 we found or we concluded that we cannot have a very low value of D for buck nor we can have a very high value of D for boost and buck boost type of converters. So, therefore I cannot have a too low a voltage in a buck converter nor I can have a very high voltage in a boost or a buck boost converter maximum ratio could be of the order of 7 to 10 not more than that. Then that was the end of my lecture I derived a transfer function by using a transformer. What was the transfer function that I derived? I connected a switch battery to a transformer in this function in this manner I did not allow both the coils to carry current simultaneously purposely we connected this diode. Now, by doing that what we achieved whatever the energy that I stored here by when I close the switch gets transferred to the secondary. So, if I connect a capacitor here it gets charged. So, on a transfer function in this case happens to be N 2 by N 1 N naught is equal to N 2 by N 1 into V dc multiplied by D into 1 minus D. So, I have in addition to the transfer function of buck boost converter I have another term which is the terms ratio N 2 by N 1. So, there is a power supply what is known as the fly back converter this diode is also known as the freewheeling diode also known as the fly wheel diode. It is very popular up to up very popular up to say 150 and 200 is the upper limit why we will see. So, this is also known as the isolated buck boost because transfer function is same as the buck boost converter, but then there is an isolation between input and output. So, isolated buck boost converter I am not we already discuss the principle of operation it is same get connected diode accordingly and polarities of V naught this is positive dot this is also positive. So, this terminal very current enters the dot current enters the dot. So, this is positive ensure that only one coil is carrying current if of course, you can put a dot here at that time current leaves the dot. So, in this case also current should leave the dot I will repeat I have put a dot here and directly connected to the positive of the battery and when I close the switch current enters the dot no need I can connected in such a way that current can leave the dot as well. Now, if the current leaves the dot here current should leave the dot in the secondary as well. So, you connect the diode and the battery accordingly how to determine the voltage rating the diode and the switches is given here current enters the dot of the teller I too should leave the dot is explained it is fairly simple go back and read if you are if you are still doubts you can send me an email no issue and switches if an flux in the core must be continuous. So, I to flow in the secondary in such a way that direction of flux due to I 1 is same as that of the primary. Now, there was a question how does the current waveform look like in the primary there are two possible conduction continuous conduction and discontinuous conduction. Now, here conduction means flux in the core has not become 0 prior to closing the switch for the second time I will repeat continuous conduction in buck boost buck boost or chook converter is the inductor current here when I say continuous conduction implies that there is a flux in the core is finite prior to closing the switch that is all flux is continuous. So, the B H loop something looks something like this the flux becomes 0 this is the residual flux whatever it starts when I close the switch current increases therefore, the flux and when I open the switch current in the current in the coil in secondary decreases and therefore, the flux in the core as well and I close it again. Now, if the flux if there is a finite flux. So, B H loop looks something like this which mode we operate we will discuss. So, if I operate in continuous conduction in other words there is finite flux or in other words when I just prior to closing the switch for the second time in the primary there was some current flowing in the secondary there was some current flowing in the secondary. Instead the moment I close the switch whatever the current that was flowing in the secondary get transferred to the primary now it is I 1 increases linearly it reaches a peak I p. Now, the moment I open the switch here this I p gets transferred to the secondary now I need to take care of the turns ratio I p is flowing in N 1 what should be it is equivalent when it is flowing in N 2. So, and current decreases similarly this is the value of current just prior to closing the switch it is flowing in N 2 turns. When I close the switch for the second time it gets transferred to the primary and it starts flowing in I 1. So, this value you need to you can determine by knowing the turns ratio and see here current in the coil I 1 increases flux also increases it starts with the finite value of flux because flux is continuous here it was flux current in the secondary was carrying some current there was some flux here it increases and therefore, decreases generally operated in discontinuous mode. In other words secondary current has become 0 therefore, the flux in the core there is some time neither the switch nor the diode are conducting after sometimes switch S is closed and current starts increasing from 0 and this is the B H loop if it is continuous if it looks something like this. Now, why am I saying that generally it is operated in discontinuous mode reason may be bit reason is quite obvious if I operate in this mode and accidentally say for example, load has changed see that load has changed output therefore, output voltage will fall the signal comes to the controller saying that output voltage is falling. So, naturally the controller will try to increase the value of D duty cycle and if there is no limit on D and if D increases. So, high value your core may get saturated core may get saturated see I will repeat I just show you the you want generally the output voltage has to be regulated V naught ref this is V naught actual now, load there is a sudden increase in load therefore, nothing has happened to the primary primary does not know the controller does not know the load has changed current capacitor starts supplying the power capacitor voltage will try to fall. So, V naught starts decreasing it is trying to fall it was almost constant now starts decreasing. So, this is error increases there is a regulator assuming that there is no limiter assuming that there is no limiter D has to in D will increase controller says increase D by increasing D I can supply more power. So, increase D now this increase in D might result in core saturation. So, might increase in D implies that switch is switch is closed for a longer time. So, flux continues to increase it I have shown you linear increase because core is not saturated now for a higher value since the D has increased it may so happen that core may get saturated and if that happens you can see may be smoke coming out of the fly back converter because if the transformer core is saturated current will be limited by the internal resistance of the coil fortunately here solar panel there is a inner and short circuit limit. So, you may have to take care so that nothing should happen to the devices transformer as well as other components if you operate the core in saturation. So, generally thumb rule is thumb rule is we operate in discontinuous core mass saturate this how we have derived transfer function increase in flux should be equal to decrease in flux that is for steady state condition same transfer function we have derived. Therefore, magnitude of it is can be significantly different D max is generally kept at 0.5 and choose N 2 by N 1 suitably now which is a very important parameter it is L 1 is a very important parameter that should answer why we are why this circuit topology is popular for low power low power why it is not suitable for high power that should answer the question now what are we doing when I close in the switch the energy is storing energy is being stored in the primary inductance L 1. So, current is the peak value of the current is given by V dc divided by L 1 into D into t provided it is operating in discontinuous mode current is starting from 0 increases linearly it is a peak value of I p is given by this this is the peak value what is the total power of this converter in the sense the p in given by V dc is the input voltage I am assuming it to be constant average of this because from d t to t current is 0 source does not supply any power from d t to t. So, average of this is area of the strangle is half I p into d by d. So, power N input power is given by the input voltage V dc into the average value of this. So, that also should be equal to power out neglecting the losses. So, neglecting the losses this is the expression for I p d is limited to 0.5 now you see what will happen to I p as the power rating increases. So, as I p increases primary inductance for a given value of d will decrease. So, your overall size of the transom are of the converter size may increase because of the peak currents. One of the advantages is output can be significantly output voltage sorry not output power output voltage can be significantly different other is isolation third is I can have multiple outputs of course, when you are doing solar this may not be very important. What I mean is I can have multiple inputs I can have multiple inputs current enters the dot current enters the dot current enters the dot current enters the dot. So, one primary large number of secondary of different voltage ratios of course, regulating this voltage is may be difficult of course, but then there are solutions. So, this will having N 2, N 3, N 4. So, I can have different voltages large number of voltages it may not be possible to have using the conventional buck boost or a boost converter. So, there is another advantages of having using a transformer, but then with large number of multiple winding. Now, one of one way to see here I said the transformer may get saturated how do I take care of the saturation I can take care of saturation by providing a small air gap because if there is an air gap it will be difficult to be difficult to saturate, but then the moment I use an air gap there is going to be a leakage coils are not tightly coupled. Now, if there is a leakage what will happen when I open the switch see here see if there is a leakage and the transformer equivalent circuit now become something like this. I am neglecting the winding resistance this is V naught, this is diode, this is L m, this is L s L. When I close the switch this is the way of the direction of current and when I open the switch switch could be anywhere I will open the switch whatever the current that was flowing through L m will flow through diode and V naught what will happen to the energy associated with this inductor. There are solutions of course we have to handle now we have to try to address or we have to address the energy associated with this inductor. My circuit overall complexity increases. So, here is one such solution. So, you close S this is how the circuit current will flow open S inductor current was flowing through this that leakage inductance. This direction of current should remain the same for the leakage inductor and it has to flow back to the source. So, what I will do I will modify the circuit something like this close the switch. So, it will flow magnetizing current increases open S now this diode leakage battery. So, that is the reason see in order to take care of saturation I provide an air gap if there is a provide an air gap leakage flux will come if there is a leakage flux my circuit complexity will increase that is what you try to address one problem it will give rise to another problem, but then solution do exist. So, these are the various possible configuration only the diode you need to take care of the diode connections you should not allow the both the cause to carry current there in the fly back converter is the time. Now, I will take few questions please come in chat mode. So, there is a question sir we know that example voltage source is a battery please tell me practically a current source yeah current source if I I do not how do I how do I construct a current source see something like this by regulating this voltage say I have a fixed voltage V dc this is L by regulating this voltage can I can I make current flowing through the inductor approximately constant something like this. See if this if V 2 if V 2 is less than V dc current increases say slowly current increases if V 2 is greater than V dc current decreases I will redraw here I will repeat if V 2 is less than V dc current flowing through the inductor increases I assume that some steady state has reached current increases the moment it touches an upper limit I will try to reduce this I will try to increase this voltage. So, if this voltage is in higher than V dc voltage across inductor is now negative current will fall when it touches a lower limit I will again decrease this voltage V 2 V dc I am not going to do anything the next question is can we have fly back converter having multiple output from different DC sources. So, it should be multiple input can we have fly back converters having multiple input from different DC sources it should not be multiple output I have type. You mean to say that you have multiple inputs from different energy sources for fly back converter. No, no, no see what you do is you have n number of sources coming n number of sources coming to one common DC link n number of sources coming to one common DC link are you with me and then and a fly back you can have n number of sources coming solar 1 solar 2 solar 3 may be a small windmill rectified whatever and you have you can have this, but then remember this is suitable for low power not more than say 100 to 150 watts in principle yes you can have one common DC link connect all the sources there why it is not it is possible this is what you have in mind, but can we have fly back converter having multiple inputs. See let me not can you can you draw the figure what you have in mind scan it and send then may be tomorrow I will try to answer multiple input fly back input I do not know see this is what you mean these are the various inputs and may be I have a selector switch connecting to various this one multiple inputs means what in the sense I have finally I have to connect this transformer to I have to finally I have to connect a transformer to a source. In the transformer there will be multiple primary coils. Multiple primary coils let me not answer straight away in the sense now this question is multiple primary coils and one core no no one transformer no no let me not let me not answer this I have to study that I have to see what will happen to the flux in the core because you are trying to energize the switches they may not be there is no are they synchronized all the switches. So, no no what I am saying is you are saying may be then I think I need to need it you have to see what exactly is happening in the flux in the core let me not answer let me think the question is how to find d max so that converter will not you know the question of converter saturating is a transformer saturating the question is how to find d max so that converter will not such converter does not saturate is a transformer saturating because you are the one who are designing the transformer you are the one who are designing that inductor you should ensure that after all it is b h loop will tell you whether it is saturates or not your n i photon should not increase beyond the saturation this is h this is b h is h is proportional to current that is flowing when the close when the close the switch. So, you should ensure that you should not go the saturation level the question is if v d c is varying continuously how can we regulate v naught see fine is a good question v d c is varying continuously what are the what are the input what are the minimum and maximum limits of v d c variation and d is generally at point five. So, you choose choose for a minimum condition for minimum input d is point five now choose n 2 by n 1 accordingly now as v d c increases v naught is the same now d has to come down it is transform may not get saturated if follow what I am saying v d c now you have to you cannot be just saying that v d c is varying continuously fine what are the upper v u and v l. So, for v l you know what is v naught you know the transfer function for this you calculate d is equal to point five. So, therefore as v increases same v naught d has to come down choose n 2 by n 1 for v l and point five as v l increases above sorry as v increases above v l d has to come down very fine you are not increasing d above point five. So, the apprehension of transformer getting saturated is not there what is the maximum reverse blocking voltage rating there is d c voltage here it has to block d c it is please try to under what is the reverse blocking question is wrong in a sense there is no reverse voltage being applied across the switch here that can happen only in a c and if you have if I am using an a c then I need to choose the devices accordingly conventional BJT, MOS, IGBT they will not block the reverse voltages. So, as I understand correctly it looks like your question is what is the voltage rating of the switch see when the switch is open when the switch is open what is the voltage it has to withstand this is v d c plus the voltage induced here that is v naught into the turns ratio now turns ratio take suitable definition v naught is being applied to n 2 what is the voltage induced in n 1 this is with. So, this is positive or equivalent circuit v d c assume the dot is here dots are negative both the dots and this is the switch s this is the ground there is no reverse voltage being applied across s. So, voltage rating is v d c plus the induced voltage in the in the primary winding that is v naught into turns ratio turns ratio now you choose your be careful while substituting the value of n it is v naught is being applied to the secondary having n 2 turns what is the voltage being induced in the primary you need to find out and that is n here. The next question is why cannot we use boost converter alone to step up the voltage from 24 to 230 you want to use only boost converter please go ahead and do my best wishes to you. Next question why fly back converter is not popular for power rating about 200 I showed you the transparency as well this is the expression for p in is equal to p out now you have to do your paper and you have to take a paper and a pencil and design and you have to find out what will happen to the transformer as the power rating increases what will happen to I p the p currents size. So, I encourage you to do a simple transformer design problem say for example, 300 watt transformer only one winding carrying current at a time fix some input d is 0.5 calculate I p calculate l 1 and try to get this l 1 what will happen to the size and the p current I p. So, can we know the power rating of buck boost and buck boost converters for design application see we discussed about this converters I told you we have a problem when if I if there is a large ratios of input and output that is all power rating we have not discussed we can power rating is not a problem only is the voltage ratings if the output voltage ratio of v naught by v d c is very high then I may not be able to use the conventional boost or buck boost it is a problem with the voltage ratios.