nanoHUB U Fundamentals of Nanoelectronics: Basic Concepts/Lecture 3.5: Landauer Formulas
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[Slide 1] Welcome back to unit 3 of our course, and this is the 5th lecture. 

[Slide 2] Now, so far we have been talking about channels which have distributed scattering processes. And that scattering process is described by something we call the "mean-free path", that is how far does an electron go before it is turned around? Now, in this lecture what we will be considering is a localized scatter, that is the way we'll think is as if the channel is basically ballistic, no scattering anywhere, except for one point where there is some scatter. So, again, in terms of cars on a highway, it's as if there's a little tollbooth, or a construction zone somewhere. Everywhere else cars just flow freely. And the question you ask then is that under these conditions, what would be the average flow, the current, and what would the potential profile look like? And what we'll show is that the conductance will be given by the ballistic conductance times this what's called "the transmission." I mean, I'm using T, but it's got nothing to do with temperature. Okay; this is what you call "the transmission coefficient." What it describes is that we have this localized scatterer, and the idea is if 100 electrons come in here per second, what fraction crosses the scatterer and gets out on the other side? So let's say if T is .2; that means if there's 100 per second coming in, only 20 second leave from the other side. And as you might expect, as a result the current flow becomes 20%, and so the conductance becomes 20%. And we'll see how it follows from our discussion. And if you compare this relation to what we have been working with so far, you'll notice that it's as if that quantity that we had for distributed scatterers is replaced by T. Indeed, we could interpret this quantity as T; that is if you have a channel of length L, with distributed scattering processes, with a mean-free path lambda, then this fraction kind of tells you what fraction of the electrons incident from the left actually come out on the right. Okay? But that's a separate thing. Right now what we want to talk about is not distributed scatterers, but a localized scatterer whose transmission probability is T. And what we'll argue is that in that case when you look at the electrochemical potential of this quasi Fermi level, if you look at the quasi Fermi level, what we'll find is the quasi Fermi level drops sharply across the scatterer, from 1 to T. And if we look at the other quasi Fermi level, it steps up from 0 to 1 minus T. Okay. Now, of course what I've really plotted here, though, and this is the point I want to stress, is actually the occupation of the states; that is the way one should think about it is, as we discussed in a previous lecture, that in this energy range, the states in this contact are all filled. So the occupation is 1. Here the states are all empty, and so the occupation is 0. Okay? Now, the question is, "What is the occupation once you cross the scatterer?" That is, once you are on this side, of course, the right moving states are all occupied, as we discussed, you know, mu plus connects up to this mu1. And, again, in terms of the highway analogy, all the northbound states are completely filled. But what happens as you cross the scatterer. Well, you know what happens on a highway. Anytime there's a construction zone, or one of these tollbooths, as you go across the occupation drops suddenly. So you'd expect that if you looked past the scatterer, what you would find is that of course the occupation instead of being 1, would be much less. You know, so if the transmission is .2, then the states would only be 20% occupied. Of course, the states below mu2, those are just occupied everywhere anyway, so that doesn't matter. It's -- in the energy range of interest the occupation would go down from 1 to T. So what does this thing represent when we draw a line here? What I really mean is it's the occupation of states. Now, of course if we do as a real electrochemical potential, if you think of it as an electrochemical potential, then the distribution should have looked something like this, because normally the way you think of a Fermi function at low temperatures, is that it changes from 0 to 1 at the electrochemical potential. But the point I wanted to stress is when I draw this curve, I'm not implying that the energy is distributed this way, because what the actual energy distribution is, that depends on inelastic processes; that is this red distribution, that's not an equilibrium situation. So there's always inelastic processes which are trying to take that red and turn it into this black. And the actual distribution could be somewhere in-between, depending on the degree of inelastic scattering. It could be like the red, or it could be like the black. We are not getting into that question at all. All we are saying is that once you cross the scatterer, the lanes will look very empty. And that just comes from common sense and experience. I mean, as you know, on a highway, as I said, you cross the construction zone, the occupation goes down. And what this mu tells us -- or this occupation tells us is just the degree of filling of the states. So what you could do is you have an f which tells you the occupation, and you could relate that f to a electrochemical potential, remembering, again, the electrochemical potential is not describing the energy distribution of electrons, rather it's just telling you the degree of filling of states. 

[Slide 3] Okay? Now, what is the current then? Well, the current is, as we discussed in an earlier lecture, it's the ballistic conductance times the separation of quasi Fermi levels. And the separation of quasi Fermi levels is constant everywhere. You can look at any point. Here one is T, the other is 0. Here one is 1, the other is 1 minus T. So the difference is always T. So if you put that in, then you get the current as proportional to the voltage, and current divided by voltage is the ballistic conductance times the transmission. Okay? Now, you could say, "Well, that's then the conductance." Or if you inverted it, it will give you the resistance. But what is the resistance of the scatterer itself; because note that the presence of the scatterer, or course, reduce the current significantly; gave rise to a lot of extra resistance. So what is the resistance of this scatterer? So then you say, "Well, what you should do is look at the voltage drop across the scatterer," you see? So if you -- you see the total voltage is that much, but only a fraction of it drops across the scatterer. So you'd say well, but you have drawn two voltages. You have drawn two quasi Fermi levels. Which one should I look at? Answer is look at either one. You'll get the same answer. If you look at the top one, it goes from 1 to T. If we look at the bottom one, it goes from 1 minus T to 0. In either case the drop across the scatterer is 1 minus T times the overall voltage. So the actual potential drop across the scatterer is the applied voltage times 1 minus T. So what is the resistance associated with that? Well, you can look at the conductance. It will be current divided by delta V, and now because of this extra 1 minus T, you will get what we had before, but now with the 1 minus T in the denominator. So these are conductances, current divided by voltage. If you write the resistances, you could -- this would give you a resistance of RB times 1 over T. But as this one would give you the resistance of RB times 1 minus T over T. And I've put an S there to indicate that this is kind of what you could call the "scatterer resistance." And back in the 1980s, you see, there was a lot of argument about what exactly represents the resistance? Is it 1 over T, or is it 1 minus T over T? You would see papers in fact whose title is just that, "1 over T or 1 minus T over T?" 

[Slide 4] Now, today of course knowing what we know, the way you would explain it is well there is a resistance which is like 1 minus T over T, which is associated with the scatterer itself, and then there is the 2 parts -- difference between those 2 that's RB, that actually is what appears across the -- at the interfaces. That's the extra part, you see? And why do we do -- put half here and half here? Well, what we could do is we could look at the average of the two. See, if you look at this average here, then you'd say, "Well, you have a drop-off -- half the extra drop is here, the other half is here." So the drop in the middle we associate with this resistance. The drop at this end we associate with this one. The drop at this end we associate with that one. Now, you might say, "Well, but then why are you looking at this black one, the average; why not that one?" Well, now that is a subtle one that kind of requires some discussion, because if you had looked at that one you would have said, "All the resistance is at this end." And if we had looked at this one, you might have said, "All the resistance is at this end." Right? But that doesn't say there's a subtle point that requires a little more discussion. But if you look at the middle one then, you could put half the resistance here, and the other half at this end. But the important thing is that the overall resistance is 1 over T, whereas the resistance associated with the scatterer itself is 1 minus T over T. And these are what you could call the Landauer formulas, in the sense that I guess Landauer had appreciated all these subtle issues, many of the subtle issues involved, although this interface resistance took much more discussion, and people credit Imry with recognizing this interface resistance in this context. But this is the original Landauer formula, whereas these days often people use this one to interpret results, because usually you are looking at the total resistance, rather than just the one associated with the scatterer. So this you could just call generally the Landauer formulas. 

[Slide 5] Now, however, this discussion often leaves a lot of people kind of uncomfortable, because it involves quasi Fermi levels. You see, because as I've been stressing from the beginning, quasi Fermi level is a subtle concept that -- and it requires you to appreciate many subtle things in statistical mechanics; and it leaves people uncomfortable. And so people say, "Well, let's do something more concrete." And the way they do it usually is one is they say, "Well, let's just look at the electrostatic potential. Let's not even talk about quasi Fermi levels." And I've mentioned before, I don't quite like that because electrostatic potentials tend to smear things out, and tend to kind of obscure the physics. The other approach people use is they say, "Let's not talk about the potential inside, but let's just ask, 'What is the potential that a probe would measure? If I actually had a scanning probe coming down here, and measuring the potential, what would it measure'?" And there the answer is usually a probe would measure the average. A weakly coupled probe would measure the average, which means the black curve. That will be one way to justify why we should be looking at the average, the black curve. Okay? But this is what we'll talk about, you know, what a probe measures. 

[Slide 6] That's what we'll be talking about in the next lecture. Thank you.