 You can follow along with this presentation using printed slides from the Nano Hub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. So to continue our discussion then, as you'll notice from the outline of my talks like today, this 1, 1B2, that's largely focused on what you might call semi-classical transport. What I mean by that is that you think of electrons as particles, essentially. Don't worry about the wave nature of things. And tomorrow I'll talk a little more about quantum transport. But this semi-classical part, it's something I don't normally talk all that much about. But what I feel is it's really very important to be clear on what you expect just from this classical picture. So that when you see something different in a quantum picture, you know that it's really just as a good background to compare to. You see? Because there is lot in transport that can be quite confusing if you don't look at it the right way. And it has nothing to do with quantum mechanics. Much of it has to do with just what you'd expect if you think of them as particles. Or as I said, like cars on a highway, how many channels you have, things of that sort. So it's very important to be very clear about the simple things before one gets confused by the harder things. You see? And what I was trying to describe to you today in general is this model that is inspired by this bottom-up view. Which actually gives you a very good, clear picture of the simple things rather easily, you see? And the concepts that I was trying to get across, you'll notice the density of states, the Fermi function, those are important things. Why does current flow? Because this one wants to fill it up, that one wants to empty it. That's really why current flows. And a new concept that I introduced, I want to say a few words about that, is this M, the number of modes. And this we'll see more of. And I tried to motivate this whole discussion with a simple statement that the transfer time is proportional to length, and then there is a term that is proportional to the diffusion coefficient. Now this is something that you are familiar with. Of course, it helps you get to the end results very quickly. On the other hand, looking at your feedback, by the way, please do give me the feedback because it helps me see which parts are harder, what's not getting across well. And one of the feedbacks I got from many people was they weren't quite sure where that came from. I mean, what was the hardest thing in the whole lecture? Some people say this one, you see? And now for the discussion section, I'll talk more about this. We can discuss this more, right? And so please do pass on your questions to Samiran. And I'll look at them and try to structure our discussion at 330 on all the different issues that came up, right? Now the result of this, of course, was finally that we had this rather simple expression for conductance, where it looked like what you had seen before, Sigma A over L, but then L plus lambda. And that tells you that when you get to small lengths, the conductance doesn't just blow up, but goes to some constant value. And I think this point will get a little clearer now. I think I'll discuss this some further, actually. We'll talk more about that. But one concept that in this context that came in, and this also we'll talk about more, is this m, which is this dVz over 2L. That's density of states times velocity. The density of states is a concept that has been around with us for a long time. That you have states in the channel and by the one of the questions I had was whether these states were in the channel or whether in the contacts. And let's be clear, this what I'm drawing here is the density of states in the channel, very much. The contacts are regions with lots of states. So the way I think of it again in that highway analogy is you have a highway running from between two cities. And then there's lots of little lanes that are feeding into it. And hundreds of things that converge at that point. That's the contact, hundreds of things coming on. And then there is this few lanes that go to the other city. And then again it breaks up into hundreds of lanes. That's the kind of picture I have always. That's what the contacts are. And so the contacts are more or less held in equilibrium. And this is what controls the current flow. Now that's density of states. The number of modes which came out of that discussion a little bit is that it's density of states times velocity. Now, and this is again a general concept that it could be any conductor. You could take even amorphous materials. One could talk about what is the velocity, what is the density of states mud like this. But towards the end of the morning's lecture, I mentioned that in crystalline semiconductor materials, you could have a model that is based on an EK relationship. Where density of states is calculated from an EK or what I wrote as EP relationship, energy momentum relationship. And based on that, I introduced a concept called N. And this was again confusing to some people, which as I said, again in the discussion session, we can talk more about it. With different EP relations, what Ns do you expect, etc. But the picture is that where does this density of states come from? The way you think about it is that as if, say, this point corresponds to zero momentum, this point corresponds to a little more momentum, a little more momentum, and so on. And the energy is related to that momentum in some way. That's the picture you have. And question is, how do you go from an EP relationship to a density of states? And that is where, what I say it is, the standard way of counting states is, to write down the total number of states that have a momentum less than some value P. And that is proportional to P. And as I said, it's usually this P to the power, the number of dimensions. It's something in front, some constant in front differs. And it's proportional to the volume of the solid. Bigger it is, the bigger that constant in front. And based on that, I tried to show a general relationship, namely that N times D is equal to D times V times P. And this, as I said, would be independent of energy momentum relationships. You could be using graphene, you could be using some semiconductor, the parabolic relation, doesn't matter. And the nice thing about it is it relates the density of states to the total number of states below it. Because as I said, conduction really depends on the density of states right here. Right around the Fermi energy, and the point I made is, that quantity then gets averaged over this del F del E type of thing, okay? And this density of states, that relationship relates that to all the states that happen to be below it. And that is how you can kind of take that conductivity and relate it to how many states you have. That's kind of what happened from here to here. And the point I tried to make is, this one is like N tau over M. That's the usual root formula that many people are quite familiar with. Because as I said, most discussions start with something like this. But the point I wanted to stress is that this is actually far more general. You could be applying it to any material you want. It doesn't even have to be a crystalline conductor. It doesn't have to have a EK relationship, right? Whereas to get here, then you need to bring in these other issues. And we can discuss these more, again, at 330, right? Now, the way I wrote this down was by saying that, given any solid, how many wavelengths can you fit inside it? And wavelength is related to the P. That was the basic argument people use. That it's the de Broglie wavelength related to the P. H over P, that's the wavelength. So how many can you fit into the box? And that's where the bigger the box, more you have. Now, interestingly, and this is the part that's not so obvious is that this M which is D times B can also be viewed in this way, except that it is the number of wavelengths that fit into the cross-section. So this is a kind of total, and this is not obvious at all from this definition. And yet, in a particular problem when I'm trying to evaluate what this is, it is much easier to think from that point of view, you see? Yes? M can be fractional, right? So what is fractional number of forces? Yeah, usually M cannot be fractional, should not be fractional, M itself. Now, when you measure something, the reason it gets fractional is because you're averaging over energy. So at a given energy, if I had to try to find M, it would be a number, 200, 201, 315, whatever. But then when I average, because of this Fermi function, then I might get 21.7. But at low temperatures, usually that averaging is not very important. And that is why at low temperatures, people see this quantized conductance in ballistic conductors. Because that M is a integer that you can actually see, right? So experimentally, that's all established. But the point that I thought I should mention is that while one way of thinking of M is density of states times velocity, and as you'll see in a minute, I mean later today I'll also talk more about this. But this is also equivalent to how many wavelengths you can fit in a cross section. And the quickest way to see that is something like this. It's like a, you see, if you again use that relationship that I wrote. That N times D is equal to D times V times P. Now, this N, as we discussed, is sort of like P to the power number of dimensions, whatever it is. And then times the volume. So let's say I'm writing A L. Now, the number of modes is like DV over L, right? So let me take DV over L, put L here. Then you see this is kind of like number of modes. And then you can kind of see that this number of modes then becomes, you see, the L goes out, one of the P's go out, and it becomes A times P to the power D minus 1. So as if you're just looking at the cross section and trying to fit wavelengths into it. You see, this was when you were trying to take the entire volume and trying to figure out how many wavelengths fit. And after you do this, it's like this cancels out and it becomes. Now this particular thing may not be clear just from my discussion right now. Again, at 330 we can go through examples and it'll be clear. But the reason I mentioned this is that this is one definition you'll see. On the other hand, if you're really trying to quickly get an idea of, okay, now I got something that's 10 microns wide, how many modes do I have? Then the quickest way is, okay, in this semiconductor for the energy involved, what is the de Broglie wavelength? Like in silicon at a carrier density, certain given carrier density, the de Broglie wavelength may be let's say 100 angstroms, 10 nanometers. So it's okay, if it's one micron, number of modes should be about 100. So you can do these quick calculations if you have that picture in your mind. On the other hand, of course, if you have all the time in the world going back there doing it properly, you'll get all the right answers. But for quick estimates, that's usually a very quick way of doing this. But this, again, let's talk more about at 330. I just thought I should mention this point. But this is an important concept that, as I said, was not well appreciated in this field or not at all present. No one used that concept till all these experiments came along looking at small conductors. Density of states has been around since the 1940s or 50s. I mean, once people started looking at solids, density of states was one of the first concepts in there. But this concept of number of modes, that became important only when you started looking at small conductors at this point of view. Now, what I wanted to go on to from here is that the points that Professor Dunström discussed about the current voltage characteristics in a MOSFET, a lot of it you could model based on this equation. And what I'd like to explain is how you do that. I think he went over it, he just wanted to say quickly how you do this. And that's what I call this point channel model. Point channel meaning that you think of this channel as a single point with some potential U. I mean, one should make this distinction. Potential is usually volts. Potential energy is like electron volts. So here when I write U, I mean potential energy. So if this U is positive, what it does is it moves this entire density of states up or down. If it's positive moves it up, negative moves it down. And this is the thing that needs to be accounted for in order to get the proper physics of MOS devices into this description. That is, this is fine as long as you're just trying to get the low bias conductance. Namely when you take current versus voltage, as Professor Dunström showed, you have something like this. And if you're just trying to get this conductance, then as we discussed, we could basically talk about this conductance. And the conductance you have these different ways of looking at it. And so you could look at density of states, you could relate it to the number of modes as we discussed, etc. So that's all about this part. On the other hand, if you're trying to model this, then one needs to take into account the electrostatics of the problem. And I just wanted to say a few words about how this electrostatics of the problem could be included into this description. And the basic idea is the following, that you see this conductance function that I've written, it is based on a particular density of states here. And if you just look at this and you say, well, when I put a positive voltage on this, how will the current increase? At first the current will increase and then once I drop below this, the current shouldn't increase anymore. Why? Because as I said, current flows because electrons come in here and get out and nothing's coming back the other way. And once I am down here, it doesn't matter how much further down I go, there's no new states here. And of course, when you get there, you'll start picking up new ones. But usually that is way out of the range of discussion. In graphene, it's not actually. In graphene, there are states down there which actually, that's one of the problems people deal with. But usually for this discussion, we don't even, let's say, worry about this. And so once you bring it down here, basically the current should saturate. Now, from this point of view, of course, you think that it should do this. And if that were so, that would be perfect, of course. Because you see, in transistors, that's the kind of IV you really want. You want it to saturate completely. Right? Rarely does. I mean, that takes a lot of work to get it to even look like a saturate. Because it's usually doing this. Why is that in this picture? Well, that's because when I pull this down, I say that when I put a voltage, all this goes down. When I pull this down, I also pull this down somewhat. So the thing is, if I take this as my zero, and I take this as one, the point is this won't stay put at zero. It will be, say, 0.1, 0.2, whatever it is. And basically the importance of design, as Professor Lundstrom said, that the art of designing a good transistor is to design the electrostatics such that when you put one here, the amount by which this goes down is as small as possible, is influenced as little as possible by that drain voltage. And so in that model I put up there, on the right hand side, you see, I've written U as, let's look at the first term, the U sub L. And I said, let's write U sub L as 1 minus alpha QVG plus alpha times QVT. So what is this U? That's the potential by which this density of states moves when I put a voltage here. And ideally what you'd like is, it should be influenced only by the gate voltage. It's the gate that should move things up and down, not the drain. And so ideally you'd like alpha to be 0. That's what it is. And good electrostatic design essentially involves making alpha as small as possible. And that's what is getting harder and harder, of course. Because these transistors are now, it's only, as I said, 50 nanometers, 60 nanometers long. So what that means is you have to make the oxide really thin. And this side is 500 atoms, but then the oxide has to be like tens of atoms. And that's, of course, the hardest part in a way of designing these things. This effect then, you could get into our model by putting a factor here, E minus U. The idea being that this quantity that conductance, it depends on the density of states. And the density of states itself is floating down. Because whichever formula you want to use, point is basically it is depending on this density of states or the number of modes. And the thing is it's floating down. Yes, please. If you make double gate, what would be the change if you make double gate? I think at this level, this is independent of the structure. I guess the question was if it is double gate. Yeah, so double gate gives you much better control from the gate. So it would effectively make your alpha smaller. So I'd say all these are clever ways of making alpha as small as possible. So that the control is entirely from the gate and not from the drain. Right? That's it. Right, so electrostatics, of course, is much the same really. Graphene, the additional thing is that this, even with perfect electrostatics. So I always say that when you have two different things, then the way to think clearly is look at them separately. So let us say we had a transistor with perfect electrostatics. Alpha is zero. Now graphene, of course, because the density of states looks like this, even a perfect transistor with perfect electrostatics would do this. It would want to go up again. Because once you have pulled this down, it again starts building up. Because then you start conducting through this. So it won't saturate even for perfect electrostatics. And then electrostatics will do what it does furthermore, right? And one of the advantages of this way of thinking is, again, that it treats all states more or less on an equal footing, this density of states. That is, it's not like you have to separate it out into conduction band, and valence band, and because filled bands don't conduct and all that, and do it separate, none of that. I mean, it's basically all depends on density of states. So if you have density of states here, density of states here, it's all part of the same model. I mean, density of states is density of states. It really doesn't matter what it is. I mean, don't have to separate out conduction valence bands. It's all the same, okay? So that's the physics behind that U of L that I've written there. It basically captures this electrostatics. Now to that, I have a second term that I add on, which has to do again with this one point that Mark mentioned. And that is, let's say you had a perfect electrostatics. As I said, when you have multiple effects, good to think of them one at a time. So let's say you had perfect electrostatics and you think that the current versus voltage would saturate somewhere there. Well, the thing is, in practice, it would actually saturate at roughly double that value, double what you're getting, what you think you'll get. So it won't saturate here, but somewhere up there. Why is that? Well, the way I think about it is that you see, you had all these electrons here. There were all these half the lanes going forwards and half the lanes coming backwards were all filled. Once you have put this bias, as Mark mentioned, only the ones going forwards are filled. The ones coming back are not filled anymore. So if you left it that way, it would have gone, yes, you'd have only half as many electrons. And that, of course, Poisson equation wouldn't like. It would say this region is getting too positive. Now all these electrons that used to be here lost half of them. It's gone positive. And if it goes positive, of course, what it means is this thing will all settle down a little more, go down a little bit. Go down so that after you're all done with half filled state, you still have just as many. And that's the point that Professor Lansdor mentioned clearly, right? That, and that is why it will saturate not here, but a little later. And so you'll get twice as much current, really. This electrostatics, essentially. Which is captured very well by saying that the total number of electrons is just given by C ox times Vg minus Vt. And in this model, this will come from adding that second term in there. That if the number of electrons changes, so let's say I do a first when I calculate it, I find that N has become much less than N0. That would give me a negative term there, U0 times N minus N0, that would be negative. And that then would tend to, again, pull these bands down and increase the current. What reason it makes it, the calculation a little harder is, now you have to do it self-consistently. Self-consistently because, in a minute, self-consistently because given a density of states, I can find N, that's the equation on the left. On the other hand, given N, I can find U, that's the equation on the right. And those two things have to be solved together. And so you usually iterate to a solution. But this kind of self-consistency, that's part of all device simulation problems. There's always transport and then there's electrostatics. Usually transport is done with drift diffusion, electrostatics with Poisson, same here. And in small devices, you may also need to correct the Poisson equation somewhat in the sense that the actual potential tends to be a little less than what you'd think from Poisson because of these exchange and correlation corrections, etc. But that, I'd say, basically goes into choosing that U0, that interaction energy. You had a question? Please. So the current also potential to the right. How do you shift these curves downwards? These, why it shifts it down? In general, whenever you apply a positive voltage, it would shift all energy levels down. So these are all electron energies. So any electron energy that used to be here, when you put a positive voltage will end up being going down by Q times the voltage that you applied. Because it's positive, so for the electron, it takes a little less energy to get there because the potential has been lowered. And because electrons have negative charge, that is why a positive voltage lowers it. It goes the other way. So in general, I said that why did I lower everything down here because there was a positive potential on the drain. And right now, the point I'm making is you cannot put a positive voltage here without also affecting the potential around here somewhat. I mean, in ideal transistor, that's what you'd do. That you'd like to put one volt here and still have this be zero. Tight to the source. That's what you'd really like it to be. But in practice, whenever you put one, you'll get some fraction of it there. And because of that some fraction that goes there, it lowers these energy levels compared to what you think it is. Because every time there's a positive voltage, it'll go down a little bit. Now, one of the consequences of this that I often stress is that, you see, there's a lot of discussion often about how one can make a rectifier. How do you make a conductor, like a molecular conductor, that conducts one way and not the other? Very well. And of course, usually when you have this discussion, what you try to do is figure out how to design this channel, that is the molecule, so that somehow you'd conduct one way and not the other. And usually the approach is something like this. The famous Aviram Ratner rectifier, that's one that everyone is motivated by, that's inspired a lot of work. Usually the idea is that on this side, you have a dense level here. On this side, you have a level here. So a molecule that has a level a little higher than on the other side. And of course, this is something that our point channel model wouldn't capture. Because in a point channel model, you think of the point as one big thing. You don't distinguish between left and right, right? And of course, one of the effects we want to talk about after this is how to talk about the left and right and take into account variations, etc. But if you have something like this, then the argument is, that when you put a voltage in one direction, you just separate them. Whereas when you put a voltage in the other direction, you'll align them. That's how the Aviram Ratner rectifier, for example, is supposed to function. And that's how usually people think of how to make molecular rectifiers. But the point I wanted to stress here is, that even if you had a perfectly simple channel with perfectly just a single density of states describing the entire channel, you could have a rectifying characteristics simply because of the electrostatics. That's the point I'm trying to make. You see, look at the current voltage characteristics of the structure I just drew. So let us say we have arranged it so that the density of states is here, more or less lined up with mu1. And so when I pull this mu2 down, no current flows, there's no states there. When I pull it up, lots of current flows. That's it, you see? So you would actually get a current voltage characteristic. When I put pulling mu2 down is like positive drain voltage, not much current in the other direction, lots of current, that's it. Now, when you look at that, of course, you say, well, that can't be. It's a perfectly symmetric structure after all, isn't it? Because it's a point channel model. So how do you even know what's left and right? You had this channel here, and you're not distinguishing between left and right. So how did you get that? Well, the asymmetry came from the electrostatics because alpha is zero. You see, because we're assuming that the electrostatics as such, that this channel is kind of held to this one. Otherwise, you see, if there was electrostatics or symmetric, then what would have happened is when you pull this down, you'd have also been pulling that down. And then you would get a symmetric IV. And if electrostatics was opposite so that alpha was one, so everything is controlled by the right-hand side, then, of course, this would have been reversed. You could have saturated on this side and gone up on that, et cetera. So the point I'm trying to make here is that electrostatics can have a major influence on the IV of small conductors, really. And so when you're trying to design a rectifier, it doesn't necessarily have to come from the asymmetry in the channel. It can just come from the asymmetry in electrostatics, thanks, et cetera. Okay, so what I'd like to talk about next is little more in depth about what happens inside the channel. That is, so far, you see, everything we did here, it kind of stuck to the ends. That is, you have F1 in one contact, F2 in the other contact. And what is the conductance? We said, well, it's density of states. That's the density of states inside. What is T? Well, it's the amount of time it takes for an electron to transfer from left to right, right? And this is very general in the sense that we don't talk about the mechanism of transfer, necessarily, or whatever time it takes. And then we said, well, if you write the T in a certain way, you'd get these various relations. But let's talk a little bit more about what goes on inside. That is, supposing in this structure, I'm trying to plot. And for this discussion, let's say we are talking about at a certain energy, that some energy electrons come in. And as I discussed before, when you look at the density of states, you have one set of lanes that are going from left to right, and other set that are going the other way. And what we ask is, what is the occupation of these? How are they occupied inside the channel? So far, we haven't talked too much about what's going on inside the channel. Let's talk about how these lanes are actually occupied inside the channel. And we're trying to plot this as a function of this direction, which I've called z. And we're talking about at a particular energy. So this occupation we'll call f z comma e. At a given energy, how does it change with z? Now, the point is that you see these lanes. And first, let's think of a ballistic conductor. So ballistic means electrons that once they get into this lane, there's no way they can switch lanes. It just goes straight. That's it. And once you're in this lane, again, you can't switch lanes. So in that case, what you'd have is, as you plot against z, you find that one of these is occupied according to the Fermi function in this contact, whatever f1 happens to be. So if you have lots of electrons coming in from here, all states are full, then what will happen is these lanes will be completely occupied. Whereas these are, of course, coming from there. So when you look at their occupation, it's more like f2. And once it's inside the contact, then, of course, everything changes, goes to the same value inside the contact. So at a given energy, you could think of it, let's say f1 here is 1, almost 1. f2 here is, say, 0. Then what would happen is all the states going from left to right would all be occupied, while these states would be completely empty. And again, this is just pure particles, just like cars going on a highway. If this is Lafayette, that's Chicago, and something important's going on out there, so that everyone's trying to get there, you'd expect all the northbound lanes to be filled. And the southbound lanes would all be completely empty. No one's coming back. That's basically it. No more physics than that. That's exactly what you'd expect. Now, if on the other hand, you have non-ballistic transport, which means electrons can actually change lanes and come back. So now you have all kinds of construction and all going on. And there are people who change their mind and decide to come back. The question is, then how do you describe what goes on inside? How do you describe the occupation of these levels here? That's what we want to discuss. Now, this is where one expression that let me write down first, which is useful, is let us say we are trying to write the current at any point inside the channel, somewhere around here. Current means the rate at which electrons are going. Now, usually you have probably seen this, that one way to write currents is q and v. That is, if you have a channel with a certain electron density, you have a certain number of electrons per unit length going along with a certain velocity v, then if I stand here, then the current I'll see will be this density of electrons times the velocity. How do you show that? Well, usually the argument is that if I stand here, then since it's in a given time t, how many electrons would cross me? Well, whatever electrons are present in a length of v times t. Because in a given time t, everyone within this length will cross me. So that tells me how many electrons cross me. And that is equal to then, if n is the number of electrons per unit length, then n times v times t is what will cross me. So the current, that is, how many are crossing me per second is n times v. And then if you want charge, electrical current to multiply by q, of course, that's it. So from that point of view, if I'm trying to write the current, I could say, well, I've got q. What is the electron density? Well, I got half the density of states and which are occupied according to f plus. And that's this, the northbound lanes. That's my f plus. And this is the f minus. So we look at the difference between the two. And of course, the way I have defined density of states is like total density of states per unit energy. For this discussion, this needs to be per unit length. So I'll divide it by 2l. And I think I missed something. What's that? The velocity, because it's q and v. So I'll put the dvz. Now you kind of see now, I mean, that's basically what this number of modes also is. So this dvz over 2l. So from this point of view, you see immediately why the current depends on that quantity. It doesn't depend just on how many states you have, but also on how fast it moves. And that's really, again, what is defined as m over h. And the only reason you define it this way is that you see, in this discussion, there is no quantum mechanics in this. It's all classical. We're just saying that you've got some states that are moving along, just cars on a highway, nothing more. But once you bring in the wave input into this, and you say, well, the way I'm going to count my states is by fitting wavelengths into this, then what you'll find is this happens to be how many wavelengths you can fit into the cross-section and is a number, et cetera, et cetera. So that's where the quantum part comes in, right? Is when you bring all that in. But till then, it's really just this, nothing more. There's nothing more to it. Yeah, that's a very naive question. No, nothing's naive, yeah, go ahead. So we have a velocity here. And I mean, very earlier, we also had mobility. So what's the difference between two and why we're using velocity here and not the mobility? Okay, now the mobility is usually related to this drift velocity. And the way you say it is, what is the drift velocity of carriers? And the way I think of it is, I've got, again, this northbound lanes and the southbound lanes. They both have basically the same velocity. That is, you know, 65 miles per hour or whatever it is. But then when you define drift velocity, what it means is you say, well, I got 100 cars per second going this way, 100 cars going this way, but then I got only 50 going this way. So what is the average velocity? That's the drift velocity. And that can be way less. In the sense, if you had 100 here and 100 there, the drift velocity would be zero. Now for this discussion, since I'm keeping the F plus and the F minus separate, I would use just the complete speed, the actual speed limit. That's what I'd use. Not the drift velocity. Drift velocity would be, after I have done this averaging, whatever I got, I could call that VD. Then it would be drift velocity. So this is how I would write the current. And in the ballistic case, what happens is, this is V1, this is F1, and that's F2. That's it. And then this becomes M over H. And that is how we ended up with Q, M over Q over H, M, F1 minus F2. That would be a ballistic case only. Now we are trying to do the diffusive case next, actually. And in the diffusive case, what happens is, this difference is much less than F1 minus F2. And that's the important point, because this curve won't look at all like this. What you'll find is, and this is what I want to explain, that if you now look at, in the diffusive case, if you look at it, it really looks more like that. This is F plus, that's F minus. In other words, you got lots of people trying to get to Chicago, but then they gradually turn back. So of all the 100 people that start out here, only about 10 get there, you see? Other 90 kind of came back, somewhere along the way. And that's what you'd call transmission, right? And that's what, as Professor Lundstrom discussed, that's this transmission, which was like lambda over L plus lambda. That if length is zero, all 100 get there. But then if length, as length goes up, fewer and fewer actually get there. And this curve will end up looking like this. And this is what I want to discuss here. So if you have a spatially varying F, so you apply this as a concentration gradient, so it doesn't mean that you have a diffusive transport. Doesn't- Doesn't that imply that you have a diffusive transport because we have a very, you know, a spatially varying- Right, so what I'll try to explain is where this comes from. For ballistic transport, as I described, say this would look like this. Just that, for ballistic transport. But if you have diffusive transport, which in my mind means that people who start out north could turn around. You have this turning-round thing. And mean-free path is a measure of how far an electron goes before it decides to turn around. What's the average distance it goes before it turns around, basically? And what I'm going to show is that once you do that, once you take that into account, the correct solution would look like that. And that's not meant to be obvious. That's what I'll try to do in the next few minutes, actually. And then you'll see where this comes from. And I'll try to relate it to the diffusion equation. That is, what exactly is missing in the diffusion equation that this then includes in it? We'll get clear here. So why aren't the two the same in the source of brain contact? Right, right. So that is what will come out. It's not obvious. But in the ballistic case, of course, they are not the same, right? I mean, oh, oh, sorry, sorry, I got your point. Sorry, I missed that. Now, so eventually, of course, it will join up. Deep inside, it will all join up. Agreed. Once you go inside the contacts, and contacts I think of where the main highway has 10 lanes. Inside the contacts, it's like thousands of lanes. It's filled up. Now, yes? In the context, there's always a line, there's always a contact. Say this again? There is a contact, there's two lanes. I guess once you get into the region where it is many, many lanes, at that point, I think it is more or less converging very quickly. So, yeah, so actually, I really shouldn't, yeah. So if, let's say, up to here is 10 lanes, and then it suddenly became a thousand lanes, I'd say it would immediately collapse. And I'll discuss this in a minute why that had to be this way. So I really should have drawn this picture more like this. Right there it drops. Yeah, yes? I still have a question about- About M. Yeah, because then this day, the chain is drawn in some sort of shape along with M. The M is the constant. Right, so we'll talk about that in a minute. But yes, at this time, let's assume these are independent of Z, but you are right. If this is a function of Z, then you should include that. And then you try to solve the problem such that current is the same everywhere. Because in a one dimensional problem, current has to be the same anywhere you look. So usually, the differential equation you would then solve would look like dI, dz equals 0, basically. Okay, so it is convenient here then to think of a current I plus, which is q this m over h dVz over 2L, et cetera, times f plus. And then there is i minus, which is q m over h times f minus, right? And the actual current is like the difference between the two, a little bit. Now the question is that inside the channel then, what is happening is as you go along, the number of things, number of electrons going in the plus direction, that's northbound, it keeps going down, just write it here. And you could say a mean free path, you almost define the mean free path as the distance it takes, but in a given distance dz, what fraction actually get turned around? It's almost like radioactive decay if you like, dn dt equals, this is the lifetime you could say. Or here you'd say, how much did the current go down? Well, whatever current you have times dz over lambda. That could be the definition of mean free path, so you could write that. Now, of course, this isn't quite right because if you took that equation, as you know the solution would be, like radioactive decay, it would be exponential. That's the first differential equation you learn about, right? Dn. And in that case, of course, this wouldn't have looked the way I've drawn here. It would have looked like an exponential going down. But what you're missing here is that people who are going north, of course, are turning around and that's why this is going down. But then there will also be people who are going south who are turning back. That's the part that you have to include. So that will lead to an increase. So the correct equation would look something like this. And always, of course, the total current stays the same because if you're not going north, you're going south, you have turned. And so this is actually equal to d dz of i minus so that if you look at the derivative of i plus minus i minus, that's zero. That doesn't change. That's fixed. So the basic equation would look something like this. This is the differential equation, actually, then that you have to solve in order to come up with this. That's the basic thing, you see? And the important point is if you had just been solving this, you'd have gotten exponential. But when you solve this one, you'll actually get a straight line with this. Okay, so let me write the equation. So this I can write it, combine the two and write it this way. Now you can kind of see now why it's a straight line. It's because, you see, the way this works, the derivative of i plus minus i minus, that's zero. That doesn't change at all. So this is actually a constant number. That's this i. That's actually the actual current, and it's a constant. And when you have something whose slope is a constant, that means it's a straight line, basically. It's going to change the straight line, that's it. And so, when you look at this i plus, as you realize, of course, f plus is proportional to i plus, f minus is proportional to i minus. So whether you're plotting f or whether you're plotting i, it's all the same thing. It's basically the same differential equation. They're just multiplied by some constant there. So instead of this equation in terms of i's, I could have put f's here in this discussion. As long as we are not talking of complicated things like the number of modes changing with distance and things like that. As long as we're not getting into that, I could take this off and write it as equation in f. That's fine. So the point is then that these things will have a straight line slope. The question is then what is the solution to this? How would you write that? So when I'm trying to draw this, then actually, yeah, let me draw it here, I've got this z. So you have, at this end, we have got this f1. And then it goes down linearly. And at this end, I've got this f2, and it goes down. What we want to figure out is, how much does it go down? What is it here? And for this, just to simplify the algebra for the moment, let me just call this 0, call this 1. Now that actually is easier to see the answers. If I call this 0 and call that 1. Of course, you could then adjust the levels accordingly. And this value then, let me call it t, that's this transmission. That tells you if an electron came in, what's the chance it will get there. And the equations I have to satisfy as df plus dz should be this. Now how do we write down what t is? So if you think about it, what we are saying is that the amount by which it went, so let's say this length is l. And this is 1, so this is a straight line. It went down from 1 to t. So the slope, that is the df dz is like 1 minus t divided by l. Because it went from 1 to t, if I write the slope of that, that's it. And the point is that that slope should be equal to the current divided by lambda. And how much is the current? Well, that's like t. Because if I look at this end, this is 0, this is t. So the actual current, which has to be the same everywhere, that is this distance has to be the same everywhere, no matter where you are. Because the difference between the two stays fixed. So whatever it is here, it will be the same. And here it is 0. Here it is t because I know it's 0. So this must be equal to t divided by lambda. So how did I do, just to go over this again, that's the basic equation. And for this discussion, I'm just, instead of i's, I'm thinking of it as f's, etc. And then I'm saying, what is f plus minus f minus? Well, that's fixed. That's the same here, here, here, anywhere I want. And here it is t, so it's t everywhere, good. So this quantity is t divided by lambda. Derivative of i plus with respect to z, well, it went from 1 to t. So it changed by 1 minus t over a length l. So the slope is 1 minus t over l, that's it, okay? So basic point is, this is constant, so I know it has to be a straight line. And all I'm trying to figure out is this end values. That's all I'm really trying to do, okay? So how do I get t? Well, when you have these two fractions like this, A over B equals C over D. It's actually equal to A plus C over B plus D. And if I add these two, I get 1. If I add these two, I get lambda plus 7. That's it. So the t is equal to, this would be it. This is the argument I use. So this would be what the shape of, this would look like. That whatever comes in here, whatever goes out there would be. This would be a fraction of this. And what fraction? Well, it would be this lambda over l plus lambda. And it all follows from just this differential equation, really. And this is this very interesting difference because, as I said, if you didn't have the i minus, you would have got an exponential solution. But together, this couple flows, one going this way, one going this way. You have this straight line solution. Now, the next point I wanted to make is that, you see, this thing about the diffusion equation that we often use. Now, you could turn this into something that would look like a diffusion equation. And the main point is this boundary condition that you are using. See, usually when you solve a differential diffusion equation, what happens is you don't distinguish between plus and minus. And you say, well, at this end, it is fixed at 1. At this end, it's fixed at 0. And I'll draw a straight line. And so you end up drawing a straight line like that. And that's the problem, you see. If you had done it right, then you would have got all these interface resistance and everything that you are talking about. You know, the basic point I tried to make, that one important insight that wasn't there before, is that the Ohm's law, instead of looking like rho times l over a, looks like l plus lambda over a. And essentially that plus lambda is basically because, you see, if this is l, usually we want to put 1 here and 0 here. But actually, you should be putting 0 somewhere out here. I mean, you should be putting a different value here, if you take into account this fact. And putting something else here is kind of equivalent to putting a 0 out there. And what you could show is that this extra length is basically the lambda. That's it. Which is why everything looks kind of like little lambda longer. That's the point I want to make. But let me show you how you can turn this into a diffusion, something that looks like a diffusion equation. Now, one important point I want to stress is, let me just restore the picture I had here with the density of states. That usually whenever you want to talk about transport, if you're not going to make any assumptions inside, then in the contacts you can always say that it looks like a Fermi function. In the contacts, those are in equilibrium. Inside, however, it's not an equilibrium. So often you do not know what it looks like inside. And so if you want to make no assumptions, then you kind of have to formulate your transport theory with an energy grid. Namely, you have to keep track of how many are here, how many are here, how many, etc. And the way we have got this simple expression for conductance that we have been using is because we neglected any transfer of electrons from one energy to another. We said that, and this I'll explain a little further later. We'll say that let's assume that electrons go from here to there without changing energy. And that is why you can calculate what is the current through this energy, that energy, and all separately and just add them all up. And that is how we essentially got that simple answer, you see, basically. And it describes a lot of things very well. So this is what we call, I call this elastic resistor. Elastic meaning it doesn't lose any energy inside. And in practice, of course, you have to, if you want to include inelastic processes, you'd have to figure out how electrons move from one to the other and so on. And that can be important, especially at high bias. Okay, now often one assumption that people make is that, well even at high bias, let's assume that inside it is given by some Fermi function. Because the idea is that if you have lots of inelastic processes, it will try to restore a Fermi function at the end of the day. Because it tries to restore a kind of local equilibrium. And so it will try to get it back into a Fermi function. Now, the point is that if you assume that you have some form of function, if you have this f of z, e, if it looks like a Fermi function, then what you can do is, instead of, the way I've written this, everything is a function of energy. But if this is true, then you can just write all your equations in terms of a chemical potential. Because that one number then more or less describes how it varies in energy, you see, because you could always write this as, you can do this Taylor series, you can write the fz of z, e, as if it is some constant plus this del f0 del e times whatever mu z is compared to some constant value mu 0. So you could look at differences in f in terms of as if it's differences in mu. That's the point. So the net result is that this equation that I wrote here, where I worked in terms of f's, I could have worked in terms of mu's. That is, I could have said that d, this is like d mu minus d z. This is also mu plus. So we could have been working in terms of chemical potentials. And then say that, well, the current is this, the actual current would be integral dE of f, f plus. And that is where, if you wanted f plus minus f minus here. So if you wanted something like f plus minus f minus, I would write it as just minus del f0 del e times u plus minus mu minus. So the bottom line is that then when I try to write the current, let me take this off, it would look something like this. So this quantity then, it is like the number of modes, which is a function of energy. But it is integrated averaged over energy in a certain way. Like I said at the beginning, that you can always think of this conductance, happening in different channels, but then it's averaged over energy, according to this function. So you could say current is proportional to mu plus minus mu minus, but then it's multiplied by this. And this number of modes, it's averaged over energy according to this function. Remember, this del f0 del e tends to look something like this, right? Around there. So the equations then would tend to look like that. And this mu plus minus mu minus then would actually become, you could write instead of this, you could write it as current divided by qm over h. Where by m, I really mean this averaged value of m. You put something there, whatever that is. And this then would be like conductivity. So that you could turn it around and write it as i is equal to minus. Just to think of this qm over h times lambda. That I think is like what we defined earlier as conductivity. Remember, right? So if you look up there, I think I had this written out here. Sigma a over lambda plus l equals q square over h, m lambda over lambda plus l. So sigma a is equal to q square over h, m lambda. So this is what it is. So the point I'm trying to make is, I guess I should. The q is simply because I'm working in terms of this electrochemical potentials, which are like voltage times q. That's why the extra q. But otherwise, this is it. So the point I'm trying to make is, this is what is usually called looks like a diffusion equation. Diffusion equation is that current is equal to conductivity times dv. Times d mu dz, sort of, right? And the main difference then, the point I'm trying to make here, is that usually you do not distinguish between mu plus and mu minus. That's the point I'm trying to make. Yes, please. No, so this tells you whether they are occupied or not, right? So, okay, yeah, let me come to that in a minute. Let's get to that in a minute, right? I'll get to that. But here, the main point I was trying to make is that either you can do a more detailed transport theory like you do in using Boltzmann equation where you keep track of electrons with all different energies and how you move around, etc. But what is commonly done, and this is not necessarily a good assumption, especially at high fields, is to assume that you can describe things locally in terms of a chemical potential. But even then, if you want to include ballistic transport, you should have two different electrochemical potentials. One for the positive carriers, one for the negative carriers. Because just as, as you know, in ordinary semiconductor devices, conduction band and valence band often have two different quasi-firmly levels. Ballistic transport means northbound and southbound are kind of way out of equilibrium. And of course, in general, as we discussed, one of them tends to be in equilibrium with this and the other wants to be in equilibrium with that. And to keep that, you need to have two different chemical potentials in the two channels for plus and minus. And so the basic equation could be written something like this. But if we ignore this difference, then it will look like an ordinary usual diffusion equation, i equals sigma d mu dz, it'll look kind of like that. But when you solve this, of course, what will happen is, as I mentioned, that if you're solving a diffusion equation, you would have put new chemical potential, let's say you put zero here and you'd put q times v on this side for a given voltage. So when you, that is how you normally try to solve it. But the important thing is that this is the boundary condition on mu plus. Whereas this is the boundary condition on mu minus. And you have to be careful about that, it's not the same mu, really. Because if you put that on the same mu, then of course, you'll just get a straight line like this. And you'll miss this extra resistance at the ends. Completely. We'll miss that roll under part. Please. If you have inelastic transport occurring, then do you mean the transport is diffusive and non-elastic? It can be diffusive without being inelastic. What I mean by that is, elastic has to do with whether energy changes. And here I'm assuming that for the moment that energy is fixed. So it's at that energy. But then it doesn't have to be ballistic in the sense it could turn around. It could turn around and come back, go in a different direction, etc. But energy is fixed, let's say. And inelastic means that it is actually changing energies as well. So it can be inelastic and ballistic or diffusive. Right, right. So far what is that there is something that's decelerized. Say this again. So far what you presented is something that's denialized and coming to control potential. And the way it starts is that it's really a fairly direct distribution. But there's really no proof that was the dynamics of the distribution function that was governed by some other equation, Boltzmann or master equation. Right, basically everything I'm saying would be based on Boltzmann equation really. So all this is kind of trying to understand what Boltzmann equation would say. So if you solve the Boltzmann equation carefully, but we're trying to simplify certain things to get insight into it. Really, right? You can't get it straight, you can't get it straight. Linearized curve, why would you do that? No, I think, oh, no. But this curve though, I didn't linearize it originally. When I drew this, remember, I drew this as f plus z comma e. So at a given energy, I said it looks like this, right? So that every energy is separate, but the assumption there was that it was all elastic transport, so that I can think of it one energy at a time, right? So the basic assumption here is that this is like a single energy, okay? So every energy is parallel. Every energy channel is parallel, and that is what that equation up there, which is very widely, can be applied to many different conductors, but not always. The top one, where the way we arrived at that one was to say, let's think of one channel, see what current you get, then add them all up, right? That's what the top equation is, right? That's what the top, and okay. So I'll try to say a few more words about the assumptions and the issues when I wind up in the next few minutes, right? Let's do that, but right. So the main point here then is that this distinction needs to be made between positive going states and negative going states in order to get this effect of the ballistic channel. And usually people think of diffusion and diffusive limit and ballistic limit as just being totally different things. But as you can see, it's not totally different, really. It's the same equation, but if you use the boundary conditions correctly, that one of them is on mu plus and the other is on mu minus, you'll get much of this, actually, you'll get all of this, okay? Now, a few points then, in the last few minutes, what I want to stress is that a lot of what I've described to you then is based on what you might call an elastic resistor model. So what do I mean by that? That is, the picture we have is that electrons go from here to there at a given energy. And how much current flows through a channel depends on F1 minus F2. And then you add them all up, that's how you get that. That's the basic picture, you see? Now, this elastic resistor, of course, the reason you do not, this is not the way you normally think even, is because 20 years ago, if we had talked about elastic resistors, people would have said, well, that makes no sense at all. You can't have an elastic resistor. Why is that? Well, because people would say that anytime you have a resistance, there's a heat that goes with it, there's I squared R. If it's elastic, that means it's not losing any energy. So where's that I squared R loss? You can't, people would say, well, elastic resistance, that's just an oxymoron. You can't, it means nothing. I mean, you can't have an elastic resistor, that's what people would say. What is now pretty clear, actually, is that in small devices, they actually function almost like elastic resistors. In the sense that an electron does go from here to there without losing any energy. And then what happens it? It loses all this extra energy in the contact and then goes out. So all the heating is in the contact. So where is the I squared R? That's all in the contacts. And a little bit at this end also, that the electron that went away gets filled up from here. So there is some heating here and some heating there. And one can talk about how much at either end. And the fact is that in small devices that is exactly what is observed. That people say that the only reason many of these carbon nanotubes can carry the amount of current that is passed through them without burning up is because the heating is not in the nanotube itself. Because the nanotubes are very small thing. If that were, all that heat were happening there, it would have burned up. And they said the reason it's not burning up is because the heating is actually in the contacts, which are of course big things which can get rid of it. So in a practical sense, of course, one could say that very small devices are really very much like elastic resistors or more, right? And with elastic resistors what is nice is it conceptually separates these two very complicated, two distinct processes. One is process by which electron at the same energy kind of goes through this. That's what I'd call the pure mechanics. Classical mechanics or wave mechanics, whatever it is, pure mechanics. And then there is this matter of giving up heat, which is like thermodynamics. And usually what makes transport such a difficult topic is because these two things are all mixed up. These are two very separate branches of physics that develop separately. You see there was, first came mechanics, that's what Newton's laws and all that. That came from observing how the planets move in the sky according to the, no friction. Much later, hundreds of years later, then there came the heat engines and people started trying to understand how heat is dissipated. That's when this concept of an equilibrium distribution, KT, the whole idea of KT, and all that comes in gradually. And then reconciling these two, that almost took all of 19th century. Culminating in the Boltzmann equation. Boltzmann equation took Newton's law and did something to it, based on which they could describe a lot of these non-equilibrium phenomena. And what he did actually was subtle enough that people still argue about it. In his time, of course, they gave him a very hard time. But even now, 100 years later, they still argue about exactly what he did, what was involved, etc. And the reason it's so subtle is again because these two things are mixed up. Mechanics in there. And what makes these small devices, conceptually a whole lot simpler than big devices, is that it's actually separated. The mechanics and the thermodynamics. These are kind of two separate things. And so, because if you ask me that this resistance that we are calculating, if you use any of these equations, where exactly is the heat dissipated? And the answer is the dissipation is here. It's all in the context. Nothing's getting dissipated here in this discussion. And so you can ask them, where is the resistance? Well, that is why this picture is kind of useful. Because one way to say it is, well, the resistance is really not in the channel. It's all in the context. Well, that doesn't make sense because we know that after all, if you took the same channel and made a big hole inside here, the current would go down. If you cut out this part, of course, the current would go down. So it makes no sense to say that all the resistance is at the end. Not really. There's obviously a big resistance right there in this case. So how do you know where the resistance is? Well, look at where the occupation drops. So for example, if you had a ballistic channel, we said it would look like this. If you had a big hole someplace, it would drop suddenly right there. So by looking at how this chemical, how this occupation of states changes, you can tell where the resistance is in a way. And that makes sense, again, and again, no rocket science here. Again, it's just like particles that if on the interstate you see a lot of cars somewhere, you know there's a construction zone in front. I mean, that's why there is this, if there is a region here where there's a construction, that's where you'll have lots of cars before it, very few after it. No question. That basically tells you, that is what you then could identify as where the voltage drop is occurring and so on. Although the conceptual point is that that has not associated with any dissipation though, nothing's getting hot out there at all in this picture. Because this scattering is all elastic. Electrons are not losing any energy, nothing got heated necessarily. All the heating is at the end. So there are very important conceptual issues involved with all this, right? And they say this model of this elastic resistor that was proposed by Landauer way back, but no one used it very much till all these new, till all these experiments on small devices came along in the 80s and 90s. Where for small devices, as I said, this model is very relevant. It is basically what describes the physics very well. What is interesting is that it even gives you a lot of insight into big devices in the sense that if you look at that expression for conductivity I have there, sigma equals q squared density of states times the diffusion coefficient. As I said, that is the standard expression for conductivity you will see derived. It's just that usually it is derived from Boltzmann equation or from Kubo formalism with a lot more work. I mean, but finally, because somewhere in there, there is this assumption of elastic scattering, or certain other related issues that assumptions are made. That's what gets you there. So the basic expression is really just like, just this, right? So the main concept is really that of this elastic resistor, which really helps you understand at the basic level what is involved in conduction. And the simplest thing I want to say that when you think of current as being driven by electric fields, it's not even clear why all these electrons aren't moving, really. And then you say, well, no, no, field bands can't conduct. Well, why can't they conduct? But when you think of it from an elastic resistor point of view, it's pretty clear why this can't conduct. Sure, I mean, after all, here are these states, this wants to fill it up, that wants to empty it, and that's why currents flow. Here, everyone keeps it filled, that's why it doesn't flow. That's it, yeah? Yeah, the reason why you can't understand that is because you left a term in the form that allows for acceleration. Which is that? In a Boltzmann transport, we have de-activation, that allows for acceleration, to fill states, right? No, I'd say that it is basic equation is really D mu, D z. You see this electrochemical potential, often people separate it out into a part that has to do with the electrostatic potential U. And a part that is, people call the electrochemical potential. What I mean by that is, yeah, this is something we can talk more about after the break. So, supposing you have a medium in which, let's say the density, you have an electric field. Then as you know, you often draw the EC. This is the band edge, you draw it this way. And let's say there's a chemical potential like this. Now, what I'd say is that basically the current is driven by D mu, D z. That's it. Now, you could then break it up and say that, well, mu is equal to EC plus mu c, where mu c is that much. And then when you have D mu, D z, you could call it, this is D EC, plus D mu c, D z. And then say that this is the electric field, and the electric field gives you a drift term, and this gives you a diffusion term, and they cancel and all that. But that's just a complicated way of saying this. That's why I'm concerned. So I'd say just as heat is driven by temperature gradient, current is driven by a gradient in the chemical potential. That is it, not electric fields, because electric fields, if you look inside a solid. There's all kinds of electric fields even at equilibrium. How do you know which ones to ignore and which ones to keep? Especially now when you get to a small device, you're talking about a hydrogen molecule. Look at the electric fields around the hydrogen atom, enormous fields. If fields drove anything, I mean, they would be a disaster, right? And if you think of p-n junctions at equilibrium, enormous electric fields. But finally, my point is what really drives currents is really the occupation. D mu d z, or actually it is really D f d z. I mean, basic point is on one side states are occupied, on one side it's not. And that's why current flows. Because this tries to fill it up, that tries to empty it. And this is the important thing that gets obscured when you start from big conductors, and try to rationalize it. But once you get to small conductors, it's really very clear. I mean, this is the point I'm really trying to make. That sure, this model doesn't capture everything, but it captures a lot. And once you understand this, then you can go on to more sophisticated things much more easily. Whereas when you think of big conductors, what happens is even the simple things get very complicated. Even, so in a way, today's lectures were really about trying to get the simple things straight, clear away. And yeah, please do feel free to ask more questions. And I guess we'll have the session at 3.30. Please feel free to raise questions. And if there are things you wanted to pass on, ask me before that, you could write it or give it to me. And then I'll try to structure the discussion a little bit at 3.30. Thank you. I had a very nice question. Sure. So does he actually calculate the current resistance? He calculated the response from the 15th. So now the concept you mentioned, your perspective seems very consistent. But I cannot be concerned with J equal to sigma E. With J equal to sigma E. Yeah, because, yeah, let's discuss this more. In my mind, the correct equation is J equals sigma del mu. That is the real equation. See, del mu or del chemical quasi-formulable, whatever you call it. That really should be the driving. That's it. And what you can show is that if you could then break this up into two parts. One part that is an electric field and the other part which is grad of the electron density. And if you have a material in which the density of states is changing or the effective mass is changing, you'll get even more terms. Things that will depend on the gradient of the mass, things that will depend on gradient of all kinds of things. But if you don't know the answer and you want to start, this is where you should start. Gradient of the mean is coming because of the potential. I'm trying to put out something. The chemical potential, however, it could come. You could, for example, just have. Usually, because they have put a voltage. But in principle, you could have shining light here, for example. You could have created it in any way you want. But finally, it is that. And I'd say, actually, it is more like gradient of the occupation, F. That's important. Because to go from F to mu, you have to assume that something is, it is almost a Fermi function. There will always be the del F0 del E in front.