Overview of Bipolar Junction Transistors

So we shall continue with our last discussion on BJT bipolar junction transistor. If you recall from the last couple of lectures we had introduced the concept of BJT which is essentially two back to back pn junction I told you they are called emitter base collector we had taken an example of n pn of type of BJT. I told you there are various current components and how a BJT works in principle how you actually control the current between two terminal. With the help of a third terminal, okay, that is what we have discussed. Today we will go ahead and we will actually give some mathematical or quantitative shape to the equations of current so that we can derive some very simple expressions of things like gain for example, okay or how to design a better BJT for example. And there are many effects associated with the working of a BJT, okay, there are various aspects of that. So all those things we will take up class by class, okay. So let us come to the white board now. If you recall from the last lecture, a BJT is two back-to-back PN junctions. So, for example, I have an N and a P and an N here. Suppose this is a BJT. I told you that this is emitter, this is base, and this is collector. And you forward bias the... base emitter junction so that you inject electrons from the n type to p type of course holes will be back injected from p to n you want most of these electrons to reach the collector and so you basically you basically reverse bias the base collector junction so the band diagrams look something like that you know this is forward bias and then this is reverse bias the base collector right so the idea is that electrons that are injected from here, some of them will recombine in the base and most of them will reach the collector edge here. Once they reach the depletion, they will be immediately swept away by the field to the other side, okay. So, electrons are injected from here because it is a forward bias junction, holes are back injected here. So, this component is called electron current diameter current because of electron, this is emitter current because of hole, this is also a base component by the way because it exists in both base and this thing. Some of the current will recombine here, we call that VBE recommendation, recombination because the current will also recombine electron hole. And whatever fraction reaches the other end here contributes to collector current, the electron that comes here will contribute to collector current and causes the output. Of course, there is some leakage here, reverse leakage, base collector reverse leakage current which we essentially ignore, but nevertheless, the voltage between base and collector is VBC that you are maintaining and between the base and emitter you are maintaining VEB forward base. I had defined a few terms if you recall you know about efficiencies for example emitter injection efficiency and base transport factor. We will come to that again again. This is how at the end of the last class I told you that this IE and the electrons that are injected from emitter to base eventually some of them will come out to the collector most of them will come out to the collector and this contributes I and C and this is what we actually is important as the output current okay. This is what is important as the output current because this actually matters in amplification or any other thing, okay. This back injected hole from base to emitter because remember this is p type, this is n type. The back injected holes that contribute to IEP, this emitter current because of hole actually comes out for the emitter only. This is a useless component of current. You want to minimize that. And in the last class, towards the end of the last class, I had mentioned that if you have n plus sort of a p junction, for example, this is 10 to the power 18 doped, this is 10 to the power 15 doped. I had shown you in the equation if you remember. The last slide from the last class, you know the current components that you are injecting from emitter to base and base to emitter can be dramatically different, okay. If you recall these expressions you will know that the electrons that are injected from here to here will far outnumber the holes that will actually be injected from here to there in terms of the current. And so this junction if you forward bias predominantly the current will be because of electrons going here and not from because of holes coming here. So, if you dope this emitter very high compared to base let there be collector, then you can make sure. So, if your emitter doping you remember emitter doping that is emitter doping right and the base doping is N A in the base. If the emitter doping is much larger than the base doping, then you make sure that your emitter current because of electron which is this is much larger than the back injected current I E P and this will give rise to a good you know transfer ratio if you recall. We had discussed many of the terms here. Gain we had discussed, base transport factor and you know we had discussed emitter injection efficiency. So, if you recall the emitter injection efficiency was what fraction of the total emitter current is because of electron. If your base component is the component because of holes coming from base to emitter is reduced, then this component will approach 1 and that as much as it approaches 1, we have a better and better transistor. And if your gamma approaches 1, of course, your you know your alpha the ion transfer ratio. gives you the ratio of the output current to the total input current also will become better okay, total amine diameter current okay. So everything becomes better with making the emitter higher doped than base to a first extent but of course in the way you make basically you know a three terminal I told you a transistor is a BJT is a transistor, it is a three terminal device actually when you have a transistor like a BJT and this is suppose the Venn diagram under operation, so this is collector which is you know you are applying V. c b that is the collector voltage between this you know the base and the collector this is p type this is n type collector and this is n type emitter of course between the base and the emitter you have a forward bias voltage. The thing is how the transistor you know the work is that this output current that comes out which is i and c this is the output current it actually does not depend on what potential difference you have this at this v b c the base collector voltage. So, you see the base collector voltage actually ideally should not affect the output current. In other words the output current should be independent of the output voltage, it should depend on input condition and what is the input? You see some of the electrons that go here they recombine here know that when it recombines then to maintain the equilibrium you will basically the base will inject some holes here. So, it is a very tiny component of current and this base current basically injects holes in order to replenish the holes that are lost here when the electrons recombine while going to collector. So, a small current I b a very small current I b when you inject essentially this is a forward bias junction a very small this will lead to a very small change in for example, a very small change in your base emitter voltage. Your base emitter forward bias that you have applied can be mildly changed by a small current that you are applying and a small change in this thing leads to an exponentially large change in. Exponentially large change in what? Large increase or change in the current that goes here. The current that goes here. Why? Because this emitter current depends exponentially. If you remember Q D by L, etc. square by L, you know, B, L N here. Exponential of Q V B E by K T minus 1. So, if this quantity the emitter base voltage is slightly changed, it leads to an exponentially large change of current that you are pushing here and most of this current actually reaches the collector, only few of them get few of the electrons will get recombine here, most of them will reach the collector. So, the collector current I and C this quantity also has a very large change. So, this is the transistor action in a way you can say a small change in the I B which changes the delta E V you know the potential the difference in a small amount. can lead to a large change in current from the emitter which then comes out to collector and that is how the output current which is INC does not depend on the output voltage but it depends on the input voltage which in turn depends on the input current you can say. So essentially by controlling the input current by a very small amount you are able to make a very large change in the output current okay. So that is how a transistor action is that is how amplification should take place ideally and so There are different modes in which you can operate these devices. Of course, I will come to the device equations very soon, but suppose you have you know this is your schematic of the BJT for example, NPN right. So, typically there are two modes in which you can operate a BJT, one is say common base ok, common base mode, common base means that the base is grounded and one mode is called the common emitter mode which means the diameter is grounded. Common base and common emitter. So, what does common base mode means? So, first if I take common base mode and we will of course, derive the generic expressions for current of the various components of current very soon. So, that we can get a quantitative feel of the things like you know injection efficiency, beta, alpha, b. If you remember these are the these are the matrices that will tell you how better you know how good the transistor is all of these depend on the on the values of the equations of current like I E n, I B and so on. Just before we come to that we will just quickly touch base on the modes of operation because to understand a BJT we need to understand how it is operated. So, a BJT can be operated in common base or common emitter mode when you want to get you know a forward active kind of operation you want to get some gain and stuff you do not want to switch it off for example you want real you know working device. So, common base for example means that the base is grounded actually ok the base is grounded. So, you what you do of course you know if you remember this base emitter junction has to be forward biased. and the base collector junction has to be reverse biased that has to be always holding true if you want to operate in the forward active region you want to have some gain so you keep the base grounded and what you do is that from the emitter side you inject some current i e When you when I say the base is grounded and I am injecting some current from the emitter it means the base emitter this is the electron current by the way I mean I am talking about electron it means automatically that the base emitter junction gets forward biased gets forward biased. So, basically I am injecting some current from the emitter and emitter current is almost equal to collector current right if you remember. So, suppose I might in in you know give 5 milliamp then I will increase it to 10 milliamp. So, on I can increase and then 15 milliamp and so on right and of course, this some of them will get recombine here, but most of the electrons will come here and they will collect out to the collector. So, you might want to measure the collector current and collector current will be almost equal to the you know the emitter current that you are putting in. So, instead of 5 amp you might get 4.9 milliamp for example, for 10 amp you might get 9.9 milliamp for example, for 15 amp it might get 14.9 milliamp. So, almost the entire you know the current is coming there little bit of. bit of current is dropping here. So, essentially what you do is that you plot the output current which is Ic versus you actually this voltage this you know if you keep changing the collector voltage here Vc nothing will change ideally right. So, I am plotting Vc, but I am plotting Vcb because this is grounded. So, with respect to the base I am measuring the collector because base is grounded. So, collector is measured with respect to base. So, Vcb and this is your you know Ic that I am putting. So, you know I am putting say emitter current of Vcb say 5 milliamp and then you know in the current of 5 milliamp, so it will basically be something like that. So the collector this is I am plotting collector current, but this will correspond to an emitter current of say 5 milliamp that I am putting and here I will get a collector current of say 4.9 milliamp that I am getting, then you know you might have something like this, the emitter current is 10 milliamp and that corresponds to a collector current of say 9.9 milliamp right. and this kind of things will come IE will be equal to 15 milliamp you know I will be equal to 20 milliamp. So, this is your output characteristics okay this is called your output characteristics of the transistor in the common emitter a common base mode sorry the base is grounded here the common base right the base is grounded here and you have this ratio of this IE and this is the total current here the ratio of course here is 19.79 milliamp This is your output current Ic, this is your Ie and the ratio of Ic by Ie gives you alpha. So from this you can actually find out what is the value of the alpha, right. So one important thing here to note is that you know when you have the base grounded, when you have the base grounded then essentially you measure the collector voltage with respect to the base. So you measure the emitter voltage with respect to the base. You know this is a common this is IIC for example this is your common base configuration and similarly I will come to the common emitter configuration very quickly. Common emitter configuration common emitter so as the name suggest in common emitter configuration you will basically ground the emitter right. So this is suppose collector this is base this is emitter so here you will basically ground the emitter it is grounded. So what you do is that you inject small amount of base current, very small amount of current say maybe 5 microamp, maybe 10 microamp, very small amount of base current 15 microamp, 20 microamp, you might be injecting some small base current which essentially make sure that your base emitter junction is forward biased and that small current will lead to a large change and because your base is getting forward biased of course because you are injecting this current and the correct emitter is 0, so it looks like you know, so this is 0. And then the base is getting forward biased here right and then the collector comes here collector is reverse biased anyways right collector is reverse biased anyways. So, whatever electrons can go here now immediately we will get you know strapped to the other side. So, if you look at this then you can plot the collector current output versus the you have to plot actually VCE because everything is ground in emitter is grounded. So, you measure the collector potential with respect to E. So, you plot VCE and at this you plot with respect to you know IB so IB could be very small say 5 microamp corresponds to that you have an IC of say 5 milliamp so the gain is basically beta is 1000 you know 1000 because it is 5 microamp this is 5 milliamp similarly you might have 10 microamp of base current you are injecting you are getting maybe say 9.8 9.9 milliamp of say for example you know the collector current okay so this is your this is your This is your common emitter configuration. So if you look into this carefully, you will see that here. Actually, your electrons that are moving from the emitter, they go down to the collector side here which is reflecting here. So here the step input is basically your base current. In the other case, your step input was basically your emitter current, okay. Please keep that in mind. So these are two common modes in which you operate and there are many things associated with this operations that we will come to that later, okay. So we should keep this in mind because when we talk about different, you know, different plots and different kind of matrices of performance in these devices, it is very good to recall how a common mode or how a common emitter or a common base configuration actually should look like okay. So this is all about that. There is also associated something called Gammel plot, we will come to that sometime soon after we discuss some of the current things but Gammel plot also will tell you about the gain and many other things that we have to plot and there is also a Gammel number okay, there is a Gammel number and the Gammel number also will become handy. So all these things will come but before that we have to discuss about the current equations actually okay. all the basic components of current we should know them, the equation should be known only then we can relate properly about the Gammel plot, Gammel number and many of the other things associated with common emitter and common base will become clear only then okay, will become clear. So now let us again come back to the main schematic of a BJT, this is N, this is P, this is N, I told you what are the current components, one component is this, emitter is going so this is I and E. One is back component IEP, this is IEN for example, most of them will reach here, so there is a collector current, some of the base will get recombined here, so that is base recombination current, this also is included in the base by the way sometimes and the base and collector junction will have some reverse leakage from both this sides and from this side which we are ignoring now but later on we can take it. So the most important thing to learn here is the base minority carrier profile, when we are injecting electrons from emitter to base. The electrons become minority in the base, so they will decay or recombine in the base, so this N of B tells you the minority carrier decay profile of electrons that are injected in the base, okay. So I will write this as X and if you see carefully here, there is a forward bias and there is a reverse bias junction between the emitter base and base collector respectively. So if you draw it little large, this is your emitter base, this is the base collector. If you recall the emitter base is forward bias, so there is a small depletion here but the base collector is reverse bias so there is a large depletion here. I told you the true depletion width the neutral base width actually is this only WBN all your base current decay and everything will happen here this red mark to red mark you know that is called metallurgical if you remember that is called metallurgical base width and that does not matter now what matters is the neutral base width from the depletion edge to the depletion edge okay. So this is a minority carrier profile and this is you know If you recall to have a more component of electrons reach the collector that is the idea one thing that we have discussed is that the emitter doping should be much much larger than the base doping that we had learnt right. The emitter doping this is emitter doping okay should be much much larger than the base doping base doping is NAB collector doping is NDC which is not important now. So the emitter doping should be much much larger than the base doping then most of the electrons will basically be injected but very few holes will be injected this side and another very important thing is that You want the electrons that are injected here most of them should reach the collector because if maximum the electrons decay and recombine in the base region very few electrons will reach the collector which means the collector current is low your gain and your many other things in the transistor will be very poor. You want maximum the electrons injected from emitter to reach there and that is possible only when the base is narrow when I say base is narrow base should be thin which means this neutral base with W bn if you recall W bn the neutral base with. over which the carriers will decay has to be very short. What will happen otherwise if the base is very long for example I am having a emitter collector and then this is base this is small depletion here this is a large depletion here and then this is the base. If the base is very very large then electrons that you are injecting from here the electrons that you are injecting from here they will get such a large base that they will recombine almost everything the current that comes out here will be very little so you do not want that so you want the base to be very short. And if you remember for short base the carrier recombination or decay carrier decay is almost linear for a very large diode it is exponential but for a very short diode it is linear okay. So you are going to get a linear minority carrier decay profile so if I plot the minority carrier profile okay I am going to plot the minority carrier profile minority carrier profile. is only for the minority that gets injected to that, the majority that becomes minority on the other side, right? So, for example, I take here You know maybe I will have this so this is your x axis okay this is your base emitter junction so there is a small depletion here and there is a large depletion here this is the base collector junction right. So injecting electrons from the emitter side to the base side so they will decay linearly like this because this is a minority this is a depletion region and this is reverse biased technically according to depletion approximation this linear bias should end. At 0 here because at the edge of the depletion region the X the total carrier has to be 0 okay the total carrier has to be 0. So this this is your N B X this should be a linear function this is how it is decaying okay at this point it should become 0 similarly at this point it should become 0 and at equally and eventually far away it should become your minority carrier electron which is a holes which is Ni square by N d C okay. Here similarly you will be injecting holes from. the base to the collector emitter side. So, this should be decaying exponentially because the emitter is very long this should be decaying exponentially far away from the junction it becomes ni square by NDE this is your minority carrier this is an excess carrier here if you remember that excess carrier you can find out exactly as ni square NDE into x e to the power whatever forward bias you are applying by kT minus 1 gives you the excess carrier which is equal to delta. the the the the hole that you are injecting here delta p and this is at x equal to this is at at this point which I can say this is at this is x e this is the you know this direction x e equal to 0. So, at the interface you will get this kind of a excess minority carrier decay that will that the carrier concentration that will decay exponential is e to the power minus x by L p. But this is a linear decay more or less but we will come to that again very quickly this is a linear decay more or less at the edge of the depletion it should come to 0. So, at this point at this point The total carrier concentration should be not the excess the total carrier concentration NB at 0 I am talking about this is X equal to 0 this is X equal to WB N because this distance is WB N. So, at X equal to 0 at this point at this point your excess the total the total carrier concentration will be the background minority carrier concentration which is N naught that is the background minority carrier concentration okay. Do you know what is my background minority carrier concentration this is equal to N naught by NAB that is your background minority carrier concentration in the base times e to the power qve by kT this is the total carrier here the total carrier here the total carrier here should be almost equal to 0 but this is one boundary condition that the carrier the total carrier here should be this and the total carrier over here actually the total carrier at x equal to W b N Total carrier at dx equal to WB n should not it should be very close to 0 but it should not be exactly 0 what it should be is that it should depend on the bias between the base and the collector. So it depends on the reverse bias so what will happen is that it will be the same ni square by nAB which is the baseline minority carrier here into exponential of the base collector reverse bias voltage which is e to the power minus Q VBC VBC is positive by kT minus 1, this is not there is no 1 here, this is the net carrier concentration here, the excess carrier concentration of course will be minus 1 here. Similarly the excess carrier concentration here if you want it will be minus here okay. So this is another boundary condition that I know because at this point, at this point please remember at this point the excess carrier concentration will depend on the voltage difference between the base and the collector. This quantity is very very small, so this quantity actually goes to almost 0 but is exactly not 0 but these are the 2 boundary conditions, this is boundary condition 1. boundary condition 2. With these two boundary condition you solved the continuity equation here which is dn d square by delta nB by dx square is almost equal to delta nB by tau that the recombination time there okay. So this is approximately the relation that will hold this is the continuity equation in which we apply this boundary condition for a sufficiently short diode we can assume it to be linear equation but if you solve it exactly. If you solve it exactly then you will get the you will get a minority carrier profile Nb of X which will have some hyperbolic functions and so on okay. But this is the expression that you will get and this will have a hyperbolic function here okay. The boundary conditions are written here one boundary condition is that at X equal to 0 this is your carrier concentration at X equal to WN this is the carrier concentration you solve the continuity equation with those the boundary condition okay. So, you will get that. So, now if you look again this is your diffusion here, depletion and this is your large depletion, this is your X, this is your WBN, so this distance is WBN. Your minority carrier profile should decay linearly but it is very linear but slightly non-linear maybe, slightly you cannot see it quickly okay. So the minority carrier profile is NBX okay. So the current that you are injecting here IEN, the electron current because of emitter, I am not talking about the whole current here. Okay this hole current is easy you are injecting holes to the n side no so that will come out to be IEP it is a p n junction basically Q A D E by L E on the emitter side n i square by n D E into exponential of Q V B E by K T minus 1 this is your back injected hole current and you this component is useless because it comes out to the emitter and one way to actually minimize this current. is to make the emitter doping here very high. If you make the emitter doping very high then this quantity goes very small because and this is in the denominator okay. So that is one thing that is okay. But this point this you know the carrier profile here can be actually obtained from the boundary condition and if you take a derivative of that this this if you take a derivative of this and you multiply it by Q a the electron that is going there right so dn by ln sort of thing at x equal to 0 if you take this if you take the slope here. If you take the slope here and you multiply by this whole thing, you actually get I and E, the electron current that you are emitting injecting here and if you take the same expression this profile here, if you take the same expression here but you take the derivative of this, it is not slightly non-linear right, you take the derivative of this at X equal to W B N which is this point. If you take the slope at this point and you multiply by this, then you actually get how much current is coming out there in the collector from the I and C okay. All the electron that you have actually injected from here what is the fraction that is coming out here is obtained by the slope here times QADN how much electrons you are putting here I mean what is the current here depends on this constant here and the slope here okay. So this is your INE this is your ICE and if you subtract this two you are supposed to get the base current there okay you are supposed to get the total base current okay you are supposed to get the total base current. That will give you an idea of you know and we will actually exactly do this also, we will exactly find out what is the expression here, we will exactly find out the expression here, we will subtract them out and we will find the base current, okay. So let us do that in the next class, so we will wrap up the class here, okay for today's here. So we have basically now in the flow of essentially deriving the expressions for current, I told you that the emitter current and collector, this collector current can be expressed from the slope of the minority carrier profile in the base. It is a very almost linear like profile but exactly not linear you obtain that expression from basically solving the continuity equation in the base with minority electrons that are injected with the two boundary condition that we know which I have written down here. So we have now the expression for emitter current and collector current and we will also extract the expression for the base current. So all of these things we will do in the next class so we will meet you then and then we will take the current expression forward okay. Thank you.

And there are many effects associated with the working of a BJT, okay, there are various aspects of that. So all those things we will take up class by class, okay. So let us come to the white board now.

If you recall from the last lecture, a BJT is two back-to-back PN junctions. So, for example, I have an N and a P and an N here. Suppose this is a BJT.

I told you that this is emitter, this is base, and this is collector. And you forward bias the... base emitter junction so that you inject electrons from the n type to p type of course holes will be back injected from p to n you want most of these electrons to reach the collector and so you basically you basically reverse bias the base collector junction so the band diagrams look something like that you know this is forward bias and then this is reverse bias the base collector right so the idea is that electrons that are injected from here, some of them will recombine in the base and most of them will reach the collector edge here. Once they reach the depletion, they will be immediately swept away by the field to the other side, okay.

So, electrons are injected from here because it is a forward bias junction, holes are back injected here. So, this component is called electron current diameter current because of electron, this is emitter current because of hole, this is also a base component by the way because it exists in both base and this thing. Some of the current will recombine here, we call that VBE recommendation, recombination because the current will also recombine electron hole. And whatever fraction reaches the other end here contributes to collector current, the electron that comes here will contribute to collector current and causes the output.

Of course, there is some leakage here, reverse leakage, base collector reverse leakage current which we essentially ignore, but nevertheless, the voltage between base and collector is VBC that you are maintaining and between the base and emitter you are maintaining VEB forward base. I had defined a few terms if you recall you know about efficiencies for example emitter injection efficiency and base transport factor. We will come to that again again. This is how at the end of the last class I told you that this IE and the electrons that are injected from emitter to base eventually some of them will come out to the collector most of them will come out to the collector and this contributes I and C and this is what we actually is important as the output current okay. This is what is important as the output current because this actually matters in amplification or any other thing, okay.

This back injected hole from base to emitter because remember this is p type, this is n type. The back injected holes that contribute to IEP, this emitter current because of hole actually comes out for the emitter only. This is a useless component of current. You want to minimize that.

And in the last class, towards the end of the last class, I had mentioned that if you have n plus sort of a p junction, for example, this is 10 to the power 18 doped, this is 10 to the power 15 doped. I had shown you in the equation if you remember. The last slide from the last class, you know the current components that you are injecting from emitter to base and base to emitter can be dramatically different, okay.

If you recall these expressions you will know that the electrons that are injected from here to here will far outnumber the holes that will actually be injected from here to there in terms of the current. And so this junction if you forward bias predominantly the current will be because of electrons going here and not from because of holes coming here. So, if you dope this emitter very high compared to base let there be collector, then you can make sure. So, if your emitter doping you remember emitter doping that is emitter doping right and the base doping is N A in the base. If the emitter doping is much larger than the base doping, then you make sure that your emitter current because of electron which is this is much larger than the back injected current I E P and this will give rise to a good you know transfer ratio if you recall.

We had discussed many of the terms here. Gain we had discussed, base transport factor and you know we had discussed emitter injection efficiency. So, if you recall the emitter injection efficiency was what fraction of the total emitter current is because of electron.

If your base component is the component because of holes coming from base to emitter is reduced, then this component will approach 1 and that as much as it approaches 1, we have a better and better transistor. And if your gamma approaches 1, of course, your you know your alpha the ion transfer ratio. gives you the ratio of the output current to the total input current also will become better okay, total amine diameter current okay. So everything becomes better with making the emitter higher doped than base to a first extent but of course in the way you make basically you know a three terminal I told you a transistor is a BJT is a transistor, it is a three terminal device actually when you have a transistor like a BJT and this is suppose the Venn diagram under operation, so this is collector which is you know you are applying V. c b that is the collector voltage between this you know the base and the collector this is p type this is n type collector and this is n type emitter of course between the base and the emitter you have a forward bias voltage.

The thing is how the transistor you know the work is that this output current that comes out which is i and c this is the output current it actually does not depend on what potential difference you have this at this v b c the base collector voltage. So, you see the base collector voltage actually ideally should not affect the output current. In other words the output current should be independent of the output voltage, it should depend on input condition and what is the input? You see some of the electrons that go here they recombine here know that when it recombines then to maintain the equilibrium you will basically the base will inject some holes here.

So, it is a very tiny component of current and this base current basically injects holes in order to replenish the holes that are lost here when the electrons recombine while going to collector. So, a small current I b a very small current I b when you inject essentially this is a forward bias junction a very small this will lead to a very small change in for example, a very small change in your base emitter voltage. Your base emitter forward bias that you have applied can be mildly changed by a small current that you are applying and a small change in this thing leads to an exponentially large change in.

Exponentially large change in what? Large increase or change in the current that goes here. The current that goes here. Why?

Because this emitter current depends exponentially. If you remember Q D by L, etc. square by L, you know, B, L N here. Exponential of Q V B E by K T minus 1. So, if this quantity the emitter base voltage is slightly changed, it leads to an exponentially large change of current that you are pushing here and most of this current actually reaches the collector, only few of them get few of the electrons will get recombine here, most of them will reach the collector.

So, the collector current I and C this quantity also has a very large change. So, this is the transistor action in a way you can say a small change in the I B which changes the delta E V you know the potential the difference in a small amount. can lead to a large change in current from the emitter which then comes out to collector and that is how the output current which is INC does not depend on the output voltage but it depends on the input voltage which in turn depends on the input current you can say.

So essentially by controlling the input current by a very small amount you are able to make a very large change in the output current okay. So that is how a transistor action is that is how amplification should take place ideally and so There are different modes in which you can operate these devices. Of course, I will come to the device equations very soon, but suppose you have you know this is your schematic of the BJT for example, NPN right.

So, typically there are two modes in which you can operate a BJT, one is say common base ok, common base mode, common base means that the base is grounded and one mode is called the common emitter mode which means the diameter is grounded. Common base and common emitter. So, what does common base mode means?

So, first if I take common base mode and we will of course, derive the generic expressions for current of the various components of current very soon. So, that we can get a quantitative feel of the things like you know injection efficiency, beta, alpha, b. If you remember these are the these are the matrices that will tell you how better you know how good the transistor is all of these depend on the on the values of the equations of current like I E n, I B and so on.

Just before we come to that we will just quickly touch base on the modes of operation because to understand a BJT we need to understand how it is operated. So, a BJT can be operated in common base or common emitter mode when you want to get you know a forward active kind of operation you want to get some gain and stuff you do not want to switch it off for example you want real you know working device. So, common base for example means that the base is grounded actually ok the base is grounded.

So, you what you do of course you know if you remember this base emitter junction has to be forward biased. and the base collector junction has to be reverse biased that has to be always holding true if you want to operate in the forward active region you want to have some gain so you keep the base grounded and what you do is that from the emitter side you inject some current i e When you when I say the base is grounded and I am injecting some current from the emitter it means the base emitter this is the electron current by the way I mean I am talking about electron it means automatically that the base emitter junction gets forward biased gets forward biased. So, basically I am injecting some current from the emitter and emitter current is almost equal to collector current right if you remember.

So, suppose I might in in you know give 5 milliamp then I will increase it to 10 milliamp. So, on I can increase and then 15 milliamp and so on right and of course, this some of them will get recombine here, but most of the electrons will come here and they will collect out to the collector. So, you might want to measure the collector current and collector current will be almost equal to the you know the emitter current that you are putting in. So, instead of 5 amp you might get 4.9 milliamp for example, for 10 amp you might get 9.9 milliamp for example, for 15 amp it might get 14.9 milliamp.

So, almost the entire you know the current is coming there little bit of. bit of current is dropping here. So, essentially what you do is that you plot the output current which is Ic versus you actually this voltage this you know if you keep changing the collector voltage here Vc nothing will change ideally right. So, I am plotting Vc, but I am plotting Vcb because this is grounded. So, with respect to the base I am measuring the collector because base is grounded.

So, collector is measured with respect to base. So, Vcb and this is your you know Ic that I am putting. So, you know I am putting say emitter current of Vcb say 5 milliamp and then you know in the current of 5 milliamp, so it will basically be something like that.

So the collector this is I am plotting collector current, but this will correspond to an emitter current of say 5 milliamp that I am putting and here I will get a collector current of say 4.9 milliamp that I am getting, then you know you might have something like this, the emitter current is 10 milliamp and that corresponds to a collector current of say 9.9 milliamp right. and this kind of things will come IE will be equal to 15 milliamp you know I will be equal to 20 milliamp. So, this is your output characteristics okay this is called your output characteristics of the transistor in the common emitter a common base mode sorry the base is grounded here the common base right the base is grounded here and you have this ratio of this IE and this is the total current here the ratio of course here is 19.79 milliamp This is your output current Ic, this is your Ie and the ratio of Ic by Ie gives you alpha. So from this you can actually find out what is the value of the alpha, right. So one important thing here to note is that you know when you have the base grounded, when you have the base grounded then essentially you measure the collector voltage with respect to the base.

So you measure the emitter voltage with respect to the base. You know this is a common this is IIC for example this is your common base configuration and similarly I will come to the common emitter configuration very quickly. Common emitter configuration common emitter so as the name suggest in common emitter configuration you will basically ground the emitter right.

So this is suppose collector this is base this is emitter so here you will basically ground the emitter it is grounded. So what you do is that you inject small amount of base current, very small amount of current say maybe 5 microamp, maybe 10 microamp, very small amount of base current 15 microamp, 20 microamp, you might be injecting some small base current which essentially make sure that your base emitter junction is forward biased and that small current will lead to a large change and because your base is getting forward biased of course because you are injecting this current and the correct emitter is 0, so it looks like you know, so this is 0. And then the base is getting forward biased here right and then the collector comes here collector is reverse biased anyways right collector is reverse biased anyways. So, whatever electrons can go here now immediately we will get you know strapped to the other side.

So, if you look at this then you can plot the collector current output versus the you have to plot actually VCE because everything is ground in emitter is grounded. So, you measure the collector potential with respect to E. So, you plot VCE and at this you plot with respect to you know IB so IB could be very small say 5 microamp corresponds to that you have an IC of say 5 milliamp so the gain is basically beta is 1000 you know 1000 because it is 5 microamp this is 5 milliamp similarly you might have 10 microamp of base current you are injecting you are getting maybe say 9.8 9.9 milliamp of say for example you know the collector current okay so this is your this is your This is your common emitter configuration. So if you look into this carefully, you will see that here.

Actually, your electrons that are moving from the emitter, they go down to the collector side here which is reflecting here. So here the step input is basically your base current. In the other case, your step input was basically your emitter current, okay.

Please keep that in mind. So these are two common modes in which you operate and there are many things associated with this operations that we will come to that later, okay. So we should keep this in mind because when we talk about different, you know, different plots and different kind of matrices of performance in these devices, it is very good to recall how a common mode or how a common emitter or a common base configuration actually should look like okay. So this is all about that.

There is also associated something called Gammel plot, we will come to that sometime soon after we discuss some of the current things but Gammel plot also will tell you about the gain and many other things that we have to plot and there is also a Gammel number okay, there is a Gammel number and the Gammel number also will become handy. So all these things will come but before that we have to discuss about the current equations actually okay. all the basic components of current we should know them, the equation should be known only then we can relate properly about the Gammel plot, Gammel number and many of the other things associated with common emitter and common base will become clear only then okay, will become clear.

So now let us again come back to the main schematic of a BJT, this is N, this is P, this is N, I told you what are the current components, one component is this, emitter is going so this is I and E. One is back component IEP, this is IEN for example, most of them will reach here, so there is a collector current, some of the base will get recombined here, so that is base recombination current, this also is included in the base by the way sometimes and the base and collector junction will have some reverse leakage from both this sides and from this side which we are ignoring now but later on we can take it. So the most important thing to learn here is the base minority carrier profile, when we are injecting electrons from emitter to base.

The electrons become minority in the base, so they will decay or recombine in the base, so this N of B tells you the minority carrier decay profile of electrons that are injected in the base, okay. So I will write this as X and if you see carefully here, there is a forward bias and there is a reverse bias junction between the emitter base and base collector respectively. So if you draw it little large, this is your emitter base, this is the base collector.

If you recall the emitter base is forward bias, so there is a small depletion here but the base collector is reverse bias so there is a large depletion here. I told you the true depletion width the neutral base width actually is this only WBN all your base current decay and everything will happen here this red mark to red mark you know that is called metallurgical if you remember that is called metallurgical base width and that does not matter now what matters is the neutral base width from the depletion edge to the depletion edge okay. So this is a minority carrier profile and this is you know If you recall to have a more component of electrons reach the collector that is the idea one thing that we have discussed is that the emitter doping should be much much larger than the base doping that we had learnt right. The emitter doping this is emitter doping okay should be much much larger than the base doping base doping is NAB collector doping is NDC which is not important now.

So the emitter doping should be much much larger than the base doping then most of the electrons will basically be injected but very few holes will be injected this side and another very important thing is that You want the electrons that are injected here most of them should reach the collector because if maximum the electrons decay and recombine in the base region very few electrons will reach the collector which means the collector current is low your gain and your many other things in the transistor will be very poor. You want maximum the electrons injected from emitter to reach there and that is possible only when the base is narrow when I say base is narrow base should be thin which means this neutral base with W bn if you recall W bn the neutral base with. over which the carriers will decay has to be very short. What will happen otherwise if the base is very long for example I am having a emitter collector and then this is base this is small depletion here this is a large depletion here and then this is the base. If the base is very very large then electrons that you are injecting from here the electrons that you are injecting from here they will get such a large base that they will recombine almost everything the current that comes out here will be very little so you do not want that so you want the base to be very short.

And if you remember for short base the carrier recombination or decay carrier decay is almost linear for a very large diode it is exponential but for a very short diode it is linear okay. So you are going to get a linear minority carrier decay profile so if I plot the minority carrier profile okay I am going to plot the minority carrier profile minority carrier profile. is only for the minority that gets injected to that, the majority that becomes minority on the other side, right? So, for example, I take here You know maybe I will have this so this is your x axis okay this is your base emitter junction so there is a small depletion here and there is a large depletion here this is the base collector junction right.

So injecting electrons from the emitter side to the base side so they will decay linearly like this because this is a minority this is a depletion region and this is reverse biased technically according to depletion approximation this linear bias should end. At 0 here because at the edge of the depletion region the X the total carrier has to be 0 okay the total carrier has to be 0. So this this is your N B X this should be a linear function this is how it is decaying okay at this point it should become 0 similarly at this point it should become 0 and at equally and eventually far away it should become your minority carrier electron which is a holes which is Ni square by N d C okay. Here similarly you will be injecting holes from.

the base to the collector emitter side. So, this should be decaying exponentially because the emitter is very long this should be decaying exponentially far away from the junction it becomes ni square by NDE this is your minority carrier this is an excess carrier here if you remember that excess carrier you can find out exactly as ni square NDE into x e to the power whatever forward bias you are applying by kT minus 1 gives you the excess carrier which is equal to delta. the the the the hole that you are injecting here delta p and this is at x equal to this is at at this point which I can say this is at this is x e this is the you know this direction x e equal to 0. So, at the interface you will get this kind of a excess minority carrier decay that will that the carrier concentration that will decay exponential is e to the power minus x by L p. But this is a linear decay more or less but we will come to that again very quickly this is a linear decay more or less at the edge of the depletion it should come to 0. So, at this point at this point The total carrier concentration should be not the excess the total carrier concentration NB at 0 I am talking about this is X equal to 0 this is X equal to WB N because this distance is WB N. So, at X equal to 0 at this point at this point your excess the total the total carrier concentration will be the background minority carrier concentration which is N naught that is the background minority carrier concentration okay.

Do you know what is my background minority carrier concentration this is equal to N naught by NAB that is your background minority carrier concentration in the base times e to the power qve by kT this is the total carrier here the total carrier here the total carrier here should be almost equal to 0 but this is one boundary condition that the carrier the total carrier here should be this and the total carrier over here actually the total carrier at x equal to W b N Total carrier at dx equal to WB n should not it should be very close to 0 but it should not be exactly 0 what it should be is that it should depend on the bias between the base and the collector. So it depends on the reverse bias so what will happen is that it will be the same ni square by nAB which is the baseline minority carrier here into exponential of the base collector reverse bias voltage which is e to the power minus Q VBC VBC is positive by kT minus 1, this is not there is no 1 here, this is the net carrier concentration here, the excess carrier concentration of course will be minus 1 here. Similarly the excess carrier concentration here if you want it will be minus here okay.

So this is another boundary condition that I know because at this point, at this point please remember at this point the excess carrier concentration will depend on the voltage difference between the base and the collector. This quantity is very very small, so this quantity actually goes to almost 0 but is exactly not 0 but these are the 2 boundary conditions, this is boundary condition 1. boundary condition 2. With these two boundary condition you solved the continuity equation here which is dn d square by delta nB by dx square is almost equal to delta nB by tau that the recombination time there okay. So this is approximately the relation that will hold this is the continuity equation in which we apply this boundary condition for a sufficiently short diode we can assume it to be linear equation but if you solve it exactly. If you solve it exactly then you will get the you will get a minority carrier profile Nb of X which will have some hyperbolic functions and so on okay. But this is the expression that you will get and this will have a hyperbolic function here okay.

The boundary conditions are written here one boundary condition is that at X equal to 0 this is your carrier concentration at X equal to WN this is the carrier concentration you solve the continuity equation with those the boundary condition okay. So, you will get that. So, now if you look again this is your diffusion here, depletion and this is your large depletion, this is your X, this is your WBN, so this distance is WBN. Your minority carrier profile should decay linearly but it is very linear but slightly non-linear maybe, slightly you cannot see it quickly okay.

So the minority carrier profile is NBX okay. So the current that you are injecting here IEN, the electron current because of emitter, I am not talking about the whole current here. Okay this hole current is easy you are injecting holes to the n side no so that will come out to be IEP it is a p n junction basically Q A D E by L E on the emitter side n i square by n D E into exponential of Q V B E by K T minus 1 this is your back injected hole current and you this component is useless because it comes out to the emitter and one way to actually minimize this current. is to make the emitter doping here very high.

If you make the emitter doping very high then this quantity goes very small because and this is in the denominator okay. So that is one thing that is okay. But this point this you know the carrier profile here can be actually obtained from the boundary condition and if you take a derivative of that this this if you take a derivative of this and you multiply it by Q a the electron that is going there right so dn by ln sort of thing at x equal to 0 if you take this if you take the slope here.

If you take the slope here and you multiply by this whole thing, you actually get I and E, the electron current that you are emitting injecting here and if you take the same expression this profile here, if you take the same expression here but you take the derivative of this, it is not slightly non-linear right, you take the derivative of this at X equal to W B N which is this point. If you take the slope at this point and you multiply by this, then you actually get how much current is coming out there in the collector from the I and C okay. All the electron that you have actually injected from here what is the fraction that is coming out here is obtained by the slope here times QADN how much electrons you are putting here I mean what is the current here depends on this constant here and the slope here okay. So this is your INE this is your ICE and if you subtract this two you are supposed to get the base current there okay you are supposed to get the total base current okay you are supposed to get the total base current.

That will give you an idea of you know and we will actually exactly do this also, we will exactly find out what is the expression here, we will exactly find out the expression here, we will subtract them out and we will find the base current, okay. So let us do that in the next class, so we will wrap up the class here, okay for today's here. So we have basically now in the flow of essentially deriving the expressions for current, I told you that the emitter current and collector, this collector current can be expressed from the slope of the minority carrier profile in the base. It is a very almost linear like profile but exactly not linear you obtain that expression from basically solving the continuity equation in the base with minority electrons that are injected with the two boundary condition that we know which I have written down here.

So we have now the expression for emitter current and collector current and we will also extract the expression for the base current. So all of these things we will do in the next class so we will meet you then and then we will take the current expression forward okay. Thank you.

Transcript for:Overview of Bipolar Junction Transistors

Transcript for:
Overview of Bipolar Junction Transistors