nanoHUB-U Fundamentals of Nanotransistors/Lecture 4.4: Transmission Theory of the MOSFET I
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[Slide 1] Well Lecture 4 and Lecture 5 are really the heart of this course. This is where we've been heading in this course, to develop a transmission theory for the nanoscale MOSFET. We're ready to do that. We have all the background and understanding that we need. I'm going to do it in two pieces and part 1 will be in this lecture and then we'll continue the discussion in the next lecture. 

[Slide 2] So our goal is to develop a transmission theory for the nanoscale MOSFET. We're going to use our Landauer Expression for the current and we're going to retain the transmission, the finite transmission, in this calculation. Now in this lecture the first part of this story about transmission theory, I'm going to focus simply on the linear regime and on the saturated regime. And then the next lecture we'll do the entire IV characteristic. 

[Slide 3] Ok so let's look at the linear regime. And I'm going to make things simple. You know in general the Mean-Free-Path can have some complicated energy dependence because the scattering mechanisms have some complicated energy dependence and the velocity, which is related to the band structure, has some complicated energy dependence. I'm going to just assume that the Mean-Free-Path is energy independant. So actually that's not an unreasonable assumption for things like parabolic bands and acoustic phonon scattering. But it greatly simplifies our calculations also. If that's not precisely true then we will interpret this lambda naught as some appropriate average Mean-Free-Path over the energy channels of interest. That means that our transmission is also energy independent and I'll label it with a T0 just like I labeled the Mean- Free-Path with a lambda 0. So the calculation, doing the integrals is greatly simplified when we can move the transmission outside the integral. So our ballistic expression for the linear region current was given by this expression, assuming Maxwell Boltzmann statistics. If you go through the calculation again and if you assume an energy independent Mean-Free-Path, you'll see that everything works out the same way and it's just what you would have expected. It's transmission times the ballistic result. So we get the linear region current in the presence of scattering very simply. 

[Slide 4] Ok now before we proceed, let's just look at this in the diffusive limit and make sure that it does the right thing. So here's our linear region current. It's simply the ballistic expression we developed earlier multiplied by the transmission, average transmission, a number between 0 and 1. The transmission is related to the Mean-Free-Path for backscattering. Let's assume that we have a long channel length and we are in the diffusive regime. The transmission simplifies to Mean-Free-Path divided by length of the channel. Ok if I do that, if I put my transmission in a diffusive limit into the first expression, I'll get an expression for the linear region, drain current, in the diffusive limit. And then if I lump some terms together and if I pull the W/L out front, I can write that diffusive region current in this form. And if I look at what's in the brackets there, I'll notice that what's in the brackets there, first of all it has the units of mobility and I can see that a little more clearly by rewriting it as thermal velocity times Mean-Free-Path divided by 2, which I recognize as my expression that relates diffusion coefficient to the Mean-Free-Path. And then I can bring the kT/q out behind. And then I see that what I have here is diffusion coefficient divided by kT/q. The Einstein Relation tells me that that's the mobility. So what we conclude is that in the diffusive limit our transmission theory for the linear region current reduces to the well-known expected answered W/L mobility times charge times drain the source voltage. And we have also developed an expression for the mobility, thermal velocity times Mean- Free-Path divided by 2kT/q. So the mobility is related to scattering. We can see it's directly proportional to the Mean-Free-Path for backscattering. So everything works out just as we expected it to in a diffusive limit. In the ballistic limit we'll simply put T equals 1. We'll get the results we got earlier in the ballistic limit. So no surprises in the linear region current. It was very easy to do. 

[Slide 5] Ok so we've done the linear region. Let's do the saturated region so we can look at things like the on-current. Now you might expect that all we have to do is what we did with the linear region current. We simply take the ballistic result, multiply by T and we're done. That turns out that's not the case. Let's talk about this a little bit and see why we have to be a little more careful with this calculation. 

[Slide 6] So let's look at you know how on on-current is related to transmission and to charge at the top of the barrier. So here's our picture. We have a high drain bias. We are injecting electrons from the source into the channel. The drain bias is high so we're really not injecting anything from the drain into the channel so we have an injected flux. I'll call it i plus at x equals 0 at the virtual source, the top of the barrier. The plus means it's the forward directed flux. Ok now what comes out at the drain is transmission times the flux that was injected. That's our on-current because we're operating at a high drain bias here. But if there is scattering we also have a backscattered flux. So what returns to the source is 1 minus what went out the drain times what was injected from the source. So we have these three fluxes to consider. 

[Slide 7] Now let's look at the ballistic case first, the one that we considered in the previous unit. We're injecting some forward flux. Now we're in a ballistic case. The ballistic on-current is just the transmission times that, transmission is 1. And there is no backscattered flux in this case. Everything just, it gets injected, goes out the drain. Ok now if I look at the charge at the top of the barrier, charge is just flux divided by velocity, I have a W, or the width, of the MOSFET that comes into there too, so I have a very simple expression for the charge in terms of the flux that's injected. And remember this charge in a well defined MOSFET, I set this with a gate voltage. Its basically c inversion times vg minus vt. 

[Slide 8] Ok now let's look at the problem that we're interested in in this unit. There is some finite transmission. So we inject the flux, i plus of 0, this comes from the source, into the top of the barrier. Some fraction tau propagates across. Now we have some fraction 1 minus transmission that backscatters. Ok now let's look at the charge at the top of the barrier. In the ballistic case, we just evaluated it, now if I'm operating at the same gate voltage, I'm going to get the same charge at the top of the barrier if I have a well-defined MOSFET. So the charge Q is not going to change but I'm no longer operating in the ballistic case. My charge now, what I have to do is to add, the forward flux contributes some charge. The backscattered flux contributes some charge. Both of these fluxes, if I divide by W times the velocity, that will now give me the total charge, the charge due to the injected flux and the charge due to the backscattered flux. These two expressions are the same. If I have a well-defined MOSFET, Q is determined by MOS electrostatics, not by transport, so these two expressions are the same. I have to equate these two expressions. And if I equate those two expressions I'll find an expression for this flux that I inject in the presence of scattering that gives me the same inversion layer charge that we had before without scattering and it will simply be the ballistic flux divided by 2 minus transmission. Transmission is always less than or equal to 1. So the flux that I inject is always less than the ballistic flux because now I have some charge contributed from the forward flux and some charge contributed from the backward flux, both of those have to add up to the same charge that I had in the ballistic case. Now what comes out is transmission times the injected flux so what comes out is my on-current, because I have a high drain bias here, so I simply multiply this expression times the transmission and what comes out is transmission divided by 2 minus the transmission times the injected flux. This injected flux was also my on-current in the ballistic case because whatever I injected propagated across and came out. So we don't simply multiply by the transmission. We multiply by the transmission divided by 2 minus the transmission. That all comes about because of the necessity of keeping the charge at the top of the barrier constant at the value that MOS electrostatics is demanding, even in the presence of scattering, Whether there is scattering or whether there isn't scattering. 

[Slide 9] So our expression for the saturated current, it's not transmission times the ballistic result, it's transmission divided by 2 minus transmission. So not what we might have guessed but easy to understand where it's coming from. 

[Slide 10] Alright let's take a look. This is some experimental data. The circles here are the measured IV  data for a 30 nanometer silicone extremely thin SOI MOSFET. And you can see the IV characteristic there or the gate voltage at the drain voltage. We also see a virtual source fit when we can see that we can do a nice virtual source fit to that data. Now we can also calculate the corresponding ballistic IV characteristic for this same device using the same series resistances, the same inversion layered gate capacitance and we can compare the ballistic calculation to the measured calculation. And we see that the measured result is below the ballistic limit. There must be some scattering going on. So we've just determined, you know, we can see that our ballistic calculation is here, our experimental result is here. The experimental result is about 70 percent of the ballistic result. We have determined that the current in the presence of scattering divided by the current without scattering is given by this ratio transmission divided by 2 minus transmission. So we can equate these two expressions and we can find out what the transmission is in this 30 nanometer silicon MOSFET. And we'll find that it's 82 percent. So that's actually a quite high number. Eighty two percent of the electrons that are injected from the source exit from the drain, ok? So modern day silicon MOSFETs operate at a significant fraction of the ballistic limit. 

[Slide 11] Now one other thing that I want to point out that's quite interesting here is that I've actually pulled the wool over your eyes a little bit. You know this transmission that we said T0, this energy independent average transmission, actually it's a little more involved. The transmission at low drain voltage is different from the transmission at high drain voltage. It is drain biased dependent. Now it also turns out that for low drain bias the transmission is less than it is for high drain bias. We will see that quantitatively when we analyze experimental data in lecture 6. But I'll just ask you to accept that for now. That is a bit surprising. That's a little counterintuitive. Because under a high drain bias we expect carriers to gain a lot of kinetic energy. If they gain a lot of kinetic energy they'll be at energies that are way above the bottom of the conduction band. The number of states there is much higher. I've mentioned that the scattering rate often is proportional to the density states. There are many more places to scatter too. We would expect the scattering time to decrease under high drain bias. And we would expect the device to have a transmission that is lower under high drain bias. We find when we analyze experimental results that exactly the opposite occurs. 

[Slide 12] Well we can discuss that and see if we can understand why that occurs by looking at the linear regime. In the linear regime we have a small electric field throughout the entire channel so we know how to compute the transmission in that case. It's just Mean-Free-Path divided by Mean-Free-Path plus the channel length. Ok what happens under high drain bias? Under high drain bias we pull everything in the drain down. If we're electrostatically well designed we maintain a small electric field near the beginning of the channel, the region under the strong control of the gate. So now there's a very small region of length L, script L, which is some small fraction of the channel length where the electric field is low. We argue that in a structure like this where we have a small electric field region followed by a large electric field region, the transmission is controlled by the small electric field region. So instead of using the entire length of the channel here, we should replace it by the length of this short bottleneck region that the carriers have to get across. So our transmission is higher because the length of this region is much less than the channel length. So in general the message is that the transmission depends on the drain voltage. There is some critical length at small drain voltage. That critical length is the entire length of the channel. If we backscatter anywhere within the channel, there's a chance that the electron could return to the source. Under high drain bias that critical length is a  small part of the channel. If we get across that small bottle neck region without scattering, then it doesn't matter if we scatter. It's bound to come out the drain. Ok so that explains, qualitatively anyway, the biased dependence, the drain biased dependence, of this transmission. 

[Slide 13] Now I also want to mention that when we say that we're operating near the ballistic limit, that doesn't necessarily mean that there isn't a lot of scattering going on. What it means is that there isn't much scattering going on in this critical region. We get across that critical region, then even if we scattered, the strong electric field ensures that we're going to by and large be swept out the drain and even if we do scatter we swept out the drain and the transmission will be quite high. So the message is that operation near the ballistic limit doesn't imply that there's no scattering, it implies that there is little scattering in the critical region that controls the current of the device. 

[Slide 14] Ok now I want to return to something that I mentioned just briefly earlier and that is, talking about the transmission, when we developed our expression for the transmission and when we inserted it in a Landauer Expression, we just had a number T for transmission. Well in principal the transmission from the source to the drain is not necessarily equal to the transmission to the drain to the source. We've been assuming that the two are equal. 

[Slide 15] How can we do that? Well if the drain to source bias is low then everything is symmetrical and we would expect that to be the case. One can also prove that if the scattering is elastic within this region, then the two transmissions are the same and we can deal with one transmission as we've been doing. 

[Slide 16] Now under high drain bias we'll pull everything down in the drain, then electrons that are injected from the drain that can get up to the top of the barrier, they have a lot of kinetic energy. They can scatter frequently. And what we'll find now is that the transmission from the drain to the source is much, much less than the transmission from the source to the drain. So how can we deal with one transmission in our transmission theory of the MOSFET? The reason is that under high drain bias, even though that assumption isn't valid, there are virtually no electrons being injected from the drain to this critical region. So the fact that we have the transmission for that part of the problem wrong, doesn't matter because there aren't any significant numbered electrons being injected to the virtual source under those conditions. So that's the reason that we can deal with one transmission and not have to differentiate between transmissions from the source to the drain versus transmission 

[Slide 17] from the drain to the source. One final point that I want to mention is this question about is mobility relevant at the nanoscale? So we think of mobility as something that applies to large bulk semiconductors that are near equilibrium. It's a material parameter that we can measure and then look up. Well that's not the situation that we have in a MOSFET. Under high drain bias especially we're very far from equilibrium, the conditions are very much different than they were in equilibrium but the point is that the mobility is related to the nearly equilibrium Mean-Free-Path. The backscattering in this critical region that controls the current, even under high bias, is related to the near equilibrium Mean-Free-Path. And since the near equilibrium Mean-Free-Path is related to the mobility, we can say that the near equilibrium Mean-Free-Path  controls the current even under strong drain bias. But it's important to note that in this region where the carriers have a lot of kinetic energy, their Mean-Free-Path might be quite a bit shorter than our near equilibrium Mean-Free-Path but it doesn't really matter because those electrons are going to go out to drain and they won't lower the transmission. The transmission will remain high as long 

[Slide 18] as the carriers can get across that critical bottleneck region. So just to summarize, we've covered a lot of ground and we will do a little bit more in the next lecture because this is really the heart of the course, this Transmission Theory of the MOSFET. The key points are scattering lowers the drain current, no surprise there. The surprise is that the transmission is actually higher under high drain bias than it is under low drain bias and we tried to explain physically why that makes sense. The reason it makes sense is that under low drain bias, scattering in the entire length of the channel matters and can lower the current. But under high drain bias scattering in just a very short bottleneck regime is what limits the current. And as long as carriers get across that short bottleneck regime, then even if they scatter, and they will scatter more under high drain bias, it doesn't matter. The transmission is still quite high. So with that understanding of the linear region current and the saturation region current, I'd like to continue this discussion in the next lecture and talk about the entire IV characteristic from low VDS to high VDS. So we'll continue this discussion in the next lecture. Thank you.