nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L2.2: Semiclassical Model
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[Slide 1] Welcome back to Unit Two of our course. This is lecture two. 

[Slide 2] Now as we discussed in the introduction, what we want to talk about is this process of inflow and outflow from the contacts into this channel, and N is the number of electrons in the channel, and you are considering a range of energies here with a total number of states given by D0, and there is this outflow that is whatever electrons are in here, they flow out at a certain rate, and the outflow can be written as nu1 times N and nu2 times N, and the inflow depends on how many states you have here. That's D0, and then there's this constant, S1 and S2, and we can write an equation for DN/Dt, the rate, at which the electrons change inside the channel in terms of this inflow and outflow. And one physical picture we want to carry in our mind that's very clear is that, you see, you have this N, and DNDt depends on current coming in and current coming in here, and you can either think in terms of number of electrons per second when you talk of current. In which case, you don't need the q. On the other hand if you'd like to think in terms of coulombs per second, that's amperes, then you should multiply by the charge carried by each electron, the q. Okay? So DN/Dt, what's written here is just the flow per second. If you want the actual current, you should multiply by q. So this term is the flow from the left, S1 D0 coming in. That's the positive. Nu1 N going out, that's the negative. S2 D0 coming in, Nu2N going out. Positive, negative. Now, at steady state, what happens is there is no further change in the number of electrons. So in the beginning, when you turn on things, of course, there'll be a period of transient flow, where things will come in, and the number of electrons will change. But after some time, everything will settle down, and there'll be a steady flow, and that's, in general, what we'll be talking about in this course, the steady-state current, and under these conditions, that's 0. What that means is I1 plus I2 is equal to 0. What comes in from here and what comes in from here, and that's kind of, if you have taken a course in circuits, what you might have called Kirchhoff's law. That if there's a node here, then the total amount of current coming in adds up to 0. So what that means if there's 1 milliamp coming in here, there must be 1 milliamp going out. So that this will be 1. This will be minus 1, and I add the two, I get 0. Now since this is equal to 0, I can solve for the number of electrons inside the channel, under steady-state conditions. So N times Nu1 plus Nu2, N times Nu1 plus Nu2 is equal to D0 times S1 plus S2. So I've kind of divided it around and written it as N over D0 is S1 plus S2 over Nu1 plus Nu2, and as I said, if you multiply these two, that's equal to those two. That's what this equation tells us. So this is then the steady state. Now we want actual current, the steady-state current. So what we could do is look either at this quantity or at this quantity but then substitute from N from here, from the steady-state value. That's what we'll be doing next. 

[Slide 3] So what we do is we write the current as this S1 D0 minus Nu1 N. S1 D0 minus Nu1 N, and I've pulled the D0 out of it. So it looks like this. And then for N over D0, I can put in the steady-state value. So that's what we'll do next. For N over D0, I put in the steady-state value. Now we can do a little algebra and, you see, you have the Nu1 plus Nu2, but when you bring this over to the numerator, the Nu1 S1 will cancel that Nu1 S1, and you will have the Nu2 S1 minus Nu1 S2. That's that one. Okay, so then you have this expression for I1, and as I said, this is the flow per second. If you want coulombs per second, you multiply by q. Now you can put this together, I1 is equal to this D0 times that quantity times q, and that's equal to minus I2, because that steady state, as we discussed, I1 plus I2 is equal to 0. 

[Slide 4] Now to proceed further, we need to look at one additional thing, and that is that you see the way this looks, the Nu that determines the outflow and the S that determines the inflow seem like independent things, but actually those are connected. There's a very basic connection between them, because if you have a good contact, electrons will flow out easily, but they'll also come in easily. Okay, so those two are kind of related things, and the way you can discover the connection between them is by saying, well, let's think of a special case where you have only one contact. Let's say we don't even have the second contact. Then what would happen is N over D0 at steady state should be equal to S1 over Nu1, because there is no S2 and no Nu2. So it's just this. But if you have a channel connected just to a contact, then the basic laws of equilibrium stat mech says that they will have to come to equilibrium, and if they come to equilibrium, they should be described by the same Fermi function. So in other words, inside this contact, the ratio of occupied states to the total number of states is given by the Fermi function. It tells you what fraction of states are occupied, and since we have only one contact, it's in equilibrium with the channel, and so under those conditions, the ratio must be equal to whatever it is inside the contact, and hence, S1 must be equal to Nu1 times F1. So note that in general we have two contacts, but in trying to discover the connection between them, we conceptually say, "Well, let's just disconnect the other contact and figure out what relation must exist in order to make sure that we are in compliance with, I guess, equilibrium stat mech, which says that when a channel is in contact with just one reservoir, then they will come to equilibrium and have the same Fermi function, have the same fraction of states that are occupied." Similarly, you could make the same argument with the second contact and get S2 equals Nu2 times F2. Now if we substitute those in here, instead of S1 I put in Nu1 F1, and instead of S2, I put in N2 F2, and I get this expression for the current. Note that this is Nu1, Nu2, appears on both terms. 

[Slide 5] So what we can do next is pull the Nu1, Nu2 out, and then have an F1 minus F2. So finally what we are getting is, the current is, proportional to density of states. This is Nu1, Nu2 over Nu1 plus Nu2 times F1 minus F2, and if you remember our discussions from part A, we talked about an inelastic resistor, you can always read the current as proportional to the difference between F1 minus F2. Indeed, you can see the significance of this quantity here if you write it in this form. You see, with a little algebra you could write it as 1 divided by that. So what's 1 over Nu1? Remember, Nu1 is like, let's say the rate at which electrons escape into the contact. Let's say 10 per second. So what that means is 1 over Nu1, that's like 1/10 of a second, that's the time an electron takes to cross that interface because it's like 10 per second, which means as if every electron takes a 10th of a second. Similarly, this is like another 10th of a second. So how much time does electron take to get through two interfaces? That's like a 10th of a second and a 10th of a second, and so the overall time is this total. That's what you could view this as, and if you remember from part A, conductance was proportional to this q squared D over t, t being the time it takes to get from one contact to another. So this result, of course, we didn't invoke any of this. We did it from a totally different point of view just to help us connect with the quantum theory that we'll be getting into, but you could connect it to whatever we have done before. 

[Slide 6] Now, the other thing I should point out before we move on is that we did this with just a small range of energy, dE, with a total number of states, D0, but you could just add these all up. That's the beauty of this elastic resistor, because all the energies conduct independently. So once you have written out the current in this energy range, you could just add them up. That is, instead of D0, I've put in dE D(E), and then you can just go ahead and integrate it, and then you have a complete expression for the current, and this is what you would call, I guess, like, the conductance function in part A. So that's what you could do to write the current. You could even write the number of electrons in the channel, same way. That is, instead of D0, put in D(E), and then integrate over energy. 

[Slide 7] Okay, anyway, so with that then we are ready to move on to what we really want to talk about, and that's the quantum model. The purpose of the semi classical model is to give us a physical picture, which we can use to understand and interpret the quantum results. The basic picture is consider this small range of energies. You have an outflow and an inflow and from which you can figure out these currents, and at steady state, there is no dN/dt. The currents kind of add up to 0. Thank you.