Lecture 13 Module 01

Now, welcome back, this is lecture 13. So, in the previous lecture, we were looking into what happens to the condition of a common source amplifier, right, when the signal at its input, input swings, right. So, by the way, let me take a moment to emphasize the fact that, so what is the configuration of the amplifier, potential amplifier that we were interested in. So, this is the gate, this is the drain and this is the source. So, the input is incrementally, what is the incremental equivalent of this? Incremental equivalent of this structure is, so you have the gate. So, what I will do is I will sketch the MOSFET, sketch this entire structure and then show you the incremental equivalent. So, what should you do in order to get the incremental equivalent? So, you should first replace the non-linear device with its linearized equivalent and what is the linearized equivalent of this MOSFET M1? So, the linearized equivalent of this MOSFET is nothing but a control source, a current control source whose controlling terminal are the gate and the source, the constant voltage sources go to 0. So, this shorts, we leave the incremental voltage source as is on the output side, the constant voltage source goes to 0 and what is the other terminal? This is the drain terminal and this connects to, what is the value of this control source? The value of this control source is gm times vgs. So, what, so incrementally if you see, the input is applied between the gate and the source and the output, where are you taking the output? The output is taken between another terminal which is drain in this case and the source, right. So, the common terminal between the input and the output is a source terminal and since the common terminal in the incremental model is a source terminal, it is called a common source configuration, right. So, this is also called common source amplifier, right. So, another way of thinking, another way to think of it is the fact that whenever we say, whenever we say common, right, you will see several configurations which that we will be dealing with in the upcoming lectures. This common term will come, you will see common source, common gate, common drain and so on. So, this common essentially in your mind, in the back of your mind, you can replace common as grounded source, right. You can say that this is almost equivalent to saying you have a grounded source configuration, ok. I find this terminology a bit more useful because it is much more general while dealing with other configurations as well, right. So, however, in the literature common source amplifier is a generic term. So, we will refer to this configuration as common source amplifier, but at the back of your mind, you should have the understanding that whenever we are saying common source amplifier, we are essentially pointing to the fact that the source is incrementally grounded, ok, great. And also it is not only the source is incrementally grounded, it should be common, the input should have been applied between the, wherever between the gate and the incrementally grounded source and the drain, the output is also between some other terminal and the grounded source, ok, fine. So, now in the previous lecture, we were delving into, we were delving into what happens when the signal actually swings, right. So, it is one matter to just apply the voltages, the quiescent voltages and set up the quiescent currents in order to keep the transistor in saturation, but it is a different ballgame altogether when we say the transistor should be in saturation even when I apply an incremental, even when I apply an incremental input, right. And when we saw that and we saw that in case of this particular configuration or if we saw that if incrementally my input V i is or even if I say that incrementally this is the V gs and assuming the transistor is in saturation, right, so subtext is transistor is in saturation, we saw that there were two conditions that we needed to maintain, right. What were the two conditions? So, this is gate, this is drain, this is source. The primary condition that we had to maintain was for saturation the total drain voltage V d had to be greater than equal to the total gate voltage V g minus the threshold voltage, right, ok. In other words, if I have to break it down, what is total drain voltage? Total drain voltage is nothing but the quiescent drain voltage V dq or capital V capital D, right, plus the incremental drain voltage should be greater than the total gate voltage minus the threshold voltage. So, this was the saturation region, this was, this condition we had to ensure in order to keep, in order to make sure that the transistor always remains in saturation. What was the second condition? The second condition was to ensure, ensure the transistor is sufficiently on, right. So, when we, when we say the transistor is sufficiently on, what are we implying? We are implying that we know that the V gs, when the V gs swings, it can go up or it can go down. When V gs goes up, right, when V gs goes up, I run into, run into a potential problem of the transistor going out of saturation because the drain voltage can, because I am closing the gap between the gate and the, and the drain, right. However, when the, when the V gs goes down, when the V gs goes in this half cycle, when the V gs goes down, what is, what is the likely cause of problem? The likely potential problematic cause is the transistor can, can get into cut off, right. And we did the analysis and we saw that we had to ensure that the incremental V gs had to be much, much less than, had to be much, much less than what? Twice the overdrive of the transistor, right, twice the quiescent overdrive of the transistor, correct. And if we, in a common source configuration, in a common source configuration, if I go to this common source configuration, what is this incremental V gs? The incremental V gs is, so in, in common source configuration, this incremental V gs is the input that has been applied directly, right. And if I say that the input is sinusoid, right, if the input is sinusoid, then V gs essentially becomes V p sin omega naught t. So, what is the condition on V p? What is the condition on V p to ensure that the transistor or the small signal approximation always holds? So, I, I, I, let me, let me take a step back and I think I should rephrase, I should rephrase this, this statement. So, if we go back to the previous lecture, we should, we saw that this condition was required in order to ensure, in order to ensure the validity of the small signal approximation, regardless of whether, whether the signal is swinging on top, swinging to the positive cycle or in the negative cycle, right. So, so, so let me, let me correct myself here. So, the condition that I am referring to here is to ensure, to ensure validity of small signal approximation, we have to ensure this V gs is much less than twice the overdrive, right. We will come to, we will come to what is the, what is the condition, what is the minimum V gs that we can apply shortly, but again, I mean, this is a correction, right, from the statement that I had made two minutes back, that this condition is to ensure the validity of the small signal approximation. So, in a common source amplifier, V gs essentially is equal to V i, which is equal to V p sin omega naught t, right. Since this is, since this is that case, so what is the maximum V p that you can apply in order to ensure, in order to ensure the small signal validity, right, the small signal approximation is still valid. Note that when you have the sinusoid V p sin omega t, the sinusoid goes on either direction and this peak to peak is twice V p, right, which means, which means V gs max is twice V p, which means you have to ensure V p is less than 1 overdrive. So, in a common source configuration, right, in a common source configuration, you have to ensure that the sinusoid that you have applied is much less than the overdrive of the, of the transistor. And if you are not particularly bothered about whether it is a common source configuration or not, right, then the question arises, how should I, let us say I have a network which has a transistor where whole bunch of stuffs are connected to the source, whole bunch of stuffs are connected to the gate and whole bunch of stuffs are connected to the drain, right, then how do I know, how do I know what is the, what is the constraint, what is the constraint for small signal approximation? That is a quite a valid question to have because it is not necessary that we will be using our transistor in a small signal configuration all the time, right. So, in that case, what you have to do is you have to find out what is the gate voltage, you have to find out what is the source voltage, right and then you will have to ensure V gs, whatever that V gs might be after solving KVL and KCL is much less than twice overdrive. And if V gs is supposed to be a sinusoid, right, if incrementally V gs is equal to some sinusoid, let us say some A sin omega naught t, A gs sin omega naught t, right, you will have to ensure, you have to ensure that A gs is much less than overdrive, right, this is for validity of small signal approximation. Now, one might say that what is this much, much less, whether I mean 10 times less is much, much less or 100 times less is much, much less that depends on the application, right. So, the whatever you will be designing a circuit for certain application, right, that application will have a specification what is this much, much less and for you will have to, you will have to design your circuit based on the requirement of that particular, of that particular application, ok. So, let us move on, let us move on. So, so in, now if I, if I ask you, if I give you, if I give you a, let us say a common source amplifier, right, the way we derived, the way we went about finding, finding out the total currents and voltages in the common source amplifier in the previous lecture was to combine the incremental and the, combine the incremental and the quiescent conditions in the same figure, right and we, I walked you through the steps, but that can be a bit of a hassle when the circuit becomes a bit more involved, when you have multiple transistors, when you have multiple elements, right, it might not be, it might not be so easy to draw or to sketch all the, all the incremental and the total currents in the same figure. So, so what we generally do is we break the problem into two parts and then combine them in order to come up with a, come up with a solution. Let me, let me demonstrate that with the same example that we have been taking till now. So, let us take, what I am asking, what I am basically trying to motivate here is that in, in the next part of this lecture, what we will do is we will see how to, we will see how can we combine the incremental and quiescent conditions to get to total currents and voltages in the network, right. So, this is what we will be focusing on in the next, next few minutes. So, so let us take the same example. So, what is step 1? Step 1 will be, step 1 will be find quiescent currents and voltages, right. So, how will you find quiescent currents and voltages? You get rid of inputs, get rid of incremental inputs. In this case, the incremental input was only one voltage source, we get rid of them. When I say get rid of them, I mean I desensitize them, right. So, voltage source desensitizing means it gets shorted. If I had a current source, I would have had to open circuit them, right, okay. So, now from this, what will I do? I will solve the full blown nonlinear equation, right. So, how do you do that? Solve for I and V using KCL and KVL using the nonlinear I-V current equations of M1, right. So, you will be getting, what you will be getting from this? You will be getting Vgsq, you will be getting all the parameters, right. You will be getting Vgsq, you will be getting Vdsq, you will be getting Idsq. So, we will combine them. What is the next step? The next step, next step is make the incremental model of the nonlinear network. What is the incremental model of the common source amplifier? We know that we have done it sufficient number of times now. The incremental model of the common source amplifier is this, where there is no, we do not have any constant voltage or current sources. We only have incremental quantities and the MOSFET has been replaced with its incremental equivalent Vi, this is gate, this is overforce, this is drain, this is gm, Vgs, right. So, what will you do now? Now, note that this is the completely linear network, right. So, what will you do? We will find, so here we will be solving, solve for I and V in the incremental network, linear network and what will be step 3? Step 3 will be combine, combine the currents and the voltages that you have gotten from step 1 and step 2, right. To get total current and voltage, combine the currents and voltages, let me also mark the figures that will be helpful. This is figure 1, this is figure 2, combine the currents and the voltages obtained from the quiescent analysis. So, this is figure 2, figure 1 and incremental analysis, this is figure 2, right. So, what will you get after, at the end of this? So, from figure 1, we will get Vgs q, Ids q, Vds q, right and from figure 2, we get small vgs, small ids, small vds. So, what will be the total? It is small v capital Gs will be Vgs q plus small vgs, similarly Ids will be Ids q plus small ids, similarly Vds will be, this will be Vds q plus small vds, okay. So, once we have, once we have this framework, we can use this for any network, right. We can essentially close our eyes, find out, I mean, find out the quiescent conditions, convert the quiescent transistor circuit into its incremental equivalent, find out the incremental conditions and combine those 2. Now, note that the incremental equivalent is related to the quiescent network because in order to figure out the incremental equivalent, you need the values of gm, right and what does gm depend on? gm depend on, depends on the quiescent condition of the primary network, right. So, the value of the gm will change if you change, if you change the quiescent condition, but the analysis procedure or the steps that we, that we go through does not really change, okay, okay.