We will now conclude our discussion on strong inversion models by deriving a quadratic strong inversion model that will reduce to a very popular so called square model used in circuit design. Here is where we are on our chart. We have already covered the complete and the body reference thrown inversion models, and now we will talk about the source reference simplified strong inversion model, that will in fact result in the very popular so called square models that people use in circuit design. So, we start from our simplified source-reference all-region model that turned out to give us this equation here. In the nonsaturation region, as we have done before, we assume this value for the surface potential next to the source and this value for the surface potential next to the drain. You plug them into this model, and you get this expression. Now, we can recognize several quantities here. For example, the difference of these two, Vgb minus Vsb is simply the voltage Vgs, the gate source voltage. Vdb minus Vsb is Vds, the drain source voltage. And it also appears here. And finally, this quantity, it, it's easy to recognize as the negative of r threshold voltage, at the value of Vsb. You will remember that the threshold, because of the body effect, depends of vsb. And we have derived the expression for the threshold, and it was simply Vfb plus phi 0 plus gamma squared over phi 0 plus Vsb. So what we have here is the negative of that threshold. So let's replace the quantities by the quantities we recognize them as being equal to, and we get this equation, which now is a second order, a second degree polynomial in terms of Vds. Then where the threshold VT, the value of this Vsp is given by this expression. Let us now discuss the value of alpha in this current equation. As you may recall from our discussion of simplified origin models, alpha resulted from an approximation of the decrease in region charge. And the nominal value of alpha was this. Now for strong inversion, CS0 is V0 plus Vsb. And therefore the nominal value of alpha is as given by this equation. But back then we had also mentioned that the better approximation is obtained if you lower the value of alpha slightly. So sometimes empirical equations are used for alpha. For example, of this form, where a quantity phi 3 greater than 0 is added in the denominator. Or the coefficient d2 < 1 is inserted. The final value of alpha that results can be significantly larger than 1, for example, 1.2. And sometimes a single constant alpha value is used. If we plugged the non saturation expression versus Vds, we see, not surprisingly anymore, that it predicts a nonsensical behavior above a certain value. This part here. And the value of Vds where this predicted I will call Vds prime. We are going to ignore this because the equation is simply not valid in that part, just because the channel and next to the drain is no longer strongly inverted, so we should not be using the equation in the first place. And instead, we are going to take the peak value and extend it using a horizontal line and this becomes our saturation model. We've already done all of this for the complete model so I'm bypassing many of the details. So what do we have? We have the nonsaturation expression given by this and the saturation happens at some point in Vds prime. And it's easy to calculate where that point is because from here, you can see graphically that this is the point where the slop of the IDSN expression becomes zero. So, you take dI dV DS from that expression, you set it equal to zero, and you find the resulting Vds value and you call that Vds prime. And Vds prime turns out to be simply Vgs minus VT over alpha. So IDS prime is the value of IDSN. The nonsaturation current had Vds equal to Vds prime, so IDS prime is simply the maximum value of the current. And from that, we have now a two part model, which I can summarize. You have this equation for nonsaturation. And when you plug in the value of Vds it occurs from here into this equation to find the value of the current at this peak point. You find this expression for such iteration. Again, of course, please go through the algebra to get some feel for all of these things. So let me summarize what we've done. We have a new, for us, strong inversion model that has two branches. It has this expression as long as Vds is less than Vds prime and has this expression as long as Vds is larger than Vds prime. Both expressions are valid at Vds equal to Vds prime. So arbitrarily we include the equal sign over here. Now this is very, very close to the so-called square model that you see in circuits books. Instead of alpha there you see a one. We'll come back to this point in a minute. Vds prime we derived to be Vgs minus V T over alpha and this is the value of Vds that separates non-saturation from saturation. Sometimes this is called the saturation. Value for Vgs. So if you plot these equations you find this behavior. Here we plotted drain source current versus the drain source voltage and Vgs is the parameter, and the larger the Vgs is, the larger the current you get for the given Vgs. The non-saturation is separated from the saturation by employing Vgs prime which is here. And of course, the value of Vds prime depends on Vgs. So the equation itself, Vds prime = Vgs- VT / alpha, corresponds to these points For each value of Vds. You can see that this Vds prime, the point where you achieve saturation is different for different Vgs values. Keep this in mind, because it is characteristic of strong inversion. We will see later on, for example, that this property does not hold in weak inverse. This is what I would call the correct square law strong inversion model, it is the basis of the so called spice level three model. If you compare this model to the complete strong inversion one, you'll find these results for a typical set of parameters given in the book. So you'll find that overall you do make some error of a few percent but the model is. Otherwise pretty decent. Now, I promised I would talk about the popular square law model. This is the popular square law model. As you can see, it is identical to our model if you assume that alpha is equal to 1. And Vds prime is simply Vds minus Vg over alpha. Again you put alpha equal to one you get this small. However, if you go back to our derivation for alpha, you find that alpha can be significantly larger than one. If you assume it is equal to one you're going to make an error and in fact this model can give you serious error. For example, it can over predict the value of the current in saturation. So this one, because alpha is missing from the denominator it over predicts the value of saturation. Current. And it is tempting to say well, if it overpredicts it, then I'm going to use a smaller value for the mobility to make the current go down to the value it should have. But that is a very, very bad thing to do, because if you make the mobility smaller, it's also going to be smaller In the nonsaturation region. And then you're going to underpredict the current in the nonsaturation region. So, this is one example of trying to cook numbers to match experiments. It's a very bad idea to arbitrarily take parameters and give them a non-physical value. It may help you In the particular thing you're trying to do, it's gonna hit you somewhere else. Going back to the complete model, there is one compact way to write the equations, which we will be using in some of the future lectures. We define the parameter Eta, which I will call the degree of non-saturation. And it is defined like this when Vds is in the non-stauration region, and it is zero when the Vds is in the saturation region. So it plots like this. For example, when Vds is equal to 0, eta is maximum, the degree of nonsaturation is maximum. Then the degree of nonsaturation goes down and in the saturation region, the degree of nonsaturation is 0, because you are in the saturation region. If you play with the equations I showed you, and use either of them, it turns out you can write them like this. And this equation is valid in both nonsaturation and saturation, the entire model we derived can be expressed as compactly as this. Now, how is it possible to have a model that has two branches, a nonsaturation branch and then saturation branch? How is it possible to express it in terms of a single equation? The trick is that inside Eta, you do have the two branches. You have one equation for Eta here, and another equation for Eta there. One final Point I would like the mention is that so far we have only looked at the channel near the source and near the drain but there will be cases where we need to look at the channel in between the two. And we would like to know what the surface potential is at any point at position X in the channel. So here is a trick on how to do this. Here is the device. The channel extends between x equals 0 and x equal to L. And we would like to know what is the surface potential of point x. If you take any of the models we have derived they're of this form, in strong inversion. The non-saturation is W/L times some function of Vgb, Vsb, and Vdb. I don't care which model we take. It can be written like that. Even if it's written in terms of Vgs, you can write Vgs equal to Vgb minus Vsb. And again, you have the correct variables there. So this is then the equation for the complete device between x equals zero and x equal to L. So it's between here and there. But now consider the following. If I take this as a new sub-transistor, sort of speak, where the source is the source that I already have, and the drain is at this point here, so this a fictitious device with length x, and drain at that point, then instead of l we are going to have x in the denominator. And instead of Vdb the reverse bias between brain and body, we will have Vcd at point x. So this Vcb is the reverse bias of the channel between this point and the body. Just like we have the Vsb near the other source and Vdb near the train. Here what I have I will call Vcb. And Vcb changes from Vsb at the axes called zero. To Vdb at x equal l. But for this particular sub-device, I will consider that this point here is the drain, and it has a drain body voltage of Vcb. From these two equations now, you can eliminate the variables you don't need and end up with this equation. This now relates Vcb of x to x, and you can solve for Vcb. For the particular square law model that we derived, the correct square law model, it turns out you get this expression. So let's plot. This is Vcb as a function of x for different values of Vds. Let me first take Vds equals 0. If Vds equals 0, this should be the same throughout. Take out the Vsb, as you see in the first curve. If I raise Vdb a little by making Vds larger than zero, making it equal to Vds1. Then the Vcb changes from Vsb near the source, to Vdb near the drain. For an even larger value of Vds, I have an even larger value of Vdb. So then the visibility changes from VSB near the source, to VBD near the drain, and so on. So as you keep praising the drain voltage, it go in this direction. An interesting thing that happens is when you make Vdb equal to the value that the channel supposedly gets pinched off. In other words, when Vds is equal to Vds prime, the saturation value. An interesting thing happens with this curve, and I will let you think about what that is at home, if you have the time. So in this video we have seen the correct square model and how it reduces to the very commonly used. Square log model for strong inversion. And at this point we'll have finished with our strong inversion models. In the next video we will derive models that are valid in weak inversion.