 Good morning. In the last class, we discussed choke converter and fly back converter, buck converter, boost, buck boost and choke converter. There is no transformer that is being used, a high frequency transformer is not being used. Hence, the name non-isolated DC to DC converter. Since, there is no transformer, they belong to a class of converters, what are known as non-isolated DC to DC power converters. What are the limitations of those converters? In boost and buck boost, theoretically, though it is possible to get V naught as I has infinity in a non-ideal converter as D increases beyond a certain value, instead of V naught increasing, V naught starts decreasing. So, the maximum ratio of output voltage to the input voltage could be of the order of 7 to 10 provided you have a very high Q inductor. And how do you get that high Q inductor using what is known as the Litz wire, using Litz wire to wind the inductor. In a buck converter, if the output voltage is too low, then we have to reduce the value of D and it will so happen that converter may operate in discontinuous conduction mode. And if it operates in discontinuous mode, V naught is not equal to D into V input. It is higher than D into V DC or V input. So, one it is possible to use all non-isolated converters provided the ratio of input to output is either it is not too low or is not too high. What if the ratio is too high? Then we have to use a transformer. By the way, I am using a transformer in DC to DC conversion. Only thing is I am not allowing the transformer to saturate and one such power supply that we discussed was a fly back converter, which is generally suitable for low power application. Why it is low power application? Yesterday we derived an expression for the output power. Then we derived an expression for the peak current that is lost to flow through the switch. These two factors may limit and therefore, the size of the transformer may limit or may not allow us to increase the power rating beyond a certain value in fly back converter. What is the operation of fly back converter? I will just repeat it. One such configuration is shown here. I will close the switch. Current enters the dot. There are two coils here which are mutually coupled. So, if current enters the dot, the current direction in the secondary is to leave the dot. But because of this diode, current cannot leave the dot. So, therefore, though I am using a transformer, only one coil is carrying current at a time. When I close switch S or in other words when I am energizing the primary, there is no current in the secondary. Therefore, current drawn from the source is equal to the magnetizing current itself. So, voltage V dc is applied to the primary inductance, primary self inductance L 1, which is equal to leakage plus the magnetizing. Since V dc is held in one switching cycle, if V dc remains constant, current flowing through the inductor that is nothing but the magnetizing current will increase linearly. If the magnetizing current increases linearly, flux also will increase linearly provided we are working in the linear region of BH curve. Here after some time, after D into T, I operate S. When I am opening S, there was just prior to opening S, there was some flux in the core. Flux in the core must be continuous. We cannot allow the flux to fall. If that happens, a large voltage will be induced and that may damage the switch. And we found that when only one coil is carrying current at a time in a mutually coupled coil, if current enters the dot in one coil, in other coil also current should enter the dot. Yes, current can enter the dot here. So, what will happen? Whatever the stored energy here is transferred to the secondary and it will charge the output capacitor. Using the volt second per turn, what does it give? Volt second per turn, we said V is equal to N d phi by dT. So, therefore, d phi is V dT divided by N. Volt second per turn is V dT divided by second per turn that is nothing but d phi. So, d phi in the rate of change of sorry d phi in primary coil should be equal to d phi in other coil as well for steady state operation using this condition. So, V dC into D into T, the time for which the switch is closed divided by N 1 should be equal to the voltage applied to the second coil that is V naught for 1 minus D into T duration and it is applied to N 2 turns. So, therefore, I can derive a transfer function V naught is given by N 2 by N 1 V dC d divided by 1 minus D. So, this is nothing but this is d divided by 1 minus D is nothing but a buck boost transfer function. I have in addition I have N 2 by N 1. So, generally I told you D is kept around 0.5 that is because though I am using a transformer only one coil is being used at a time. So, therefore, if I see the BH characteristics my operation is in the first quadrant itself BH. So, increase in flux decrease in flux. So, this is for discontinuous conduction sorry this is for discontinuous mode of conduction. What exactly mean by discontinuous conduction? It implies that just prior to closing the switch S for the second time flux has become 0. In other words current flowing in the secondary winding has become 0. So, that is known as a discontinuous conduction and if there is a finite current in the secondary winding when you just prior to closing the switch in the primary that is known as a continuous conduction under that condition I will have a BH loop something like this. So, if I operate the fly back converter in continuous mode there could be a chance that the convert or transformer may get saturated when that is if there is I told you that it suppose there is a sudden change in the load this is the reference voltage output voltage has to be regulated at certain value. I am sensing the actual voltage I told you there is a sudden change increase in load V naught as V naught tries to fall because controller does not know action on duty cycle as in taken place V naught starts decreasing error increases and it will so happen that it is so happen that D may increase to a high value and that high value of D may saturate the transformer. So, in order to eliminate this possibility generally this converters are operated in sorry discontinuous conduction mode operated in discontinuous mode generally a small air gap is provided during the fabrication of a transformer. So, therefore, the chances of saturating the transformers are reduced, but then the moment I use an air gap there is going to be a leakage flux. So, if there is a leakage flux when I open the switch I need to provide a path for the energy associated or to dissipate the energy associated with the leakage inductance. So, I have to modify the power circuit configuration those power circuit configuration I did discuss yesterday. So, another advantage of fly back is multiple outputs are possible and this closed loop operation is a must closed loop operation is a must because why is closed loop operation is a must assume that there is no load at all for some reasons there is a open circuit is created at the output the controller that does not know that there is an open circuit there is an open circuit. So, there is no closed loop at all the input does not know what is happening at the output in the input side you are going on storing the energy and dumping to the capacitor and there is no load here. So, the capacitor voltage will go on increasing and after a point your diode will fail capacitor may the voltage rate the voltage across the capacitor may exceed this rating and then for capacitor may get damage or when the switch is open and when the diode is conducting there is an equivalent voltage will be induced in the secondary and that voltage plus the supply voltage appears across the switch S and that voltage and switch S has to block that voltage. So, that may exceed the voltage rating. So, in order to protect a closed loop operation is a must. So, what is the voltage rating of the diode or the switch S see diode D 2 has to block when the switch is closed when switch is conducting there is no current flowing and diode D 2 has to block that what is the voltage that it has to block when I close S voltage induced in this sorry voltage applied to this coil is V dc current is increasing linearly therefore, flux is also increasing linearly there is a change in flux in the core. So, therefore, voltage is induced in the secondary and this voltage is voltage induced in the secondary with dot as positive voltage induced in the secondary with dot as positive why dot as positive in the primary when I close the switch dot is connected to the positive of the battery therefore, dot is in the secondary is also positive. So, if I multiple winding all dots or of the same polarity. So, the primary it is positive even in the secondary it is positive. So, equivalent circuit is something like this this V dc into N turns ratio V dc is being applied to N 1 turns the equivalent voltage induced in the secondary is how much that you need to calculate. So, this is V naught. So, voltage across the diode is V dc to N plus V naught this voltage diode is V naught. Voltage diode has to block similarly when you open the switch in the secondary diode starts conducting voltage applied to the secondary voltage applied to the secondary with dot was positive when I close the switch now when I open dot is negative now. So, this polarity is positive voltage applied to the secondary with the dot as negative is V naught that coil as N 2 turns the equivalent voltage in the primary will be multiplied by the turns ratio. Now, the turns ratio is the inverse of whatever that you have taken in the first case plus V dc is a voltage that switch has to block. What is the peak current rating of S? Peak current rating of S is the volt the current equation is I 1 is equal to V dc V dc is equal to L 1 d I 1 by d t. So, rate of change of primary current is given by V dc by L 1 and this is switch is on for d into t. So, therefore you can calculate I peak provided we are operating in discontinuous mode in the sense when I close this just prior to closing the switch S there was no current flowing in the secondary. So, current starts from 0 this is the value of I p this is the slope d I by d t and this is d into t. So, you calculate what is the value of I p if we are operating in continuous mode or just prior to closing the switch there was some current flowing in the secondary the equivalent current now I need to take care of I need to do the magnitude of the current that is flowing in the winding as well as the number of turns this is the equivalent current flowing that was flowing in the secondary just prior to closing the switch and from there it starts increasing slope is the same. So, when I open the switch the peak current into the turns ratio that this peak current was flowing in N 1 turns it gets transferred to the secondary with having N 2 turns we calculate what exactly. So, this current starts flowing through a diode. So, one has to choose the device current accordingly forward converter I am not going to discuss an ideal forward converter it does not make sense a non ideal forward converter ideal forward converter is the one which as mu r is equal to infinity the moment I say when mu r is equal to infinity you do not require any ampere turns to establish the flux in the core. So, that is all the difference then we will see I will tell you what modification do we need to make or which part is absent if I consider the ideal ideal forward converter in non ideal forward converter the transomer is non ideal non ideal in the sense mu r is finite. So, source has to supply some ampere turns to establish the flux in the core. So, magnetizing current is finite therefore, when I 2 is equal to 0 when there is no current in the secondary primary will not be 0 when I energize it that current is nothing but the magnetizing current. Now, see the power circuit configuration do not worry about this winding as of now we will see some time later right now you worry about N 1 and N 2 do not worry about N 3 why N 3 is required do not ask me now we will see N 1 and N 2 primary and secondary. Let us see we will close the switch S for d into T a duration current enter the dot. So, right direction is in the mutually coupled coil is here is current to leave the dot yes current can leave see the diode connection current can leave in fly back it is not possible current enter the dot current should leave the dot because of this d 2 current cannot leave the dot only 1 winding is carrying current at a time that is the magnetizing current itself here current enter the dot current can leave the dot current can leave the dot here see the diode connection. If diode If diode starts conducting, what will happen to D f? D f is reverse bias. Why? Dot is positive. Therefore, this dot is also positive. So, this polarity, sorry, this terminal is negative. So, voltage, what is the voltage induced in the secondary? The voltage applied to the primary is V dc. So, there will be an equivalent voltage induced in the secondary with the dot as positive. Divided is on, diode can conduct. So, this terminal gets connected to this point. In other words, this dot or entire voltage induced in the secondary is applied to or appears across D f. I will repeat, what is the voltage that is applied, whatever the voltage induced in the secondary is applied across D f. Dot is positive. So, D f is reverse biased. D 2 is reverse biased. D 2 is conducting and see the remaining part. There is l, c and r. Where did we see this? See, if I draw the equivalent circuit, it looks something like this in the secondary. V dc into n, voltage induced with the dot as the secondary. Here, second diode, D f is open. This is the output stage. A load is connected across this. Where did we see this circuit before? There is a source voltage. Diode or a switch does not matter. It is applied to L f and C f. Looks like this is something similar to a buck converter. The equivalent circuit when the buck converter, when the switch is on in buck converter is same as that of forward converter when I close the main switch. There, the input voltage is V dc. Here, it is V dc into n. Now, let us see what happens. Current enters the dot, current leaves the dot, current should leave the dot. Here also, current cannot leave the dot because of this diode. Current cannot leave the dot in the tertiary winding. There is no current flowing in the tertiary in n3 because of this diode. Only two windings are, when the primary is energized, there is a equivalent current that is flowing. When I close, energize the primary current that is flowing in the primary coil is magnetizing current plus the equivalent secondary current. See, you have to choose the cross sectional area of the conductor accordingly. In fly back converter, though I am using it as a transformer, only one coil is carrying current at a time. So, this is only the magnetizing current. The peak value is I p. That I p depends on the self inductance, supply voltage, the value of the duty cycle into the switching frequency d max. Whereas, in the forward converter, when I close the switch, in addition to the magnetizing current, there is an equivalent secondary current is flowing through the primary coil. That is something similar to the transform action. So, I 1 is I m plus I 2 prime. I 1 is equal to I m, the magnetizing current plus I 2 prime. No current in the tertiary because current enters the dot, current should leave the dot, current should leave the dot. It is not possible. After sometime, I would into t open s. What will happen? What happened in fly back? Flux in the core must be continuous. Flux in the core must be continuous. Even here also, flux in the core must be continuous. So, we have to take care of I m and we have to worry about I 2 as well. I 2, the current that is flowing in the secondary. The moment I open the switch, d phi by d t is now negative. When I close the switch flux in the core increases, when I open the switch flux has to fall. So, d phi by d t is negative. So, all dots are now negative, voltage induced. So, dot is negative diode d 2 cannot conduct. But there was some current, there was some current flowing in the inductor. I l was increasing linearly. In this equivalent circuit, if you see, this circuit v d c by v d c into n is applied to L f and C f. v naught is regulated. So, in one switching cycle, this voltage is expected to remain constant. A constant voltage is being applied to L f. Therefore, I f increases linearly. Something like this is the current I f. Why am I starting from some finite value? I will tell you. As of now, I am starting from a finite value. It reaches some peak. At this point, switch is opened. So, d into t. So, inductor current must be continuous. So, what will happen? That current will flow through d f. The equivalent circuit in the secondary is something like this. This is v naught. So, capacitor voltage, the output voltage is applied across L f with the reverse polarity. When switch is open, this terminal was positive. Now, when the switch is opened, this polarity is now negative. So, if the current was increasing, when I close the switch, the current will fall. Current will fall. This is t. Now, just see these two equivalent circuits. So, these two equivalent circuits I think we got while analyzing the buck converter. Buck converter input voltage we had taken as v d c. Here, the input voltage looks like, why it looks like it is v d c into n. So, the same transfer function should hold good. So, the transfer function is v naught is equal to n 2 by n 1 into v d c divided by d. Sorry, into d. v naught v d c into d. So, this part is nothing but the buck converter transfer function v d c into d. n 2 by n 1 is the Tons ratio. So, this buck converter is also known as the isolated, sorry, this forward converter is also known as the isolated buck converter, isolated buck. Transfer function is same as that of the buck converter. In addition, I have n 2 by n 1. This isolation is provided by the transformer. So, inductor current, there was some finite current when I had just prior to closing, just prior to closing the switch for the second time in the primary. So, when I close the switch again, this current will increase from here at steady state. So, at steady state these two peaks are same. So, we have derived the transfer function. We have taken care of the secondary, the inductor current that was flowing, but we have not taken care of the magnetizing current. What happens to the flux when I open n 1? What happens to the flux when I open the switch S? Flux must be continuous. It was increasing when I close the switch S, now it should decrease. We have to provide some path for the magnetizing current to flow and repeat, we have to provide some path for the magnetizing current to flow. By the way, what did we do in the fly back converter? When I close the switch, the magnetizing current increases linearly, therefore, the flux. How do we provide the path for the magnetizing current? We used, we did not allow the secondary coil to carry the current when I close the switch and we connected the diode in the secondary in such a way that it does not allow the flow of current when the switch is closed and it allows the flow of current when the switch is opened. So, therefore, when if the current enters the dot, when I open the switch, the right direction of the current to flow in some other coil is to enter the dot as well. So, what do I need to do? With the only the two winding, it is not possible. The two winding is not possible. Current enters the dot here, current leaves the dot. Now, when I open the switch, right direction for the magnetizing current to flow is, if I enter the dot, it can enter the dot in some other winding. It can enter the dot or the right direction of the current to the magnetizing current to flow in some other coil is to enter the dot. So, we need to have an additional winding, third winding. I have connected across the supply here. I do not need to connect it across the supply. I can do something like this, something similar to the fly back. What did we do? Redot, this is N 1. So, N 1 is to N 2. Current enters the dot, current leaves the dot. So, current in this coil is I 1 plus, sorry I m plus I 2 prime. When I open the switch, the right direction for the magnetizing current is to enter the dot in some other coil. So, what I will do is, I will have another coil. Current enters the dot, current should enter the dot here with this. Either I will have a capacitor or a source here. This is the direction of current. So, I need to have another winding which is something. So, this is nothing but I can have a separate capacitor here to charge. It will charge 2 may be, why may be? It is N 3, N 3 by N 1 V dc d divided by 1 minus d and this is N 2 by N 1 V dc d divided by 1 minus d and this is N 2 by N 1 V dc into d. Forward, fly back. Instead of using a separate capacitor and therefore, having a separate voltage, what you can do is, might as well connect this winding across the same source. Third winding, a diode here, current enters the dot, current can enter the dot. So, this is the direction of the current for the magnetizing current. So, magnetizing current flows in N 3, N 2 is the load current equivalent secondary current. I 1 carries when the switch is closed, equivalent secondary plus the magnetizing current. So, I 3 has to carry only the peak value of current in N 3 is the equivalent peak value of the current of magnetizing current. See the current waveform in I 1, if I have to draw, it looks something like this. Current waveform primary current has two components. One is a flux component, this is I m magnetizing. Now, there is an equivalent secondary current. Now, equivalent secondary current what we had assumed, because we have L f there and we have assumed a continuous conduction. This is the secondary current that we had assumed, current through L f. This is the current through L f something like this. So, when I close the switch, this current is starts from this value, some finite value. So, there is an equivalent current flowing in the primary. Whatever the current that is flowing here, equivalent current has to come from the primary. So, this value is finite. So, definitely I 2 prime means also finite. I 2 at t is equal to 0 is finite. Therefore, I 2 prime in the primary when t is equal to 0 is finite, not the same value into the turns ratio. So, this was the current that is flowing in the secondary or through L f. This is I L. So, the equivalent current is something like this secondary. This is I m, this is the inductor current, this is the magnetizing current, I m, this is I 2 prime. Some of these two currents is the primary current, some of these two. At any given time, I have to add this and the sum of these two, sorry, this is I L magnetizing current, I m is the magnetizing current. This is the equivalent secondary current. At this point, you are turning off the switch d into t. So, the primary current is something like this sum of these two. This is I 2 prime and this is I 1. So, the tertiary current through the tertiary is now or I 3 is look something like this. Somewhere here it is the next cycle starts. Why it is so? I will explain to you. I had assumed almost a constant current or constant current in the inductor L f, but then magnetizing current becomes 0 much before the switch is closed for the second time. The reason for this I will explain to you. So, this is the equivalent peak current into the turns ratio here. Terns ratio is number of turns in the tertiary to number of turns in the secondary. Please while determine the current ratio, current peak value of the current and equivalent current you have to take care of the number of turns in those two coils whether it is primary secondary primary tertiary. I will not continue now. I will take up few questions. Feel free if you have any questions on the topic that I have covered. Why B H curve is a closed curve with same B values in the beginning and end of the cycle? That is the steady state. The question is why the B H curve is closed curve with same B value in the beginning and end of the cycle. Temporarily I will assume the residual flux in the core is 0. So, when I and I told you that most we generally operate in discontinuous mode of operation. So, when I close the switch magnetizing current starts from 0. Therefore, flux also starts from 0. It increases linearly flux increases linearly. It reaches a peak value phi p correction corresponding to I peak in the primary. When I close the switch, when I open the switch flux in the core starts falling and it becomes 0 much before the switch is closed for the second time provided if I operate in discontinuous mode. And this I am again I am closing for the second time. So, this is T this is D into T for discontinuous. For continuous flux, there is some value of flux increases and decreases at steady state these two values are same. This is the steady state is not it? This is for continuous conduction. This is for discontinuous conduction. This is I this is T. So, same as I and flux both are same. Now, you plot B H you will get the same thing no B H square H as I increases H increases therefore, flux also increases almost linearly. But then when I decrease it, it will not take the same path and it will come in that is all. There is a very good question where is the question is will the transformer saturate if I operate the if I operate the fly back converter in continuous mode. I have no answer it may saturate it may not saturate it depends on your transformer. Have you provided an air gap there? When I close the switch current increases linearly if the if it is a continuous conduction it increases in this fashion. What is the value of H corresponding H here and therefore, in the B H curve? Will it saturate or no? If there is a sufficient air gap flux will not increase because of the air gap reluctance flux will not increase to a very high value. It will be much less than the saturation value. So, it is entirely depends on the transformer design if there is a sufficient air gap it may not. If there is no air gap it may what is the condition needs that to be satisfied for D 3 to conduct as V D C is reverse biasing it V D C is not reverse biasing it. The question is what is the condition that needs to be satisfied for D 3 to conduct as V D C is reverse biasing it? Are you talking about forward converter? Now, I will ask you the same question why did you accept my argument that D 2 in fly back will conduct when the switch is open? See the fly back converter see the polarity of the output voltage V naught polarity of the output voltage V naught with this as positive what happens to the diode connection? Whether the diode is forward biasing or reverse biased? You just cannot comment on the biasing condition of the diode just by looking at the battery you need to see the voltage induced in the coil as well. What is the equivalent circuit here? When I open the switch I told you when I close the switch dot is positive when I open the switch dot becomes negative d phi by dt is negative induced voltage in the secondary induced voltage in the secondary now flux is decreasing therefore, phi is negative now. So, equivalent circuit is something like this equivalent circuit is something like this this is V naught this is this is this is the equivalent circuit you accepted this then why why aren't you accepting the operation of the fly back converter? What is the difference what is the difference? I told you that I do not have to connect it to V d c I can have a separate a capacitor and that capacitor gets charged by the energy stored in the magnetizing inductance there is no condition at all it will this is N 1 and N 3 will form a fly back N 1 and N 3 is similar to fly back there is no difference at all you do not need to connect it to V d c the advantages of connecting to V d c I will tell you you do not have to connect it to V d c I can have a separate capacitor and that capacitor gets charged corresponding to a value N 3 by N 1 into V d c into d divided by 1 minus d. Now, instead of connecting to a capacitor I have connected to a d c source that is all assume that tell me now assume that there is no path there is no path for the current to flow what will happen now assume that diodes your connection of diodes your reverse or whatever that is happened there is no path for the flux to flow. So, d phi by d t is going to flux in the core is going to collapse is not it when I open the switch when I close the switch flux in the core increases it has attained some value now we are opening the switch there is no path so what will happen to d phi by d t now there is no path what will happen to the d phi by d t or what will happen to the current that is flowing through the so called the magnetizing inductance there is no current there is no path. So, that what will happen to d l d f by d t the voltage induced hello N i t c why fly back convert is not popular about 200 watts I had calculated I had derived an expression for the peak current I p in terms of supply voltage and the primary self inductance L 1 and the primary self and I had also derived an expression for I p in terms of p in that is approximately equal to p naught. Now, you find out what will happen to the transformer what will happen to the peak current that will that has to flow through the switch as the power rating increases the question is how fly back convert operates in boost mode v naught is equal to N 2 by N 1 v d c d divided by 1 minus d yes I told you maximum values d is equal to less than 0.5, but then N 2 by N 1 I have if d is less than 0.5 it is of course it is the transfer function is a buck boost converter, but then output voltage is if there is d is less than 0.5 ratio is less than 1, but then I have N 2 by N 1 to take care of whatever the voltage that you want at the output. Sir whether N 1 and N 3 are electrically as well as mutually connected the question is whether N 1 and N 3 are electrically as well as mutually connected. Now, I told you there is no need for or need to connect N 3 to v d c why are we connecting N 3 to v d c I will discuss it after during my next lecture, but then we need to have N 3 to provide a path for the magnetizing flux whether you connected to v d c or you connected to a separate capacitor it does not matter. There are certain advantages of connecting to v d c we will see in my next lecture. How can we fix the dot transformer one minute how can we fix the dot in transformer winding is it in any relation with the transformer winding rotation what is transformer winding rotation G C T S G C T S I do not know what exactly you mean by transformer winding rotation how can we fix the dot I think yesterday around 10 30 I did explain to you how to place a dot though this topic is covered while discussing mutual inductance in the first level of circuit course. I did explain to you how to place a dot generally the winder this winder in the shop floor you places a dot there if you do not know the winding arrangement it is not possible to determine or it is not possible to place the dots the only the winder can determine or place the dots having wound the coil and if you do not know the sense of winding it is not possible to place the dots. And I did explain to you yesterday what is the what is the range of q can get what is the range of q to q you can get v i v t v what is this question what is the range of q to get high gain in buck boost converter higher the q the better it is higher the q the better it is q factor is the ratio of inductance to the inherent winding resistance higher the ratio the better it is that is the reason I told you to use to get a higher q you have to use a lids wire or a series of a large number of strands instead of having a single conductor thereby reduce the effective value of r. So, what is that f switch is suddenly open and in transient state what will be the roll of tertiary coil with respect to the primary coil switch is suddenly open that is in transient state of course when you are suddenly opening when you are opening the switch is a transient state is not a steady state what will be the roll of tertiary coil with respect to the primary coil when I open the coil when the open the switch s I told you flux must be continuous see here I will give a simple circuit v d c yeah this should answer the question the very first question that I answered when I close the switch what happens current in this inductor increases linearly what happens when I open the switch what happens if I open the switch if I have a very fast diode I need to have a very fast diode t on of this diode should be very small as I open the switch there should be the diode should turn on and this and therefore whatever the current that was flowing through the inductor continues to flow through a diode if the diode is slow diode the so called rectifier diode and if you are using a MOS here you may see a spike across the device because you are opening the switch, but then diode is not responding it taking its own time to turn on so current may be abruptly or trying to reduce so you may see a spike across the switch to avoid that you need to have a very fast diode t on of this should be as small as possible you need to have a very fast diode you cannot use a rectifier diode now there was a question that was asked in a forward converter what should be the condition see v i t v i t v the question is what is the condition that needs to be satisfied for the d 3 to conducts as v d c is reverse biasing it so baisab you tell me here in this circuit come back when I open the switch whatever the current that was flowing through it will flow through the diode, diode cathode is connected to the positive of the supply diode cathode is connected the positive of the supply you need to see where the anode is anode is connected across the coil it is the voltage induced in the coil that will forward bias the diode it is voltage induced in the coil that will forward bias the diode that is all if this is positive V dc. So, this should be V dc plus 1.5 volts for a power diode that is all if this is positive V dc the anode potential should be V dc plus 1.5 volts. So, this voltage is clamped at this voltage is clamped at V dc plus 1.5 volts when the diode is conducting. And if the diode is does not conduct a huge spike because of abrupt breaking of inductor current will appear across the switch and that will damage. So, this voltage sorry the potential of this point is clamped at V dc plus 1.5 volts. So, that should answer the question first as well as as well as when I open the switch suddenly what should happen to the tertiary coil. I need to have a very fast diode if there is a slow diode you make an experiment you conduct the experiment in the lab and see a put of very slow rectifier diode some sort of I n 500 sorry I n 4001 small a slow rectifier diode and you try to open the switch and you see the voltage waveform across the switch you may see a spike that spike is because diode is not turning on. The magnetizing current flows all the time in the third winding or only when the switch is open or that you need to answer that in the magnetizing current flows all the time. Didn't I tell you when I close the switch primary current in the primary enters the dot. So, what is the right direction of current that has to flow in other coils current leaving the dot. So, there is no current flowing in the tertiary winding there is no current flowing in the fly back transformer when switch is closed and whether the current continues to flow during the entire off period of the switch in the tertiary winding we will see in continuous conduction mode of forward converter there will be steady state core loss what is the allowable rate of change of core loss. So, a b v c your assumption itself is wrong we cannot operate the fly forward converter in continuous conduction mode why I will discuss now after the during my next lecture it has to be flux has to become 0 prior to closing the switch for the next time in the existing this one otherwise we may the transformer may saturate core loss. Yes, if you have a core if the flux is changing sorry if you have if there is a core with the finite sorry I will repeat you have a core it has a hysteresis loop the area is finite and if the flux is changing there is core loss you cannot do anything about it. Yes, MITK what is your question? Hello sir, in the normal buck bus converter the voltage gain will be order of around maximum around we can say practical limit around 5 7 times what is the voltage ratio we can obtain in this fly back converter and the power handling capacity means I would like to highlight because buck bus connector may be efficient at the range of 100 watts level or at the kilowatt level or other higher kilowatt level can it be applied this principle your fly back converter what you are speaking what is the voltage gain maximum we can achieve and the power handling capacity this principle is applicable in the watts range or in the can we apply go power in the kilowatt power handling capacity or a much higher level. No sir, I told you in the beginning fly back converter is popular can be used only if the power rating is less than less than 150 watts less than 150 watts if the power rating from 150 to around 500 to 600 watts then you go for forward as the power rating increases circuit configurations do change we will slowly we will cover this what is the what are the possible circuit configurations as the power rating increases we have to do a forward converter after what till around 1 kilowatt range there is what is known as the push pull converter then there is a half bridge full bridge there are whole lot of converters for low power application less than 150 watts or so you go for fly back and if there is a significant voltage ratio you go for fly back that is all low power only less than 150 forward is up the order of 500 to 700 above 700 what is known as the push pull we will see we will discuss why did I use fly back for low power application I want around 150 watts but then there is a significant difference between the output voltage ratios then yes I need to boost the voltages see what I am saying is let me do the inverters as well let us discuss the final application then what we will do we will see see if your application requires 230 volts AC either I will whatever I do is I will invert a low voltage DC it is a 12 volt DC to 12 volts AC then I will use a step of transformer at that time I require a 50 hertz step of transformer 50 hertz another way is by the way 50 hertz run the moment I say 50 hertz magnetics they are heavy weight size so you may not be able to sell this in the competitive market now if you have to reduce the size in the weight I cannot use I cannot use I will repeat I cannot use 50 hertz magnetics so I need to reduce the size as well as the weight so another way to do is boost DC so I will boost the DC voltage from 12 volts to the required DC voltage to get a 230 volts AC then I will invert it thereby avoid a 50 hertz transformer I am using a high frequency transformer okay so if you have an application if you have a solar panel of 12 volts 200 volts solar panel now you want and you want to invert it to 230 and you want to do it with a bit of sophisticated way you may have to boost the DC voltage how do I boost the DC voltage one way to use is since the power level is low use fly back as the power rating increases use forward and if you have a still a higher panel you have to do other modifications or use large number of fly backs in cascade or whatever.