We will now consider one of the types of noise we have discussed in the previous video, specifically thermal noise. Let us assume strong inverse operation for now. We have a transistor showing here, stripped out of the features that are not of interest to us at this point. So we have a source, a drain and channel in between them. we assume we are in steady-state, so there is a current IDS flowing all through the channel. If you plot the voltage between channel and body versus distance, it varies from VSB at the source to VDB at the drain. We will concentrate on one particular point, that x equals to x1, and we will discuss noise due to the resistance of the channel. We start with the well known drift current equation. I remind you that drift is dominant in strong inversion. The equation, we have seen it before, is given by this, and it involves the inversion layer charge at position x, which is a function of the chan, channel body voltage at point x. And it also involves the gradient of the voltage with respect to position which corresponds to the electric field. Now we have seen how we can take this equation and multiply by the x both sides and integrate then divide by L and we get this relation for the drain source current. Where the integration on the right is between VSB and VDB. Now think of the, the point, the channel here, magnified. And let's go to point x1, let's say x1 is right in the middle between these two lines, we'll take a very short chunk of the channel of length delta x. And therefore, on the, on one side of this, on the left side you have a channel body voltage, VCB, and on the other side you have the channel voltage VCB plus delta VCB. This means that from the right side to the left side there is a voltage difference dVCB. And if you have a current and the voltage, you can define a resistance as the ratio between the two. And that will give you a small resistance, delta R, for the chuck. So to find delta R, we take delta VCB and divide by IDS. Now for delta VCB from here you, you let the, the d's become deltas and you solve for delta VCB and you're replacing it here and you get this result. So this is the resistance of this chunk of the inversion layer centered around point x1. And of course if it is a resistance that you expect to be noisy, for reasons we have explained in the previous video. So we'll represent it with the noiseless resistance in series with some thermal noise voltage. And for this thermal noise voltage, we have mentioned in the previous video that we have a corresponding power spectral density 4kT times the resistance value. And since this noise is white, we can get the total mean square value of the value over the bandwidth B by simply multiplying by that bandwidth B. So we need to understand better how this noise voltage influences the current. So that we can find the noise in that current. this one if you replace delta R by the expression we've got here, it gives you this result. So this is the mean square value of the thermal noise voltage at position x1 in the channel over the bandwidth B. Here is a repetition of the previous figure, we have seen that for a chunk of the invertially point x1, there will be some voltage variation across it. The question is now how does it influence the voltage in the channel. And then from that, we will infer how it influences the current. So let us for a moment assume that there is a voltage difference at one point, x equal x1 in the channel. And I will represent that difference for a moment. By a DC source,a tiny DC source which I will call delta v. Of course this is a thought experiment, there is no voltage source over there. But if you do have a delta v source from left to right, at one point in the channel, it follows that as VCB, the channel body pool teach rises from VSB. When it gets to x1, it would jump by an amount delta v, which is this voltage. And then it will continue, it will come VDB at the drain. So let us assume that at x1 right before the jump, VCB is equals to V1, that means that right after the jump, will be V1 plus delta v. Now, we can represent this situation as follows. We split the channel into two parts. The left right part, right before the delta v source, is one sub-transistor, and in series with it you have another sub-transistor, and in between them you have a delta v. The two transitions have a common gate and a common body. Now, V1 is the voltage right before, to the left rather, of the delta v source. This is the voltage of V1 that we said here, and therefore the voltage here as we already mentioned is V1 plus delta v. The first subtransistor, which extends from 0 to x1 has a channel length x1. And, therefore, the second transistor has a length L minus x1, if L is the total channel length. And, because of the presence of delta v, you're not going to get exactly IDS, but you'll get IDS plus a small change, delta i. So now we need to calculate what delta i is. Okay, we know that the current throughout the channel is the same, it's IDS plus delta I. Looking at the first transistor, we applied the relation that we showed on the previous slide. Which had minus W over l, and so on, but instead of l now we have the effective channel length of the first transistor, which is x 1. The rest is the same. And we integrate from VSB up to the drain of the sub-transistors which has a VCB equal to V1. As you can see from here. So it goes from VSB to V1. Now we can write the similar relation for the second transistor. The current is still the same because the two transistors are in series but the channel length is L minus x1. And now we integrate it from the voltage at the source of this sub transistor which is V1 plus delta v. So the lower limit of the integral is V1 plus delta v, up to the drain voltage which is VDB. So we have one equation for each of these transistors and both give you the same current because the two channels are in series. So between these two equations you can eliminate x1. And then you allow delta v to go to 0, and the result, which I'll bypass, but you can check it, is this one. So the current plus the small change delta i due to the presence of the delta v source is given by this. But this first term over here can be recognized as being exactly the train source current. We've seen this term on this previous slide. so now crossing out IDS from both sides you end up with delta i, which is this part of the expression. So we have found then, a way to find how much the drain source current is influenced by the presence at delta v as position x1. It is influenced by this much. So now on the next slide I will write that delta i is equal to this part of the expression, and we have this result over here. Now, if instead of a DC source delta v we have a noise voltage, nothing really changes, but instead of delta v here, we have delta v of t. This is the thermal voltage noise, in, at position in a chunk, at position x1 of length delta x. You can square both of these sides, and take the mean value and you get the mean square value of the current noise. Here you get the mean square value of the voltage noise of the chuck, and here you get a multiplying quantity. And this one here can be found as we showed two slides ago. It is given by this. I remind you, delta x is the length of the chunk, B is the bandwidth over which we are calculating the noise. So now plug this into here, and you get this result. [NOISE] Now this, I remind you, is the contribution to the current noise, because of the thermal noise in chunk of length delta x around, the point x equal to x1. Of course, all of the chunks in the channel will have noise, so to get the total noise, you have to integrate from x equal 0 to x equal L. And now you get the total mean square value of the noise current over bandwidth B. It is given by this expression. This one we recognize is the inversion layer charge per unit are times an elemental area of the chancel integrated over all of the channels. So this is Qi, this is the total inversion layer charge. And therefore if you now divide both sides by the bandwidth, you get the power spectral density, which is mean squared value over the bandwidth. And it is given by this. I remind you, we're talking about thermal noise, so the bandwidth doesn't have to be very small, simply because the prospect of density is constant. So, we have finally a formula, a very simple formula in fact, that gives us the power spectral density of the current noise as a function of the total inversion layer charge in the channel. Although we will not do this, we can show that in all regions, you can have a similar formula if you use quasi-Fermi potentials rather than voltages. If you're not express qi in terms of terminal voltages. You find this expression. We already have Qi expressions in terms of terminal voltages. So this is the expression that we get for eta is the degree of non saturation. Being 1 when VGS equals 0 and then linearly going down to 0 at the point where you reach saturation. So let's look at two special cases. In non-saturation, where VDS equals 0, you replace eta by 1 here, and you get this expression. This one here, if you go back to the small signal parameters lectures, can be recognized as gsd, the source drain small signal conductance. Our VDS equals 0. We can also find the power spectral density in saturation by replacing eta by zero over here, and we get this result. And again, you can recognize here the transconductence of the transistor multiplied by the parameter alpha, so this can be written like this. So we have one expression for VDS equals 0, a different expression in saturation. I would like to warn you that in some books, and early programs, this formula has been used throughout, both in saturation and non-saturation, by mistake. And so, instead of using the correct formula for non-saturation this one was used and of course that is a big problem. Because when VDS equals 0, gm goes to 0 and this would tend to predict that there is no noise when you make, VDS equals to 0. It would seem as if you have a noiseless resistance which of course is wrong. You do get noise with VDS equals 0 and it is given by this value. In between, you can plot this expression here. So you find that as you increase VDS for a given value of VGS, the power spectral density of the noise current goes like this. Okay. Now, let's go to weak inversion. There are two theories for white noise in weak inversion. One assumes thermal noise, and using the weak inversion Qi in the previous formulas gives you this result. And when VDS is large, the exponential goes to zero, so the whole thing reduces to that. The other theory assumes shot noise, and of course from the general expression for shot noise that I showed you in the previous video, you get this result. So, if you look at them, they're exactly the same. And in fact, you can show that even in between the saturation point and VDS equal zero. Even if you use shot noise, you get exactly the same result, as you get if you assume thermal noise. So two different theories that give you the same result which has generated some controversy. And you can find references to that in the book. So, here we show the power spectral density versus VDS, in general has the shape that we also show in weak inversion. But it saturates at this value here. For, for comparison, I have the drain current versus VDS. In weak inversion you will recall that the current saturates within three to five [UNKNOWN], in other words thermal voltages, and so does the power spectral density. Now, small-dimension effects, of course if I don't mention small-dimension effects like I didn't in the previous slides, it is implied that we're talking about long channel transistors only. But if we have small dimension affects, the power spectral density versus VDS, with VGS, as a parameter, may look something like this. And we have to take into account channel length modulation, which can be handled, through the pinch-off region length. We need to take a talk on velocity saturation, hot electron effects possibly if they are present, and drain-body, body current shot noise. [INAUDIBLE] you will recall that there is a leakage current between drain and body. This shows shot noise and because the drain-body current is part of the drain current, it will have a direct effect on it. In addition, because the drain-body current passes through the body resistance Rb, It modulates the internal VSB. The internal source to body volt. It's, and through the body transconductance it modulates the current as well. So both of these effects will contribute to current noise in the drain current. You also get gate tunneling current shot noise, you have very thin oxides. people define an excess noise factor like this. You take the power spectral density predicted by the long channel theory and the factor which you must multiply that in order to get the actually observed short channel noise power spectral density, is called the excess noise factor. This excess noise factor has been claimed to be large, for example, 2 to 5 but this has been questioned. and it actually depends on whether you have hot electron effects, or you don't. Now, elaborate computer aided design models do include noise due to these effects, but unfortunately the results are very complicated analytically. So I will not present them and I will refer you to references in the book instead. In this video, we have discussed thermal noise and shot noise in MS transistors. We talked about the main formula and how it is derived for thermal noise. In the next video, we will concentrate on very high frequencies. And see what additional considerations we need to take into account then, as far as thermal noise is concerned.
