nanoHUB U Fundamentals of Nanoelectronics: Basic Concepts/Lecture 3.2: A New Boundary Condition
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[Slide 1] Welcome back to Unit 3 of our course. And this is the second lecture. 

[Slide 2] Now, as I explained in our introductory lecture, our purpose here is to show that, when you look at the resistance, there is a part that's proportional to the length. And a part that's constant. And what we'd like to argue is that the part that's proportional to the length is associated with the channel. Whereas, the ballistic part is actually associated with the two interfaces. And the key point I tried to stress in the introductory lecture is that, if you want to locate where the resistance is, it's very important not to think of the heat associated with the resistance. That won't get you anywhere. What you want to do is follow the voltage drop. That is, you ask the question that when a current flows, how does the voltage drop? And a very important part of answering that question is also recognizing what voltage means. And I argue that you should be looking at the electrochemical potential and not the electrostatic potential. So how do you locate this electrochemical potential? In the sense, how does it drop? How does it vary across the device? Well, you start by noting that, in the two contacts, it has these two different values. Mu1 and mu2. And because you have put a negative voltage on the source relative to the drain, this side is higher up. And they are separated by the applied voltage. And the question is, how does the electrochemical potential vary spatially inside the channel? Now, what you could do is start from this diffusion equation. Okay, and we discussed this in the introductory lecture. That it's actually the drift-diffusion equation really. It includes both drift and diffusion. But this is the basic equation which says that current is proportional to the slope of the electrochemical potential. Now, there's one more equation actually you need in order to be able to solve this. And that's what's called the continuity equation. So what's this continuity equation? What it says is that in a one-dimensional conductor like this, the current has to be the same everywhere, at steady state. And you can kind of see why. That is, you have this channel, electrons are flowing. And let's say I look around here. Then you have, say, 100 electrons flowing per second. Well, if I look at a different point, you still must have 100 electrons per second flowing. Because, if they were different, let's say you had only 50 coming in from this end and 100 leaving at that end. Well then, inside the number of electrons would be going down with time. Because only 50 come in and 100 leave every second. And if it keeps changing with time, that's not steady state. So at steady state what must happen is, whatever comes in per second must be leaving per second. And hence, dI/dz must be zero. So that's what's generally called a continuity equation. Now, this tells you then that the current has to be a constant spatially. And if the current is a constant, then, as long as what's out front is constant, it means the slope of the electrochemical potential must also be a constant. Now, the curve with slope as a constant is a straight line. Straight line has constant slope. So what that means is that the electrochemical potential must vary like so across the channel. So what's the current? Well, we can find the slope.  00:04:08,046 --> 00:04:13,716 You know, it goes from mu1 to mu2 over a distance L. And it's a straight line. So the slope is just mu1 minus mu2 divided by L. Of course, it's going down, which means the slope is negative. So that's why minus dmu/dz is mu1 minus mu2 divided by L. So I could use this to calculate the current now. So I could put that back in there. For minus dmu/dz, I put in this value. Mu1 minus mu2 divided by L. And so we have an expression for the current. Now, what you'll notice, of course, is that this is not what we got in the earlier part of this course. What we have been discussing for the last couple of units, actually, is that the current is really given by, not this in the denominator, not L, but L plus the mean-free path. In other words, if you made the device very small, if L were getting really small, it's not like the current would blow up, you know, go get very high. Once L is very small, you just drop the L here. And you'd have a certain constant value of the current for a given voltage. So this is, of course, the result we know. That's what we have been discussing. And the point is what we just did is not getting us this. It's getting us something else, without this plus lambda part. So your first reaction would be, well, that makes sense. We are talking about ballistic transport, and we use the diffusion equation. That can't be right. Basically, you're using the wrong equation. You should be looking for a different equation. Now, the point is what I'd like to argue is, that's not the case really. The equation is fine. The boundary condition is wrong. What we need to do is change the boundary condition. That is, what happens at these interfaces. 

[Slide 3] So what we'd like to argue is, that the boundary condition we used is that the electrochemical potential at z equal to zero must be mu1. And that z equals L must be mu2. And then we drew a straight line in between, as required by the diffusion equation and the continuity equation. Now, what we'd like to argue is that the correct boundary condition is the electrochemical potential at z equals zero is not mu1, but a little less than that, by an amount that is proportional to the current. And at z equals L, it isn't mu2, but a little more than that. Again, by an amount proportional to the current. So when we draw this line inside the channel, the point is it's still a straight line. Because the straight line is required by the diffusion equation. So it's still a straight line. But it doesn't start at mu1 and end at mu2. It starts a little below, and how much below depends on the current. And it ends a little above, how much above depends, again, on the current. See. So this is the new boundary condition that we are arguing we should be using. And once we use that, what I'd like to show you is you do get the right answers. 

[Slide 4] That is, if you now calculate the slope again, this minus dmu/dz. Previously, we took mu1 minus mu2 and divided by L. That's what we had before. But now you see you start a little below mu1 and end a little above mu2. And so the slope is also a little less. How much less? Well, by qIRB because mu1 is less by half of that. Mu2 is increased by half of that. So when I look at the difference, there is minus q I RB that appears here. So what I can do next is take this and substitute it back into our current equation.  00:08:22,056 --> 00:08:25,796 So when I put that in, I get what I had before. But in addition, it's not just mu1 minus mu2, but mu1 minus mu2 minus this qIRB. RB being the ballistic resistance. So now you see, well, we are trying to calculate current. We've got current on this side. But I also have a current on the other side. Because, of course, these discontinuities are proportional to current. That's why we have that on this side. But what we could do is take the current over to the other side and just solve for the current. And if you do that, you could, you know, bringing that term over, you'll get something like this. I times 1 plus something is equal to that. And then, if you make use of this identity that we had before, that sigma A is GB lambda. And you see, you can cancel the lambda from there. I mean, rather, replace this with lambda, and then you'll get the correct answer. So it's a little bit of algebra, but it's straightforward. You could check this out and the point is this gets you just the right answer. So the point now that I'm trying to go show is that, if you modify the boundary conditions, as I've shown here, you'll get the right answer, the one we have been talking about, out of the diffusion equation. Equation doesn't need to be changed. Just the boundary condition that needs to be changed. See. And this is the correct new boundary condition. Okay. 

[Slide 5] Now, how do we interpret this boundary condition? Well, what we are saying is that the electrochemical potential at this end is not mu1, but a little less, by the amount proportional to the current. What that means is there is a potential drop here. And that is proportional to the current. And that, of course, corresponds to having a resistance there. Because in any series circuit of resistors, when a current flows, the drop is current times the resistance. So if you have a drop of I times RB over 2, it means there's a resistance RB over 2. Similarly, there's a drop here, and that's another RB over 2. See. So that's how you could pictorially represent this boundary condition. So what we have shown you is that this boundary condition gets you to the right answer. But what you might ask is, well, where did that boundary condition come from? I can see that it gets me the right answer, but you just put it in there. How can you justify this? Well, that's what we'll do in the next lecture. 

[Slide 6] Because this requires us to bring in this concept of Quasi-Fermi levels, the idea that left moving carriers have one Quasi-Fermi level and right moving carriers have another Quasi-Fermi level. But I'm getting ahead of myself. That's for the next lecture. Thank you.