 DC to DC conversion basically these are building blocks. We may be able to use this building box directly or some modification we may have to do while integration. So, right now we will just study this building blocks and may be towards the end depending upon the power level when we integrate various blocks we will study the modification that we need to make in this building blocks. Basically the entire DC to DC converters can be broadly classified into two groups with transformer without transformer. Again remember I am using DC to DC conversion, DC to DC conversion and they are classified into two groups with transformer and without transformer. I have to how the transformer can be used in DC to DC conversion that question we will answer sometime later. But then yes we may be we are using a transformer in DC to DC converter or we are using an inductor in DC to DC conversion and the frequency generally the depending upon the power the frequency does change. Let us see what is the effect of using this or what is the effect of switching frequency on the size of the converter. So, if I say the V a rating of the transformer is given by this equation 4.44 f into 5 into n this is nothing but the voltage induced in the coil into the current. For a same V a rating number of turns come down number of turns reduces as the frequency increases. So, therefore as the frequency increases n come down but then as the frequency increases if the core has a hysteresis loop my core losses increases. So, therefore I may not be able to use the core that is used in 50 hertz conversion in high frequency power conversion. So, in high frequency power conversion I need to use a core which is having a very small hysteresis loop. Generally a ferrite core is used the problem with the ferrite core is it is very brittle and the operating flux then it is very low it could be of the order of run 0.2 tesla 0.2 tesla of course. Then now amorphous alloy cores are also available average B average is of the order of 1.1 tesla. So, these are the two cores are popularly used in DC to DC conversion. Coming back to a buck converter may be all of us know I will quickly cover this the circuit configuration is given here switch S is switched at a very high frequency L f L is the filter inductor, it is written L here is the filter inductor, C f is the filter capacitor, V naught is assumed I have to regulate the output voltage. I will repeat I have to regulate this output voltage this voltage may be directly coming from the solar cell and it could vary. Now how do I analyze this circuit switch is on turned on off at a very high frequency it is on for D into T where T is the switching frequency D is the duty cycle and it is off for 1 minus D into T. When the switch is on the equivalent circuit look something like the equivalent circuit of this buck converter look something like this diode is reverse biased. I told you V naught we need to regulate V naught and in one switching cycle I will assume that even this voltage remains constant. So, under that condition when the switch is on voltage across L voltage across L is V dc minus V naught I will repeat when the switch is on voltage across L is V dc minus V naught. I told you in one switching cycle V dc is assumed to remain constant V naught will remain constant because that has to be regulated. So, therefore V dc minus V naught is a constant. So, voltage across the inductor is constant during this time D into T. So, if the voltage remains constant inductor current will increase linearly and at the output stage I need to apply k c l at this node it is given by I L is equal to C d V naught by d T plus V naught by R V naught by R the k c l at this node will give I L is equal to current flowing in resistor R the load that is V naught by R. It does not matter output stage could be anything I can have a inverter as well, but then I will represent that inverter feeding a load as a current source. See I will repeat here I have connected a resistor we can have an inverter feeding some other load as well I will represent that inverter along with the load as a current source which is equal to I naught in this case it is V naught by R see d V by d T is a capacitor current. So, when the switch is on voltage across the inductor is V dc minus V naught what happens or what is the voltage when the switch is open. When the switch was closed current flowing this is the direction of current that is flowing through the inductor we know that inductor current has to be continuous when I open the switch direction of current will still remain the same and therefore it will flow through the diode d. So, when the switch is open voltage across the inductor is now is the output voltage V naught itself, but then with respect to the negative dc bus voltage across the inductor is now minus V naught minus V naught plus is here negative of the output or negative terminal or the ground potential is the reference terminal is connected to this point. So, voltage across the inductor is now minus V naught I told you V naught is regulated suppose to remain constant therefore voltage across the inductor is constant current decreases linearly the same voltage equation sorry current equation at this node I L is equal to C d V by d t plus. So, how do I determine the transfer function I will use the condition that average voltage drop across the inductor at steady state should be 0. So, in other words V dc minus V naught into d t that is voltage across the inductor during t on should be equal to V naught into 1 minus d into t. So, average voltage across the inductor is 0 this is the voltage condition. So, therefore, if I simplify I will get V naught is equal to d into V dc very popular anyway. Now, this is the average current I will average you can see d V by d t average for a V naught by R at steady state at steady state average value of the current through the capacitor should be 0 otherwise capacitor voltage will change I told you the V naught is regulated. So, therefore, this current has to be 0. So, I L average is equal to V naught average by R these are the various wave forms voltage wave forms switch on switch off inductor current increases linearly decreases linearly how does the source current look like I s switch current. See the moment I close the switch whatever the current that is flowing through the diode and inductor now starts flowing through s I have assumed a continuous conduction I am saying continuous conduction in the sense current through inductor is continuous remember that in other words at any given time there is some finite current flowing through the inductor. So, the moment I turn on the switch that current will start flowing through this switch. So, instantaneously the source current sorry or current flowing through the switch look something like this. So, this is the inductor current varies linearly this part of the current source supplies. So, this is the current wave form. So, if you see here remember this is one of the limitations of this buck converter source current instantaneously has to change from 0 to whatever the current that is flowing through the inductor in other words even the source even the device has to carry this current when the switch is turned off whatever the current that is flowing through the switch or whatever the current that is flowing through the inductor it starts flowing through the diode this is the diode current. So, at this instant in the second cycle switch is turned on again. So, diode current instantaneously will become 0 instantaneously will become 0 and this current this cycle will repeat and this is the voltage or voltage wave form across the capacitor V naught. By the way see there is a flaw in our analysis in one point I am saying that V naught should be regulated V naught is supposed to remain constant, but then here I am showing capacitor voltage does change yes. Now, you need to understand in what context I am saying that V naught will suppose to remain constant for all practical purposes I am assuming V naught will remain constant for all practical purposes. In other words output voltage ripple has to be regulated within less than 1 percent or so. So, I can assume that V naught will remain constant, but then there is some current when the switch is close some current will flow through the capacitor therefore, capacitor voltage will increase and during sometime capacitor will discharge and therefore, the capacitor voltage will fall. When does the capacitor charge and when does the capacitor discharge this condition is given by the Kirchhoff's current law. So, if I use KCL here I L is equal to current flowing through the capacitor plus current flowing through the load. If I represent current load as a constant current source I will repeat I will assume that load is a current source having a constant magnitude. If the inductor current is I will repeat and if the inductor current is less if the inductor current is less than the this current source the remaining current capacitor supplies. If the inductor current is more than the value of I naught the additional the difference between these two currents will flow through the capacitor. So, therefore, KCL will tell me whether the capacitor is charging or not. If I L is higher than I naught C will charge if I L is less than I naught C will discharge that is all. See in this case this is the dotted line shows the value of the current source or value of the current that is flowing through the load. This is the capacitor this is the inductor current in this period inductor current is less than I naught. So, the remaining current has to come from the capacitor. So, therefore, capacitor is discharging output voltage C v naught is reducing. In this zone see in this period in this period inductor current is higher than the load current the I naught. So, the remaining current has to flow through the capacitor. So, therefore, capacitor voltage will increase. So, remember just because you have turned on the switch here it does not mean that capacitor will start charging. No KCL will tell you whether the capacitor will charging or not not the conduction state of the switch. Similarly, here just because you have turned off the switch does not mean that capacitor is discharging. Again the KCL will tell you whether the case whether the capacitor is charging or discharging in due to apply it stops voltage it stops current law. So, see here switch is open in this period since the inductor current is higher than I naught the remaining current is flowing through the capacitor and therefore, is charging here. Now, how do I choose the voltage current voltage and current rating of the devices? So, what is the what is the voltage rating of this diode? Voltage rating of the diode is nothing but the supply voltage itself here VDC itself or the whatever the maximum solar voltage whatever that happens when the switch is closed here. So, all the voltage appears across the diode and diode has to block and what is the voltage rating of the switch S that I encourage you to find out draw these two equivalent circuits and find out what should be the voltage rating of the switch. This equivalent circuit will tell you the voltage rating of the switch. See here the entire supply voltage comes across the switch. So, therefore, voltage rating of the switch is determined by the supply voltage itself. This is a input output losses input output sorry input output power condition I am assuming the ideal condition no losses are 0. So, this is the input power VDC into I S should be equal to V naught into I naught average values. So, I S is D into I naught by the way this is the average current rating. So, weak current rating you need to determine from you need to determine from this waveform. So, what could be the what should be the current rating see the see this I S has to flow through the device at this point device is turned off. Whereas, I naught is this is the average value of I naught. So, you need to take while choosing the current rating and voltage rating you should be bit extremely careful. Now, how do I choose the value of L and how do I choose the value of C? The value of C will determine the ripple in the output voltage and value of L will determine the ripple in the current. See now, let us let us we have an expression for the current ripple in the I L. So, D L by D T T D L by D T is given by VDC minus V naught divided by L when the switch is closed. So, therefore, I max is given by of course, this is a this equation of this straight line with this I mean as the y axis intercept I mean plus. So, I can determine the value of I max is given by this equation I max minus I mean will give the total ripple this is when the switch is open current falls linearly this is the voltage this is the this equation will give you the rate of change of current. And therefore, current ripple is given by where does the ripple is maximum? You find out how do I find out D L is given by this. Now, D L by this L maximum and D is equal to 0.5 that maximum ripple is given by VDC by T divided by 4 L. Now, you put a limit on this T is the switching frequency you know the DC link or you know the supply voltage. And therefore, you can calculate L I will repeat this is the maximum ripple that a designer has to select T is the switching frequency VDC is the supply voltage. In this case you can take an appropriate voltage and that will give you the value of L. Now, capacitor voltage ripple this is the variation of inductor current and this is the variation of capacitor current I told you KCL will tell you whether the capacitor is charging or discharging in this zone delta T by 2. So, this is entire period is delta T this is delta T by 2. In delta T by 2 capacitor is discharging because I naught is less than I naught is less than the inductor current. So, capacitor has to supply that current at this point at T is equal to DT by 2 capacitor current is 0 because inductor current is equal to 0. I naught and beyond that beyond DT by 2 capacitor inductor current increases therefore, capacitor is charging. So, this is the capacitor current waveform and this capacitor if I know the capacitor waveform I can draw the capacitor voltage waveform. I c is given by this is a voltage the current equation for the current flowing through the capacitor I c is given by this equation this is delta I L by 2 this is entire period is DT this change is linear. So, this is the equation for the capacitor current and from DT to T capacitor current is given by now capacitor voltage is V naught max minus V naught min. So, 1 over C you integrate I c DT therefore, delta V naught is equal to 1 over C delta I L T by 8 this is the expression for delta V naught. Now, you put a limit on delta V naught delta I L you chose the previous case yes it is a iterative process because delta I L will see if you see this value here inductor value will the moment I choose L delta I L gets fixed, but then here if you see delta I L will determine delta V naught as well. So, you have to may have to go back and forth iteratively and calculate the value of C and L to get the required ripple output voltage and required ripple in inductor current this about the buck converter voltage and current ratings I can determine from by seeing the waveforms the value of L and C I can determine by fixing the value of the current ripple and the voltage ripple T is the T corresponds to the switching frequency. So, this what it is G is equal to 0.5 maximum notice I do not want to discuss about the discontinuous conduction not very important may be I will discuss the boost converter where do you want to use boost converter let us not discuss now we will see sometime later integrating then various power electronic equipment right now we will just study the subgroups the circuit topology for a boost converter is given by is shown here the switch is turned on off at a very high switching frequency close S when I close S D F gets reverse biased because sorry this diode gets reverse biased because output voltage appears across D naught sorry appear across D. So, diode is reverse biased in the input side supply voltages or whatever the solar energy voltage appears across L when the diode is off capacitor is supplying power to the load in other words capacitor is discharging. So, this is the equivalent circuit voltage across voltage applied to the inductor is constant because of course, it may change in real life it will change, but then since the switch is turned on off at a very high frequency in one switching cycle V dc is expected to remain constant. So, therefore, in one switching cycle voltage across the inductor will remain constant therefore, current increases linearly capacitor is supplying power to the load so capacitor will discharge see here V naught is expected we are expected to regular to V naught, but then in this case what happens is capacitor is supplying power. So, capacitor voltage will fall. So, now you need to fix a limit on delta V naught and therefore, and thereby calculate the value of C. So, remember insert some applications or while deriving the transfer function I have to assume that V naught will remain constant and I have to change I have to take this change in V naught into account while calculating the value of L and C. So, remember that I will repeat while calculating or while determining the transfer function I will assume V naught will remain constant, but then when I am determining the value of C and L I have to take the change in V naught and change in I L into account. Now, open S in this case prior to opening the switch switch was closed current was flowing through the inductor inductor current has to be continuous. So, therefore, this inductor current starts flowing through will flow through the capacitor and arc. Now, voltage applied to the inductor here voltage across the inductor is V d c minus V naught here it is V d c. When I close the switch current flowing through the inductor increases. Therefore, when I open the switch current flowing through the inductor should fall only then I can have a steady state. What is the steady state? Say I will repeat this is I will take T is equal to T 1 this is T is equal to T 2. If the both the values of current at T is equal to T 1 and T 2 are same then I can say that circuit has attained almost a steady state. Now, this can happen only if only if current falls here because when I close the switch current increases here also current increases. So, that can happen only when voltage across the inductor is negative voltage across the inductor is negative and voltage across the inductor can be negative only if V naught is higher than V d c. If V naught is higher than V d c voltage across the inductor will be negative. So, in other words for steady state operation V naught should be higher than V d c. I will repeat for steady state operation V naught should be higher than V d c. So, hence the name boost converter. What am I doing after all? I am storing the energy in inductor transferring the capacitor. So, therefore, capacitor voltage has to increase. So, it will increase hence the name boost converter. These are the voltage equations they are very simple you can I am going to do to discuss them. If I equate them average voltage across the inductor should be 0 V d c into d into T is equal to V d c minus V into V d c minus V naught into 1 minus d into T. If I equate I will get this transfer function V d c divided by 1 minus d. So, for any value of d V naught is higher than V d c. What happens to V naught as d tends to 1? As d tends to 1 V naught tends to infinity see here when d tends to 1 V naught tends to infinity. But then what is the reality? d tends to 1 it implies the switch is permanently closed and this is the equivalent circuit. So, in this case when the switch is permanently closed inductor is permanently connected across this d c supply and output capacitor is continuously supplying power to this load. So, what will happen? Three things can happen inductor will get saturated if inductor get saturated s switch can fail inductor can fail the source can fail. Because if the source is saturated current is limited by the internal resistance itself, but in this case it happens to be I s c the short circuit current of the solar panel and if it is less than the current rating of s nothing may happen. But at the output stage what will happen? Output stage capacitor will continuously discharge and this voltage will tend to 0. So, where is the flow? In real life if I analyze this circuit it will tell me by inspection which if it tells me that V naught tends to 0 when d tends to 1, but the transfer function tells me when d tends to 1 V naught tends to infinity. So, how can there be such a huge difference between an ideal case and a non-ideal case? If I just see the circuit here V naught tends to 0, if I see the transfer function V naught tends to infinity where is the flow? The flow is that how do we derive this transfer function? We derive the transfer function making certain assumptions. What are the assumptions? We assume that V naught and V d c are constant and ripple free. Can V naught be ripple free for high values of d? V naught cannot be ripple free for high values of d. Why? As d increases capacitor is continuously supplying power to the load. So, that is for capacitor voltage has to change. So, our assumptions are not valid for high values of d. For low values of d they are valid. For high values of d I cannot take the ideal conditions such as see I have represented this inductor. I assume that L is ideal. This is true for low values of d. For high values of d I have to take all the non-idealities into account. I have to take the resistance also into account. Now, if I take the resistance into account or non-idealities into account what sort of a transfer function will I get? See here I have taken this R into account and I will write the equation same equation of V d c is equal to R into I L plus L d A by d t. This is the K c l at the output stage and again this is K v l and K c l. Using K v l and K c l I get these two equations and if I simplify them now taking the average values average of this is 0, average of this is 0 at steady state I will get this function. This is the transfer function taking the binding resistance of the inductor into account, binding resistance. Now, see if I put d is equal to 1, v naught tends to 0. If d tends to 1 even v naught tends to 0. Now, if I substitute R is equal to 0 then there is nothing, but an ideal case. I have a transfer function for an ideal and boost v d c divided by 1 minus d. I will repeat see I have just taken this R into account in a non-ideal case. If I equate it to 0 I will get a transfer function for ideal. This is the transfer function for non-ideal when d tends to 0 v naught tends to 0. So, if I plot both the voltage and functions it looks something like this. This tends to infinity for ideal boost and this is for non-ideal boost. When d tends to 1 output is equal to 0 for d is equal to 0 ratio is 1. So, it increases for slow values of d reaches a peak and falls down and this peak depends on see here. I have differentiated this equation and equated to 0. I get v naught max is equal to v d c by 2 divided by R by R. So, this maximum value that I can get depends on the ratio of the load resistance to the inductor resistance. Smaller the resistance higher is the value. Now, how do I reduce the value of R? I will show you two different inductors. Same value of L and the technique that is used to reduce the winding effective winding resistance. How we have tried to reduce this value of R? If I try to reduce the value of R my v naught the maximum voltage that I can get can be increased. Here lower the value of R higher is the value of but thumb rule is that the ratio we can get a maximum ratio that we can get is 10 maximum 8 to 10. So, if there if you want to boost the existing voltage by a factor 8 or so you can use the conventional boost. Otherwise we have to really think of a alternative solution. What is that alternative solution? We will see sometime later. This is the filtering requirement same. I will quickly I will do. This is the source current waveform for a buck converter. I said the moment as soon as in a buck converter the moment you close the switch see here. The moment I close the switch whatever the current that was flowing instantaneously it has to flow through the switch S or to the source here. So, there is a step jump in the source current. So, the so called d i by d t that is flowing through the source is really high very high. So, I may have to use a special filtering technique or special filters may be required at the input whereas, what about for a boost converter? Boost converter is source current is how do I represent this see the equivalent circuit of the boost converter when the switch is on and switch is off. This inductor is always there in the circuit always the same current see here the current increases linearly and current decreases linearly. Again I close the switch again it starts increasing. So, I can represent this source current input stage this stage as a current source. Yes any electrical engineer will when he sees this part a DC voltage an inductor you can represent this by a current source. So, I have almost excellent input characteristics a current source, but then see the filter output stage the filtering requirement here the capacitor is continuously discharging during when the switch is open this capacitor is continuously discharging. Definitely capacitor voltage will fall see here in buck converter since at any given time some current is flowing through the inductor. So, ripple here for same current may be less compared to boost converter because here capacitor has to continuously discharging capacity continuously discharging this sort of equivalent circuit I do not get in buck converter. So, these are the various wave forms and using these wave forms I need to calculate the value of L and C say ripple in V naught and ripple in I naught neglect R and delta V naught to determine I delta I L I am neglect in a ripple in the output voltage to determine I L this is the voltage simplify solve you will get this equation. So, delta I L is given by V dc by L into delta into t proportional to d similarly now I calculate the ripple in V naught for that I will neglect the ripple in I L this is the voltage equation and delta V naught is given by this equation. I will just show you this is a capacitor current wave form continuously supplying see here when the switch is open KCL will tell me KCL will tell me when the capacitor is charging and discharging. Here when the switch is open all the capacitor all the load current has to flow the capacitor it will discharge here when the switch is open KCL will tell me whether the capacitor is charging or not if this inductor current is higher than I naught capacitor will charge if inductor current is less than I naught capacitor will discharge this is what I have shown here capacitor is discharging and in this case for some time capacitor is charging this is true only for certain conditions otherwise you need to apply KCL and only from there you need to determine whether the capacitor is charging or not this is the expression for V naught and delta I L and suitably give the suitable values for delta V naught and delta I L and calculate the value of C and L. Third one is the buck boost see all these are quite popular it does see the circuit verification close S VDC applies gets applied to L at the output stage capacitor is supplying power something similar to boost something similar to boost open S whatever the current that is flowing through inductor will continue to flow through S something similar to buck but then when the switch is open the equivalent circuit is not something similar to buck but something similar to boost here when close S inductor is charging similar to boost output is capacitor is supplying power similar to boost when I open switch S this is the voltage and current equation same procedure that you need to follow and equate it transfer function is given by VDC into D divided by 1 minus D this is the transfer function. So, if a D is less than 0.5 V naught is less than VDC when D is equal to 0.5 V naught is equal to VDC and D is greater than 0.5 V naught is higher than VDC. So, I can get all three values less equal or more but then only difference is see the reference point the capacitor output voltage this terminal is negative with respect to this point if I take this as the reference capacitor voltage is negative with respect to that. So, this issue has to be taken into account while designing the solar I will take up few questions as a very popular converter cook converter may be I will start in the next class. Yes, NIT Kurukshetra go ahead with your question. Sir, which is regarding your previous picture my question is that for I D D T regarding S O A is this different from MOSFET S O A and if it is different then why this difference is there. No, no, no, same S O A is the same same S O A as MOS. S KG SOMAYA, Mumbai. Sir, normally when we go for a converter the requirement is that output voltage should be controlled. When we go for any converter DC to DC conversion requirement is the output voltage at the load should be controlled according. So, in converter the output voltage has to be controlled depending upon the availability of the input DC voltage. Here what exactly is the nature of the load because is it a voltage that we want to apply somewhere or it is a voltage from a solar photovoltaic which we want to control. So, the output load condition is not so clear. No, no, no, see depending upon your end use application now you may be charging the battery or you may be converting this AC to DC. You can either charge the battery or you may be feeding any other load does not matter does not matter does not matter my input voltage I said input voltage does change V DC is just I represent V DC that V DC actually represent the solar voltage does change as the insulation changes but then in one switching cycle it is expected to remain constant. See my sincere request to is these are all building blocks when we integrate we will see what happens whether DC to DC converter has to regulate this voltage or whether inverter has to regulate the voltage we will see who will do this regulating to whom this task is given whether to DC to DC converter has to regulate or whether inverter has to regulate we will see when we integrate it. See when I have to feed the power to the grid looks like inverter may have to do the voltage regulation see the things do change all these issues cannot be explained at this stage because I said one student ask how do I synchronize the inverter to the grid see this question may be answered towards the end. So, voltage regulation who does as of now buck converter is doing or boost converter is doing or buck boost is doing when we if when we are feeding power to the grid looks like this control may have to be taken over by the inverter. So, as of now yes inverter the buck boost or buck boost converter has to do it when we integrate it things may change things may change because see this is an interfacing stage between solar panel and the inverter I will I will just repeat I will tell you this is a this is a one block between solar panel and the inverter this buck converter this DC converter may have to just to extract the maximum power from the solar panel that may be the task that is all who has to do the regulate who has to regulate the DC link voltage we will worry sometime later. V I T weller go ahead with your question sir in the case of this buck converter or in the case of buck converter sir first this when the switch is open in the switch is open the capacitor and the inductor will come in parallel they will act as a voltage source for the load am I right. When the switch is open the equivalent circuit is something like this this is L yeah. If the inductor value and the capacitor value is increased we get a better regulation but the problem is the source should be should be able to provide. Provide. Provide what? Current in a very short time to the capacitor so for better regulation it is not just the inductor and capacitor values but the source ability to provide the power. Now I agree now value of L value is chosen such that see you have to first you have to fix the value of delta I L the variation in the inductor current ripple that will determine the value of L capacitor value will determine the value of V naught the change in V naught C is delta V naught L is delta I L. Now it is expected that source will be is response to whatever the in whatever the output wants it is expected that. Right thank you sir. NIT Varangal. Yes sir I want to know the maximum ratings of IGBT that is presently available sir and one more things are in do the switching frequency should also be considered while you are designing the DC DC converter. Yes switching frequency depends on the power rating smaller the power rating higher can be the switching frequency higher is the switching frequency. I told you see just now we did we have discussed three types of converters as the as the voltage rating as the see there is we derived a transfer function V naught is equal to D into V DC V naught is equal to V D V naught V DC into 1 minus D and third one is V DC into D divided by 1 minus D. I told you maximum ratio in this case I can get is 7 to 10 what do I need to do if this ratio is higher similarly what happens as the power rating increases can I use this conventional topologies or do I need to make any alternative changes we will see in detail as the power rating changes the configurations will change as the voltage ratios changes we may have to change this we may not be able to use this converters directly. So, right now we discuss the converters without transformer if there is a large voltage ratio we may have to go in for converters using transformers. Again transformers in DC to DC not DC to AC yeah right now we use only internal capacitor if they if I want a higher ratio you may have to use a transformer as well in DC to DC all these issues depending upon the issues like what could be the switching frequency what is the power circuit configuration depending upon power level voltage level we will discuss at the appropriate time may be sometime the next lecture definitely I will take up if I have a large voltage ratio what do I need to do yes we will see take for example if the voltage solar panel voltage is 24 volts. I wanted to convert into 230 AC RMS can I use this boost I may not be able to use this boost directly something else you may have to do what we have to do we will see. Manipal Institute please go ahead with your question. You are telling about this boost converter whatever we have discussed in theory VDC input voltage limitation if that is up to volt voltage level this is applied whether it is a power transfer at what level L or kilo watt level or mega watt level. What is the limitation there what I told you was V naught is equal to VDC divided by 1 minus D this is for ideal whatever and this maximum ratio I can have maximum ratio I can have is of the order of 7 to 10 maximum ratio V naught by VDC sorry V naught by VDC maximum ratio I can have is of the order of 7 to 10 provided I design a very good inductor very good inductor in the sense it has a large Q factor it has a very high Q factor if I have I may be able to get of the order of 7 to 10 how to get that a good Q factor I will show you the inductor may be in the next class I will forgot to bring I will bring it it depends on R by R R is the winding resistance of the wire that you are winding resistance of the coil smaller the resistance higher is the Q quality factor. So, it depends on the quality factor of the of the inductor higher the better how to increase this we will see what you may have to use the leads wire leads wire not a just an ordinary wire conductor you can you may not if you use a just an ordinary conductor to make an inductor you may not be able to get a high Q and you may not be able to get this factor 7 to 10 may not you may not be able to get this factor if instead if you use a leads wire wherein you can get it is possible to get a high values of Q you may be able to achieve this ratio 7 to 10 it is a strong function of the quality factor of the inductor with this ratio for a higher power transmission input voltage VDC only should be increased for example if it is in a kilowatt range VDC that means the solar power generation to be connected in series and we should add up to the voltage level VDC input to be higher to this boost converter so 10 times it can make the implication that it can be applied in that way you circuit my my sincere suggestion to you is actually apply only voltage level if for example 220 volts VDC we could able to connect for example when the power transfer SV2 versus say the grid connection kilowatt level or megawatt level power has to be transferred can we apply this principle of VDC increase that means SVV connected in series so that we can have input VDC of a 1000 volts and 10 times so you can go up to the 1000 volts and that way can we do in this type of circuits. Sir I have got a limitations on the voltage Sir be with this question for some time we will discuss this when we discuss the grid connection the things will change completely when we are doing grid connection grid feeding but I sincerely request you is be with this question you ask this question when I am doing grid connection what exactly we need to do because most of the concepts may remain the same circuit apologies and our school of thought might change might so let me not let me not answer this question what we have to do when it comes to grid feeding I think I will answer this question when we do the grid feeding topics at that appropriate time please be with this question for some time sir since we are discussing about the high frequency switching is it not required to consider the switching losses yes definitely definitely whatever that you do in life you have to pay a price okay if you want to reduce the size of the power electronic converter okay I have to reduce I have to increase the switching frequency higher the switching frequency is smaller the value of L higher the switching frequency higher will be the switching losses fortunately there are techniques there are techniques to minimize these losses what is known as the soft switching techniques the moment I use the soft switching techniques complexity increases so you remember to reduce the size of the power electronic equipment I have to increase the switching frequency if I increase the switching frequency my losses increases okay so therefore losses increase of my efficiency will come down my heatsink requirement may increase fortunately I said there are techniques where I can try to minimize these losses by using soft switching technique the moment I use the soft switching technique my complexity increases so that is why I said whatever that you do in life you have to pay a price switching losses will be there that they can be minimize using soft switching what are the soft switching technique if the time permit I will discuss towards the end right now we are doing what is known as hard switching that my question is what is the role of the parasitic resistance connected in series with no parasitic role they are inherently present this can I have an ideal inductor I cannot have an ideal inductor the moment I take a conductor to make an inductor there is definitely there is resistance rho L by A higher the value of resistance smaller the value of Q quality factor and therefore I may not be able to boost it to a higher value this resistance the moment I use I take a conductor to make an inductor there is going to be resistance now the question is how to reduce this resistance there are techniques in the next class I will bring a two inductors and I will show you how to reduce this value of R thereby increase in the quality factor. Good morning sir there is another question Can we connect two converters in parallel any DC DC converters in parallel and what are the issues behind that? Yes why not in the sense interlude interlude back there are there are techniques as the power rating increases or you know to reduce the current ripple flowing there are techniques interlude back you may be knowing them to two buck converters are connected in parallel that my switching my for a given switching frequency the current ripple reduces significantly the control issues I do not know whether I will be able to discuss them or not you have asked that question if time permits I will I will discuss it sometime later as of now I have no plans of discussing interlude buck and all those issues but then I am telling you I can I can tell you the there are there is enough literature available I can put it on the module and you can refer interlude buck buck converter all those converters are well established anything else thank you