BJT: Small Signal Circuit Model-I PDF

Microelectronics: Devices to Circuits Professor Sudeb Dasgupta Department of Electronics and Communication Engineering Indian Institute of Technology Roorkee Lecture−08...

Microelectronics: Devices to Circuits Professor Sudeb Dasgupta Department of Electronics and Communication Engineering Indian Institute of Technology Roorkee Lecture−08 BJT: Small Signal Circuit Model-I Welcome to the NPTEL online certification course on Microelectronics: Devices and Circuits. This module we will be taking care of a Small Signal Circuit Model for BJT. The obvious question asked is, why should we require this? Well, BJT as a stand-alone device has got no usage or application, so this has to be integrated onto a circuit for example BJT as an amplifier, BJT as a switch, BJT as a current source and so on and so forth. Therefore, if you have understood how BJT works which we have already done in our previous interactions, we will understand or we will together learn how this BJT the device can be broken down into sets of active and passive elements, so that it is very easily integrated with the circuit analysis. And it is easier with circuits and it is easy to do the analysis in circuits in a much detail manner. (Refer Slide Time: 1:45) Why small signal? We will discuss why small signal now. What is the meaning of small signal? Let me first make it clear to you. A small signal primarily means that the Q-point does excursion of approximately plus minus 10 percent from its quiescent value, right? So if you have biased your device then we define this to be as a small signal, right? So if you have biased your device say at 1 volt, VCE, right? 1 volt then if the input signal changes it from 1.1 volt to 0.9 volt peak-to-peak, I would expect to see that, we can define this to be as a small signal. This is a rather not a very decent definition but nonetheless works fine for all circuit applications assuming that this much amount of variation in the voltage, input voltage will not force the device to go into either the saturation or cut-off region. So we have biased Q somewhere in the middle of active region, a 10 percent high rise or fall will result in approximately a linear change in the value of your amplification, right? So you are still in the linear region of amplification as far as Q point is concerned. So two things one is a small signal circuit model primarily means that I am assuming that my input signal is basically small signal which means that is just a plus minus 10 percent of quiescent value and the second point which I am assuming is that there is no nonlinearity into the picture by virtue of a large signal, right? So and the third point is that obviously we require to have a corresponding circuit elements being inserted, so that whenever we plug those BJT and use it as a switch or any device we can easily do that by doing small amount of manipulation in terms of circuit application. So let’s see what the issues are. (Refer Slide Time: 3:54) So let me come to that point in a detailed manner. I will take one by one as we move along. If you see clearly again this is, as you can see this circuit is basically a BJT in an emitter grounded, so this is basically your emitter grounded configuration or even a common emitter configuration, emitter grounded configuration or common configuration. And what you see from here is that VBE is the applied AC bias which you are giving and this VBE is responsible for biasing it in the appropriate Q-point. We define therefore IC, the output current to be equals to gm times VBE. So if you understand VBE is the voltage which you have applied here, right? And this VBE is varying voltage because that is what the input voltage is all about. So one time it is 1 volt, 2 volt, 3 volt then again comes to 2, 1, so on and so forth. So it is a rising and falling maybe a sine wave maybe a pulse waveform or maybe any other wave forms available to you. And therefore this base to emitter voltage which you give that converts into a corresponding current on the collector side, how? By mathematically multiplying with a value which is gm. gm is referred to as a transconductance, right? And it is defined as, in this case as 𝜕𝑖𝑐 , 𝜕𝑉𝐵𝐸 , right? So it is defined as rate of change of collector current with the base of voltage. So once you get this to be as gm, right? I simply get 𝜕𝑖𝑐 to be equal to gm times 𝜕𝑉𝐵𝐸 , so higher the value of gm of a transistor, I would expect to see a larger amplification appearing to me, right? Because higher the value of gm will primarily mean that the transistor is more sensitive to variations in the input voltage, right? So gm is basically a notation or a parameter which gives me how sensitive your BJT is all about. So if your BJT is quite sensitive, I would expect to see a large value of transconductance 𝑖 available to me. So I define 𝑔𝑚 = 𝑉 𝑐. So if you see it is ic by VBE and this is replaced by VT 𝐵𝐸 for the lowest case because this is thermal energy which you get which is 25 mVs at 300 K, 𝜕𝑖 right? You can also write down, therefore 𝑔𝑚 = 𝜕𝑉 𝑐 for the fixed value of IC, right? 𝐵𝐸 So at a fixed value of IC what is the change in this thing? But you need not forget about all these things, this you just forget. gm is basically 𝜕𝑖𝑐 , 𝜕𝑉𝐵𝐸 which you see, right? Now if you plot IC vs VBE, this versus this, let’s see why and how we get IC vs VBE in this graph. (Refer Slide Time: 7:07) So for common emitter configuration if you plot say VBE, right? And you plot say, maybe an iB, let us suppose, right? Let us see how you plot iB. If you plot iB it will look something like this, right? We have discussed this in our previous discussion, this is the threshold voltage or the knee voltage for the PN junction diode which is the 0.7, right? So for iB if you have a common emitter configuration in this case then, this is the value of VBE, right? And when this VBE crosses 0.7, right? This fire this junction, switches on this PN junction and as a result the current starts to rise, right? Below that particular point the current is very -very low or very small and drops very drastically. So iB with VBE is in this manner. We have also seen from our previous discussion that if you bias your device in such a manner that your Q- point is here, so this is your active region. If you bias your device in the Q-point here VCE let us suppose, right? Then I can safely say that for the same change in the value of iB, I would expect to see same change in the value of iC. So remember when we are plotting the graphs, right? For various values of iB when we are plotting IC vs VCE, right? The common emitter configuration, for various values of iB. So it is iB1, iB2, iB3, so you see this is iB1, this is iC1, this is iB2, iC2, this is iB3 , iC3, so this iB is of the order of few µA and this is of the order of few mA, so thousand times increases the approximately between base current….. but..but please understand that for the same increase, so if this is say iB2 is say 1 micron, this is 2 micron let us suppose, then this current will be actually if suppose 100 times larger, so I get 100 microampere (µA) here or 0.1 mA and I get 200 µA. So you see for the same change in 𝜕𝑖𝐵 , right? I get the same change in 𝜕𝑖𝐶 , so 𝜕𝑖𝐵 , right? And 𝜕𝑖𝐶 if you want to find out the rate of change is always the same. So for the same increment in 𝜕𝑖𝐵 , I get equal increments in 𝜕𝑖𝐶 , right? The increment themselves might be different obviously 100 times larger so quantitatively they are different but I am 100 percent sure that if I increase the base current by once my collector current will also increase by once. If I increase the base current twice my collector current will also increase by twice, why? Because I am in the linear region of operations, right? And therefore I wanted to tell you is, that this is iB vs VBE exactly the same graph is for ic vs VCE. So if you plot iC vs VBE, you will get the same curve again, right? Only thing will be that this iC will be mA which was here it was in µA or may be mA as well but this will be more on mA, right? (Refer Slide Time: 10:49) So let me show you therefore if you plot iC vs VBE here and then in input side if you vary this is your input curve VBE is varying, it is a triangular waveform which you have given, this triangular waveform again small signal assuming so that, the Q is just shifting from this side and coming to this side. So the maximum value is Q is this and this. So we are still in the linear region of operation of the transistor. I get this to be the output, right? Triangular output waveform will be there, right? Now, the slope of this curve which you see, this curve, the slope of this curve, sorry, the slope of this curve, right? Which is 𝜕𝑖𝐶 , 𝜕𝑉𝐵𝐸 , this is basically your gm transconductance. So if this slope is not steeper, I would expect to see a much larger variation in the output waveform, for the same change in the value of input waveform, fine. So that’s the small signal model and in the small signal model therefore we have learned one important point and that important characteristics is transconductance (gm) and gm is therefore defined as the transconductance which in this case happens to be 𝜕𝑖𝐶 , 𝜕𝑉𝐵𝐸 , right? And it gives me an idea about sensitivity. So if I know the base to emitter voltage I can predict the value of the collector current directly by multiplying gm with the rate of VBE, right? And that’s quite an interesting phenomena which is there with us. (Refer Slide Time: 12:21) And now we define another resistance known as the small signal input resistance and it is referred to as rπ, right? And it is nothing but the resistance looking from the base side where between base and emitter, right? So looking from the base side, if you’re looking from the base side the resistance offered by this is basically Vbe by iB. So Vbe is the voltage which you see between these two points and iB is the current by virtue of this voltage. So Vbe by iB is equals to rπ, right? So it is basically the forward biased on resistance of a BJT, right? It is an on resistance which means that the base emitter junction is forward biased it is getting on and therefore the resistance offered is defined as rπ, right? Sorry! But plus understand, this you are looking from the base side, right? So you are looking from this base side and you are trying to find out. So your emitter is grounded, please understand. Another is re which is basically the base to emitter resistance looking from emitter side. So you give a potential on the emitter side and you ground the base side and then looking at that, you will get the value of re to be equals to Vbe by ie, right? So it is Vbe by ie, right? Obviously as you can find out ie is much-much larger as compared to iB. So I would expect to see re to be much-much smaller as compared to rπ. So rπ will be relatively larger as compared to re, fine. So I get rπ is equals to therefore Vbe by iB this is also equals to Vbe by re, right? So rπ can be written as, so what we can write down is that, it is equal to β/gm, right? How β/gm we will just show it to you. If you remember gm, 𝜕𝑖𝑐 𝜕𝑖 what was its definition? It was, , right? So if you put 𝜕𝑉 𝑐 here, so I get what? 𝜕𝑉𝐵𝐸 𝐵𝐸 So I get 𝛽 divided by 𝜕𝑖𝑐 by 𝜕𝑉𝐵𝐸 , so this VBE goes to the numerator I get 𝜕𝑉𝐵𝐸 at the top, right? Divided by into 𝛽 divided by 𝜕𝑖𝑐 , this you get, right? This you get. So this Vbe is coming from here, so now you see remember the definition of 𝛽 from a previous discussion, 𝛽 was equal to your ic by ib, remember. So if you put ic by ib here I get ic, so I replace 𝛽 and make it ic by ib, right? This ic , ic gets cancelled out I get VBE by ib, so this is equals to rπ. So rπ equals to 𝛽 by gm also we can write. (Refer Slide Time: 15:52) So I will do it once again if you want to do. rπ is given as Vbe by ib, remember. So we also know rπ has been also given as beta by gm. We know gm to be equals to what? gm to be equals to 𝜕𝑖𝑐 , 𝜕𝑉𝐵𝐸. So if you put the value of gm I get rπ. If you at place this equation here, I get 𝛽 times 𝜕𝑉𝐵𝐸 divided by 𝜕𝑖𝑐 , right? Now we know the definition of 𝛽 as the current gain in common base configuration and therefore I can write 𝛽 to be equals to ic by ib. So 𝜕𝑖𝑐 , 𝜕𝑖𝑏 , let me write down. Place this value of 𝛽 here and therefore this 𝜕𝑖𝑐 and 𝜕𝑖𝑐 gets cancelled out, I am left with 𝜕𝑉𝐵𝐸 upon 𝜕𝑖𝑏 , right? And that’s what you are getting here, fine. Since it is a small signal therefore 𝜕𝑉𝐵𝐸 equals to Vbe and 𝜕𝑖𝑏 equals to ib and therefore this is valid assumption which we get. So I get 𝛽 by gm as the rπ value, fine. So 𝛽 is relatively very high value and therefore rπ will be very-very high, so this is what we get from our discussion as far as this is there. (Refer Slide Time: 17:16) Similarly re therefore can be written as ie is the emitter current and VT is the base to emitter voltage at 300 mV we will write it as VT thermal equivalent voltage. Therefore I can write down that rπ could be written as 𝛽 plus 1 re, right? And you can do it yourself to find out this value that means rπ is approximately 𝛽 times larger as compared to re, right? So…So… Which means that, and it is expected also, if you look very carefully, the reason being, the base is, so looking from the base side the resistance is very large as compared to looking from the emitter side and you can understand why is it? Because base has got relatively low doping, right? And the area is also very small. So in both will result in very - very high resistivity. Resistance will be typically very high and that’s the reason rπ is much larger than beta re approximately 100 to 200 times larger, right? And that the reason why you get it. Therefore rπ can also be written as VT by IB in case you want to do it and re can be written as 1 by gm as well, as you mean 𝛽 is approximately equals to 1, I can get to be approximately these. These approximations can be taken care of or it can be handled easily. So we have understood therefore that for small signal model if you’re looking at the base side, emitter side resistance is not equal from the base at resistance is much higher as compared to that from the emitter side. (Refer Slide Time: 18:40) We come to the models, various models of BJT which is available. Small signal model is, these are again small signal models, these are all small signal model which is available, right? And they are quite interesting and very easy to handle from circuit point of view once we know the basic fundamental principles of working of these devices. Look at, so we have already learned that BJT is basically a three terminal device. That everybody knows that BJT is obviously a three terminal device, so what are those three terminals? Base, collector and emitter, right? So the arrow head which you see, the blue arrow here, the blue arrow here, the blue arrow here shows you the direction of the conventional current, right? And therefore we say that looking from the base side the amount of resistance offered is rπ and across it we have a VBE as the voltage drop available to me. So therefore ib can be directly written as VBE divided by rπ, right? VBE is always fixed approximately 0.7 is supposed the approximately voltage. If I give you therefore the rπ value you can predict the value of ib directly, right? Now this is therefore the base emitter junction, right? Let’s look at the base collector junction. We have already discussed that base collector junction, the collector starts to behave like an ideal current source, right? So it gives us ideal current source. How much is the value of current? Will be given by gm times VBE. Remember gm was equal to 𝜕𝑖𝑐 , 𝜕𝑉𝐵𝐸. So 𝜕𝑖𝑐 , 𝜕𝑉𝐵𝐸 multiplied by VBE, so VBE gets cancelled out and I get the current available to me, right? So this is the current source with me and this is the direction, the direction of the arrowhead of the current tells me that the current is directed inwards which means, it is basically an NPN transistor and therefore it is directed inwards and therefore you see this directed inwards whereas emitted current is directed outwards, right? So by the previous definition we know that gm is equal to Ic by VT which is available here and this is 𝛽 by gm is rπ, right? So it’s completely analogous to see on the right-hand side, this one, right? It is also a hybrid model, hybrid-π model. Now, it looks like π, so that is the reason it is hybrid-π model. Moreover it is made up of rπ and therefore it is a π model. So you remember, rπ is the resistance offered by the base side, base emitter junction and therefore we have rπ as the resistance here. Whereas on the collector side it is the current source, wherein current source ic will be equal to β times ib, right? So if you have a large value of ib, I will get a large value of ic and I can therefore have a constant current source towards the collector side, right? Therefore this model will give you a very first instance the input impedance and output impedance of the device, right? (Refer Slide Time: 21:50) So if you look at the BJT therefore the output impedance is basically mine, Zout if you want to find out, is looking from the collector side this is the output. How much amount of resistance is offered? Of course you can see, from the output side if you look very carefully it is being terminated across a constant current source and therefore Zout will be approximately equal to infinity in idealized condition. So if you want to follow a hybridized pi model, then Zout will be equal to infinity and Zout will be equal to infinity primarily meaning, the reason being that, we are looking into the current source here, an ideal current source, right? And assuming it to be an ideal current source, we are assuming that the Zout will be the infinity. So whereas looking from the input side Zin, input impedance will be approximately equals to rπ, fine. So this is for the common emitter configuration, right? I leave as an exercise for you to find out common base and common collector, please do it, right? Please do the necessary, right? π model for CB and CC, common base and common collector, right? In common base, you will see it will be re which will be coming into picture as the input impedance, right? Therefore just be able to handle this hybrid-π model in a much better manner in this case, right? So this is one model. The second model which is most prevalent but most of the time we use of hybrid-π model for most practical purposes but sometimes we also use what is known as a T model, right? This is a transmission model or T model which you see, this is the T model which is there. (Refer Slide Time: 23:31) And the T model, still again three terminal device so base, collector, and emitter. We have also seen that the resistance between emitter and base is re looking from emitter side. Looking from the base side, it would have been here rπ but since we are looking from the emitter side at this case we are giving re, you can also have rπ here but they will be in parallel. So being parallel it will be always less than the least, remember? And therefore this has been neglected. So you see rπ and re will be in parallel. Since they will be in parallel the maximum, the highest value will get neglected. So highest is nothing but rπ and therefore rπ is not shown in this diagram, right? (Refer Slide Time: 24:16) So therefore understand that why rπ is not here or here or here, because it is so large that it does not play a major role, re plays a major role. So I get re here, the voltage across re is basically my VBE, base to emitter voltage and this is the applied voltage which you see the current direction is of course, the conventional current direction as we have been discussing quite often. And therefore, the collector current will be equal to gm times vbe, right? We have again discussed this point as the input point. So... So… you see, what I’m trying to tell you, is your base emitter junction or your base emitter voltage, right? vbe starts to change the value of your current flowing through collector, right? And that is basically therefore a control current source with you, right? So basically, it is a current control current source because ib will change the value of ic and therefore it is basically known as the current control current source. So I get from here that it is α by gm is the value of re which we get. α by gm you can prove it yourself to get the value of re. α is basically the common emitter current gain, right? Common emitter current gain will mean that you will have re approximately just to α by gm, right? And you can get the value of α and gm, it is typically very low because, α will be hardly of the order of 0.99 or 0.98 and this gm will be typically large because sensitivity is very large, so your re will be typically very small quantity. So your resistance looking from the emitter side will be very small, right? And that makes it go to a very small state or a very small region. And therefore as you can see on the right-hand side if you look very carefully from the common base configuration I get re I get α times ie as the collector current and therefore, for a common emitter configuration, so this is common emitter consideration CE and this is for the CB configuration and therefore I can get the value of this thing. So I will recommend that for both hybrid-π model as well as, T model for all the three configuration CB, CC and CE which means common base, common collector and common emitter please try to find out the value and the model itself please try to get it from any standard books which are available in the market, right? Right? (Refer Slide Time: 26:34) Let me therefore now come to the basic, we will finish with this common emitter configuration. Let me come to the common emitter configuration now and explain to you what common emitter configuration how it works, as far as, how we are inserting the small signal model in a common emitter configuration, right? So this is the application of the small signal model, right? So what is the application? The application is something like this that I have a signal input resistance, I have an input signal, I have BJT which is there, RC is the resistance, so if you look rn is basically, which is this one, is the input resistance looking from the base side and output resistance looking from the collector side is Ro. So what we are doing, we will replace this. So if you look out of the blue box. This blue box, out if you look, this RC is this RC, so this RC is this RC, right? This Rsig which you see is this Rsig, right? Signal. So looking from this side this can be broken down into an rπ, right? Because rπ is already known to you and this will be gm times rπ, right? And Ro will be the output impedance offered by the device which is there for the device itself, not for the circuit point of view. So therefore I get R0 always parallel to my current source here, right? Howsoever large or small we don’t know but it will be the value which will be there R0 here. So we get Rin to be equals to rπ, no need to understand why is it, very simple, because looking from this side from the base side I only get rπ as my impedance, so rπ is the Zin value. V0 is gm times Vπ, Vπ is nothing but the voltage across this VBE or Vπ whatever you want to name, it is basically the output current. This multiplied by Rc, so this Rc and Ro in parallel, right? So the affective resistance seen by the current is basically Rc parallel to Ro, this if you multiply with ic, I will get V0, right? And I get V0 here. So R0 is nothing but Rc parallel to r0. So the output impedance is basically Rc parallel to r0, right? So this is the output impedance which you see. So for a common emitter configuration we are able to see the output impedance, right? Okay, with this knowledge, with this basic knowledge will stop here today and what we have learned is basically the small signal input model for the BJT and in the next turn we will look into how we can apply further for the common base as well as common character configuration design. So we have learned how to bias a device, DC bias, super impose AC signal over it and we have also learned how to use those for the purpose of common emitter configuration using the small signal model, fine. I hope you have understood the issues taken up in this lecture, thank you for your patient hearing.