 I stopped at discussing the gate drive characteristics. What are the important characteristics that we need to incorporate while designing a gate drive? Please try to understand the issues that are involved for a BJT. Only then you will be able to appreciate for a MOS and IGBT. I said it requires a faster turn on, faster turn off. Why do I require faster turn on, faster turn off? They are required because to reduce the turn on and the turn off losses. Faster the turn on smaller is the losses. Then I need to operate the device in quasi saturation. I should not operate the device in saturation. Why? I will tell you and I should have a inherent short circuit protection. The moment the collector current or the moment the drain current in a MOSFET increases, I need to turn it off by using the gate drive itself not by using external means. These are the characteristics. See here the base drive current in the sense it does not remain constant in the entire on period. There is a small high pulse of current. It falls say by 30 to 40 percent then remains constant and again say again it does not change instantaneously. There is a certain rate, the positive IB becomes negative IB and then it becomes 0. See the profile of collector current. Collector current slowly increases, remains constant and see here IB from positive to negative it has become but then collector current is still approximately the same. See the IB as from positive to negative it has become IB is approximately the same and after sometime it starts falling and see the VCE profile. VCE from rated voltage it was blocking, you have turned on, it falls towards saturation value and after what is known as a TS, TS is the storage time. What exactly this we mean I will tell you, TD is the delay time. You have applied, you have applied the positive gate drive, collector current is still 0, it start slowly rising and it around 0.1 of the rated. This delay is known as the delay time or this is the rise time TS is known as the storage time. You have applied the negative, you want to turn off the device here. So, you applied the negative gate drive, current has remained, current remains constant for some time provided you operate the device in saturation, it may not happen if you operate the transistor in active by the way we do not operate in active and after sometime the current becomes current starts falling. So, this time is known as TS storage and this is the fall time TF. So, TD corresponding to these are the various definitions is there in any one of the text books is there in the slide you can read and understand. What is the storage time in the sense when it happens, when the transistor is driven deep into saturation, when the transistor is deep into saturation storage time see here when excessive number of charges are injected into the base. See that see here as you go on pumping in more and more charges into base, let me tell junction see what happens to this layer N plus and N minus layer is there because of the voltage rating as the voltage rating increases N minus the thickness of N minus layer increases. So, as the degree of saturation increases see here this layer completely vanishes. See the difference in these layers as the degree of saturation increase the active quasi saturation and device driven deep into saturation. So, if I operate the device into deep into saturation this TS increases storage time increases. Now if I in addition what will happen, if I apply a instantaneous positive negative I B here negative I B here what will happen is negative I B means I am pumping out the charges from the base emitter junction what will happen is I just show you the slide see. If prior to turn off the transistor is over saturated rate and if a large I B is applied rapid evacuation of the carriers at the base will occur you are rapidly evacuating the charges. So, it results in a rapid cut off of base emitter junction base emitter junction goes to cut off. But then see here if you see the this figure entire N minus region as vanishes now. So, what all the minority charges that are trapped here they take finite time to recombine at that time base emitter junction is already in cut off mode. So, there are minority carriers they are trapped they take finite time to recombine in other words a small current continues to flow voltage across the device starts increasing. So, what will happen is the voltage the power dissipated during this time increases. Since the current that is flowing is very small it is known as the current tail current tail that has to be avoided. So, I will repeat if you operate that device into deep into saturation and you apply a minus I B sharp minus I B you will have a current tail and therefore, device and during current tail voltage across the device is increasing. So, you may have substantial power loss and device may fail. So, that has to be avoided. So, first precaution that you need to take care is never operate the device deep into saturation and in case if you operate the device deep into saturation is never apply a minus I B instantaneously it has to be gradual. So, here is it holes in the collector region requires certain time to recombine and negative I B has a negligible effect on this. So, base emitter junction cut off base collector current continues to flow operation is equivalent to a diode during T R R it is also known as the current tail during this period in most cases V C is already high high losses risk of thermal run away. So, there is a reason you need to apply a gradual the I B has to be gradually applied in the negative direction. So, that you need to ensure now question to you is how do I ensure the BJT to operate in quasi saturation. See the problem here is you have designed a gate drive circuit or you have designed the base drive circuit taking or for rated IC or rated collector current you have designed the base drive collector current is approximately equal to emitter current that is nothing but that is approximately equal to the load current itself. Just give an example here see how I have a circuit load is variable is a independent variable you have no choice. You have designed I B for a rated load at that time current is the rated current assume that now the load has reduced IC as reduced to 0.1 IC no load to approximate sorry rated load to almost no load I B is still remains the same now what will happen. Now the device is driven deep into saturation one way to avoid is as IC changes you reduce I B also. If you do not do anything to I B IC has reduced because of the load nothing has happened to I B. So, therefore device will now go into saturation. Now if the device goes deep into saturation my turn of time increases. Now the turn of time is made up of there are two components one is T S the storage time it depends on the degree of saturation as the degree of saturation increases T S increases therefore T of increases if T of increases turn of losses increases turn of losses increases. So, you need to ensure that device should not operate into deep into saturation how do I ensure that one way is to reduce I B accordingly. Second one is to use what is known as anti saturation network is known as a becker clamp see the arrangement here there are two diodes are connected in series the base and one more diode between collector and this point and there is another D 4 in this direction why D 4 is required D 4 is required because to apply minus I B. So, minus I B will flow through because now I have connected two diodes in series here. So, this is the direction for positive I B. So, negative I B can flow through D 4 why I connected two whether do I require two or three we will see and why D 1 is required by the way how do I here how do I infer that device transistor is going into saturation or how do I stop the transistor going into saturation what do I need to do the answer is somehow ensure that or somehow ensure that V c does not attain a value which is equal to V c is at somewhere you maintain a V c the value V c which should be higher than V c is at that you need to ensure how do I how do I ensure that I should not allow V c e to attain its saturation value see this circuit now what is V c e V c e is V B e plus voltage across D 2 plus voltage across D 3 plus voltage minus voltage across D 1 V c e is equal to V B e plus these two diode drop plus this D 1 I am here. So, if they are same identical diodes V D 3 cancels with V D 1. So, now V c e is V B e plus V D 2 at any time V c is always clamped at V B e plus V D 2 when there are two diodes now if you want to increase V c further you connect two three diodes in series if there are three diodes. So, V c e is V B e plus V D 2 plus one more diode drop that is ensure V c is not V c sat at all it cannot attain that value what this D 1 does D 1 acts like a tap or some sorry in other words it operating like a regulating valve how does it operate a regulating valve just see here see for some other reasons this transistor is trying to enter the saturation. So, when I am saying that when it is trying to enter saturation V c e is trying to fall it is trying to attain V c sat see the moment this voltage V c draw to a certain value this D 1 starts conducting now V c is always clamped at potential at this point minus the diode. So, I will repeat initially all the current assume that V c is high all the current starts flowing through D 2 and D 3 all the current starts flowing from D 2 to D 3 and for some reasons load has reduced I c has fallen. So, V c therefore transistor is trying to enter the saturation the moment it enters the saturation what happens V c starts decreasing the moment V c start decreasing D 1 gets forward bias and some current base current gets diverted to D 1 and starts flowing through here. So, this is acts like a valve it will not allow this voltage to fall below a certain value a part of I b starts flowing through D 1. So, part of I b starts flowing through D 1. So, automatically there is a reduction in the current flowing through the base. So, transistor comes out of saturation it is explained here see assume the transistor is off I b positive I b is applied to point x till V c is equal to V x plus V d 1 till V c is equal to V x plus V d 1 all the current will flow through D 2 and D 3. Sorry this should be D 1 D 1 is off please correct this should be D 1 is off this is D 1 sorry all I b will flow through it D 2 D 3 all the current will flow through this. Now, assume the load has fallen I b is held constant corresponding to the rated load transistor might get saturated V c falls when V c is equal to V x plus V d 1 see here when V c is equal to V x plus V d 1 diode D 1 gets forward biased all the current starts flowing through sorry V c a part of I b starts flowing through D 1. So, therefore, a part of I b which was flowing through the base starts flowing through diode D 1 and the collector to emitter back. So, therefore there is automatic reduction in the base drive current transistor comes out of saturation something of a negative feedback or a control wall coming to isolation the gate drive circuit by the way what is isolation I will just explain to you why do you require isolation I will bring back the same circuit back this could be either a BJT or a MOS or IGVD does not matter I have shown a BJT it could be MOS or IGVD or anything any device all the signals are applied with respect to the emitter in a BJT source in a MOSFET all the signals are with respect to emitter if I use it as a MOS if it is a MOS. Now, what is the potential of this emitter that is floating when this device is on emitter potential is the DC link voltage this point gets connected to E when this is off and when this is on this E gets connected to the negative DC bus. In other words potential is continuously changing or what we say is potential of E is floating by the way how do I generate this gating signals generally a controller is used to generate this gating signals a microprocessor or a microcontroller or a DSP. DSP has its own ground a 3.3 volts DSP means VCC supply voltage is 3.3 volts with respect to its ground a flexed ground. Now, can I connect the DSP ground to this because the reference point for this for this BJT is emitter and the emitter is continuous the potential of emitter is continuously changing. So, it is we cannot connect the so called the so called control ground to this ground absolutely there is no absolute ground in this power circuit this point is continuously changing. So, therefore, you cannot connect the control ground and this reference point see the see for this point emitter of this switch is sorry the potential of this potential of this switch emitter is fixed is always negative bus whereas potential of this emitter is floating. So, we cannot use one controller and control and connect that ground to emitter here and here it is not possible. So, we have to isolate it. So, we need to have an isolation between the control and the power this is one of the reasons there are other reasons as well we will see at the appropriate time what are the reasons why isolation between control and power is required. What sort of isolation to use what sort of isolation we use in SCR circuits by the way I am not covering SCR here, but then it is an ideal switch almost sorry it is almost like an ideal switch why it is almost like an ideal switch see it just requires only a sharp pulse to turn on it does not require a continuous gate drive MOS, IGBT or a BJT they require a continuous gate drive continuous gate drive whereas SCR requires just a sharp pulse that is all. Now, I said this devices fully controlled devices require a continuous gate drive. So, what sort of isolation that is required isolation is a mass now what sort of isolation what sort of isolation that we used in SCR circuit we use a pulse transformer or we use a transformer. Now, what will happen if I use now the question is can I use a transformer in BJT or a MOS or IGBT the question one if I can use what modification can I make the second question. The first question let me try to answer what if I use a transformer in gate drive circuit of a BJT or a MOS which requires continuous. See I have shown a figure here a transformer I am giving a square wave what do I get at the output of a transformer basically is a differentiator transformer is a differentiator I get a sharp pulse at this point and another sharp pulse at this point in this region I expect the transformer will get saturated. So, I do not get any output at this. So, the faithful reproduction of the primary signal the secondary is not happening here I get this. So, if I and this signal may not be able to turn on the BJT here IB has become 0 this sharp pulse I can use to turn on a SCR because I need only to turn on I can turn on unfortunately I cannot turn it off see that is only the problem. So, it may not be possible I am saying it may not be now you may say that I will design a transformer such that it will not get it will not get saturated at all. If you say that and if you are able to do it go ahead very good. So, I am just saying that it may not be possible to use that pulse transformer in isolation please may not be possible do not say that it is not possible. What is the second level of second method of isolation may be use an opto isolator or optocoupler sorry optocoupler opto isolation optical isolation what is the structure it has a diode photo diode and a BJT when this diode and it is light on the base of a BJT this BJT turns on what price am I paying for this the moment I use a BJT here it requires a biasing voltage it requires a biasing voltage transformer does not require whereas, here it requires since I there is a BJT I require a biasing voltage you may say. So, what you will use the biasing I will not answer that question right now we will see what happens what are the issues when we discuss a 3 phase inverter 3 phase inverter not now you may say what I will use the biasing voltage is biasing voltage. So, what I have no answer at this stage sometime later we will see at this stage I am saying that as the number of devices increase a base day circuit may become bulky may be if you are feeling bit restless I will answer this question right now see here one leg of an inverter I said I am using opto isolation second it is a secondary is a BJT there it requires a biasing voltage. Now, there are two transistors in one leg here can I use one power supply or the common power supply for these two answer is now why because reference point for this BJT and this BJT is not the same or this MOS or this MOS is not the same. So, I require a separate power supply for this separate power supply to bias the opto coupler BJT I require a separate supply to bias the BJT of the opto coupler of this switch. Now, what if I have more switches something like this. So, number of gate number of power supply that are required to design this gate drive increases which are completely absent in SCR circuit SCR requires only one power supply that is all that at the at the low voltage side or then in the control side and these two control and power circuit are isolated by isolated by a transformer and you do not require you do not need to bias a transformer you need you have to bias a transformer that is the reason you require a large number of large quote unquote large number of power supplies and therefore, power circuit may become bulky. I will just discuss the characteristics now and maybe I will give a small hints to design a gate drive circuit for a BJT mind here this is an assignment you have to solve it during the tutorial please try to understand see I used BU 208D VCEO 700 volts VCEO see what is the various O stands for base emitter with IB is equal to 0 with IB is equal to 0 that is the reason they have put O if I this blocking voltage increases if I negative if I apply a negative gate drive for a BJT if I apply a negative gate drive for a BJT this voltage increases that the time they will put VCEX see here VBEO or reverse voltage is 10 volts VCEO it can block voltage of 700 volts when it is in the forward bias mode it cannot block negative voltage see why base emitter junction is highly doped. So, break down voltage reverse break down voltage of the base emitter junction is very low very low. So, that is the reason you cannot use a BJT in AC supply. So, remember that I am using I told you transistor is used as a switch and switch when I say switch for an electrical engineer I can put it in AC circuit and it should be able to block both positive as well as negative peaks BJT will not be able to do it why because base emitter junction is highly doped and therefore, it can block a very small voltage at 10 volts only it can block collector current is 8 amperes. See here VCE sat maximum maximum could be of the order of 1 volt you should not allow VCE sat to attain this value by connecting suitable number of diodes there. See storage time typical 7 microseconds fault time 500 nanoseconds see here storage time is much higher than the fault time. So, the total turn of time is of the order of 7.5 microseconds. So, therefore, there is significant losses that will occur when you turn of the BJT if there is no storage time your total turn of if there is no storage time turn of time is very small. So, therefore, you can operate the transistor at a much higher frequency. So, at what price are you paying I have saturated the transistor thereby on state losses have come down on state losses mean the losses that are occurring in BJT when it is conducting they have come down at what cost storage time has increased. Now, storage time increases means turn of time increases and therefore, it has two consequences one is turn of losses increase and maximum switching frequency comes down T of itself 7.5 microseconds that was the question that was asked how to reduce TRR I had no answer nor I have, but here again T s and T f I can control T s T s depends on entirely on your gate drive circuit entirely on a gate drive circuit T s T storage and see the safe operating area at any given point it should lie in this zone and see here H f collector gain I c versus H f e beta it is a very strong function of the junction temperature T just stands for junction, junction temperature 25 degrees and 125 degrees. Now, see collector gain for a gain of 8 amperes see here 5 amperes also gain is of the order of if I am operating at the junction temperature is 125 degrees gain is of the order of 5 of 4 5 or it is somewhere here around 5 if the junction temperature is 25 degrees gain is of the order of 8 or so it is quite interesting if the junction temperature is 125 degrees for a 5 ampere collector current gain is of the order of 5. So, base current is required is 1 ampere steady state base drive current see the problem issues of issue that I involve steady state base current is 1 ampere 1 ampere I will emphasize 1 ampere because why my emphasizing I will see here all 3 I mentioned base current approximately 1 5 1.5 ampere see by the way now what sort of a gate drive you should design why it is why did I supply a high pulse of base drive current to the gate this I am doing because to reduce the turn on time to reduce the turn on time. So, to reduce the turn on time if you give a high pulse of current your turn on time decreases, but then can I maintain the same current if I maintain the same current transistor will go into saturation. So, there what I am doing automatically I am trying to I am reducing this current. So, what are the basis characteristics one is a high pulse of current during starting after T on after T on automatically reduce it automatically reduce it and when you want to turn off gradually reduce. So, in that example I said 5 ampere collector current gain is 5. So, steady state current is 1 not to reduce the turn on time I have to give a high pulse of current say 50 percent. So, I have to supply a current of 1.5 amperes 1.5 amperes. So, remember there is a very interesting value these are 1.5 amperes. Now, can I use a small transistor to drive this transistor SL under a rating is around 500 milli amperes SL under which transistor the transistor which we use in the lab. So, definitely I require a small power transistor to drive or another PT a small PT see 1.5 ampere is a small signal transistor whose voltage rating and current rating are small to drive another PT. So, that is the problem in a BJT. So, the this is about overload protection let me let me summarize my gate drive requirement increases in case of BJT it is a basically a current fed device current fed device in order to drive a 5 ampere through 208 D transistor I require another PT and last one is overload protection you cannot provide protect using a fuse, fuse is not fast enough. So, necessary to detect an over current and immediately base the base drive. So, how do you how do you provide a overload protection it is C A detect the over current connection detect the overload over current condition and immediately remove the base drive the question is. So, you need to your base your gate drive circuit has to do 2 functions first is it has to detect and immediate and it has to act. Now, how do I detect how do I detect this overload condition how do I what will happen I will just wind up soon do not worry I B safe for example, I B some I B is flowing and I C is the rated current. So, therefore rated I B was flowing now because of some fault on the load side load current is increasing now load current is higher than I C rated I B is the same. So, what will happen now this transistor starts coming out of saturation sorry out of quasi saturation initially we were operating in quasi saturation. Now, this I L is higher than the rated. So, this transistor starts coming out of quasi saturation what does that what happens at that time V C is starts increasing. So, so if you can detect or if you monitor V C E when the transistor is on you can you can or you may be able to detect the fault under normal condition V C is expected to remain in quasi the value corresponding to V C is set plus 1 or 2 diode drops and if this value increases during on time it implies that there is a fault in the load side. So, what do we need to do detect V C E when V J T is on and if it increases immediately apply a negative I B. So, steps I will give you during on time detector measure I B sorry during on time measure V C E if this V C E is higher than the prescribed value there is a fault immediately apply I B negative I B and send an information to the main controller to block the other pulses. So, these are the characteristics your gate drive circuit should have. Now, I will take up few questions. Agartala sir why we do not consider the reverse recovery current for normal signal drive? No, no why are we not you are supposed to consider this when if it is if you are switching at a very high frequency if you are see if you are switching at a high frequency and you know while designing a power semiconductor devices you have to consider reverse recovery current you have to consider the reverse recovery current you have to in a normal signal normal electronic circuits this current may be very small in a normal electronic circuit this current may be very small in power electronic circuit in high frequency diode this current is substantial in a what is the normal signal diode that you are using 1 and 4 0 0 1 what is the current you just go and find out it is very very small it is very small as the voltage rating increases this current increases your normal signal this I R R is very small when the voltage ratings are very small low let me emphasize what is the voltage that you are applying for in your electronic circuits 5 volts at that time current is very small. But in power electronic circuits when the currents are in amperes you have to consider I will see what else you have any questions sir my question is related to the BJT why BJT is called minority carrier device it is specially for N P N transistor N P N transistor only generally they are very popular they are being used P N P transistor not popular they are not being used in the inverter generally N P N transistors are used N P N transistors are being used they were being used they are not even they are not being used now I am doing BJT because only then you may be able to appreciate MOSFET and IGBT and only generally N P N transistors are used if you see my very first sentence N P N I have mentioned there my very first sentence in the very first sentence in my slide is N P N not P N P in inverter and BJT in DC to DC only N P N are used not P N P.