nanoHUB U Fundamentals of Nanoelectronics: Basic Concepts/Lecture 2.7: Conductivity versus Electron Density (n)
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[Slide 1] Welcome back to Unit 2 of our course. This is the seventh lecture. 

[Slide 2] Now, as you know, here we have been talking about this expression for conductivity, this new perspective in which the ballistic conductance kind of plays a central role. It's like the starting point. You start from ballistic conductors, and then you obtain an expression for conductivity that's like ballistic conductance times mean free path. On the other hand, old perspective, or the one that you have seen in all text books in freshmen physics texts. Most books will start with it, is the Drude formula, and the Drude formula relates conductivity to electron density, effective mass, and this mean-free time. As I had mentioned earlier, that in a way this part here that depends on scattering, this is a mean-free path, this is mean-free time. Those correspond to each other, and then there is this ballistic conductance, and here you have electron density and mass, and these are kind of corresponding quantities, and what I'd like to do is start to relate the two because this, of course, is widely used and it's fundamentally correct, and it usually works fine. It's just that there are special cases where you have to be careful when you use this. Like if you don't have a parabolic band, what does mass mean? Or, in general, which electrons are free? What does this free electron mean? Whereas in this view, as you know, this ballistic conductance, it just depends on this density of states and velocity or this number of modes, and so it's more straightforward in that sense. On the other hand, this is what is widely used. This is what people in the field usually carry in their heads. This is what they use to interpret data on a daily basis, and so what I'd really like to do is connect these two. Now, in order to connect those two, first thing is, we need an expression for electron density, and that's what we did in the last lecture. We had this N of p that we had from a few lectures back. It's rules for counting states, Lecture 3 I believe, and we said if you want electron density per unit volume, then we drop the L or the WL or the AL, so this would be electron density per unit length, or per unit area, or per unit volume. Now here we have this quantity, ballistic conductance, and that depends on the number of modes, and again, a couple of lectures ago, we obtained an expression for the number of modes, and as I had mentioned, that looks a lot like the N, except that this kind of tells you how many half wavelengths fits into the volume. This tells you how many half wavelengths fits into the cross section. So you'll get that, and from here, you can deduce an expression for the ballistic conductance, because ballistic conductance is q squared over h, times the mass. You see? And what I've done is, you want ballistic conductance per unit area, and so I've dropped the area. The area is gone. I only have the pi left. If it was two dimensions, of course, it would be GB over W, and that's why I've dropped the W here, and if it's one dimensions, then it's just GB. That's it. So the expression for the ballistic conductance per unit area then looks something like this, and the expression for the electron density looks something like this. Now what we'd like to do is look at the ratio of the two, because, I guess, what you're trying to do is find the ratio of this quantity to this quantity and to see if they're equal or not. So how does this compare to that? 

[Slide 3] So for that, let's collect the results we had, ballistic conductance. This is the electron density, and divide one by the other. So what we have here is the ratio. GB over A, that's the ballistic conductance per unit area. That's that quantity. Divided by this q square n over m. That's that quantity appearing in the Drude formula. So you look at the ratio of the two, and you'll notice q squared cancels out. p over h, d minus 1, this is p over hd. So you kind of have a P over h left. Except that there's an h there, so the h cancels out. So finally, you'll see what is left is one p and that m. So when you take the ratio, all that is left is m over p, and as far as the dimensional factors are concerned, I guess it's straightforward. 1 divided by 2, that's half. 2 divided by pi in two dimensions, that's 2 over pi, pi divided by that, that's 3/4. See, two dimensions-- in 1-D, 2-D, and 3-D, but in every case, then the ratio of this quantity to that quantity is this m over p times that. Now there is another factor here which is lambda over tau, and lambda is mean-free path. That's kind of like velocity times tau. So when you look at the ratio of the two, lambda over tau, if you look back again, I think this is in Unit 1, we had a special lecture on angular averaging where we obtained an expression for the mean-free path, after averaging over angles in 1-D, -2D, and 3-D, and the answer we got was mean-free path is v tau times 2 in 1-D, pi over 2 in 2-D, 4 over 3 in 3-D, and this is tau from here. So that tau just cancels out. So this ratio is velocity times, again, a numerical factor depending on the number of dimensions. So that is the ratio of this part to this part. That is the ratio of this part to that part. 

[Slide 4] So if you collect those two things together, now what you want is the product of the two, because conductivity is this times this, or this times this, and we have a ratio of this to this and a ratio of that to that. So we could take the product, and what you'd find is you get mv over p, and you say, "Oh, that's 1." After all, the momentum is m times v, but I think when we discussed the energy momentum relations, I did make a point that that's only true if you have a parabolic band, where energy goes quadratically as through momentum. In general, that's not necessarily true, and you have to think through carefully, in terms of how you define the mass. Because one of the things you'll see in a lot of the solid-state physics texts is the way they define mass is as dv/dp. That rather, inverse masses dv/dp, and v is dE/dp, so this amounts to the second derivative of the energy with respect to mass, with respect to momentum. So that's what you'll see in a lot of solid-state texts, in a different context. But in this context, that is when you're trying to discuss the Drude formula, that's not really not the correct relation to use at all. The correct relation to use is that 1 over mass is v over p, not dv/dp. See, and the thing is, in parabolic bands, those two are the same thing. So it really doesn't matter, but in general, those two are not the same thing, and what we want in this context is really this quantity and not that one. So why do I say that? Well, for that you have to kind of look back at the derivation of the Drude formula. That how is the Drude formula actually derived? And this we went through in Unit 1, so I'd just like to remind you how that was done. You can go back and look at the details again. The point is we have these two quantities. There's the current, the expression for current, which depends on the velocity. And then there is the expression for momentum, which comes from, like, Newton's Laws, which says dp/dt is equal to the force, and then you say, "Well, it's not dp/dt equals to a force. You also have a frictional term because you are in the solid with viscous drag." And so you put in a p over tau to it. And you can use that then to get an expression for p. But in order to calculate current, you really need an expression for velocity, and that is where you usually make use of this idea that mass relates the p to the v, and so the point I'm trying to make is when the Drude formula is actually derived, the way the mass gets introduced is as the ratio of momentum to velocity, not as dv/dp or anything else. It's really this ratio of the momentum to the mass is velocity. That's the relation that we are using, and so, the point is, that is what we should be looking at here, and so mass should really be viewed as p over v, and in general, for nonparabolic bands, that answer can depend on, I guess, at what energy you are, because the mass will keep changing with energy, because velocity to momentum ratio will not be a constant, and when using the Drude formula, then, you see, you have to worry about that. So what I just tried to show you is the formula we talked about kind of, I guess, correlates well with the Drude formula, the ratio is one, but then, when you use the formula we have been using, you need things like density of states, number of modes, et cetera. Which are, again, well-defined things. Whereas, when you are using the Drude formula, you have to worry about this number of free electrons and effective mass, where you have to be careful about what they mean. And in general, if they don't agree, it's really this result, the new result that we have been presenting. That is really more general. It is what you'd get straight from the Boltzmann equation, which is cornerstone of all semi-classical transport. If so, why doesn't everyone talk about that? Why is it everyone carries this in their head and not other one? Well, just because normally, this is very easy to derive. This requires Boltzmann equation, and so this the one you usually carry in your head, but using this new approach that we are talking about, see, this expression comes out in a relatively simple and straightforward way. That's what I guess I've been trying to convey to you. 

[Slide 5] So in the remaining two lectures then of this unit, what we want to do introduce this third terminal. That is, you see, so far, we have been talking about devices as if there's two contacts and a channel, but in practice, of course, a transistor has a third terminal, and that third terminal is all important. That's what's called a gate, and what it does is it controls the conductance of the channel. How does it control it? Well, basically, by moving the electrochemical potential up and down. Now how it does that, how much gate voltage is needed to change it, that's what we'll talk about in the next lecture.