We have now come to the point where we can start discussing models that are valid in particular regions of inversion. We begin with a discussion of a basic strong inversion model. Here is our chart. So far we have developed a variety of models that are valid in all regions. And now we are ready to start with strong inversion models. These will be simplified equations that are valued only in strong inversion. We will begin with a complete strong model that will be derived directly from that corresponding complete all-region model. This is the structure we have been analyzing several times already. We're going to neglect diffusion because as I have shown you, the diffusion component is negligible in strong-inversion. So then we take only the drift component of the current, as we derived it for the complete origin model. Which is shown again here. And for now, I will assume that I'm in the triad region, or rather, in the non saturation. Sometimes it is also called the triad region. The non saturation part of the characteristics. Now because we are in strong inversion, we can use for the surface potential right next to the source. Here, this is CSO, we can use an approximation that is similar to the one we had used for the European junction, just like for European junction the total potential variation across it is the built-in potential plus the externally applied bias. Here it will be phi zero. The pint surface potential for the two thermonamic structure, plus the externally applied bias, VSB. So we can write this expression, shown here. Now, next to the drain end of the channel, we're going to have a surface potential CSL. That will be given by similar expression phi 0, the pint value of the surface potential for two terminal structure, plus the externally applied reverse bias, which in this case is VDB. So we can write CSL is equal to phi 0 is equal to VDB. If you do not see these relations, you may have to review the material for the three terminal and more structure. So now if you replace PS1 and CSL by these quantities in the top equation. You end up with this equation. And this becomes now our complete strong-inversion model. It applies to the nonsaturation region because we made this assumption before and you will see why this model is not valid in saturation shortly. What is remarkably about this equation compared to the ones we have shown is that it gives you the current as an explicit function of the externally applied voltages, V GB, V DB and V SB. No longer do you have to solve implicitly for the surface potential. Because we assumed very simple expressions for the surface potentials, but remember, these are only valid in strong inversion. This now, first of all, is a symmetric equation, just like the model where it came from is. If you interchange the role of source and drain, you're going to get the same magnitude for the current, but in the opposite direction. Just interchange Vdb and Vsb in the equation, and you get it. This is a very old classical strong inversion model that formed the basis for the so-called Spice level 2 model. Now let me say something about the assumption of nonsaturation that we made here. Nonsaturation basically assumes that you have strong inversion throughout the channel, including next to the source and next to the drain. Because if you didn't have it, you wouldn't be able to write these equations here. These are. Approximate expressions for the surface potential in strong inversion. Now if the channel is strongly inverted next to the source and next to the drain, then it is also strongly inverted throughout. Okay. But if you keep increasing VDB, what will happen? As you recall from our discussion of the three terminal [INAUDIBLE] structure, the level of inversion in the channel next to the train will be come lighter and lighter. In fact, this will happen for two reasons. First of all. If you increase VDB, you're reducing the potential of the oxide over here. So you have a smaller field in the oxide and, therefore, it attracts fewer electrons underneath it. And, in addition, the larger you make VDB, the wider the depletion region becomes. So more and more of the gate charge is being balanced by ionized acceptor atoms. And you need fewer electrons in the channel. So if you keep increasing VDB, the inversion level will become lighter and lighter until eventually you will not be strongly inverted anymore. You will be moderately inverted there, then weakly inverted, then eventually depleted. And then of course, the equation we just derived does not apply, because that equation, again, had assumed strong inversion at the drain end. Okay, let's take this equation, and plug it, and see what we get. This is what we get. This might be surprising in the beginning. So here we're plotting the current that we calculated, versus VDB, the drain body voltage. And when VDB is equal to VSB, there's no potential difference across the channel, and therefore there's no carbons, so this passes through zero and it goes up as expected but eventually it goes down again. Why does it go down? Well, it goes down because over there, VDB's so large that we do not have strong inversion anymore. Next to the drain. So VDB has increased so much that over here, we don't have strong inversion anymore. So we don't have the right to be using the equation we developed on the assumption that we do have strong inversion there. So then of course, this equation predicts some non-sensical behavior over here, because you're not supposed to be using it in the first place. What is the point where this happens? Where is the maximum of this? If you calculate it, and it is done in the book, you find that this is nothing but the so-called pinch off voltage. I remind you that pinch off voltage is the value of the externally applied voltage, in this case VDB, that you need to have in order to pinch off the channel right next to this region. Pinch off means that the strong inversion equation predicts zero charge for it. At this point, you may be wondering how come there is current flowing near the drain if the channel is pinched off. But the fact is the fact is not pinched off, it is the strong inversion equation that predicts the channel is pinched off when the electron density is low. We do not have the right to use that equation. So the fact that it predicts it is pinched off is immaterial to us. What happens is that the density of electrons becomes low near the drain. But the electrons travel at very high speeds, and they can still carry the required current. We will explain this in more detail later on. So we will using the non-saturation equation all the way to V sup P, in lack of anything better for the strong inversion approximation. And we will denote the peak value of the current by IDS prime. So since we know that this part is some non-sensical behavior because we're using an equation in a region where we're not supposed to be using it. What we will do instead is we will extend the current and assume it is constant at the peak values, so we take the maximum point in non-saturation and then extend the current like that. So the combination of these two branches, this branch and this branch, forms our model for strong inversion, according to this model. So now we can say that the overall drain source current consists of two branches. The curve for the current consists of two branches. It is equal to IDSN which is the equation we derived provided that VDB is less than VP so we are somewhere here. And this is the non-saturation region. And it is equal to IDS prime, the maximal value in the saturation region. And IDS prime is defined as IDSN when VDB is equal to the pinch off value, VP. Of course, all of this assuming we have a long channel. Later on we will see that this curve in the saturation region is not exactly horizontal so we will keep refining our model to take into account more and more effects. But for now, we assume a very long channel. So you can assume that this is a horizontal curve. So this is a classical representation of strong inversion currents. You divide the region into two, non-saturation and saturation, and you use different equations in each region. We will see this practice again later on. So we have seen a simple, strong inversion model that gives you the drain-source current explicitly as a function of the externally applied voltages, VDB VSB and VDB, the model contains three half powers and which are not very efficient computationally. We're going to get rid of those powers in the next video.