 Good morning. Now, I will continue my lecture on DC to DC conversion. Here, let me starting, let me tell you one thing. I am just doing the basic blocks. Now, as the power rating increases, you may have to parallel them and our control strategy may change. So, all these issues we will discuss when we finally integrate the entire solar system. Right now, I am just discussing the basic building blocks. So, there was a question that can I can be parallel? Yes, how to parallel? What is the control strategy? What modification do we need to make? We will discuss in detail while integrating the entire power electronic interfaces to solar and to the grid. Last class, we discussed the basic, one of the basic building blocks is the buck converter. If the inductor current is continuous, if the inductor current is continuous, output voltage is given by D into the input voltage. And the value of L will determine the current ripple and value of the output capacitor will determine the voltage ripple. And this ripple is maximum at D is equal to 0.5. When we discuss the ideal boost, we found that output voltage tends to infinity as D tends to 1 for an ideal boost. But in a non-ideal boost, the output voltage tends to 0 as D tends to 1. Why there is such a significant difference? It is because the assumptions that we made while deriving the transfer function are not valid for high values of D. We had assumed that output voltage will remain constant and ripple free, so as the input voltage. But then if you see the equivalent circuit and for D tends to 1, we will see that capacitor will continuously discharging, it will continuously supply power and at the input side the sources or the inductor is connected across a DC source. Now, here it so happens that if the solar cell, yes current may be limited to ISC. But then if there is a capacitor, we have a problem. At the output side, capacitor voltage tends to 0. So, our assumptions are not valid for high values of D. Now, for high values of D, we need to take the non-idealities into account and we found that if I take the inductor resistance into account V naught max, the maximum output voltage that I can get is given by this equation. The maximum ratio that is generally obtained in a boost converter could be of the order of 7 to 10, that too using a inductor, which is having a very high Q, the quality factor, the strong function of the quality factor of the inductor. So, then we discuss the buck boost converter, the transfer function is given by D into V DC divided by 1 minus D, again this is for an ideal buck boost. For D to 0 to D to 0.5, it is in buck mode and from for D greater than 0.5, it is in boost mode and here and here the transfer function says that at D tends to 1, V naught tends to infinity. But then same whatever that happens in a boost converter, it happens in buck boost as well as D tends to 1, V naught tends to 0. The maximum ratio here also we can get is of the order of 7 to 10. So, we have discussed 3 converters, buck converter boost, buck boost. See here in the buck converter, the moment I close the switch, whatever the current that was flowing through the inductor starts flowing through the switch as well as the source. So, if I assume the inductor current to be continuous, the source current looks something like this, almost instantaneous it will jump from 0 to the current that was flowing through the inductor at t is equal to 0 minus. I will assume that the switch is closed at t is equal to 0 and there was some current flowing at t is equal to 0 minus that that current will start flowing through the switch as well as through the source. So, this how it looks. So, D I B D T at this point is very high whereas, at the output stage we have a reasonably a good circuit in the sense I have a inductor. So, I can represent this as a some sort of a current source. At the output I have a current source. Now, for a boost converter if you see at the input stage I have a current source, I can represent this L and V dc by a current source. This current if I plot it looks something like this in a boost converter, source current looks something like this. It is approximately constant does not change by choosing a suitable value of L I can control this ripple. So, current is approximately a current source sorry the input stage is almost like a current source, but then at the output if you see during on period capacitor is supplying the full power. So, this current this is approximately V naught by R. So, my size of the capacitor to regulate the voltage may increase. What about buck boost? The output stage is something similar to buck converter and the input stage is something similar to buck. See the source current is the same as buck here and the output stage. So, it is always desirable to have a current source at the input as well as the output always desirable to have a current source both at the input as well as the output. So, how do we get? So, that is what the cook converter become very popular. The circuit configuration is shown here. First I will explain you the principle of operation then we will see whether we have what sort of a stage we have at the input as well as at the output. I will close S for d into t period the equivalent circuit is look something like this closes S gets connected to this point. So, at the input stage we have a source inductor and the switch at the output stage if you see whatever that the current that was flowing through it will continue to what could be the path here at when the switch is closed this point is gets connected here. Now, after some time I will open S whatever the current that was flowing through the inductor will continue to flow through the capacitor and diode D. So, it looks something like this L C 1 and D here and at the output stage whatever the current that was flowing through the inductor will continue to flow through D. When I close switch at the input stage V D C L 1 and S and torque capacitor voltage appears across D. So, D is reverse biased see this point is connected to this point. So, V C 1 appears across D that reverse biases this diode. So, so output stage is powered by the capacitor C 1. I will repeat the output stage is powered by the capacitor C 1 when you open S the stored energy in the inductor continues to flow through C 1 or it will charge C 1 and whatever the current that was flowing through L 2 will flow through D. So, when the switch is on this is the equivalent circuit and when the switch is off these are two equivalent circuits. How do I derive the transfer function? This is the input stage of a boost this is a buck converter see what is the relationship between these two and what is the relationship between these two or what could be the relationship between V D C and V C 1 and what could be the relationship between V C 1 and V naught. These two circuits look similar to that of the boost converter closes stored the energy in the inductor open S transfer it to the output stage open S it transfer to V C 1. So, the relationship between V D C and V C 1 could be I say the relationship between V D C 1 and V C 1 could be is given by V C 1 is equal to V D C divided by 1 minus D the boost converter see these two this is the equivalent circuit when the switch is closed this is the equivalent circuit when the switch is open. So, the relationship between V D C and V C 1 is similar to that of a boost converter see here these are two equivalent circuit that I get for a buck converter when I close the switch V C 1 supplies power to the circuit when I open S this is the equivalent circuit the relationship between V C 1 and V naught could be could be V naught is equal to D into V C 1 V naught is equal to D into V C 1. So, what do I get V naught is equal to D into V C 1 now you substitute for V C 1 what do I get here. So, V C 1 is equal to V naught by D if you substitute you may get V naught is equal to V D C into D divided by 1 minus D 1 minus D something similar to is nothing but a buck boost converter. So, chip converter is nothing but a buck boost converter, but then only difference between the buck boost converter and chip converter is at the input stage as well as the output stage I have almost a current source I have a current source. See buck converter has an excellent output stage I am maintaining it here boost converter has an excellent input stage that is being maintained here. So, combining these two I get a chip converter the transfer function is something similar to that of why similar is the same as that of a buck boost converter. By the way I told you yesterday I will show you an inductor see I have not used a single wire to make this inductor I could have used one single conductor to wind around across the ferrite core instead I took thin strands join them together and I used it to wind this inductor. Now, thereby I reduce the effective resistance of this inductor this inductor seem to have a very high Q very high Q see here some other example is here the ferrite core inductor ferrite core inductor I have thin strands are joined together thereby reducing the effective R effective R effective R. So, if you want to have a high Q inductor one has to go in for what is known as a leads wire leads wire is bit expensive that, but then if you want to do it wind an inductor in the lab what you can do is thin conductors twist them together and wind it around the either across the ferrite core if you are switching at a very high frequency then you may have to go in for a air core inductor. So, analysis of converter is given here you can refer same equations differential equations and finally, you can derive the waveform both input and output current waveforms are very smooth. So, far we discussed four converters buck converter boost buck boost and choke what is the transfer function for a buck converter for a buck converter this is D into V d c boost converter V d c divided by 1 minus D and for a buck boost V d c into D divided by 1 minus D these all these three expressions are valid only if the inductor current is continuous. I will repeat these expressions are valid only if the inductor current is continuous point number one. Second is see the expression for boost and buck boost that is same as well for choke converter as well the ratio of the maximum ratio of V naught by V d c that I can get is not more than 7 to 10. So, if I want a higher ratio looks like I may not be able to use one boost say for example, I have a 24 volts solar panel V o c is 24 volts and I want to generate 230 volt d c sorry 230 volts a c r m s. Now, as of now I do not know how to convert this d c to this 230 volts a c, but then the ratio here 240 to 230 looks this is r m s again. Now, if I just do a square wave inversion I may get say full bridge inverter if I use now you do not ask me what is the full bridge inverter I will discuss sometime later I will get something like this. So, this is 24 this is 24 I am neglecting the drop of course and moreover this is V o c. So, the moment I start drawing some current this voltage will fall and if I want to have a 230 volt r m s. So, that corresponding the peak is of the order of 300 and odd 310 volts also. One way to use is use a step up transformer that may not be an very elegant way of converting this d c to a c. The second way could be this low voltage d c you boost it to a high voltage d c of an appropriate value what is that appropriate value we will see it could be around 300 volts also more than 300 volts depending upon the control strategy. So, let us assume that to get a 230 volt r m s we may require around 350 volt d c for an inverter I may not be able to use one boost one stage. Now, you may say that use a two stage boost that could be one of the possibilities. Now, in that case now two boost they come in series or cascade. So, overall efficiency is eta 1 into eta 2 and plus that device count may increase. So, reliability may come down. So, is there a other way out instead of using a just an ordinary boost or a buck boost can I use some of the circuit configuration ok. So, that is the question that I ask in the beginning or same question that I am asking now we will try to find an answer in this lecture. So, one of the limitations of this converters is a high ratio V naught by V d c is just not possible to get though ideally it says that V naught can tends to infinity. So, what is the problem with the buck converter now buck converter is the transfer function is if the current is continuous it is d into V d c. Now, for a low values of d for low values of d you may it may happen that inductor current may become discontinuous. See the equivalent circuit for low values of d see this is the equivalent circuit for a buck converter when the switch is closed switch closed switch open. So, for low values of d in the means the switch is closed for a very short duration compared to compared to the off position or off state. So, I am storing the energy for a very shorter duration compared to this period. So, it may so happen that inductor current may become discontinuous and if the inductor current become discontinuous V naught is not equal to d into V d c it could be slightly higher V naught is not equal to d into V d c. So, therefore, the conclusion that we can draw is buck converter boost buck boost or chook converter d cannot be too small or d cannot be too high. If d is tends to towards one we have a problem with boost converter buck boost and chook and if d is too small we have a problem with the buck converter current may become discontinuous and I may not be able to regulate the output voltage. So, is there a way out yeah there is we will see what it is before going directly into that topic I will see I will just discuss briefly the recovery of trapped energy. I will just take an inductor a switch close s what will happen current through the inductor increases linearly because voltage a constant d c voltage is applied l d a by d t is equal to V. So, I is V by by l into t I am neglecting r can I open the switch I cannot open the switch because inductor current has to be continuous. So, I have to provide a path. So, one way to provide a path is connector a freewheeling connector a diode across diode across this inductor something like this. If I assume diode to be ideal and inductor to be ideal current will not decay because at close the switch current increases at this point current switch is opened whatever the current that is flowing through the inductor will freewheel through this path. So, inductor current will remain constant. Now, if I take the non ideal it is an account the on state resistance of a diode on state resistance of sorry the internal resistance of the inductor the current will slowly decay current will slowly decay whatever the stored energy in the inductor is dissipated as heat. Now, if I want to have a faster decay of current I have to connect an external resistance in this path current decay is faster. So, what are we doing here we stored energy in the inductor and whatever the stored energy in that inductor is dissipated as heat. The question that I am asking is instead of dissipating it as heat can we transfer it to either to the load or can we transfer it back to some of the circuit or to the source. Now, that is what the topic says recovery of recovery of trapped energy before getting into the recovery I will just briefly discuss the transformer theory. I have a two winding transformer first coil is connected to the source AC source and second is to the load these are the two dotted terminals what do they imply the dotted terminals imply that at a given time these are of the same polarity if this is plus this is also plus. What is the right direction of current if current enters the dot here current will leave the dot that is a simple transformer theory if current enters the dot current will leave the dot this happens only when or this can happen only when both the coils are carrying current simultaneously both of them are carrying current simultaneously are connected to AC source voltage alternating flux is produced voltage is induced if the circuit is complete some current will flow. So, both of them are carrying current at the same time. So, therefore, if the current enters the dot here it will leave the dot here. Now, what I will do is I will modify this circuit something like this shown here in this secondary I will connect a diode in this fashion. Now, let us see how it how it works close S current enters the dot here. Now, if there is no inductor here and just the load current will leave the dot something similar to this two winding current enters the dot current will leave the dot if there is no diode here and if there is just a passive element of a current will leave the dot. Now, because of this diode current cannot leave the dot diode is connected in the reverse fashion current cannot leave the dot. So, therefore, when the switch is closed here when the coil is carrying current there is no current in the secondary there is no current in the secondary. In other words I have a mutually coupled circuit there is a coupling between these two circuits. So, they are mutually coupled, but then only one coil is carrying current at a time only one coil is carrying current at a time. But then what should happen to the current when I open this switch here see when I close the switch there was some current flowing in the coil this coil will produce its own flux just prior to closing the switch it has attained some value of flux and some value of current I am trying to open this switch flux in the core must be continuous I have to provide some path for the flux to flow or the inductor current to flow what should be the direction of current in the second loop. Now, before coming to this the tell me how do we place this dots how do we place this or what should be the direction of flux produced by the primary current and the secondary current when the both the coils are carrying current at the same time current produce the sorry the flux produced by I s opposes the flux produced by I p flux produced by I s opposes the flux cause and effect this will I s the flux produced by I s opposes the parent flux that is the primary flux why and it happens because both of them are carrying current the same time a simultaneously, but then in this circuit only one coil is carrying current at a time when I close the switch s only the primary is carrying current secondary does not carry any current there is no path for the inductor current to flow in the secondary. By the way I have one more question when I close this there is no current in the secondary only primary is carrying current. So, if I recall my transformer theory if there is no current in the secondary only primary is current current that current is only the magnetizing current magnetizing current if the secondary is carrying current the primary current is has two components one is magnetizing current and the equivalent equivalent secondary current. So, here secondary is open I am saying open because there is no current is flowing here. So, the current that is flowing here is only the magnetizing current. So, a constant voltage is being applied to this inductor or to this magnetizing inductance or to the primary inductance L 1 current increases linearly and therefore, flux in the core also increases linearly what happens when I close open the switch see I have explained current starts flowing current enters the dot current direction of I 2 is to leave the dot current direction of I 2 is to leave the dot only if there is a path in the secondary and that should happen the same time it is not possible due to due to no current flows in the secondary when I 1 is flowing in the primary. So, here is a voltage applied to the primary winding is N 1 trans is V d c voltage induced in the secondary winding is N 2 type V d c into N 2 by N 1 I am saying voltage induced voltage induced only if there is flux in the core changing though I am applying a d c voltage here remember though I am applying d c voltage here I am not allowing the course to saturate. So, L d a by d t is equal to V d c inductor is still in the linear region or it is not saturated. So, current does change with time if current does change with time therefore, the flux also will increase. So, flux increases linearly in the core. So, that linearly increasing flux is seen by this coil and therefore, the voltage is induced in the secondary and that is given by V d c N 2 by N 1 with dot as negative why dot as negative dot is positive this also will be positive because when I close the switch positive terminal of the battery is connected to this dot. So, these are of the same polarity dot is positive dot is positive voltage induced here is V d c into turns ratio that is required because diode is blocking that voltage. So, what is the equivalent circuit in the secondary dot is positive dot is positive and the magnitude is V d c into the turns ratio dot is positive and that is connected to the negative of the output or negative of this battery V 2. So, these two induce voltage and V 2 they come in series. So, the diode voltage rating is V d c into N 2 by N 1 or V d c into the turns ratio plus V 2 is the voltage that is appearing across the diode D 2 and diode D 2 should be able to block that voltage. So, I will repeat the voltage rating of diode D 2 is voltage induce here that is nothing but supply voltage V d c into the turns ratio plus V 2 that should be the voltage rating of D 2. What is the voltage rating of this switch S we will see sometime later. So, voltage V 2 across D 2 is given by this open S. So, flux is established in the core I am. So, the magnetizing current why am I calling magnetizing current because secondary is open. So, I 1 is nothing but the magnetizing current itself core in the core increases linearly. So, I have to open switch S after sometime I cannot keep it permanently closed. So, when I open switch S what happens flux in the core must be continuous flux in the flux cannot collapse. If it collapses a large voltage is induced and that will affect or that will damage your switches flux cannot become 0 instantaneously. So, when I closed S current increases therefore flux in the core increases. So, when I open S flux in the core should fall or should reduce it tries to reduce. So, if the flux in the core tries to reduce d phi by d t is negative. So, when d phi by d t is positive dotted polarity was positive now if when d phi by d t is negative. So, d phi the dotted polarity is becomes negative see here in this circuit when I close the switch S current increases therefore flux increases d phi by d t is positive this polarity is also positive this is also positive. When I open S switch flux in the core tries to decrease tries to fall. So, this will become negative. So, this polarity will become positive. So, if this becomes positive that will for bias diode d 2 and current starts flowing in this circuit. When I open S sorry when I close S if current enters the dot when I open S in some other winding current should enter the dot. Remember in this circuit when current enters the dot current leaves the dot that should happen when both the coils are carrying current simultaneously flux produced by the secondary current should oppose the flux produced by primary. If both of them are carrying current simultaneously now only one coil is carrying current we found the flux produced if the current enters the dot in one winding current should enter the dot in either the second or a third or some other winding when you open the switch S. So, that is the right direction of the current d 2 starts conducting d 2 starts conducting whatever the stored energy or current starts flowing in d current flow is current starts flowing in the secondary circuit see here current starts flowing in this direction and it charges a stored energy is transferred to the output stage v 2. Explain to you using the simple transomer equivalent circuit see I had a circuit something like this plus v d c current enters the dot. Now, current can enter the dot current cannot so diode direction should be something like this current enters the dot. So, current can enter the dot here so direction of diode is something like this. So, if I am trying to explain to you using the transomer equivalent circuit. So, right now I will neglect the winding resistance and leakage inductances and let me take for simplicity 1 is to 1 transformer. So, how does it look like I have an equivalent circuit something like this only l m let this switch here if this is v 2 this could be a capacitor as well no problem no issues right now I am taking 1 is to 1. So, this also is v 2 this also diode drop v d itself when I close s this is the direction of current no current can flow in this circuit no current can flow in the secondary open s whatever the current that was flowing through the inductor will flow this fashion whatever the current that was flowing through the inductor will flow in the secondary that is one way to explain. So, I have explained you two different ways that there is a third way as well may be sometime one has to be extremely careful while connecting the diode in the secondary you need to ensure such that only one coil is allowed to carry the current at a time. So, when I close the switch current enters the dot if current enters the dot here you have to connect the diode such a way that it should enter the dot here in the secondary as well. So, when I open the switch sorry when I close the switch we determine the voltage rating of the diode now we will see the what is the voltage rating of the switch when the diode is conducting dots are negative why d phi by d t is negative. So, when the diode is conducting the voltage applied to the secondary is v 2 with dot as negative. So, voltage induced therefore, the voltage induced in the primary is now v 2 into the turns ratio with the dot as negative I will repeat when I close the switch voltage applied to the secondary is v 2 with dot as negative the flux in the core is changing it decreases linearly. So, voltage induced in the primary is now v 2 into turns ratio with dot as negative. So, the voltage that is appearing across s is when it is open v d c plus see dot is negative now v d c plus v 2 into the turns ratio that appears across s. So, while determining the voltage rating of diode and the switch s we need to know the turns ratio as well. So, voltage rating of s is v d c plus v 2 into the turns ratio here similarly the voltage rating of the diode is v d c into the turns ratio by the turns ratio definition changes here v 1 v d c is applied to n 1. So, what is the voltage induced in the secondary you need to find out plus v 2 is the diode voltage rating similarly when the diode is conducting voltage applied to n 2 is v 2 correspondingly what is the voltage induced in n 1 I need to calculate plus v d c is the voltage rating of the switch s. I will derive the transfer function of this converter and then I will take some question answers how do I derive the transfer function yeah one way is when I close the switch flux in the primary increases linearly and that is d phi is given by what could be the transfer function between v d c and v 2 in terms of the turns ratio when I close the switch s is closed in the primary v d c is applied to n 1 turns. So, what is the rate of or d phi by d t so n 1 d phi by d t is v d c and d t is for d into t seconds this is for d into t seconds now when I open the switch s flux in the core must fall. So, what the voltage applied to the secondary is now v 2 n 2 d phi by d t is v 2 or v naught whatever for 1 minus d into t. So, here d phi is v d c into d into t divided by n 1 volt second per turn that is nothing, but in the primary what is volt second per turn in the secondary that is d phi in the secondary is v 2 into 1 minus d into t divided by divided by n 2. This is positive d phi by d t of this is the increase in flux that is v d c by n 1 into d t volt second per turn in the primary this is volt second per turn in the secondary they should be equal. So, if I equate these two I get as get the relationship between v d c and v 2 or v d c and v naught what it could be. So, v d c into v d c divided by n 1 into d into t should be equal to v 2 into 1 minus d into t divided by n 2 t gets cancelled. Therefore, v 2 is equal to n 2 by n 1 into v d c into 1 minus d divided by 1 minus d divided by n minus d v 2 is equal to n 2 by n 1 into v d c d divided by what is this this is the transfer function of buck boost converter. So, this is nothing, but the voltage the value of the voltage source connected at the secondary if I connect a capacitor here at the secondary and I will call that as v naught v naught is equal to n 2 by n 1 into v d c divided by d into 1 minus d. So, this is buck boost converter transfer function in addition I have a term a turns ratio. Now, if there is a significant difference between or if I want a higher voltage a higher v naught v d c is low or the solar panel voltage is low here I want a high voltage which I may not be able to get using a ordinary buck sorry ordinary boost or a buck boost converter I can use this topology the name of the topology I will tell you sometime later I can use a transformer having a suitable turns ratio and I can get any v naught now. So, suitably by change by suitably using or by choosing a transformer of suitable number of turns ratio I can get any voltage at the output. So, looks like I have or we have find an solution to the problem that we faced in boost or a buck boost converter or a chook converter. I will try to take few questions here Saint Joseph Kerala see I the question says that the transformer core is magnetizing in only one direction after a long time is there any problem with the transformer yes definitely. By the way why did why did or when we teaching when we are teaching machines in the for undergraduate why are we saying that do not connect transformer to DC and here I have showed you that we have to use a transformer if there is a if you want a much higher voltage than at the input there is no way out. If I want to boost the DC voltage I have to use a transformer yeah we should not allow the transformer to saturate I need to choose the value of D in such a way that I should not allow the transformer to transformer to saturate. So, when I see the circuit equation when I closed as the equivalent circuit is something like this I may be using a transformer, but then only one coil is carrying current only one coil is carrying that is nothing, but a inductor itself nothing, but I should not allow the core to saturate I should not allow the core to saturate how to or take care of this issues I will discuss it during the next lecture yes if I if I use if I close the switch for a longer time definitely core will saturate I should not allow that to happen by the way I may be using a using it I have used a transformer, but then I am using it as an inductor. So, if I plot the B H characteristics we are operating only in the first quadrant see when I close the switch current increases therefore, the flux increases when I open the switch current decreases and therefore, the flux. So, B H loop is only in only in the we are operating only in the first quadrant entire B H loop characteristics we are not utilizing that is the limitation of fly back and other limitations I will discuss during this lecture yeah that is about the first question what is the second can I go to that NID Trichy which transfer function I do not have an expression for power output and problem no I am not able to I have no answer to your question in the sense you have a solar panel and you are connecting to the load you have you have made various combination as the input series and parallel or is there a power electron interface in between as I understand he says that he has some solar panel some of them are connected in series some of them in parallel and here is a power electronic interface and here is the output and he wants to derive a transfer function now you need to know here what is the now how do I derive the transfer function for this for me I will say that basically no I have no answer to this question how to derive a transfer function for such a combination I am taking this as there is some voltage here some voltage source here which is supplying power and I can derive a transfer function for input and output now you have some solar cells in series some in parallel. So, you find out what is the total power and what is the voltage rating and current rating and therefore, you may be able to find out what is the output power honestly I have I do not know the question is sir how do how to decide on the diode rating in a boost converter conventional what is the voltage rating of the boost converter you mean the voltage rating of the boost converter is nothing, but the output voltage itself this is the equivalent circuit this is V naught here is the diode if I plot if this is positive negative of V naught is connected to the anode. So, voltage rating of this diode is V naught itself V naught itself diode has to block V naught what are the advantages of current source at the both the input and the output of the DC to DC here as a good question see here say for example, a buck converter the moment I close the switch as the current this is what the current and if the current is continuous source current has to change from or change instantaneously to 0 to whatever the current that was flowing through the inductor at just prior to closing the switch. So, this current the source has to supply this current as a source has to supply see the d i by d t current that is flowing through the flowing through the source. So, what is being done is in the in the conventional buck you have to put some sort of a filter at in order to protect the course a small inductor and a small capacitor to take care of this d i by d t a small filter to take care of the source this d i by d t this filter is not required in case of a boost because input boost itself I have a current source current source. Now, see at the output side of the boost entire period the capacitor is discharging or capacitor is supplying power definitely V naught ripple will be much higher ripple in V naught, but if you see the output stage of a boost output stage of a buck converter see the inductor current inductor is supplying sourcing some current when when the switch is open. So, I have almost a current source at the output my filtering requirement comes down V naught size of the capacitor that is required may come down see here entire power is being supplied by the capacitor when the switch is closed the entire power is being supplied by the capacitor for d into t in boost period. So, if the if the entire power is being sourced by this capacitor voltage will fall. So, my filtering requirement comes down in the buck converter itself I told you to take care of d i by d t in the source I put a small filter small filter. So, initially the entire current is being supplied by the capacitor filter and slowly the source the inductor current source are supplying current if I understand this question correctly it is surat if s is open what is the potential of diode in secondary when s is open diode d is conducting. So, voltage across the diode is 0 or it is only drop when the switch is open in the secondary sorry when the switch is open in the primary I assume that current is entering the dot current can enter the dot diode is conducting diode is on diode is on diode is on ok when a current is entering. So, current is entering the dot current can and current should enter the dot. So, this is the positive of V naught positive of V naught see be careful while connecting the diode and the positive of the battery of the source this is a diode. So, voltage across the diode is 0 or whatever is on straight drop what will be the normal flux density in powder core while designing the inductor or transformer. Now, are you talking about a ferrite core ferrite core is of the order of it could be it should be less than 0.2 less than 0.2 if I using a ferrite core less than 0.2 m r first core you can go as I as 1 Tesla I have 1 question you know that solar radiation in north east India is so poor even in known time we are getting very very very poor radiation. So, I would like to design one DC to DC boost type converter which converts voltage 3 volts to 16 volts 3 volts to 16 volts to supply to my battery is it possible. So, which type of diagram will you prefer for getting better efficiency if the solar radiation itself is very low I do not know how much of energy that you can extract I would like to design one DC to DC boost type converter which converts voltage to 3 volts to 16 volts I will assume that I have a voltage source of the. Sir, can you derive the expression for power output of a solar terminal which is connected in series in terms of transfer function. Single stretch boost is should be enough transfer means actually suppose if a solar channel is where connected in series and parallel and connected to a load I just want to have an expression for power output and for this I want to derive a transfer function. 3 volts to 16 volts. Over to you sir. What is the operating frequency as function of transformer size? See the yesterday I did explain to you voltage rating V into A is 4.44 f into 5 5 m into m. Yes sir. Number of turns voltage induced into the current I said as f increases n decreases. So, size of the transformer comes down as the frequency increases, but then as the frequency increases you may have to choose different type of course. For a very high efficiency you may have to use the air core then your coupling is very poor the leakage flux will increase there are various issues. Let me tell you one thing often I will not be able to answer this question you try to understand the issues. This is the basic equation as f increases n comes down, but then as f increases I may not be able to I may have to go in for either ferrite, amorphous or air core because amorphous ferrite core and amorphous core they have their own hysteresis loop. So, my losses will increase core losses will increase efficiency will come down. Now, if I use a ferrite air core my core losses are become 0, but then my leakage flux will now increase coupling will be poor. So, you know there is a tradeoff how will the wave form of current in the primary and secondary of the transformer do not worry I will discuss it now how will the wave form of current in primary and secondary of the transformer why though I have not done I will discuss it in the next lecture, but then why there should be a doubt a DC is being applied to primary primary inductance is l 1 a constant DC is voltage constant DC voltage is being applied to l 1 current increases linearly provided the inductor is not saturated V DC is applied to n 1 turns which is having an inductance self inductance not leakage self inductance l 1 d i 1 by d t constant constant if it is not saturated. So, if I say that current increases linearly if I assume a discontinuous conduction now what is discontinuous I will discuss it in my next lecture current increases linearly it is triangular when I open the switch current become 0 in the primary immediately this is the current flowing in the primary this current will start flowing in the secondary the peaks will change now I need to take care of the turns ratio do not worry I will discuss it in my next lecture in high frequency transformer core loss increases how can we select a proper design I asked I already answered in high frequency transformer core loss increases yes it has hysteresis loop now one way is to use the material which has no hysteresis loop air core leakage increases. So, that is the reason I told you yesterday whatever that you do in life you have to pay a price what do you mean by in air core what about the flux density range the B H characteristics of air core is straight line is not it I cannot saturate it, but then what is the current that is required to establish this flux you need to see because because my leakage is going to be very high now there is no coupling there is no coupling there is no coupling there is no coupling there is no fear of getting the core getting saturated, but then but then I may be able to see the value of inductance also that I can get is now very small or I may have to use a large number of turns to get large number of turns to get in order to get a suitable value of L my size of the inductor definitely will increase. So, if S is closed what will be the current in the secondary current enters the dot current enters the dot the right direction if for the current in the secondary if it is carrying simultaneously is current to leave the dot it cannot leave because of this diode. So, when the primary is current in this topology in this topology secondary current is 0. Now, you may say that why not why not reverse the direction now if you just see here if I put something like this if I do something like this current enters the dot when I close the switch current can leave the dot. So, both the coils are carrying current simultaneously there was some current here there was some sorry when I close that the current increases. So, there is voltage induced some current will flow. So, this current is in the very first cycle in the very first cycle magnetizing current plus the equivalent secondary current there was some current flowing here I will open the switch after some time what will happen can I open the switch I may not be able to open the switch because there is no path for the magnetizing current to flow there is no path for the magnetizing current to flow in this configuration. So, there will be a huge spike and circuit will get damaged by the way in this configuration. So, here I need to ensure that only one coil is carrying current at a time and I have to collect and accordingly I need to connect this diode, but in this configuration if I connect a diode in such a way that both of them are carrying current simultaneously I have a problem the problem is see I if I use the equivalent circuit approach both the coils are carrying currents simultaneously this is something like this this is a transformer equivalent circuit I have neglect winding resistances leakage flux I will use one is to one transformer I have something like this both the coils are carrying currents simultaneously magnetizing current is also flowing some equivalent secondary current is also flowing I will open the switch now here there is a switch I will open the switch S what will happen what will happen to the current that was flowing in the inductor there is no path here. Let us see as we go along what modification can we make in this path circuit do not worry no the question is output ratio depends on N 2 by N 1 how much maximum ratio is allowed you can have what is the problem V naught I have derived that N 2 by N 1 V dc divided by d divided by 1 minus d now generally value of d is limited to 0.5 why I will tell you d is limited to 0.5 now if you want a higher voltage you choose the circuits circuit components accordingly and choose N 2 by N 1 d is generally limited to 0.5 for various reasons we will see do not worry. So, you choose N 2 by N 1 select select the voltage rating of the diodes and switches accordingly circuit should work no I do not know if the input voltage is boost convertible 12 volt dc is it possible to obtain the voltage up to 600 volts answer is no what sort of a boost I told you the ratio cannot be 7 to 10 that to maximum ratio that I can get if I using a very good inductor I might have told this at least 3 to 4 times what is the range the boost converter can give the boost output voltage by using a voltage double or tripler or combinations of this we can we run an ac from 100 watt solar panel no I do not know how can you run an ac from 100 watt power there should be a power balance you may be able to boost the voltage but then input power should be equal to output power provided efficiency is 100 percent what will be the output power if the input is 100 watts and you do not know how are you going to convert what is the overall conversion efficiency assuming it is to be high I do not think you can get more than say 85 to 90 watts I do not know with this 95 watts if you able to run an ac there should be a power balance or input output power is equal to input power minus the losses I will stop here.