nanoHUB-U Fundamentals of Nanotransistors/Lecture 4.5: Transmission Theory of the MOSFET II
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[Slide 1]  Hello again. So let's continue our discussion of the Transmission Theory of the MOSFET. 

[Slide 2] In the last lecture, we derived the expressions for the linear region current and for the saturation region current in the presence of scattering, which reduced the transmission below one, and the question we have now is can we derive an expression that smoothly goes from small-drain voltage to large drain-voltage. 

[Slide 3] That's what we'll attempt to do in this lecture. So were going to begin with the Transmission Theory. The Landauer Approach. Were going to make the same assumption that we made in the last lecture, that we're going to assume that the transmission is energy independent, so that we can pull it out of the integral. So if that's not the case, we'll have to interpret what we pull out as some appropriately average transmission across the energy channels of interest. So once we pull it out,we're left with an integral 

[Slide 4] that we can do, and that's the approach that we will take. So we need to work out this integral. Remember that we know everything inside the integral. We know the number of channels per unit width. The total number of channels is W times M sub 2 D. We know the Fermi functions. We know everything. We can work out those integrals. In fact, we worked out that integral when we did the ballistic MOFSET. Because if you look at this here, this part describes our ballistic MOSFET. We've simply multiplied it by the transmission. Okay, so the result of that integration, and it's good practice for you to do this yourself, but the result of that integration is given here. It's just transmission times this expression, and that expression is exactly what we got in the ballistic case. So you can see that it's just transmission times what we got in the ballistic case. Now in the last lecture, I told you we had to be very careful about multiplying our ballistic results by the transmission. That was okay in the linear region, but it wasn't okay in the saturated region. This expression is presumably valid everywhere, and it simply multiplies the ballistic result by the transmission. So something is happening here that we need to discuss a little bit. 

[Slide 5] Okay, now the reason that things get a little more involved is that what we want to do is to relate the current to the charge at the top of the barrier, because we know how to relate that to the gate voltage. We don't know how to relate things like the location of the Fermi energy to the gate voltage. That's a little indirect. So that's why we would prefer to re-express this in terms of charge. Now when we did that in the ballistic case, for the small VDS result, we just said that the charge at the top of the barrier was Q times the sheet carrier density. That was Q times the effective density of states times E to the location of the Fermi level with respect to the bottom of the conduction band at the top of the barrier. Actually, there were two equal Terms here. Half of the charge was coming in From the source. Half of the charge was coming in from the drain, but under low bias, the drain Fermi level and the source Fermi level were almost in the same location. So we just lumped them together. This was our expression. We could solve for E to the eta  sub FS, and we could then express the current in terms of the charge in the linear regime. For the large drain-to-source voltage case in the ballistic limit, we only have electrons injected from the source. Only half of the states can be filled with those positive-velocity electrons. So we ended up with a factor of 2 here that we had to be careful about, and when we did that, we could relate this expression to the charge at the top of the barrier for large VDS, and we got our saturated region expression. Okay, while things are a little more involved now when there is scattering, we just have to do this calculation valid from the small to the large VDS regime a little more carefully. 

[Slide 6] So to do that, Let's look at the fluxes again. We inject the flux from the source to the top of the barrier. That's the I plus of 0. That's the first component of the flux at the virtual source. Some fraction of that flux back scatters and returns to the source. That's the second component of the fluxes at the top of the Barrier, and some flux gets injected from the drain, transmits all the way across, and makes it to the top of the barrier. That's the third flux component at the top of the barrier, and we need to consider each of their contributions to the charge at the top of the barrier. 

[Slide 7] So we have the charge injected from the source, and that's just Q. We can fill half the density of states up, and we fill them up according to the Fermi level of the source. The charge that injected from the source that back scatters, well, the negative velocity states are half of the states. So there's a factor of 2 there. We also fill those states up with the Fermi level from the source, because they were injected from the source, and there's some fraction 1 minus transmission that was back scattered. So that gives us the magnitude of that back scattered flux. And then our third component are the electrons that were injected from the drain with negative velocities, so half of the states can be filled, but we have to fill them according to the Fermi level of the drain, because they were injected from the drain. So those three components   together added up give us the total charge at the top of the barrier. 

[Slide 8] So this was our drain current expression that we got simply by integrating the Landauer Approach. We simply had transmission times the ballistic result. Our charge now is a little different. If we add up those three components, we get a 2 minus transmission times something that depends on the source Fermi level plus transmission times something that depends on the drain Fermi level. The total charge now is a bit more complicated than it was in the ballistic case. All right, but if we use that, what we want to do is to write the drain current expression in terms of this charge. So I'll multiply by the magnitude of the charge. I'll divide by the magnitude of the charge. I'll use this expression for the magnitude of the charge downstairs. I'll do a little bit of algebra. You can refer to the lecture notes for the details, and I will re-express this term, this expression, in terms of the charge at the top of the barrier, and here's the result. 

[Slide 9] So this is our result For the drain current at arbitrary drain to source voltages. Width times charge times some function times the injection velocity. The injection velocity is the ballistic injection Velocity that we've seen before, multiplied by the fraction transmission divided by 2 minus transmission. That is some number that is less than one. The ballistic injection velocity, if we're assuming Maxwell Boltzmann statistics, is just our friend the unidirectional thermal velocity. So this is our expression now for arbitrary drain-to-source voltage. One thing that were missing that we will have to talk about a little bit is that we know that this transmission is different for low drain voltage, and for high-drain voltage. Presumably it varies continuously from low-drain voltage to high-drain voltage, and we don't have an expression for that. We'll have to talk about that a little bit. 

[Slide 10] But first, let's check the ballistic limit and make sure that this sort of complicated expression gives us the right answer. In the ballistic limit, the transmission is 1. In the ballistic limit, then, the injection velocity is the ballistic injection velocity or the unidirectional thermal velocity. That's the correct answer, and this expression in terms of the brackets, when the transmission is one, just reduces to 1 minus this exponential over 1 plus this exponential, that's what we got in the ballistic case. So this expression gives us the correct result in the ballistic case. 

[Slide 11] Okay, we can also check the small VDS limit that we did in the previous lecture. So if we take this expression, now were going to say that the transmission in the linear regimen is T sub lin. We're going to expand the exponentials for small argument, and when you do that expansion, you'll find that you'll get this expression, which is exactly what we got in the previous lecture when we did the linear region current in the presence of scattering. 

[Slide 12] If we check the large VDS limit. Then we'll have a large-drain voltage. The exponentials will approach 0. This factor will approach one, and our drain current in the large-drain voltage limit will reduce to this expression. That expression is exactly what we got before. Again, we're taking the energy independent transmission and we're saying we have to call that the transmission in saturation now because it depends on drain voltage. 

[Slide 13] So this drain bias-dependent transmission is what makes things a little more complicated. In general, the transmission depends both on the gate voltage and on the drain voltage. When the drain voltage approaches 0, it approaches this quantity T lin, which we understand, mean free path over mean free path plus channel length. In the limit that the drain voltage is much bigger than the drain saturation voltage. It approaches the saturation transmission, which we understand as being mean free path over mean free path plus the length of the critical bottleneck regime. We can actually write it in general. The transmission is mean free path over mean free path plus some critical length, and that critical length goes smoothly from the channel length to some small fraction of the channel length. You can actually compute it. Computing it rigorously presents some challenges. There have been a number of approximate calculations. This is one way to get approximately the right answer, and what this expression tells us is that if we know the potential profile throughout the channel, which can be a little bit challenging to determine, then we can compute what that critical length is. 

[Slide 14] Okay, so our model is this. Current is width times charge times average velocity. Average velocity is some drain saturation function times injection velocity. The drain saturation function is given by this expression that we just derived, and the injection velocity is some fraction of the ballistic injection velocity. The major challenge in using this model is getting an adequate representation for how this transmission changes from T lin to T sat as we go from low to high drain voltages. That's the reason that this empirical virtual source model still finds wide use, because in that case, we're simply empirically fitting that expression rather than trying to smoothly develop a theory for that. 

[Slide 15] There has, however, been some recent work, and if you're interested in seeing how one can take an estimate for what the channel potential profile is from the source to the drain, compute at any drain bias this critical length, and see how that works out, there have been some promising results that were reported a couple of years ago at the Electron Devices Meeting, and I encourage those of you that are interested to have a look, you can see how that's done. 

[Slide 16] So then, let's wrap up just by quoting the results for Fermi-Dirac statistics. The ideas are most easily illustrated from Maxwell-Boltzmann statistics, which is why I've done so throughout this lecture. Actually, Maxwell-Boltzmann statistics frequently work rather well in practice, but from time to time, we will want to use the Fermi-Dirac results also. So when we use Fermi-Dirac statistics, current is still width times charge times velocity. Velocity is still some drain saturation function times injection velocity. The drain saturation function now involves ratios of Fermi-Dirac integrals. This is the result for 2-D carriers. If I were treating a 1-D nanowire MOSFET, I'd have Fermi-Dirac integrals of different orders, but the equation would look similar. The ballistic injection velocity now, in general is gate voltage dependent, because the location of the Fermi level depends on the gate voltage. So it's the unidirectional thermal velocity times some ratio of Fermi-Dirac integrals. The actual injection velocity in this expression is some fraction of that ballistic injection velocity, and the fraction depends on the transmission. Okay, so we have developed, in this lecture, and in the previous lecture, the transmission theory that we set out to do in this course. What I would like to do in the next lecture is to take another look at our virtual source model to relate this transmission model to the virtual source model, and we'll show how we can interpret the parameters in that virtual source model in a way that is fully consistent with this transmission model of the MOSFET that we have developed. So I'll look forward to continuing this discussion in the next lecture.