nanoHUB U Fundamentals of Nanoelectronics: Basic Concepts/Lecture 3.4: Current fromn QFL's
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[Slide 1] Welcome back to Unit 3 of our course and this is fourth lecture. 

[Slide 2] Now as you may recall in the second lecture we tried to show that if you use the diffusion equation and use this new set of boundary conditions you'll get results that we have been discussing in this course that includes the ballistic resistance. What we did in the last lecture is we introduced this concept of quasi-Fermi levels the idea that right-moving and left-moving carriers have different electrochemical potentials or quasi-Fermi levels and we just argued what the correct boundary conditions are for these quasi-Fermi levels and what we want to do, one of the things in this lecture is to connect up the two. What I'll show is this implies that. From this you can deduce that. Now in order to get from here to here we need a relationship which is very important in its own right. That's this relation that relates the current to the separation of the quasi-Fermi levels and this is ballistic conductance in between. Now often there's some misunderstanding because when you see the ballistic conductance you think oh this only applies to ballistic transport but not really. This is a general relation that applies to ballistic or diffusive transport and we'll come back to it later. And if you take this you could turn it around and write it in this form. So this follows algebraically from here and what it says is that the separation of quasi-Fermi levels is proportional to the current. That is, if there's no current the two quasi-Fermi levels will collapse into one and that of course makes sense because that tells you how well the right-moving states are filled. This tells you how well the left-moving states are filled and if there's no current then they're all equally filled and the current flow moves right-moving states are more filled than the left-moving states and this is this relation that expresses it quantitatively, relates the separation of quasi-Fermi levels to the current that is flowing and this is valid regardless of whether it's ballistic transport or diffusive transport. So we'll try to explain where this comes from in the next slide but before I go on I just wanted to show that if you accept this then these quasi-Fermi level boundary conditions give you the boundary conditions for mu quite straight forwardly. It just takes a couple of algebraic steps. So let me just show that and then we can get on to discussing where this came from. So what you do is you first note that we call this electrochemical potential that appears here is actually the average of the two quasi-Fermi levels. That is mu plus, this is mu minus. Mu is the average of the two. Now with a little algebra I can just take this mu plus, plus mu minus over 2 and write it as mu plus minus the difference over 2. So this you can check it out. It's just a little straightforward algebra. This can be written like this and of course once you have this you could take this quasi-Fermi level difference and express it in terms of the current that makes user relation we just talked about which we haven't yet proved but I will come to that. But if you accept it then you say that could be written in this the form. And now you see we immediately have the boundary condition we are looking for. Why? Because if you look at z equals zero the quasi-Fermi level boundary condition tells us that mu plus at z equals zero must be mu1 so that mu plus becomes mu1 and this mu, that's this mu z equals to zero and minus qIRB over 2 from there. So this boundary condition automatically gives you that. Just one line. Now the other one, to get the other one what you do is take mu and instead of writing it as mu plus minus this you write it as mu minus plus this and that's also again something you can do, straightforward algebra. From here you could write it this way and then this difference of quasi-Fermi levels you could replace in terms of the current, again make use of this relation and once you're here the other boundary conditions follows automatically. Why? Because at z equals L mu minus is mu2, according to the second quasi-Fermi level boundary condition. So mu minus is mu2 and then the plus q I RB over 2. So the main point is what you obtained in the last lecture gives these boundary conditions that we discussed in the earlier lecture as long as you accept this relation. So what we want to do next is talk about where this comes from. 

[Slide 3] Now this relation in order to obtain this we first write the current of electrons that is flowing to the right. See in general when we look here there's a flow to the right and then there's a flow to the left and the actual current is the difference between the two but let's just consider the flow to the right. So that's this I plus so how much is the flow to the right? Well, there's the q which is the charge on an electron so what is out here tells you the number of electrons per second and if you multiply it by the charge on an electron then you get the amount of charge per second. So what is this quantity? Well it's got two parts to it. One is you look at the total number of right-moving electrons because we're only interested in the right-moving current. That is we're trying to figure out how many cars are on the northbound lanes. You're not worried about the southbound just the right, northbound lanes. Well what's the number? Well first look at how many states are there and then look at whether they're occupied or empty. So this f plus is kind of like a Fermi function but the thing is it just describes the occupation of the right-moving states and these are how many states there are which is the density of states times the energy range of interest. Why did I divide by two? Because again I'm only looking at the right-moving states so if there are a total of eight lanes, there are only four lanes that are northbound. The other four are southbound so dE divided by 2 times F. So that's the number of right-moving electrons and then I multiply it by the amount of time that each electron spends on the highway going from one end to the other. That time is L divided by the average velocity of right-moving electrons. Now why is there an average? Well, that's what we have discussed in Unit 1 of the course. That's because there's actually some electrons going straight, some electrons going at an angle, some electrons going at another angle and you need an average which in the two-dimensional case I think amounts to 2 over pi, some small number but basically, I mean a number closed to one. But the point is this is just the average of all the electrons that are headed to the right. Of course the ones that are headed to the left have a negative component of the velocity and those you are not considering, only those going to the right. So that's this L over u-bar here. Now why is the number times the time? Why does that give you the current? That's actually also not obvious unless you have seen this before. And here they often use the analogy that if you had a graduate program where you have say a hundred students graduate every year. Then what you can convince yourself is that if you ask how many students are there in the graduate program, just steady state? Every year a hundred come in and hundred leave. The question is how many students in your program? Well, the answer is let's say every student say spends five years in the program then the steady state number of students is 500. So if a hundred students per year come in and leave each student spends five years then the number of students is 500. So similarly here you want the flow of electrons. It's equal to the steady state number of electrons on the highway, that's like the graduate program, times the time each electron spends on the highway which is the length divided by velocity. So this then with a little algebra we could just rearrange it a little bit and write it in this form. Du bar over 2L times f so I'm just straightforward algebra getting from here to here and then we note that this quantity is what in Unit 1 we had defined as the modes as the number of modes, M. See that's this M so you have q over h times M times f plus so that then tells you the total current on the states that are moving to the right. 

[Slide 4] Now if you want the net current then what you have to do is subtract off the current flowing to the left and that will be exactly the same except that now the occupation instead of being f plus will be f minus. So overall the current in the energy range dE is given by this quantity. Now how do you get from occupation to quasi-Fermi level? Well, that's where we use this general Taylor Series Expansion idea that we have used many times in this course. That is you assume this Fermi function can be written, that this occupation can be written in the form of Fermi function and then use the Taylor series and so f plus minus f minus becomes this derivative of f then times mu plus minus mu minus and if this is the current over an energy dE and if you want the total current you have to integrate this over energy and that then gives you this current is q over h times this part integrated over energy. That's what you call M zero times mu plus minus mu minus and the point I want to make is this again is the ballistic conductance basically and so you then have the net current is the ballistic conductance times the separation of quasi-Fermi levels and note that this again holds at any point. That is if you looked here the current would be that. If you look somewhere else the current would be the same. So for example, the continuity equation requires that the current be the same everywhere. That immediately tells you that the separation of quasi-Fermi levels must also be the same everywhere and as you can see this is a straight line. That's a straight line. They're parallel. There's a difference but the difference is the same everywhere. And so the current is the same and at any point the current is related to the separation by this relation. 

[Slide 5] Now how does this go with the conductance? You see usually what we write is conductance time mu1 minus mu2. And this mu1 minus mu2 is of course the separation between the two contacts. Note that the contacts are separated by that much but the quasi-Fermi levels are separated by an amount that is less. So current you can write either as the conductance times the total separation or you can write it as the ballistic conductance times the quasi-Fermi level separation. This ballistic conductance is bigger than the actual conductance but then the quasi-Fermi level separation is smaller than this overall separation between the two contacts. So you could connect the two and say the quasi-Fermi level separation must be equal to the separation between the two contacts times G0 over GB and G0 over GB you could write as lambda divided by L plus lambda where lambda is the mean-free path. So the point to note is if you have a ballistic conductor that means L is much less than the mean-free path so you could drop the L and then the quasi-Fermi levels separation is exactly equal to the applied voltage, the separation between the contacts. On the other hand if you had a diffusive conductor then L is big and so the quasi-Fermi level separation is much smaller than the applied potential difference. In fact that's why in diffusive conductors people tend to forget that there is a quasi-Fermi level separation. They tend to say that there's only one mu. You don't need to worry about mu plus and mu minus. Forget about that one but as I say in a ballistic conductor that can be as big as the actual applied potential. So that then completes our discussion of this boundary condition and what we'll do next 

[Slide 6] in the next lecture let's talk about this Landauer formulas which were historically very important in developing this whole point of view and I want to show you how it connects up to what we have discussed. So far we have been discussing a channel with a continous scattering processes everywhere. That is you have cars that are continually turning around as they go along the highway. For if Landauer formula we'll consider a slightly different situation, you see one where the channel is largely ballistic but there's a localized scatterer someplace. You know something like a little construction zone or a toll booth in the middle of the highway and that helps clarify certain things. Anyway so that's what we'll do in the next lecture. Thank you.