Lecutre 18 Module 02

Welcome back, so we are now at this stage where our biasing of the transistor using a constant current source where the constant current source has been connected at the source of the transistor, it seems like we do not have to do anything and the transistor is getting biased automatically, right, okay, so far so good, I mean looks like best of all worlds but yet we have not yet figured out, we have not yet figured out what the gate voltage Vgq needs to be because ultimately it is, when you have an ideal current source one might say that okay, I mean the voltage at the source, voltage at this node can be anything and the current source will not get affected but ultimately everything is a real element, the current source is also a real element which essentially means that this voltage, voltage at the source cannot go below ground, right. So, since in practice this I naught, if this is I naught and this is the voltage across I naught, let us just say that Vi naught, since in practice Vi naught minimum has to be positive, some positive value, we need to ensure the gate of M1 is biased appropriately, why did I make that statement? I made this statement because if let us say Vgq for some reason is, if let us say R2 is much, much less than R1, if R2 is much, much less than R1, Vgq will be very close to ground, right. If Vgq let us say is almost close to ground, right and you want, and you want I naught to flow through M1, what do you expect this voltage to be? You would expect some finite overdrive voltage and you would expect that overdrive voltage to be on top of threshold voltage, you expect a positive Vgs, right. If you expect a positive Vgs and the gate is grounded, what will be the source voltage? The source voltage will be negative. If the source voltage is negative, do you think a real current source, a practical current source will work? It will not work because for a practical current source to work, you need to have a, you need to have a positive voltage in the direction shown, right. This I mean in an, in an ideal world where the current source can have any voltage across it while still, still operating like a current source, then it is not a problem. But we do not live in an ideal world and as it turns out current sources are also made out of transistors, right, so, so which, which essence means that we need to put a minimum voltage Vi naught across the, across the current source, right. So, if that is the case, if that is the case, we cannot have any arbitrary value of Vgq. So, then the question begs what value of Vgq we need and the answer to that we will start from, the analysis to that we will start from where? This analysis to that we will start from what is the minimum voltage that the current source needs, right. So, essentially I would need to know what is Vi naught min, let us assume Vi naught min is known, right. If Vi naught min is known, then Vgq min or rather if Vi naught min is known, then I know that the Vgs of the transistor, Vgs of m1 is threshold voltage plus under root 2 I naught by nu n C ox W by L of the transistor. Now, what is Vgs, Vgs is Vgq minus the source voltage, right, Vsq which is equal to threshold voltage plus under root blah, blah, blah, ok, which means what? Which means Vgq is the source voltage plus threshold voltage plus this additional stuff, whatever this additional stuff this is, what is this? This is essentially the overdrive of the transistor, right. So, I will just simply write it as the overdrive of the transistor, I am sorry. So, this is the V overdrive of m1 for a current of I naught, right. Moment you say overdrive, overdrive can change with respect to current, so we need to also be cautious and ensure that we specify the current, right. Now, what is Vsq? What is the minimum Vsq that we need? The minimum Vsq is Vi naught min, right. So, this is the necessary, I mean Vi naught min can be 100 millivolt, can be 200 millivolt depending upon how the current source has been made, if that is the case, what will be Vgq? Vgq will be Vi naught min plus the threshold voltage plus the V overdrive for the transistor and we know that Vgq is Vdd times R2 by R1 plus R2 and then you decide the ratio of R1 and R2 based on whatever Vgq, ok. So, that is as far as biasing the transistor is concerned, right. So, let us now assume that this transistor has been biased properly, let us assume the transistor has been biased properly and I want to use this like an amplifier, right. Ultimately, we want to use this as an amplifier, I mean that is what we have been doing all along, we want to use this like a common source amplifier because we can get a gain of gm times some resistance, right. So, if we have to use this as a common source amplifier, what do I need to do? So, firstly we have a input and we have a output or rather input source and output load, they need to be connected, right, ok. We can take one at a time, however what is the common source amplifier, incremental model of a common source amplifier? The incremental model, ideal incremental model for a common source amplifier is, this is Vi, this is Rs, this has to be open circuited, this is V1, right or V1, instead of V1 I should say that this is Vg, this is Vs, Vs is grounded, incrementally grounded and this output goes into Rf, right. So, this is gm Vgs, this is the incremental model of a common source amplifier, right. So we want, we want the incremental model or incremental picture to look like this, to look like this, but the biasing picture should be, should be the one in the left, right. So, I hope the problem statement is clear, we want, we want to modify the biasing picture in the left in such a way that its incremental picture looks like the one on the right, ok. So, let us do the easy thing first, in a common source amplifier where is the input connected, input is connected directly to the, to the gate, right. So, here we also need to connect the input to the gate, how can I connect, I mean we have done that before in case of a voltage biased common source amplifier, we will do the same thing, what should we do? We will put a, we will put a capacitor, we will AC couple the input, right, we put C1 and we will assume C1 tends to infinity or C1 is infinitely large, now we know that infinitely large means it is infinitely large with respect to, I mean we can, we can figure out the time constant of the network and then find out the maximum value or the constant on C1 that is necessary for this capacitor C1 to act as a short circuit at the frequency of interest, we know that, I mean this is, this is a done deal, we have done that before. But what about the output side, so in a common source amplifier, what does this common source mean? So, you can say that this common source, this terminology common essentially means grounded, right, so common means grounded and note that this grounded does not mean absolutely grounded, this grounded means incrementally grounded, so this means incrementally, so this means incrementally grounded. So, in this case as we saw the source is open circuited, right, in the incremental picture we saw that the source is open circuited because we had a current source connected at the source of the MOSFET and in the incremental picture the current source opens and the source is incrementally, incrementally open circuited, okay. And what should we do, what should we do to incrementally ground the source, again what is the, if this voltage quiescent wise is Vgq, what is this voltage, this voltage will be Vgq minus the Vgs of M1 for a current of I naught and what is incrementally grounded, what does incrementally grounded mean? Incrementally grounded mean, if some node is incrementally grounded all it means is that it is not changing with time, right, it is not changing, it is a small signal excursions is 0, right. For example, if you have a battery, if you have a battery, a battery of a certain value is incrementally grounded, right, the terminals are shorted between the battery because they are not incrementally changing, right. So, in this case in order to honour the biasing picture, we need to ensure that this voltage is equal, voltage at the source is equal to Vgq minus Vgs times Vgs of I naught. So, what can we do, we can put a battery of value Vsq which is exactly equal to Vgq minus Vgs of I naught, so nothing changes, right. What will be the current through the battery in the biasing picture? There will be no current through the battery because before I connected the battery, the voltage at the source was Vgq minus, was Vsq essentially which is V, let me call this Vsq, the voltage at the source was Vsq, after I connected the battery, the voltage at the source is still Vsq, okay, fine. But you might say that again, from where can I get a, get a battery because ultimately I have, I have made a point in the previous few lectures that you cannot have indefinite number of batteries, you cannot have one, you will only have one master battery, so what was the, what was the hack? The hack was instead of a battery, I can use a infinitely large capacitor, right, so which essentially means that I can replace this battery with a very large capacitor. So let us do that, so we can replace this battery with a very large capacitor, let me call this C2 and assume C2 tends to infinity, okay. So in the incremental picture, how does, I mean till now how has the incremental picture gotten affected, right, how has the incremental picture gotten affected? So incrementally at the input side what has happened, at the input side we have Vi, Rs, input capacitance is shorted, this is R1 parallel R2, again we have to ensure that R1 parallel R2 is much, much greater than Rs, if this is ensured, this voltage will be approximately equal to Vi, right, okay, so this is the gate voltage, right, so this is the gate voltage, this is the Vg, it should be approximately equal to Vi. What is happening to the other side, this is grounded, this is the drain, what is drain connected to? Drain is connected to a short circuit incrementally because it is connected to a, it is connected to a voltage source Vdd, now what is connected to a source? What is connected to a source? The source is, note that source is now connected to a ideal current source and an infinite capacitor, the ideal current source is open and infinite capacitor is a short, right, in a incremental picture, so incrementally this becomes, the source becomes a short circuit, right. Notice common source amplifier, grounded source using that infinite capacitor, we are able to sort out the source, which means that, which means that, which means that if there is an increment, increment at the gate voltage, this, the current here will be Vgm times Vgs but s is grounded, so the current is gm times Vg, Vg is almost equal to Vi, which means this current becomes gm times Vi, right. So, let me write it fully, so this is gm times Vgs, which is gm times Vg, which is gm times Vi, ok, now all the, now we are able to generate the incremental current gm times Vi. Now, if I ask you, what would have happened if I had not put this capacitor C2, infinite capacitor? If I had not put that infinite capacitor, so let me not connect this, if I had not put that infinite capacitor, this source terminal would have been open, right, this source terminal would have been open, which means there was no way the gm times Vgs current would have been able to, right, so if gm times Vgs current would not have been able to flow, which means this current had to be 0, right, if gm times Vgs has to be 0 and we know that gm is not equal to 0, then what is, what is likely to happen, which means that Vgs is 0, incremental Vgs is 0, which means that incremental Vg is equal to incremental Vs, right. So, if this capacitor were not there, any change in Vgq, right, any change in Vgq would have been accompanied with a change in Vsq, right, but in the presence of a capacitor, the presence of a infinite capacitor is not allowing the source voltage to move at all. Since it is not allowing the source voltage to move at all, which essentially means that this is incrementally grounded, hence we are able to, hence we are able to short the, incrementally short the source, but note that we have, we have still not arrived at a common source amplifier because what, we have not yet looked into the load picture, right. What is the load picture? In the load picture of it is, you have to ensure that the load is incremental current, right. So, now where is this incremental current flowing, this incremental current is now flowing into this short, into this, into this loop, right, right, we have to ensure that this incremental current does not flow into the loop, we have to ensure that the incremental current flows into the resistance RL, our load RL, right. So, we can connect the load RL somewhere here, so let us say we connect the load RL somewhere in the loop and then we should be, we should be done, right, so if this is load RL. So, this is in the incremental picture, but what happens in the total picture, because ultimately incremental picture is a model, that is not, you cannot give your customer an incremental picture, an incremental design, you have to give your customer a total design, right. So, what will be the total design that, that we will end up with, so again, you have Vi, Rs, capacitor C1, R1, R2, MOSFET, constant current source, very large capacitor, now we have an RL, where should RL be connected? RL obviously has to be connected at the, at the drain, but what was the issue? The issue is drain needed to be biased at VDD, if drain needs to be biased at VDD to maintain, to maintain saturation condition of the MOSFET, then we land into trouble, but we have solved this issue in case of a common source amplifier, right. What was the solution? The solution was do not connect it directly, connect it through a resistance RD and, and then use another capacitor, let me call this C3 and let us say this C3 is infinitely large also, let C2 is also infinitely large, so if that is the case, it seems like the incremental current will now flow into, the incremental current will now flow into a parallel combination of RD and RL, right, so this will get modified as parallel combination of RD and RL and this incrementally is the, is the common source amplifier, incrementally this picture is exactly equal to the common source amplifier that we initially had, but what is the difference? Why would you want to go through all this trouble, if the original common source amplifier was serving the purpose, you would want to go through this trouble because it gives you, it gives you a configuration where the gm is invariant of any threshold voltage variation and less variant with respect to mobility variation or in other words, this configuration has, is robust, much more robust towards change in ambient conditions rather than the original common source amplifier topology that we had studied, right, with respect to, with respect to change in, change in ambient conditions, ok, ok, so let us stop here.