 The power to convertor there are lot of questions as to V dc is reverse biasing this diode. Now, do not come jump to those conclusions. So, I will explain to you why what the I will explain to you the advantages that we can get by connecting tertiary winding to this source. You do not have to connect this tertiary winding to the source. You can connect it to a separate winding something similar to the fly back, but then you need to have a third winding. I will repeat you need to have a third winding and current should flow in that third winding when you open the switch s for the magnetizing current to flow. For the magnetizing current to flow we have to provide a path in the third winding second winding it is not possible here. One made to have a fly back connection something similar have a separate capacitor that capacitor gets charged or you can connect it directly to V dc. Now, what are the advantages of connecting to V dc? You just see here if I connect this diode if I connect this tertiary winding to the supply V dc what is the voltage rating of the what is the voltage rating of of this winding and what is the current rating of this winding by the what is the current rating this coil has to carry it has to carry only the magnetizing current the peak value corresponds to the corresponding value that was flowing in the primary just prior to opening the switch and multiplied by the turns ratio turns ratio that current was flowing in i n 1 now it has to flow through n 3. What is the when the when this diode is on what is the voltage that is being applied to this winding when I when I when the diode is on voltage applied to this winding is V dc. So, if I assume that l 3 is the self inductance of this coil voltage applied is V dc having n 3 number of turns what is the duration what is the maximum value of d that I can have I can determine and what should be the conductor size and the insulation that I need to wind this transformer we will discuss in detail here is the equivalent analysis I will quickly I will go through i 1 is equal to i 2 prime plus i m i m is the magnetizing current V is applied to l m l m is the magnetizing inductance of the transformer i m increases linearly with time V induced in n 2 supplies current to the load what it means is voltage induced in the secondary will supply the current to the load that forward bias is d 2 and whatever the voltage induced here is V dc into the turns ratio the equivalent circuit in the tertiary is V dc what is the voltage rating of d 3 see here voltage applied to the primary is with the n 1 turns is V dc with the dot as positive. So, this terminal is positive voltage induced in this coil is V dc into the turns ratio now turns ratio is n 3 and n 1. So, positive of the battery is connected to the negative of the voltage source a induced voltage source something it is here positive of the battery is connected to the negative of the voltage induced or negative of the voltage source which is induced in the winding V dc and that is and this voltage d 3 has to this voltage d 3 has to voltage rating I mean the d 3 I can determine from supply voltage and the turns ratio. So, comes up to be this one current rating I know the equivalent peak current of the magnetizing peak value of the magnetizing current when i 1 is 0 i 2 is equal to 0 i m and i l should be continuous i l can flow through d f now voltage across d 2 is voltage induced in n 2 itself what is the voltage across d 2 here voltage across d 2 is voltage induced in n 2 see this when diode d f is conducting this point gets connected here. So, voltage induced in d 2 is the voltage rating of sorry voltage induced in n 2 is the voltage rating of d 2, but then what is the voltage induced in d 2 when 1 is see there are 2 values you just find out here when primary switch 1 is called switch 1 switch s is closed it is V dc. Similarly, when diode d 3 is conducting voltage applied to this coil is V dc voltage applied to this coil is V dc having n 3 turns what will be the voltage induced in what will be the voltage induced in n 2 now voltage applied to n 3 is V dc what will be the voltage induced in n 2 that voltage d 2 has to block when the switch s is closed d 2 gets forward biased when s is open when d 3 starts conducting d 2 gets reverse biased and the voltage that it has to block is voltage induced in n 2 that is proportional to V dc turns ratio between n 3 and n 2 turns ratio between n 3 and n 2. Similarly, I can determine the voltage rating of the switch as well because voltage applied to n 3 is V dc there will be voltage induced in n 1 in addition there is V dc as well the total voltage the s has to block the sum of these 2 voltages procedure is the same. So, voltage across s is V dc plus n 1 plus n 3 voltage across in n 2 is V dc into n 2 into n 3 that is the voltage rating the d 2 itself what is the value of n 3. So, increase in flux V dc d t into n 1 V dc d 2 into n 1 is the flux n 1 d phi by d t is equal to V dc these are increase in flux. Now, how does the flux decrease it is because of the current that is flowing in n 3 same voltage V dc divided by n 3 into some time. Now, let us see how much is that now if I equate increase in flux should be equal to decrease in flux for the steady state t a is given by n 3 by n 1 into d t n 3 by n 1 into d t if both of them have see for the core flux to become 0 for the 4 core flux become to 0 t a should be less than 1 minus d into t. Now, we will see why the core flux has to become 0 for the core flux to become 0 core flux has to become 0 d flux has to become 0 before closing the switch for the next time. So, t a should be less than 1 minus d into t t a should be less than 1 minus d into t. So, that is a condition that I am applying here. So, d must be limited to d max such that n 3 by n 1 d max t should be equal to 1 minus d into t. So, max d max is half d max is half if n 3 is equal to n 1 if n 3 is equal to n 1 d max is equal to 1. So, number of turns is the same now. So, may be and n 3 cross sectional area of the conductor that is used in the tertiary coil is very small compared to the primary and secondary why secondary has to carry the equivalent load current primary has to carry magnetizing current as well as the load equivalent load current whereas, the tertiary should carry or will carry only the magnetizing current. So, I need to have a very thin conductor number of turns in the primary and tertiary I am keeping it same. So, that my winding of the transformer or during fabrication it becomes easier one thick conductor one thin conductor I will take them together and I will wind it it becomes easier number of turns is the same. So, it becomes easy. So, it is not only the design we need to worry about the fabrication the effort that is required to fabricate the transformer as well. So, that is the one of the reasons of connecting this tertiary across the supply. Now, if I choose n 3 is equal to n 1 and if d max is greater than 0.5 what happens see I have chosen n 3 is equal to n 1 and d is greater than 0.5 what will happen for the first time I will close the switch d max is greater than 0.5 controller decided the switch is on for a longer time. So, the flux see this equation here d phi is given by v d t into d v d c into d t into n v d c into n v d c into d t into n 1 this is the rate of the flux. So, it has started from 0 I am neglecting the residual flux the slope here is what is the equation says slope here is v d c divided by n 1 this is see d phi by d t is v d c divided by n 1. So, v d c divided by n 1 this is d into t. Now, n 1 and n 3 have the same number of turns voltage applied to the coil is also the same n 3 is connected across v d c n 1 is also connected across v d c. So, d phi is minus d phi by d t is v d c divided by n 1. Now, d into t d into d into d into d into d into d is greater than 0.5. So, what will happen when I close the switch flux falls, but then slope of the line is the same it is again v d c divided by n 3 that is equal to n 1 this slope is the same, but then switch is on for a longer time this is for a shorter time definitely flux will not become 0 flux will not become 0 at t there is some finite flux at the end of the first cycle. In the second cycle again I will close what will happen now same condition slope is the same v d c divided by n 1 the switch is on for a longer time this is d into t for the second cycle it has reached a much higher value after sometime I will open the switch again it is v d c divided by n 3 or n 1. So, end the flux in the core at the end of the second cycle is I will repeat the value of the flux at the end of the second cycle is much higher than the flux the value of the flux at the end of the first cycle because this period is smaller than this period slope is the same slope is the same this period is much higher than this. So, definitely it will not become 0 again in the second cycle I will increase now this value is higher than the previous and the value of flux at the end of 2 t is higher than the value of flux at t in the third cycle what will happen it will go further. So, what may happen of flux sorry transformer will saturate and once the core saturates you know there is no d phi by d t current is limited by only current is limited by its internal resistance. Now, in order to have high q you are used leads wire or whatever which has a very small resistance and current will be too high may damage your source you may damage your switch or a transformer may damage everything. So, if I connect the tertiary winding to the coil sorry if I connect the tertiary winding to the source and if n 1 is equal to n 3 d cannot be greater than 0.5 d cannot be greater than 0.5 n 1 is equal to n 3 my fabrication becomes designing and fabrication if I become relatively simpler. So, this what it is given the equivalent circuits here this is negative d phi by d t this is the equivalent circuit when I close the switch current in l m increase at the magnetizing current when I open the switch it was through source here if n 1 is equal to n 3 it is the same voltage itself. So, increase in current should be equal to decrease in current l m is the same source value is the same. So, duration should be the same if one is if off period is smaller than the on period flux will saturate sorry core will saturate if on period is less than the off period flux in the core becomes 0 no problem flux core transformer will not saturate core will saturate. Therefore, d is less than 5 discontinuous flux operation it is something like this slopes are the same because number of turns is the same voltage is also the same these are the various voltage wave forms primary current wave form through the tertiary this is the current through d 2 current through d 2 current through l f is this this current is almost like a current source I will say almost constant current I am trying to maintain increase or decreases increase or decreases. So, d 2 current peak current rating is the peak current that is flowing through l f these are the various this is about forward converter there are various issues as well these are some special cases see here I have not connected to the supply tertiary winding could be a fly back the capacitor I do not know whether to explain this or not I will just anyway I will explain. Suppose, what if I do not have a tertiary winding see if I do not have a tertiary winding and if there is d 2 here then the circuit cannot work because there is no path for the magnetizing current assume that there is no d 2 assume there is no d 2 I will just explain to you circuit here there is no d 2. Now, you see current enters the dot current leaves the dot no problem this is the path when I open the switch current enter the dot. So, the magnetizing current should enter the dot can it enter the in the secondary yes no problem. So, the secondary whatever the current that is flowing magnetizing current can flow through this path. So, now d f has to carry in addition to l f sorry in addition to whatever the current inductor in addition to inductor current it has to carry the magnetizing current as well will this work my gut feeling is it may not work why see what happens here when I close this when I open the switch the equivalent circuit is something like this equivalent circuit is something like this. It is actually what is the voltage equation here for this coil what is the voltage applied is 0 I am applying a 0 voltage here voltage. So, will the flux decay it entirely depends on the winding resistance entirely depends on the winding resistance if the winding resistance is very small flux will not decay flux will not decay something similar to inductor current flowing like this. In the sense if the resistance is very small current will remain approximately constant if the current remains approximately constant flux in the core also will remain approximately constant. So, when I had open the switch sorry when I had close the switch flux increases when I open the switch magnetizing current trying to tries to flow through this path yes there is a path in the very first cycle. Since the winding resistance is very small your current may not decay the equation is l d a by d t plus r into i is equal to 0 if r is very small. So, l d a by d t is equal to 0 in other words i remains constant i is constant is not it l d a by d t is equal to 0 l cannot be 0. So, d a by d t is 0. So, the I am I will remain constant. So, magnetizing current will not decay in the second cycle it the flux will start from a finite value and therefore, the core may saturate. So, you require three windings there is no way about, but then there are modifications may be as we go along if time permits we will see otherwise I will encourage you to go through the literature both again even in forward converter as well as fly back converter. If I see the B H low operation in the first quadrant only operation in the quadrant only now how do I increase in other words I am using a transformer I am using a core, but I am not using the other half of the B H curve can I improve the utilization the core. So, use bidirectional core excitation that is something similar to AC current excitation current becomes negative. So, B H loop is transverse in the negative direction as well can I use it yes let us see in other words I may have to use a two forward converters may be this converter is known as the push pull converter push pull converter why both converters deliver power to the load in each half cycle both of them are pushing power to the load both halfs name should have been push push converter, but then it is known as a push pull converter. I do not think I can start this push pull converter look something like this as the power rating increases one has to go from forward to push pull forward converter is suitable till say 500 to 700 watts or so the reasons are same as what we give for fly back the forward as the power rating increases definitely I need to increase the I have to use the magnetics I have to improve the utilization of the magnetics. So, definitely I need to use both halfs of the B H curve then I so what is known as the push pull converter the question is l m and l 1 are same or is there any other significance l m is the magnetizing inductance l 1 is the self inductance. Self inductance is leakage plus l m over here we have neglected the leakage definitely however tightly coupled coils there is a small leakage will be there l m is the magnetizing l 1 is a self leakage is always there you can reduce it by tightly coupling by improving the coupling between them. In a fly back converter when the air gap is introduced in the transformer to reduce saturation how can it fall output voltage equation? Yeah I agree I understand your question. If there is an air gap see whatever the expression that we have derived first of all there are lot of approximation that we have made then it is ok see V R deriving an expression what is the voltage equation how did you derive the transfer of equation by equating volt second per turn volt second per turn. So what is the voltage applied to the primary? I said V dc but it is wrong it is not V dc see there are one is what are the errors that are introduced one is this coil has its own resistance R this device has its own voltage drop. So therefore actually it is V dc minus i into R minus V switch is the voltage applied to the coil if the winding resistance is very small if V s is very small compared to this then yes I can neglect otherwise you have to take the device drop especially if I am using a very low solar voltage panel then I have to switch now one volt may be comparable with your 12 volt solar panel then I have to take this into account this drops this drops. So that has to be taken care while designing the number of turns is suitable values of N 1 and N 2 take now it is not definitely not V dc we are not applying V dc to the coil voltage actual voltage is resistance drop minus the saturation for example, see you know AC what is the voltage induced in this what is the voltage here definitely not V 1 N 1 into N 2 definitely not V 1 into it is V 1 by V 2 is equal to N 1 by N 2 no this is not equal it is true for an ideal transformer it is true for an ideal transformer actually it is E 1 is to E 2 is equal to N 1 by N 2 E 1 is E 2 is equal to N 1 by N 2 not V 1 by V 2 is equal to N 1 by N 2 I need to the winding resistance leakage everything here in addition there is a switch drop as well and that may be comparable with or comparable to the solar panel itself. So, 1 volt or 2 volts you may not be able to neglect. So, take care while choosing the suitable turns ratio it is not V 1 by V 2 is equal to N 1 by N 2 please. Sir, in a discontinuous conduction mode the current is always going to be in 0 level. So, for operating in a converter in a discontinuous mode it reduce our efficiency. So, how can we. Which converter are you talking about? Sir, what the forward converter? Now, what do you mean by conduction and discontinuous conduction? Sir, continuous. No, no, what exactly. Sir, because if we operate our. No, please see there are I am not talking about load current being discontinuous the flux in the core is discontinuous please it is not the load current see if I take a just an ordinary buck discontinuous conduction means the inductor current see there is a difference see ordinary buck converter when I am saying discontinuous conduction it is this inductor current becoming discontinuous becoming 0 prior to just prior to closing the switch it is something like this I L for a buck converter when in fly back and forward or using a transformer or in isolated DC to DC converter in isolated DC to DC they come fly back forward and push pull discontinuous means flux in the core flux in the core flux in the core see if you see the current waveforms this is the inductor current is almost a current source almost a current source, but the flux has become 0 see here flux has become 0 much before closing the switch for the second time. So, this is discontinuous conduction please do not mix up two issues in the sense in non isolated converters discontinuous conduction implies inductor current in isolated DC to DC converter discontinuous implies flux in the core this current is approximately current source, but then flux in the core has become 0 here and therefore, the discontinuous conduction. So, another question so in that case what is our condition of the load current so whether we are supplying by a discontinuous mode to a load. So, what is the load current condition whether it is continuous or discontinuous where have you connected the load you tell me you are connecting the load across the capacitor is not it this voltage is constant we are regulating this voltage at a constant value. If the circuit is complete current will flow there is nothing like discontinuous load current you can have discontinuity in load current only if there is an open circuit because v naught is regulated at a particular value it is it is supposed to remain constant v naught is supposed to remain constant. So, if the circuit is complete current has to flow so load current will remain constant as long as load resistance remains constant that has nothing to do with discontinuity in flux or whatever please. So, do not try to make seen drives for that matter when you said discontinuous conduction current in the armature is 0 in non isolated power supplies discontinuous implies inductor current becomes 0 not the load current you are connecting the load across the capacitor and that voltage is supposed to remain constant if there is if the circuit is complete current has to flow discontinuity in current can happen only when there is an open circuit is that ok. Paramati go ahead with your question. My question is what is the frequency range of fly back converter when we are using ferrite core transformer? Frequency depends on the power ratings or power and voltage ratings I cannot give any unique answer it all depends on the power rating see there are two types of switching one is the hard switching and a soft switching if you are doing a very hard switching in order to reduce the size you may try to increase the switching frequency your losses will increase you may not be able to go for a higher value it all depends on the power rating the way you switch the way you switch are you doing a soft switching or a hard switching what do you mean by hard switching soft switching I will tell you sometime later towards the end of may be next lecture it all see there is no unique answer to your question it depends on the amount of power the device has to handle the voltage and current rating now please see there is a question from Schupper regarding load current one parameter we have to consider that boundary between two modes what will be decided the value of inductor hello sir why forward converter can be designed for a higher rating compared to fly back converter you try to understand the operation of fly back and forward what are you doing in fly back you are storing the magnetizing energy you know you are storing energy in the magnetizing inductance and transferring it to the output stage what are you doing in forward you are simultaneously drawing the current primary and secondary both of them are carrying current simultaneously so you have to design the transformer accordingly cross sectional area of the conductor that cross sectional area of the conductor that is used to wind the primary coil is much thicker than the one that is used in fly back fly back it has to carry only the magnetizing current whereas in forward the primary current the primary coil has to carry magnetizing current as well as the equivalent secondary current you are using a thicker conductor see the difference in operation itself what is the effect of leakage used in fly back converter what do you mean by what do you mean by what is the effect of leakage you have to take care of the energy associated with the leakage inductance I showed you the circuit configuration in the last class I will show you the equivalent circuit transformer equivalent circuit see here this is the leakage magnetizing this is the fly back when I close the switch yes this is the path when I open the switch there is a path for the magnetizing current there is no path for the energy that is stored in a leakage inductance so if you do not have any path definitely there will be a voltage spike appearing across this switch might damage of course there are ways either you put a snubber across the switch you put a snubber across the switch or you modify the circuit itself the circuit modification I had shown in the last class please refer to yesterday's lecture note it is there is one power circuit configuration I have why again I have to draw when I close both the switches yes this is the path when I open the switch whatever the current that is flowing the magnetizing inductance it will try to flow through this path sorry do not see it so this is the path when you close s 1 and s 2 this is the direction of current this direction current has to be maintained is there a path yes there is a path that current now can flow through bold lines I will draw this is to take care of leakage energy associated with leakage inductance there are other way circuits also