2 V D C N 2 divided by N 1 into D. So, this is I said this is nothing, but a forward buck converter when T 1 is on D 1 is on. So, this is a buck converter with the forcing function of V D C into N 2 1 N 2 1 divided by N 1 1 when I open the switch we proved that this voltage is 0. So, this is nothing, but the output stage or this is nothing, but the equivalent circuit of the buck converter when you open the switch. So, the transfer function is only for one winding is V D C into N 2 1 divided by N 1 1 into D something similar happens for these two winding as well. So, a factor 2 2 V D C N 2 divided by N 1 into D nothing, but an isolated buck converter isolated buck converter quite popular if the power rating is around 1 to 1.5 kilowatt or 2 kilowatt or so. So, above 700 to 800 watts one has to go in for the push pull converter wherein I am using both of the magnetic circuit. So, what are the limitations of the push pull converter see I said I need I told you both windings are number of turns in both the halves is the same N 1 1 is equal to N 1 2 N 2 1 is equal to N 2 2 what will happen if they are not the same they change by a fraction primary winding may differ by a fraction of a turn what will happen here switches may have a fine circuit switches may have a slightly different saturation voltage. So, this is the equivalent circuit when I close T 1 a dot is positive. So, may be we are in the first half of the V H loop increasing flux, but then I said N 1 1 is not equal to N 1 2 there could be if both the there could the number of both the windings could change by a fraction or device drop may not be the same if device drop is may not be the same voltage applied to N 1 1 is equal to V D C minus the drop across T 1 that is being applied to the coil which is having a small or a small resistance plus the leakage, but then N 1 2 voltage applied to N 1 2 is V D C minus the voltage drop across T 2. So, if there is a difference in the magnitude of the voltage applied to the coil. So, therefore if there is a difference in the voltage applied to this coil. So, what will happen d phi by D T I will write this I have something like this magnitude of voltages are not the same they change there is a small difference plus there is a number of turns also may not be the same if they could change in a small by a small fraction if that is the case what will happen when I am here I am the first half here I am the third quadrant. So, what will happen B H curve may not be travels symmetrically B H curve will not travels symmetrically if that is the case a D C flux will exist in the core average is going to be finite average is going to be finite the positive half is not equal to the negative half now there is average is finite average is nothing but the D C flux. Now, if there is a D C flux what may happen assume that there is a small d by D C flux and in second or third cycle load has changed also load has increased now the control is trying to increase D. So, as D increases time for which T 1 is on also increases now instead of starting from 0 since there is a D C flux it starts from a finite value switch is on for a longer time. Therefore, a D C voltage is applied to a winding for a longer time flux and current increases they cannot go on increasing linearly after some time core may get saturated or so that is one core may get saturated second is I have a core imbalance and what is known as a flux walking in one direction there could be gradual increase in flux. So, that is also known as a flux walking in one direction. So, these are the limitations of of push ball you need to ensure that both the winding should have both the winding are same number of turns and third one is if there is no dead time between these two dead time in a sense there is before T 1 turns of you are turned on D T 2 what may happen the flux produced by I 1 opposes the flux produced by I 2 number of turns is the same. So, the current is so the number of turns is the same L 1 is equal to L 2 I am neglecting the leakage of that is equal to the mutual they cancel out and now I is limited by the internal resistance itself that is a dead short circuit current is limited by a small r. So, there has to be a dead time between T 1 and T 2. So, these are the limitations of the push pull converter load and I am assuming the current in the inductor is something like this. So, under this condition how does the current through S look like D into T this is T at this point again you are increasing. So, whatever the current that was flowing through the diode just prior to T starts flowing through S. So, current through S looks something like this see the D I by D T of the current that is flowing through S there is a large D I by D T we all know that in the initial when the device is getting turned on the entire area is not available for conduction. So, you if the D I by D T exceeds a rating because the hot spot device may fail. So, a D I by D T is very high also if I see the voltage across the device when I turn on it gradually falls. So, therefore, during on time voltage and current flowing through the device voltage across the device and current flowing through the device are finite and that is good the product is nothing, but the power loss that is taking place. So, as the switching frequency increases losses that are occurring the device also increases. So, therefore, the efficiency point number 1. So, as the losses increases your junction temperature also increases and in a safe operation setting area there is one area is always is limited by or one of the boundaries is the junction temperature that one has to consider. Second is the D I by D T and D I by D T through the device is very high therefore, there will be a E M I electromagnetic interference it will produce its own field try to effect the nearby the control circuit. Can I try to address this limitations? So, this set of a switching is also known as hard switching why hard I am just I am not bothering about what is the current flowing through the device at the instant of turn on. The whatever the pulses that are coming out from the controller are just passed on to this switch. So, depending upon the current that is flowing through the inductor the switch starts taking the current starts carrying the current. So, therefore, D I by D T is very high plus the voltage the product of V and I during conduction is during the turn on is high losses are high. So, that is so losses that are taking place in the device is one of the it puts one of the limit on the switching frequency. So, if I am doing a hard switching if I am doing a hard switching it may not be possible to switch this device at a very high frequency though it is capable of switching at a very high frequency. See I will repeat if I am doing hard switching in other words I am just not bothered about what is the current that is flowing through the device at the instant of turn on due to the power loss that is taking place and other effects I may not be able to increase the switching frequency beyond a certain value. So, if I am in a hard switch converter there is always in a hard switch converter switching frequency is generally low. See one of the question that was asked in one of the centers sir what is the switching frequency in a fly back converter? How am I supposed to answer that? I cannot answer I do not know what sort of a switching is being used are you doing this hard switching? What is the power rating? What is the voltage rating? What sort of a device based on that you can decide? So, if you are using a fly back converter power is of the order of 100 watt also safely for hard switch converter I can say around 25 kilo hertz. First you try to operate at 25 kilo hertz you may not be able to do because all the non idealities that parasitic capacitors inductance all those issues will come. It is very easy to make the circuit to work on a breadboard or on a table the moment you start integrating it the moment you start integrating it the because of this EMI it will affect the control circuit. So, you may not be able to increase beyond a certain value. So, please when I say that I do not know you need to try to understand it may not be possible to often tell you what could be the switching frequency? It all depends on what sort of a switching that you do what sort of a device that you use what is the voltage and power. Fine if you want to start off with any design take one assemble the circuit start with a very conservative figure if you are able to operate it is fine. There are no problem then go on increasing the switching frequency do not have to take that value from me I am no god please that is about it. Second is now coming to the hard switching because of this hard switching there are losses that are taking place. If the losses that are taking place since I have to control the junction temperature I have to use a bigger heat sink therefore, my size increases plus my cooling requirement I may have to use a force cooling as well. Because of this hard switching I cannot increase the switching frequency though in principle this device is capable of operating at a high frequency. I will repeat because of this hard switching we may not be able to it may not be possible to switch this device at a high frequency though this device is capable of switching at a high frequency. Switching frequency may not be governed by this switch switching frequency may be limited by the way you are switching the way you are switching. Can I try to address this issue? Yes it is possible yeah I will try to cover in my next lecture is 9.50 for 10 minutes or so I will take up few questions. Last three minutes you Tanjavur over to you. Secondary of the push pull amplifier behave like say full wave midpoint converter. In midpoint converter at a time the two switches are not conduct but after that you told here at a time d 1 and d 2 conducts how its possible sir please explain. Please try to understand in what context that I have said. See what is this tell me secondary what is this? It is the rectifier circuit sir. Why am I doing this? Why am I using rectifier circuit in DC to DC conversion? So, I am here when I turn on both t 1 and t 2 what happens direction of flux here increase in the positive direction this was positive now the dots are negative I am using the both after the magnetic circuit. So, what will happen? Voltage induced in the secondary has a positive half as well as a negative half. Finally, I want to have a DC to DC I want to have a DC voltage here what do I do filter it? Now your question is yes both the diodes are not conducting in full wave rectifier. Yes here also it happens in the sense when I open when I close when switch s 1 is on only one diode is on when both are off both are off primary is completely cut off do not come do not directly relate to what happens in full what center type transformer in AC to DC converter that you use in the lab. Now when both the diodes are conducting here neither t 1 and t 2 are on both these two are off and if it so happens that in this case if d 1 starts conducting d 2 also will start conducting you need to write the circuit I wrote the circuit equation and that showed you please in the sense I explained to you because this why did I why did I say that this looks like a center type transformer is yes it looks does in the sense secondary is there are it is a center type a full wave rectifier center type transformer why am I using this looks like looks like voltage induced here is AC I want to have DC looks like. Now I have told you the principle of operation now you go back understand in your own way is that ok. Sir thank you sir no problem. Government college Salem over to you. Sir good morning say if I say what do you mean if you can say sir what do you mean by high power density converter high power density DC to DC converter if it means what what it means sir high power density converter. What is a what is a foot area a foot print area foot print area to the other rating I have a huge converter power rating is very small density is very small 75 ampere device full bridge 3 3 legs of a voltage source 30 ampere 30 ampere this is 75 ampere ok 3 legs I need to have a separate now this I do not have 4 ampere now you convert now you try to make a converter with where in you require 75 ampere device you design the converter using this this and this now you try to find out the foot print area the entire area of the size of the converter bit will have a smaller size. So, therefore, do not you think the power density in this case will be higher. So, the overall size the foot print area the ratio of the output power to the foot print area I can say the power density. Thank you sir. Thank you. Hello sir can we have normal primary without tapped primary normal primary in the sense without center tap where will I connect where will I connect T 1 and where will I connect T 2 and where is the supply voltage you tell me can we have a normal primary without center tap somehow I have to reverse the direction of flux both the halves of the entire BH loop in the both half positive first quadrant both I need to use. If you do not have a center tap then I may not be able to use with two switches I may have to change the device configuration. I may have to use the device configuration so with normal winding in principle yes, but not with two switches alone not with two switches alone you have to change the input stage because if dot is positive after sometime I have to make dot as negative I have one source. So, what will you do here it can be done think it over R C P S the question is fine the what is method to reduce I think dead time of switches in circuit method to use a dead time of switches I think I expect this what you mean if I have a input as a voltage source something similar happens in the so on. So, if I these two switches are complementary only one of them can be on if both are on yes there could be a short circuit. So, in other words having turned off having turned off T 1 T 1 is turned off T 2 should be turned on after sometime this is known as a dead time this depends on the type of device. Once upon a time I have to use a dead time when we are using d j t this could be of the order of 10 microsecond. Now, if I am using I G B T of a much higher rating then it could be of the order of the type of device it depends on the type of device that is in a multi stage is boosting if you are increasing the number of stages the efficiency becomes efficiency reduces what is the maximum number of stages we can apply for boosting the voltage level. I think this is what he means I have a d c source boost 1 boost 2 and boost n and finally, he gets a higher voltage this is V d c this could be 7 to 7 7 times V d c this is a given by V d c this could be again whatever getting boosted finally, he will get the required voltage. Now, one is not only the efficiency one is the number of component count as the component count increases reliability may come down reliability one fails everything gone there reliability comes down space size cost and efficiency. Now, instead you can use a transformer and depending upon the power levels choose a suitable power configuration. Now, which one will you use how many number of stages frankly I have no answer you what is the efficiency that you can get here what is the efficiency that you efficiency are not the only issue the component count reliability cost I would like to refrain from answering this question of how many stages I would encourage you to try yourself it is not only the efficiency there are other issues as well as as well size should be as small as as low as small as possible hello sir which range of inductor gets quickly saturated what does it what do you mean by which range of inductor gets quickly saturated g c t s which range of inductor gets quickly saturated what does it mean what do you how to select the range of inductor how to select the range of inductor what do you mean by how to select the range means what in converter, what do you want to ask? So, whether low range or high range. Value of L you mean? Yes sir, value of L. Yes sir. So, it all depends on the ripple. You have to say ripple on the current and the flux and the voltage, that will determine the value of C and value of L. These are the design parameters, sir. These are the design parameters, I will say 5 percent ripple. So, depending upon the voltage and current gets fixed, D i ripple in the current, that is allowed determines the value of L, delta i ripple. Current ripple will determine the value of L and again of course, please in flyback value of L 1 will determine the peak current. I p is given by V dc divided by L 1 into t. This is what we derived in expression. Manipal, what do you? Sir, about the switching frequency I already answered sir. I think few minutes I spent when I was talking about hard switching. Switching frequency to adopted. And as it got bearing on the transform on core material selection in this. Sir, say may be around depending upon depends on the power level sir, up to say 100 kilo hertz also you can go for ferrite core. As the frequency increases, it all depends on what sort of a what type of switching that you apply. Are you doing hard switching or soft switching? I just discussed about hard switching. We will see what happens in short switching, what exactly you win by short switching and can or is it possible to increase the switching frequency as well we will see in the next lecture. But often if you ask me what is the switching frequency that I can choose? Sir, I am bit hesitant to answer this question. So, the simple reason that if I say some figure it may be possible, it may not be possible. It all depends on how well you going to integrate. See if you keep the devices I told that is how I told you when you assemble the converter or when you want to operate when you are working on a table everything works fine. As you try to integrate things are not as straight forward whatever the due to EMI your base drive may get effected. It will show some spikes here and there. So, product integration is not very simple. So, my best my my my advice is you start from a low frequency say a conservative figure of 10 kilo hertz. If you can work at 10 kilo hertz gain sufficient confidence then gone increasing the frequency. All of a sudden if you say start with a 25 kilo hertz it may not be you may not be able to design and integrate it. Let us start with a conservative figure of 10 kilo hertz find your product one cannot be say in a sense your prototype one cannot be the final product. There will be three four stages. So, first you gain enough confidence by fabricating the first prototype start with a conservative figure of 10 to 15 kilo hertz not more than that then slowly slowly you increase. I cannot I can tell you from my experience we have achieved as high as 75 kilo hertz. Now, if I say operate at 75 kilo hertz and you go back to your lab saying that once serve from IITS told me to operate at 75 kilo hertz I will operate it you may not be. So, start with a very conservative even we have started sometime back when we started we started with the 10 kilo hertz also we learnt from our experience. So, that is my sincere request to you please I do not want to throw some figure it just does not make sense. Sir, you want to design a control circuit sir. So, what is the procedure sir because power circuit it is not much difficult to understand sir, but how to design a control circuit easy there in this particular topic or what is and one more doubt regarding control circuit sir. Suppose, if we design for a let us say power circuit can we use the same control circuit for higher rating the power circuits also sir. Sir, I would think that designing a control circuit is simpler than power circuit anyway what exactly even by control circuit is the driver circuit or overall closed loop control. Yes sir, to turn on the switches MOSFET and IGBT what to turn on and turn on the switches. You need to have a control circuit and gate drive circuit I am talking about this is a gate drive circuit BJT gate the philosophy that I used to sir fine I can understand your problem my I will tell you two steps one is the philosophy that we use to design the gate drive circuit for a BJT I would explain almost the same philosophy for MOSFET and IGBT. I will give you I will tell you the design procedure I will tell you the design procedure for a gate drive circuit for MOSFET as well as IGBT I will tell you they are not very difficult making a digital and analog circuit work is not very difficult I am telling you you can incorporate all the protections my worry is here you may be able to may be able to operate it or you may be able to work on a breadboard or on a simple PCB when you try to integrate it you may face some problems my sincere request there are dedicated chips available gate drive circuits use them wherein they are worried about taking care of all this EMI issues interference issues. So, main problem comes in operating DC to DC converters as not the device or not something like that the external interference to the control circuit if your gate drive is very stable take it from me you have achieved 90 percent success 10 percent only remains the only problem is your gate drive works sufficiently reasonably at low frequencies as to increase because of the external interference gate drive starts misbehaving if there are no hardened parts so it comes through experience the integration. So, best thing is I encourage you need to know the philosophy you need to know the philosophy what is being done there are people have spent lot of time on designing and addressing all these issues why not use that same expertise. I would encourage you to use that gate drive circuits that are available I can suggest the suppliers please I do not use the various vendors you can get them and try to operate and I will assure you that it will work, but then you need to know what what may be inside that is all that is all close look control I will try to explain see close to the structure will remain the same structure will remain the same only the gains will change gains will change and the question to another question that you are asked is I have designed the gate drive circuit control circuit for low power what will happen for high power. See as the power rating increases your device also will change now if there is a MOSFET and gate to source voltage is the same in principle you may be able to use the same thing you may be able to use the same thing. So, it all one has to first choose the devices what sort of a control circuit you want from based on the parameters you have to choose the control is not very difficult sir, but one we have to take first step then things go on one or twice device may fail let them fail we have to learn from a failure that is my sincere request to you and my suggestion I would encourage you to instead of using discrete components to make a gate drive circuit I will repeat instead of using discrete components to make a gate drive circuit there are dedicated chips are available s x which is s x circuits are available use them, but understand or know the know the principle or whatever things are there you may use it as a black box, but you need to know what exactly is there inside the black box that is all yeah. Sir where can we get this control circuit sir and suppose if suppose if I have my own problem related to the power circuit suppose I have own design my own control power circuit. So, will be dedicated if I have say my problem will the vendors can design the control circuit accordingly sir. Which vendors which vendor no no sorry sir I did not get your question you say that again. I think you are going to suggest some vendors to move. No no no these are dedicated chips dedicated chips are available dedicated chips are available already existing power existing power circuits. Existing gate drivers gate drivers see gate drivers I said if you use say for IPM you do not need to use external gate driver IPM you do not need to use external everything is built in now for this also there is no I need to have external gate drive, but then dedicated dedicated circuits are available I will bring it those dedicated circuits may be in the next class not today on Monday I will bring and I will show you what I mean. I can directly use them to drive this or use transistors, diodes, resistors and other components capacitors to drive them. So, gate drive is very important if a gate drive is stable I am telling you you have crossed a major hurdle and in principle do not get intimidated take a first step and start assembling things may not work things will not work, but they have to work they have to work take it from me. How will you ensure a symmetrical how will you ensure a symmetrical flux in the core? No no no I will you need to ensure you need to ensure number of turns is the same almost same device drops see small changes here and there should not affect see even if there is a DC flux there is there are issues there are ways to address them there are ways to address them flux working can be addressed it is there in half bridge by putting a capacitor all those things can be addressed they are not a problem at all are you with me if there is a DC flux or the flux the so called flux working can be addressed literature is available I just told you the limitations I may not have suggested the solutions solutions are there see the problem here today I see major the big thing is to find out the problem having found the problem solution can be found out not a issue not a issue. Super what are you? Sir regarding that dead time of the circuit is it any current flowing through the circuit hello. Dead time of the circuit any. Sir please go to the slide sir the practical aspects of the push pull amplifier. What is the push pull? Dead time I mean T 1 and T 2 closing simultaneously or T 1 is not turned off T 2 is on that is what I am trying to tell. Since we have to close the T 1 sorry we have to open the T 1. Open T 1 wait for some time. Sir the case when the both switches T 1 and T 2 will be on. If they are on no no no no no if they are on you are in trouble I am not telling you that when they are on that is known as a dead time no no if both are on you are in trouble current will be limited by the internal resistance that is very small because these are tightly coupled I need to you operate a very high frequency and ensure that T 1 and there is both of them do not they both of them are not on and see these are the timings D T by 2 to T by 2 both T 1 and T 2 are off. At that time these two diodes D 2 starts conducting this I D 2 is the correct direction for the magnetizing current because magnetizing current enter the dot here. So, magnetizing current has to enter the dot N 2 2. So, magnetizing current will flow through magnetizing current will flow through entire magnetizing current through will flow through D 2, but then the remaining I L by 2 will flow through D 1 and the other I L by 2 will flow through D 2 entire magnetizing current has to flow through D 2 because that magnetizing current had had enter the dot. So, magnetizing current has to enter the dot here. So, at that time both T 1 and T 2 are off now it is so happened that D 2 as you are switching for duty cycle has increased and accidentally if you close T 1 and T 2 you are in trouble current is limited by the internal resistance R is that ok see I do not sir it is possible to use a full bridge circuit in the primary coil of a push pull converter that is what I meant yes. See this is the push pull the so called the original push pull converter see the question is instead of using a center tap and two devices in this fashion can we use only one winding and modify the power circuit answer is yes. That is what I told you in the beginning answer is yes, but I did not name the type of power circuit. So, in the literature if you see the conventional push pull this what it is now instead of having two switches and a center tap winding I can have I can have a bridge and connect them accordingly now you require four switches now you require a four switches of course you can further modify it go back and check can I use half bridge it is possible, but then now you require four switches whether now you decide what you want a center tap two switches or four switches see that you need to decide I am just telling you the various possibilities the final decision is going to be yours in principle yes it is possible I can still call this as a push pull I do not know, but then original push pull if you see the text books or something they name push pull as center tap with two switches whatever people have modified it this is one such modification.