 So, this is essentially in continuation with what we had discussed yesterday of what is the difference between linear and digital design. So, my basic contention is that there are no linear circuits and there are no digital circuits. We often say linear circuit, this is an analog circuit, this is a digital circuit and so on. There is no such thing as a linear circuit or a digital circuit. There is a linear mode of operation and a digital mode of operation for all circuits and if the importance to your application is in the digital mode then you will use it in the digital mode. For example, an op-amp if you use it in a mode that it is always saturated then its output is digital as a comparator. So, you can think of it as a digital circuit on the same line as we have discussed that day, that an inverter can be thought of as a linear circuit in this regime. So, this is what we had discussed yesterday so I will not repeat, but just that the analog circuits require the output voltage to be sensitive to the input voltage, essentially they want dv0 by dvi to be high. Now, on the other hand digital circuits require the output to be insensitive to the exact input voltage. Now, if you are designing a circuit for this regime then there is an important conclusion from this that is that circuits need to be biased for operation in the linear regime. What it means is that if you have a low level signal, if the signal is referred to ground or to be dv then it is not going to be in the linear regime, you will not be able to amplify it because dv0 by dvi is 0 there and those regimes as we had discussed yesterday are in fact designed to diminish this small variation. So, in order for this signal to be amplified we have to bring it to this level and that involves that we must add a bias to this voltage to bring it. So, that is the additional consideration in case of linear design right, the circuit naturally does not find itself. So, therefore in the analysis we all the time talk of a DC analysis which is about bending the circuit here and then AC analysis in which the input is not the actual voltage, but the deviation from this bias voltage. So, we are always concerned about the behavior of the circuit to DC because we want to know what its vi characteristic is to that bias voltage and then what is its response to the deviation around that. Now, in fact even the terminology reflects this we typically use capital v's for DC voltages and small v's for fluctuations around that DC voltage. So, please be very careful about the terminology I am not sure whether I have been equally careful, but in general whenever there is a small v it means a fluctuation around a bias voltage whereas wherever there is a capital v it means this is the DC response and unfortunately in analog we have to be equally worried about both of them and therefore we will often be talking about both of these. Now, in fact the second lecture that madam Shooja is going to take is how to control the capital v, how to generate circuits which will provide the appropriate bias. Right now, however before that we want to discuss how to make good amplifying circuits that is going to be the basic intent of this lecture. So, before I go I do not assume any knowledge on your part for analog design. Let us start absolutely from basics and let us find out what we actually mean by an analog amplifier. Now, here I take a very simple example. I take a transistor here. However, I actually do not need this to be simple. Any linear operation circuit can be placed here as long as we have ways of characterizing its current voltage characteristics. So, later you will see that we will put a cascode here or we will put a differential amplifier here and so on. All of them will follow essentially the same relationship. But to begin with let us start with something simple that we can appreciate very easily. So, let us put a single transistor here. So, what do I have here? I have a small transistor here and it is being fed by a current source. Notice this is the small v i and this is the capital V G. So, for bias we have applied a gate voltage here and superimposed on top of that V G is a fluctuation which is small v i. Now, let us start from absolute basic principles. So, this is not analog design. This is elementary calculus. What is the differential of total differential of the current flowing through this circuit? This is the partial differential with respect to the gate times delta V G and partial differential with respect to the drain voltage times delta V D. So, this is not rocket science. This is just expanding. After all, the current through this depends only on two voltages. It depends only on the gate voltage and the drain voltage. Therefore, the total variation in I D is the partial variation with respect to V G and the partial variation with respect to V D multiplied by the corresponding deltas. So, this is the starting equation for amplification. Now, notice that we are not assuming any transistor equations and so on. Eventually, the calculation of these partial derivatives will be from the IV characteristics of whatever device we place here. In this particular place, we have placed a single transistor. But this equation applies to anything that you can place there. Now, we define D I D by D V G to be the trans conductance, G M. So, D I D by D V G should be evaluated at this capital V G. So, this D I D by D V G is the trans conductance. It is so named because it links an output characteristic which is the drain current to an input characteristic which is the gate voltage. Hence, it transforms the input to the output. So, we will call D I D by D V G as G M or the trans conductance. Similarly, D I D by D V D, both of these are output characteristics. The voltage at the output, the current through the output. So, therefore, D I D by D V D we call G 0. Then, this equation can be rewritten as D I D is equal to G M. This is G M times D V I which is small V I. The variation in V G is the small V I. So, G M times small V I plus G 0 times V 0 that is D I D. And D I D is 0 because it is a current source. So, there is no variation in current. It is a constant current source. So, this is 0. So, this is our defining equation. Notice, we have never got in specific characteristics of a transistor here. Therefore, it will apply to whatever we place here. So, therefore, 0 is equal to G M times the input voltage plus G O O times the output voltage. And these are fluctuations. These are not absolute DC values in volts there. This is the amount of fluctuation at the input and amount of fluctuation at the output. And therefore, the voltage gain is 0 which is V 0 by V I is minus G M by G 0 as simple as that. Very often, instead of talking of G's which are conductances, we talk of a resistance. And instead of G 0, we talk of an R 0. So, the gain is minus G M times R 0. Now, this equation is very general. It applies to any device connected here. And we will be using it throughout this lecture. And whenever you do any analog stuff, you will find that this thing props up. So, this is the fundamental equation of amplification. Now, let us not just reduce this to algebra. Let us understand what is going, what is happening here. So, what is R 0? R 0 is this. The rate of change of current with respect to the rate of change of the output voltage. So, we want R 0 to be large or G 0 to be small. This is G 0. So, for getting good gain, we want G 0 to be small. When is it small? When this differential is small? And if this differential is small, what does it mean? That means, ID does not change very much with V D. Correct? That is when this differential is small. And what it means is that we should be operating in the flat region of the transistor characteristic. Where the drain current does not change a lot with the drain voltage. It is for this reason that most analog design requires the transistors to be in the saturation region. So, that the current does not change with respect to the output voltage. This part is clear. This is fundamental to good analog design. And it comes from here. And it is applies to almost anything that you amplify. This is the defining equation of amplification from which all things are derived. Now, let us make it specific to a transistor. So, suppose that device is actually a single transistor. So, in that case we can write. Now, this is a somewhat more advanced equation than we want to handle right now. So, just assume alpha to be equal to 1. Alpha takes care of a second order effect, which says that the turn on voltage of these transistors is not necessarily constant. It depends on the source voltage. So, if you take care of the dependence on the source voltage of the VT, this alpha factor is introduced. But currently just to keep things simple. Assume throughout when you see these equations that alpha is actually 1. So, then what was ID? Half K into VG minus VT square right. So, therefore, DID by DVG which is what GM is right. So, that is K by 2 this is this is ID half K K by 2 and VG minus VS is a VGS minus VT right. What we are going to actually this alpha what it does is that it says that VT is some constant VT 0 plus something which depends on VS that is what this alpha was supposed to take care of but right now we say we will ignore. We will assume that VT is constant to first order right. So, then it is nothing but VGS VG minus VS is the VGS right minus VT whole square. So, this is just the saturation equation the simple equation that we had seen last time. Therefore, if you take its derivative and this K is that mu C ox W by L in the transistor equations. Now, what I am trying to do is to introduce something here which we will try to respect as we go through this discussion and that is a somewhat qualitative fact that we should understand. What is a designer's life? What is expected of you? There are certain things that you have control over and there are certain things that you do not have a control over. So, therefore, all the equations that I have rather than writing them plainly algebraically I am going to divide them in a particular way. So, that you will be able to appreciate their impact on the process of design. So, there are certain constants which will which are given to you as technology. These are technology dependent constants the VT a designer cannot change VT. The technology has given you a value of VT saying this is the transistor I give you you work with it. There is a dependence on geometry and that you decide as a designer you decide whether you will use a fact transistor or a weak transistor and the operating condition. So, you decide what bias etcetera the combination of this is the circuit the combination of technological parameters geometrical parameters and the operating condition that is the circuit. So, therefore, wherever possible in this discussion I will try to separate out these factors and we will see the dependence on all three of them separately. That way you will know what is in the hands of the designer and what is not. So, this is what I have done this K I would like to wake up into this part which is mu C ox which is not in my hands that is the mobility of electrons as a designer I do not control it C ox I do not control. So, I would like to take this out. However, W by L is in my control. So, I would like to retain that separately. So, therefore, this conductance factor K I write as K prime times W by L where K prime is mu C ox not in my control it is some number which is given to me by the technology, but W by L is my concern. So, that I would like to see the dependence of W by L etcetera etcetera. So, this is what I am going to do with all the equations which follow. So, what I do is that I define V G minus V T in this simple case if you take alpha equal to 1. I define V G S minus V T to be V G T. So, what is V G T? It is the amount by which the gate voltage is above the conduction threshold. V G S minus V T. So, how much above the conduction threshold is the gate voltage that is what V G T is. So, V G T is. So, this is the definition for that. So, now my equation becomes very simple. Remember we are putting alpha equal to 1 in this simple discussion. So, it is just K by 2 V G T square. Also notice that V G T is linear in V G other things are constants. So, then what is G M? G M was partial derivative of I D with respect to V G and because V G and V G T are related linearly. So, this is K by alpha into V G T itself into D V G T by D V, but D V G T by D V G is just 1 because of this. So, rather than taking the derivative with respect to V G, I have taken with respect to V G T because the derivative of V G T with respect to V G is just 1. So, therefore, this gives me K by 2 alpha into V G T square. Now, I am going to play a trick on you. So, be careful and catch me if I do something wrong. So, follow the development from this point onwards quite carefully because I am going to raise a question and I will ask an answer from that because there is an apparent mismatch in things beyond this point. So, let us start. All of you agree that I D is K by 2 V G T square. By the way, this is not the catch. So, there is no disservice in this and G M is there for its derivative. So, that is that number and notice that I have again divided it like this. So, K prime is not in my hands. In fact, naught is alpha in my hand. So, but we are taking alpha equal to 1. So, K prime times W by L times V G T that is G M. So, notice G M is directly controlled by how much above the threshold voltage I am in the voltage because V G T is the amount of forward bias I put on this transistor. However, look at this equation. I can evaluate V G T in terms of current and it is nothing but root 2 I by K. Again alpha is 1. So, this is root 2 I D by K. No trick in that. This is the equation we started with. Therefore, V G T is nothing but root 2 I by K. So, I can substitute for V G T in this equation. So, this is K into root 2 I by K. So, that gives me root 2 I into K. K into square root of root 2 I by K that square root will that K when it comes in will become K square. So, it will be you will get root 2 K into I and again I am splitting this K into K prime and W by L. So, this there is no trick in the derivations by the way the derivations are accurate. So, what do I get G M as 2 K prime times this geometry factor times I D. This is another way of writing G M and the third way is that I can eliminate K. So, from this equation again this is the starting defining equation. So, what is K? K is 2 I by V G T square from here. From this equation I can say K is nothing alpha we are taking as 1. So, it is 2 I by V G T square agreed. So, if I put the substitute this value 2 I by V G T square in this then what I get is 2 I D this is K 2 I D by V G T square G M comes out simply as 2 I D by V G T by putting this value whatever I done I have created 3 separate equations for G M in which I am interested in terms of either I D or V G T or K etcetera etcetera. All these equations are derived from each other and must be consistent agreed that is when I come to a problem these are the 3 equations that we just now derived. Now, I do not care about algebra you may have derived it I do not care what I care is that for high gain I need high G M my gain is G M times R L right. So, the question I ask myself is that suppose I want to increase gain in want to increase G M what should I do. So, what is in my control geometry and operating point technology is not in under my control right. So, what should I do I want to increase G M right. So, what should I do what this equation tells me is that increase V G T right I get good G M if I increase V G T everybody agrees. However, this equation says that I should decrease V G T right and both are correct both are derived from each other. So, I have a problem similarly the other thing I can control is geometry. So, I say all right you know something has happened there too confusing we will come back to it later the other thing I can control is geometry ok. So, G M directly proportional to the geometry linearly dependent on W by L or is it dependent on the square root of W by L or is it independent of W by L ok. Now, all these equations are derived from each other they must be consistent and yet here I am a poor old designer who just has derived these equations and I am totally confused I am asked to design a single transistor circuit and now I want a high value of G M and I do not know whether to operate it at high V G T or may V G T I do not know whether how to adjust the geometry suppose I want to change the G M will the G M change proportional to the geometry square proportional to square root of geometry or it will totally ignore any changes in geometry and G M will remain the same ok. So, what is going wrong here that is the question I wanted to ask you these equations there is no trick in them they are derived from each other by eliminating one or the other. So, they must be consistent and yet here I am as soon as I want to use these equations I find that I get contradictory results. So, what is happening what is happening is this and that is why you should have paid attention when I put that last bit that in an analog circuit you need to bias it and how do you bias is very very important and depending on the way that you bias the circuit one or the other of these equations will apply other equations will not apply ok. So, let us see how that determines for example, to bring the circuit into the analog regime you may give it a voltage bias V G in that case V G is fixed V G is a known parameter right. So, V G as and transistor geometry is given to you then the current is not a free variable if V G and geometry voltages and geometry are given to you then the current is not a free variable therefore, you must not use an equation which is in terms of current. So, you must use an equation which is free of current because current you have no control over right. Similarly, there are many circuits which are current biased like a differential amplifier you have a tail current etcetera they are current biased in that case the current is given, but the voltage adjusts till that much current flows then you must not use an expression which shows voltage as one of the variables ok. So, therefore, depending on how you have biased this circuit and that is why biasing is so important that depending on how you have biased this circuit you must use an appropriate equation here ok. Have you appreciated this this this part ok because it is possible to be let us say you know algebra is very easy just find the value substitute etcetera you got three equations, but to actually use them in design you may be totally confused therefore, it is very important to appreciate that what is a fixed variable and what is something which will adjust this is because the transistor equation bring in a constraint not all three variables are independent of each other the voltage geometry and current are not independent of each other you give me any two the third is fixed ok. So, the equation must be in terms of what is an independent variable the dependent variable should not appear in that and only when you do that will you do properly will you do the designs properly otherwise it is very easy oh somebody brings around a circuit right and the manager or say sir my gain is 10 percent low oh increase increase VGT by 10 percent everything will become ok that guy comes and says sir the gain has gone even lower ok. So, therefore, you have to be very careful of what is the parameter which is eliminated from here right sometimes you may do the analysis and you may want to fix both voltage and current in that case geometry is not a choice you say ok this transistor should draw so much current at this much voltage and therefore, find the geometry which will meet that then that is not a design variable anymore and therefore, you must not be using design equations which are in terms of that variable ok that is not a free variable. So, if this is conceptually very very important otherwise you know you may look say I saw it in that textbook actually or I read it in that paper and that equation must be correct of course, all three equations are correct ok and you take some decision based on that you may end up taking the wrong decision all right. So, this is that is why it is in red bold letters it must register in your mind and equally when you teach this in the class it must register in the minds of the students because students one great weakness they are it is that they try not to understand and stuff formula with values. So, there is a formula which is given value which is given just stuff it put it into the calculator something comes up then they do not care whether it works or not ok. So, it is very important to not only tell them that this is important but also why this is important ok essentially what is happening is that as I increase vgt here there is something else which is going to change because of the current for example, here if I decrease vgt the current will actually decrease by vgt square. So, here that dependence is not visible at all ok. So, if I decrease vgt then id will decrease by vgt whole square and g n will actually decrease at that particular point if the if current is a dependent variable ok. So, I must be very careful about these things all right once I have made it then g m I know I know what is the bias that I have used I know which equation to use and now I know how to get a particular value of g m the other thing of importance is g 0 the output impedance ok what is it the the this is a slope of the transistor characteristics right. So, now let us look at the output conductance g m we have discussed the output conductance you know maybe I will draw it here now the characteristics are like this and all these lines this is the portion that I am worried about and the assumption is that all these lines will join at some point called the early voltage on the negative side ok this is the transistor characteristics we know. So, what does it talk about this slope remember this is like a hinge. So, the slope becomes larger as the current becomes larger ok. So, the slope is proportional to the absolute amount of current that is one thing the second thing is that the slope is inversely proportional to v e the closure is v e the more will be the slope the further away in fact if v is at infinity then this thing will be absolutely flat ok. And what causes this slope actually this slope is because of the channel length modulation the transistor has a particular physical channel length, but because of the drain voltage the depletion layer comes into this and that subtracts the channel length and therefore, the current increases that is why the current is dependent at all on the drain voltage otherwise physics says that the current should have been constant with drain voltage, but what happens is that as you increase the drain voltage the depletion width associated with the drain junction gets into the channel and reduces the effective channel length. So, instead of w by l you get w by l minus delta and therefore, the effective w by l factor goes up therefore, the current goes up right. So, therefore, smaller the channel length more important is delta with respect to l right. So, therefore, for long channel effects you will not get a lot of slope, but for short channel devices you will get very steep slope ok. So, let us have a very simple model here which is a linear model we will say that g 0 is directly proportional to i d and inversely proportional to l ok shorter the channel length right current is w by l. So, it should be inversely proportional to l right. So, g 0 is inversely proportional to l and directly proportional to i d larger the value of i d as you had seen larger is the slope and this is the proportionality constant lambda prime ok. I call it lambda prime and not lambda because there is another variable which spice simulation programs use called lambda which is also occurs in the same context and I do not want I do not want any confusion I want to avoid confusion. So, we are just calling this proportionality constant lambda lambda prime ok not lambda because lambda is used by transistor parameters in spice or circuit simulation programs it is not that lambda it is related, but it is not that lambda ok. So, then g 0 is lambda prime i d divided by l ok this is bias this is geometry this is technology ok. Now, I know g 0 I can now play the same trick from the transistor equations I can put i d in terms of v g t and the geometry in terms of v g t and so on. So, similarly I can find various equations ok and the early voltage is nothing but l by lambda prime this whole thing lambda prime by l is nothing but 1 by early voltage ok. So, now of course because I have various combinations 3 expressions for g m 3 expressions for g 0 I can now write down g m by g 0 which is the gain right in terms of those 3 constraints that this is a 1 of the equations should be free of v g t 1 of the equations should be free of i d and 1 of the equations should be free essentially of geometry ok. So, that I can write it down. So, the voltage gain in terms of geometry and v g t free of i d is a 0 equal to 2 l upon lambda prime by v g t in terms of grain current and geometry this is a 0 equal to this etcetera ok. So, now g m by g 0 can be found out I have done g m I have done g 0, but I observe that if I use 1 equation there which is free of i d I also use the corresponding equation here which is free of i d. So, that eventually I come out free of i d ok. This is for d c d c gain that is to say given any fluctuation I am not talking of the rate of fluctuation right. So, that is d c, but what happens if you have a c that means this fluctuation is occurring at some rate and therefore, the capacitive effects cannot be ignored ok. So, again I will just bring back that original fundamental thing is very important and it is also very important that you make your students connect to that fundamental frame of analog amplification. So, if you recall what have we used this was the current current source right and here was all device and this had bias v g and superimposed on that the small v i ok and this was my v 0 at some v d right. Now in a c I cannot ignore the capacitances right. So, notice that now there is a capacitor here which I must charge and discharge right. That is what we were discussing in some sense yesterday in the in Prasachand Rukas lecture right on logical effort it is the analog version of the same thing. So, I must now worry about this capacitor. So, why is it that I have to change my original theory which appeared so fundamental why do I have to bring in a new thing in that theory and the answer to that is that what did we do we put delta a d equal to 0 remember which is true here also delta a d is 0 that means the fluctuation through the current source is 0, but that does not mean that the fluctuation through this current is 0 anymore because now there are two parts of this current and the sum of the two is 0 right overall it is made by a current source, but now delta a d of the transistor is not necessarily 0 because it must provide the charging discharging current ok. So, therefore, it is the sum of these two which is 0 that means there is an effective R C circuit a first pole circuit which I must analyze at AC. So, what will happen is that at very high voltages my delta a being 0 my a will become less because earlier I just got g m by g 0 by putting v 0, but now this current will subtract from that the amount of current available for amplification is now lower it will subtract from that and my gain will come down as the frequency goes higher that is to be expected you are putting a capacitor across from the output the gain will come down right. So, this g 0 is essentially equivalent to an R L here and this R L times C provides a first order pole across this on the output and therefore, it will have the usual decreasing gain and with a slope of 6 dB per octave ok. So, what we do is that alright we have already done the hardware though finding the gain for this in DC we will say that the actual amplifier is that same amplifier with an R C circuit as a first order pole ok and then that is the AC picture of this right these are my gate and source terminals ok. Now, there is this CG across the input this is the gate capacitance and this VI is the same as the previous one the ideal VI, but rather than taking the output directly like that in parallel across this is the output resistance because this is a current source. So, the resistance will come in parallel and this is the load capacitance from drain to source ok essentially what what you mean is that this slope is because of this R 0 right otherwise the current would have been constant suppose this R 0 was not there then the output current would have been constant because this is just g m times VI right it is this resistance which provides the drain independence one doubt sir. So, the g m VI is the current through the transistor now through the this is the this is the current caused by yeah through the transistor and this is the current through the capacitor and the sum of these two must be must equal whatever 0 in terms of deltas ok. Now, this remember is an incremental circuit this is not a DC circuit. So, it says what happens to the fluctuations of voltages ok. So, the fluctuating part of the current is g m times VI the actual current is determined by the transistor characteristics ok. So, this is an equivalent circuit for fluctuations alright. Now, it is very easy to do the algebra you just sum the current here ok. So, the current coming here this is VI this is V 0 ok. So, VI minus V 0 times the conductance of this. So, VI minus V 0 times the conductance of this CGD right that is the incoming current this is the outgoing current g m times VI. So, you put that this is outgoing current V 0 by R 0. So, you put that with the minus sign and this is outgoing current through the capacitor right. So, that is minus S V 0 times V 0 right. Once you do that you collect all terms in VI and V 0 separately and calculate the new expression for V 0 by VI which is the old expression minus g m R 0 multiplied by this factor which is the new gain ok and notice that it is like 1 plus R C 1 plus something R C right. So, this is a first pole circuit as expected right. So, this part is clear algebra is not that important it is important to understand that now my gain is less because of this R C circuit and this is what it is also telling us ok and this CGD and this C 0 are coming in parallel here and this CGD is also giving a 0 here and that is that simply says that because of this series connection there is some output which will come because of this feed through which will increase with frequency ok, but in reality this CGD is very very small it is a parasitic capacitance between gain and drain it is because of that small overlap capacitance. So, this parasitic capacitance is very very small and initially we will ignore this. Could you just explain what that S is actually? S is j omega this is the Laplace transform. So, this is effectively j omega. So, the conductance of this is j omega c ok the resistance impedance is 1 by j omega c the conductance is j omega c. So, j omega c j omega c times the voltage ok that is the current fluctuating current you just write this ok alright. So, then we finally can write a a 1 as the dc current divided by 1 plus j omega c r ok this S is j omega and this is c times r. So, it has a time constant c r right and once omega becomes greater than c r this current will start decreasing because of that 1 plus j omega c r ok. So, that is what we expect a first order pole effect. So, essentially this is how it looks ok. Now, we define various factors here which is standard with the first pole analysis nothing new here. So, at the point where the gain has dropped by 3 dB ok that means the voltage gain has become lower by square root of 2 that point is called the bandwidth. So, this frequency is called the bandwidth of this amplifier and the point where the gain reduces to 1 that means it stops amplifying completely that is the cut off frequency ok and because it drops linearly on a long scale here that is called the gain bandwidth product ok. Any point here right the gain times bandwidth remains constant along this line ok. So, that line is called gain bandwidth product when gain becomes 1 the frequency is gain bandwidth product. So, this we will be calling GBW alright. Now, GBW is then this is the gain GMR0 and the bandwidth is 1 by r c right. So, this is GMR0 into 1 by r c which simply gives you Gn by c total ok. Notice one very important point the frequency at which this amplifier stops amplifying that is independent of the slope of the output characteristic ok it is independent of r. The gain depends on r the bandwidth depends on r when the gain bandwidth frequency that gain bandwidth product is independent of r because it occurs here in the DC gain as well as in the roll off factor. So, what it means is that as you change r suppose your r is much higher that means your DC gain is much higher, but it starts rolling off much earlier ok. So, essentially what it means is that you will start here, but you will start rolling off much earlier and eventually come down on the same straight line ending at the same GBW ok as you change r0 your bandwidth will become less and you will start rolling off earlier though you would start out with a new with a higher DC gain ok. This part is clear now I say suppose I do not put any load at all what is the maximum possible AC gain that this object can have right suppose I put no load then it gives me a kind of upper bound on the capability of this right. Suppose I do not put any external load only the self loading is there ok. Now how much is the self loading the self loading is proportional to the width of this transistor the wider the transistor larger is the capacitance on it right. So, I assume that the load capacitance is proportional to some chi times W where chi is another proportionality constant ok. So, if the C load is chi times W then I can calculate the GBW max and I get this simple equation which is 2 ID by chi times whatever ok. This is absolutely the highest frequency to which you can go with this technology ok alright. Now, we can make a little table do not worry about if this table is in your notes you do not have to worry about the actual things, but in terms of WLVGT this is GMG0A0 gain bandwidth product and GBW max ok. We calculate all of these and put down in this nice little table and we have a particular purpose in doing this ok and what is the purpose that I if I further multiplied gain with GBW max ok. So, gain bandwidth product is already the product of gain and bandwidth. So, if I further multiplied by gain that means I have gain squared times the bandwidth if I calculate that what I get are only technological factors that means the designer has no contribution to this quantity this quantity is invariant you give me a technology then this is all I can do do not ask me for more because as a designer there is no geometry here there is no operating point here ok. So, this is the this is a measure of the technology that you have given me if you give me this technology this is the gain squared times bandwidth which is possible ok. So, one thing is that you give me a technology I will look at these all these technological parameters multiply them and if you come to me saying please design an amplifier which has this much gain and this much bandwidth ok. If it exceeds this gain squared times bandwidth exceeds this I will say sorry this technology will not do bring me a better technology ok. Everybody agrees with this whenever I ask this question some rings to should start ringing in your mind ok because when I normally ask when the answer is unexpected well as far as this is concerned that is actually true ok that is called the technological constant of this amplifier ok this is the best I can do unloaded amplifier no demands of this amplifier this is the highest frequency to which it will go given this technology you want better give me a better technology ok. So, consider this scenario that you are the designer go to your boss and your boss says I do not care this is the gain I want this is the bandwidth I want and this is the technology I have are you saying that nobody can make an amplifier whose gain band which will give me this gain bandwidth ok. So, now you are a little worried whether all this analysis is correct or not. So, here is the bet that your boss places with you saying that if you give me an amplifier which gives me this much gain and bandwidth which is beyond this ok then you get a raise and otherwise I will kick you out and hire somebody who can give me this ok because I need this and this is the technology I have. So, now that such a sinking is it really are we really limiting limited to this from technology factors the designer can really do nothing I do not see anything here which designer can control ok and then suddenly like Kekilay's dream I say look I am assuming that only one transistor is used can I not use two transistors and use this transistor pair to generate better gain bandwidth product ok the answer happens to be yes. Now, there are two transistors how many possibilities are there that I connect them I can either put them in series or put them in parallel right there is no other choice that I have if I put them in parallel effectively I get a single transistor which is wider and that I already know will not help you get this take two transistors connect to gain to gain source to source right put them in parallel and now obviously that will not do because that is equivalent to a single transistor and I know that a single transistor will not give me these characteristics. So, the only possibility is to put them one on top of the other and go to the Hadovan Mandir and pray that may be this configuration will give me the characteristics that are required ok. So, that is called a cascode and that is what we will look at next ok. So, this is the cascode so, whatever I done I have put a transistor here and put another transistor on top I have given it a bias so, now I have capital VG 1 and a capital VG 2 no signal is applied here only DC is applied here ok and this is my compound transistor both of these together right which I will replace that signal transistor with rest remain the same there is a current source here this is the output this is the input voltage everything else remains the same ok and now I am hoping that my gain will be better ok. So, you got it I do not have two signal sources to connect only one signal source therefore, this guy should be connected like a single transistor and that is possible because I am just supplying a DC bias here now why do I hope to succeed ok there is a qualitative reason quantitative stuff will see later, but there is a qualitative reason why I hope to succeed in getting higher than what I was getting earlier ok the reason for that is the following what is our enemy what is limiting us in gain because the output current changes with respect to output voltage if the output current did not change at all if I had ideal saturation then I will have infinite gain right G 0 will be 0 and G m by G 0 will be infinite then I can see my boss in the eye and say tell me what do you want and I will give it to you unfortunately the real transistor the G 0 is finite. So, my attempt is to reduce G 0 right what I want to do is that as the voltage changes on the drain the current through this transistor pair should not change very much that is my hope and that is what this pair will do because if I fluctuate this voltage remember that this is a DC voltage and this voltage is like a source follower with respect to that right. So, if there is no fluctuation here there is no fluctuation here this is a source follower it will follow the input voltage and there is no fluctuation here there should be no fluctuation here right if there is no fluctuation here then this guy actually sees a constant drain voltage even though the voltage here is fluctuating okay. So, this guy sees a fluctuating is a he sees a constant drain voltage therefore the current is constant and therefore the overall current will be constant okay of course this is a somewhat rough theory in reality the current through this will fluctuate as the current through this fluctuates this VGS will change and therefore there will be a small amount of fluctuation here it will not be 0 but it will be small however I do expect that it will help me improve the output characteristics of my transistor okay this is one way of looking at it there is another way of looking and that is this think of this as an amplifier okay now the gate is grounded for AC right so it is like a common base amplifier and if this is its output voltage this fluctuation is its output voltage fluctuation here will be that voltage divided by the common base gain of this stage so that will be much smaller because we know that in common gate the voltage gain is very high right so therefore the fluctuation here by either argument will be is going to be much smaller and as a result the current fluctuation produced due to drain voltage variation will be very small that is why this becomes better okay once you have understood it the rest is just mechanical detail okay so have we understood this what have we done essentially you have put this transistor to isolate the voltage fluctuation at the output terminal from the voltage fluctuation at the amplifier transistor okay all that extra voltage fluctuation is absorbed across this okay so its source voltage is changing very little and the drain voltage is doing all the fluctuations and very little of the drain voltage fluctuation here is passed on to the drain voltage fluctuation here that is what is happening that is the advantage of the cascode and finally what do i come up with that the effective gm of this two transistor pair is the same as the gm of the lower transistor okay that is to be expected because the current is current in response to the signal is controlled by the lower transistor and because these two are in series whatever current flows through the lower transistor that is the current which will flow through both the transistors right so the gm is controlled is the same as the gm of the lower transistor right