Lecutre 16 Module 02

So, up until now we have been discussing the architecture of common source amplifier and what did we see? We saw that we saw that we after lots of discussion and soul searching we came up with this architecture which was able to give us some gain, what was the value of the gain? The value of the gain is supposed to be minus gm times, so the value of the gain is supposed to be minus gm times Rd parallel Rl under ideal condition, right? What is what I mean by ideal condition? What I essentially mean by that is when C1 and C2 are acting like infinite capacitors at signal frequencies when R1 parallel R2 combination is not loading your input, right? So, if these conditions are met then the maximum gain that you can get out of this configuration is gm minus gm times Rd parallel Rl, ok. So, now this we saw, so what, so now the issue is, so the gain that I am getting out of this common source amplifier, the voltage gain is minus gm times Rd parallel Rl, right? If we could make, if somehow we could make Rd to be much much greater than Rl then the gain for a common source amplifier would have been minus gm times Rl and that makes sense because if you look at the incremental, if you look at the incremental network what do we see? In the incremental network the current I naught, right, the current I naught that was getting divided between the branch on the top in Rd and the branch on the right, that current would prefer to go to the branch on the right if Rd was much much greater than Rl. I am obviously assuming C2 is infinitely large, right? So, the current always tries to go towards the path of minimum resistance, if Rd is much much larger than Rl then the current would definitely prefer to go to the right and hence the output would have been minus gm times Vg times Rl and hence the gain would have been minus gm times Rl, right? So, this is the maximum possible gain that we could get from the common source amplifier, ok. Now, how does this gain relate to our, to our quiescent point, right? Because ultimately the incremental characteristics of an amplifier always depends on where you have biased the transistor, how you have biased the transistor, right, ultimately it all depends on that. In this case clearly gm is the incremental parameter that depends on, that depends on our quiescent point, right? So, what is gm? gm is mu n C ox W over L Vgsq minus threshold voltage, right? If I sketch only the input side, so what is Vgsq? Vgsq is R1, R2 Vdd, so Vgsq is equal to Vdd times R2 by R1 plus R2, correct and this is, this is fixed. Why do I say this is fixed? This is fixed because once you have decided the power supply, you have chosen the power supply, once you have biased R2 and R1, R2 and R1, then whatever you do, the value of Vgsq is not going to change, correct? The value of Vgsq is not going to change, then clearly, clearly the Vgsq is, is fixed. So, so I mean you might be wondering why am I harping on this, the reason I am harping on this is the following. As it turns out, as it turns out, mu n and threshold voltage are temperature dependent variables, right? So what is, what is mu n? Mu n is a mobility, mobility of the electron in the channel, in, in, in plain English, mu n essentially gives, gives an estimate of how easily an electron is moving through, moving through, moving through the channel and clearly if you raise the temperature of the device, right, all the lattice structures and all the stuff that you learn in your device physics courses will tell you that the mobility of the electron changes, most likely the mobility of the electron goes down. Similarly, the threshold voltage of the device also as it turns out is a function of temperature, it also changes with temperature, which means what? This essentially means that the value of gm, right, will change with temperature, which essentially means the gain that you have obtained, right, the gain that you have obtained will also change with temperature, right? So, this is something that, that we do not want because ultimately when you want to deploy an amplifier in the field, right, in any product and you say that I guarantee a gain of 10, right, minus 10 or minus 2 or whatever you want to guarantee, you should, it should not be such that I guarantee you a gain of minus 10 when you make your temperature equal to 27 degrees centigrade, right, it should not be that, it should be, you should be able to guarantee the gain of minus 10 regardless of, I mean not, even not, I mean over a range of temperature, right. Let us say you, you take your cell phone from Kashmir to, to the third desert, right, so it should, I mean your amplifier should still work, which means that it should work from let us say minus 20, 30 degree centigrade to almost plus 50 degree centigrade, it should, it should work. If you are designing something for automotive applications, where electronics goes inside, inside the, inside the car, depending on where you are putting your automotive chips, the car, the temperature can go as high as, as high as 150 degree centigrade, right. So in a sense, you will have to ensure that your, your electronics, right, the ICs that you are designing should be working for a wide temperature range. But clearly, we see that if we, if we design an amplifier like the way we have done this common source amplifier, it seems like it has, it will have a lot of temperature variable, right. So, so Av, since mu n and threshold voltage are temperature dependent variable, Av that is minus gm times RL will be changing with temperature. So, this is something that will not be acceptable even to us when we buy a product, right, we want reliability. So what is gm? So again, gm is mu n C ox W over L Vgsq minus threshold voltage, ok. So how much does, I mean let us quantify the change of gm with respect to change in mobility and threshold voltage, right. So how much will gm will change? So if mobility mu n changes from mu n 0 to mu n 0 plus delta mu n, right, how much gm will change? The delta gm will clearly be equal to del gm del mu n times del mu n, correct. So this is, this is equal to delta mu n C ox W by L Vgsq minus threshold voltage, correct. So this is equivalent to if I write in terms of gm, this will becomes delta mu n by gm, sorry delta mu n by mu n 0 times gm, right. So this will be the change in gm when the mobility changes by a certain amount, right and clearly this change will get reflected in your, in the change in gain. Similarly, what will be your change, change in gm if the threshold voltage changes? So if, so again let us write down the gm expression, gm is mu n C ox W over L Vgsq minus threshold voltage. So if threshold voltage changes from some nominal Vth 0 to Vth 0 plus delta Vth, what will be the change in gm? Del gm will be again, so gm del Vth times delta Vth which will be, which will be minus mu n C ox W over L times Vth, right. So why minus? Because if threshold voltage increases, the gm, the value of the gm decreases, right. So I mean negative sign essentially is testament to that. So del gm essentially becomes, if I write again in terms of gm, so if I multiply with Vgsq minus threshold voltage, so this becomes voltage by the overdrive times the original gm, right, okay. So one might argue, one might argue that if I want to reduce, if I want to reduce the effect of change of gm due to change in threshold voltage, what should I do? I should, I should bias the transistor with as high, as high overdrive as possible, right. But now we know that what are the side effects of biasing a transistor with high overdrive? The bias, we already know, we already saw in one of the earlier lectures that if you bias your transistor with, with a high overdrive, then you, you lose on the available swing at the output, right, because you would also want to ensure that the voltage at the drain, the quiescent voltage of the drain is not very close to the overdrive of the transistor because otherwise you will, you will lose the available signal swing, right. So there is a trade-off. But if I say that if I have, I mean what can I do to reduce del gm, I mean change in gm because of change in mobility, seems like we cannot do anything because mu n 0 is first place is not in your control, correct. So this type of biasing is called voltage biasing, like whatever we did here is called voltage biasing. Why is it called voltage biasing? It is called voltage biasing because we were fixing, we were fixing Vgsq, right, we were fixing Vgsq, we were fixing the overdrive, right. So this fixing the overdrive is often referred to as, as voltage biased, voltage biased transistors, okay, okay. So I mean as you clearly can see if you voltage bias a transistor, there seems to be a, seems to be an issue with the variability, right. So but I mean let us, let us, let us look at it from another perspective. So why are we trying to fix the overdrive, ultimately what decides the, I mean why do we need to fix the overdrive, right. So if I fix this overdrive using some Vgsq, what is the purpose? The purpose is I am trying to get some Idsq, right. So this is, this is the cause and this is the effect, right. So another way of saying is that this is, this is the desired, we desire certain Idsq to establish certain gm to get certain amount of gain, hence we are applying a voltage Vgsq between gate and source to generate the current Idsq, right. We want the Idsq, that is why we are tweaking Vgsq, right. If we did not want, then we did not have to tweak it at all. So now let me tell you that what if I give you a current source, how you make a current source is a different question, right, but let us assume I give you a current source, we have been using voltage sources right until now, but let us assume I give you a current source, right. So let us say somehow I have given you a current source, right, and somehow I have ensured, right, so let us say somehow we, so let us say assume somehow, right, assume somehow we have ensured fixed current through M1, right. So I am saying that I have not shown you how, but let us assume that the current Idsq is fixed. When I say Idsq fixed, is fixed, what I mean is that Idsq will not change even if temperature changes. Anything happens to the rest of the universe, it does not matter, Idsq will not change, right. So, which means that Idsq is fixed. In the, note that this is different from the previous way of biasing, a previous example that we took, there Vgsq was fixed, the overdrive was fixed. In this case, we are making Idsq fixed, right. So let us assume that we have made Idsq fixed, we have not really, I have not really told you how we have fixed the Idsq, right. So let, if we make this assumption, let us see what are the side effects of this, right. So if Idsq is fixed, what is the value of gm? So again, you might say, you might tell me that gm is mu n C ox W over L Vgsq minus threshold voltage. I mean, it is always true, yes, it is always true, but this expression does not capture that your, that your Idsq is fixed, correct. So in order to capture that effect of Idsq, right, fixed Idsq, right, so what I will do is I will use the other expression of gm, which has Idsq inbuilt, right. So what was the other expression? This is, that was 2 mu n C ox W over L Idsq, okay. And now let us see what happens if Idsq got fixed. So if Idsq got fixed, now tell me what is going to happen if threshold voltage of this transistor change, right. So if Idsq is fixed and threshold voltage changes by delta Vth, what will happen to gm? Clearly, you see that change of gm will be 0, because gm does not depend on threshold voltage anymore, right. So delta gm will be equal to 0, right. What will happen if, if, if mobility changes, right, if mu n changes by delta mu n, what will be the, what will be the effect? So again, delta gm is dou gm dou mu n times delta mu n, which is what my expression for gm is 2 mu n C ox W over L Idsq under root, which means this becomes 1 over 2 under root mu n times under root C ox W over L Idsq, right, which means if I, if I incorporate that, that mu n terms, what would I get, times delta, delta mu n, right. So what would I get? So essentially I have to multiply with, this multiply with mu n here and divide by under root mu n here. So what do I get? I get delta mu n by 2 mu n times gm, right. So note that this is, this is better, this is better than the previous case where we did voltage biasing. In case of voltage biasing, in the case of voltage biasing, it was delta mu n by mu n times gm. In this case, it will be delta mu n by 2 mu n times gm, right. So the variability of gm with respect to mobility will also be better, but more interesting thing is to see that variability of gm with respect to the threshold voltage has completely vanished, right. So now, if you are looking at it for the first time, you might feel a bit bamboozled because of what is happening. I mean, just because I have used a different equation, does it mean I will get, I am supposed to get different result, right. What happened if I had used this expression? If I had used this expression, I mean, if I had used that expression, the del gm was coming out to be something different. How come just because I shifted to this expression, del gm with respect to threshold voltage became 0, right. So the answer to that is, the answer to that is you have to find out, you have to appreciate what is happening in this, in this network. When I say, ok, so let us take a deep dive into this. So when we say the current is fixed, right, so let us say the current is fixed, what am I implying? I am implying that probably there is a current source somewhere which is pushing current, which is pushing current into the device, right, ok. So there is probably a current source which is pushing current into the device, somehow fixing it, right, but, so let me not put the current source right now, ok. But we all know that the drain current is, is the effect, it is not the cause, right. We have to apply certain voltage at the gate, between the gate and the source in order to get a drain current, right. So again, mind you, drain current is the output, is the effect, cause is the gate to source voltage, correct. So then the natural question is, if drain current is, is the effect, then how can I guarantee a constant drain current? We can guarantee, what I am telling you is that we can guarantee a constant drain current if we already have a constant current source, right. Just like you have a constant voltage source, let me assume that we have a constant current source and as it turns out, you can build constant current sources, it is not a part of this course, but trust me and take it as a face value that you can build, build constant current sources. So if you can build constant current sources, what I can do is I can take a constant current source of value I d s q and connect it to the drain of the transistor, I can always do that, right, ok. So what will happen? A constant current source is a, is an ideal, is an ideal source, right. So if, if we have a constant current source and we, we connect it like this, what is going to happen to this, to this node? Can I leave the transistor as is? If I leave the transistor as is, what is going to happen? So the, what is the voltage between V g s is undefined, it can be anything, right, it can be 0, it can be 1 volt, it can be 9 volt, I do not know, right, it is not in my control, so it can be anything. If this is anything, if this is anything, what can you comment on, what comment can you make on the voltage at the drain of the transistor? One might say that if I do not know anything about, if I, if the voltage at the, between the gate and the source of the transistor itself is undefined, how can you ask me to make a comment between the voltage of the, between the drain and the source and that is the right assertion to make. There is no way we can tell what will be the voltage of the drain. So now if we cannot tell what is the voltage at the gate and what is the voltage at the drain, we cannot also tell what is the operating condition of the, of the transistor, correct. Now if we cannot tell the operating condition of the transistor or in other words, we cannot establish the operating condition of the transistor, then obviously we cannot even use it for a, for our amplification purposes, right. So what is the first thing we need to do? We need to establish the voltages at the gate and the drain properly, so that we will be able to establish, we will be able to establish the operating condition of the transistor, ok. So let us, let us see. So let us say I, let us say, let us do a thought experiment, so I give you a transistor and I give you a constant current source of value I naught, right, right. So let us say, let us say I tell you that I naught is 1 milliamp, let us say I tell you I naught is 1 milliamp, here is your transistor with certain mu n, C ox and threshold voltage, go and bias it with a constant current source, so that I d s q will be equal to 1 milliamp also so that the transistor is in saturation, right. So what will you do? And I also tell you that you have voltage sources available, obviously you need voltage sources in order to bias a transistor, so let us, let us assume you have voltage sources available. So what will you do? I will tell you what I will do. I know that the gate source voltage is the, is the cause and the, and the output drain current is the, is the effect. But if I have to stick the drain current into, into the, if I have to stick this I naught into the drain, what should I do? I will first go and see, plug in some arbitrary voltage, some arbitrary battery between gate and source, right? And let me call this some arbitrary battery V bat, ok, ok, call this V bat. So if I, if I connect this arbitrary battery V bat and let us say I put some arbitrary voltage 2 volt and I did my calculation, so let us say after calculation I figured that if M1, this is after, after my calculation I figured that if M1 is in saturation, right, a current of 1 milliamp will require 2 volt Vgsq, correct? So, so let us say I did all my calculations and I have figured it out, so I stick a battery in, right? So then, I have a Vgsq of 2 volt, I have a current of 1 milliamp, then do you think this will be fine? Of course, this will be fine, right? But now, let us assume that the threshold voltage changed, right? So let us, let us say the threshold voltage was Vth was 1 volt and Vth changed from 1 volt to 0.9 volt, let us say. So Vth changed from 1 volt to, let us say changed by delta Vth is minus 0.1 volt, right? The threshold voltage dropped, if the threshold voltage drops, what will happen to Idsq? Idsq will increase, this will increase by delta Idsq, correct? But note that I have a current source from the, from the top, since I have a current source from the top, the current in the top will not change, correct? So this Idsq under this initial condition was 1 milliamp, but now I am drawing more current than 1 milliamp, right? So what will happen to the node voltage Vd? I am drawing, so the node voltage Vd is now seeing there is a current source from the top, which is the current source from the top, which is pushing 1 milliamp current and a current source in, and a current source from the bottom, which is pulling out 1 milliamp plus some extra current, ok? So if you are, if you are pulling more current out of a node than you are pushing in, what is going to happen to that node voltage? That node voltage is going to drop, right? So this voltage, this drain voltage is going to drop, so Vd will, so consequence of this is Vd will decrease. And how long will Vd decrease, will it go, will it keep on decreasing and go to minus infinity or 0? No, it will not, because it will decrease till the transistor goes out of saturation, right? If the Vd decreases, decreases, decreases and it will eventually go into linear region, when? What is the problem here? The problem here is, the problem here is the, the, the current in the MOSFET in saturation is not dependent on the drain to source voltage, right? It is like an ideal current source, right? It is not dependent on the drain to source voltage, since it is not dependent on the drain to source voltage, right? You, no solution exists if, if you have a difference of current, right? You are pushing in 1 milliamp, you are pulling out more than 1 milliamp and, and the Vd will keep on dropping. But moment that transistor enters linear region, it is now dependent on, the current of the transistor in linear region is dependent on the drain to source voltage. Since that current is dependent on the drain to source voltage, a solution will exist and the solution will be such that the transistor is actually in linear region, right? So Vd will drop till, till M1 reaches or enters or not till, I mean Vd will drop and M1 will enter linear region of operation. So naturally, your, all your good properties of a transistor are gone simply because of threshold voltage changed by, changed by certain amount, ok? So what is the solution? So this I cannot do. So what is the solution? So let us, let us brainstorm a bit more. So let us say I, I still want to use this current source and I still want to connect it at the drain, right? And let us say I have a variable voltage source and I have stuck it here. So what is the problem? The problem is if this voltage source is fixed, if this voltage source is fixed and this is 1 milliamp and because of change in threshold voltage, this current became 1 milliamp plus delta, plus delta I, right? The transistor Vgs, the transistor overdrive changed, right? So if I had the ability to change the overdrive also, right? If I had the ability to change the Vgs also, then it is possible that I could have changed the capacity of the transistor to, to try to, to sink in different values of current, right? Let me explain it again. So what I am essentially saying is that, what is the goal? The goal is to match the current from the top to the current from the, from the bottom, right? The goal is to match this 1 milliamp current with the current that the transistor is sucking out. So how do I, what is the key variable or the key variable to, to focus on? Let us say I tell you that I will only try to focus on this voltage Vd, on the, on the drain voltage Vd and I will try to conclude if the transistor is biased properly or not. Is it possible? It is possible, right? So, from the example that we just saw, if we only observe Vd, right? I take it to the lab and I only observe Vd, what will I see, I will, what will I observe? If the transistor is biased such that it is trying to pull out more current than 1 milliamp, Vd is going to, going to drop, right? So if Ids, let me, let me call this Ids of the transistor. If Ids is greater than 1 milliamp, right, or rather if Ids, yeah, if Ids is greater than 1 milliamp or rather if expected Ids, I should say, not Ids because if expected Ids is greater than 1 milliamp, then Vd will decrease and transistor will go into linear region. Similarly, if expected Ids is less than 1 milliamp, Vd will, Vd will increase, right? Vd will increase and something bad might happen to the current source, let us not get into that, right? So, let us say we take the first case where, where the Vd has, Vd is decreasing. So, what, so I am in the lab and I am, I am observing this phenomena that Vd is decreasing. So, what is the only conclusion that I can draw? The only conclusion, the immediate conclusion that I can draw is the transistor is stronger than it is supposed to be. When I say stronger, what I mean, the transistor is expecting, the Vgs that the transistor has is, has been tuned in such a way that, that transistor is expecting more current, right? And again, Vgs is a controlling variable for me. So, what should I do? I should go and tune the Vgs, right? So, if I take this example, if I take this example, what is the, this is, this is observation, right? So, this, so if, so if Vd is decreasing, so, so observation is, observation is your Vgs Q is higher than required, right? It is only because Vgs Q is higher than required, the current that the transistor is expecting to pull out is higher and because the current source is limiting the current, your Vd is as a consequence of it is dropping, right? So, what is the action? So, the action required, so implies required action, action is to, is to decrease Vgs Q, correct? So, if I am in the lab, what will go? I will tune the Vgs Q knob, I will tune the Vgs Q knob till, till I see Vd is not decreasing anymore, right? So, what is again quickly, what is the algorithm? The algorithm is, if Vd decreases, then decrease Vgs Q, right? Alternatively, if Vd increases, then increase Vgs Q, right? So, this is what I will do in the lab. So, can you think of an architecture which does this automatically? Then what I am wanting to do, if Vd increases, I want to increase Vgs, Vgs Q, in this case only Vg because source is grounded, if Vd decreases, I want to decrease Vgs Q. So, what is the simplest thing that you can think of? I can simply connect the gate and the drain, right? So, some of you might be familiar with this circuit, this is often called a diode connected configuration. We will get to that later on, but the first thing to notice here is that if I do this, what is going to happen? Firstly, the thing is that we need to notice is, is the transistor in saturation? That is the first thing to notice, right? Is the transistor in saturation? If threshold voltage changes for example, right or even without threshold voltage change. So, is the transistor in saturation? What should I see? So, the only thing that you should see is for saturation, Vd should be greater than equal to Vg minus threshold voltage. In our case, Vd is exactly equal to Vg, hence it is in saturation. So, in this case for M1, for M1 Vd is equal to Vg, hence saturation condition is ensured. Now, let us assume that the threshold voltage changed, right? If threshold voltage change, is the transistor in saturation? Of course, transistor is in saturation, because the gate and the drain are connected, Vd is always equal to Vg. So, the transistor is in saturation, but what happens to the, what happens to the gm or let me ask you if threshold voltage changes, right, let us say this is 1 milliamp, right? So, threshold voltage was 1 volt, it, threshold voltage was 1 volt and Vgsq, let us say under this condition I matched my, I did the calculation and I found 1 milliamp equal to certain half unc ox W over L and let us assume we knew, we know some value of W over L, then what is the value of Vgsq that we will get? So, we will get Vgsq, let me call this I naught instead of 1 milliamp, let us keep in general. So, Vgsq will be threshold voltage plus under root 2 I naught by mu n C ox W over L, right? I basically did the algebra. So, if threshold voltage decreases, right? So, this is Vgsq, right? So, this is the Vgsq. So, if threshold voltage decreases, let me use the same color, right, for Poisson let us use black, so if threshold voltage decreases by let us say 0.1 volt, what is going to happen to Vgsq? Note that current is constant, so Vgsq will, will adjust, right, will adjust itself so as to keep the overdrive Vgsq minus threshold voltage same, right? So, Vgsq will also change by minus 0.1 volt, right? So, if you stare at this equation, you will see that, you will see that if threshold voltage changes, the Vgsq will also decrease by an equivalent amount because I have, because we have biased it with a constant current source, right? So, in this case, we have somehow figured out a way of making the current the independent variable, the current we have made the, made the constant current source the independent variable and I am asking the constant current source to develop an equivalent amount of Vgs, right? I am asking the transistor to develop the equivalent amount of Vgs that is required to sink in that current of, of, of 1 milliamp and because I am doing that, whatever happens to mu n, whatever happens to threshold voltage, right, right, your transistor will always, the transistor will always be able to sink in that 1 milliamp current and because it is always able to sink in 1 milliamp current and if we, if we, if we go back to our equations of transistor currents in, in saturation, if the current is constant, then naturally Vgs minus threshold voltage is also constant and since Vgs minus threshold voltage is constant, which means if threshold voltage changes, Vgs will have to change, right? So naturally, the transistor Vgs auto adjusts itself, right, so transistor Vgs auto corrects under the condition of a constant current source biasing, ok. So that is the beauty of using constant current source biasing instead of constant voltage source biasing, ok. So, we will talk about this in more detail in the next lecture.