nanoHUB U Fundamentals of Nanoelectronics: Basic Concepts/Lecture 4.4: One-level Device
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[Slide 1] Welcome back to Unit 4 of our course. This is the fourth lecture. 

[Slide 2] Now, just to remind you what you have done so far. Just the last two lectures, is considered a device with a continuous distribution of levels, and we re-introduced this heat current in the last lecture, and we then went to low bias, low temperature difference, this linear response regime, and had this linear equations of thermal electricity. The charge current and the heat current in terms of the voltage, and the temperature. And, you obtained expressions for these thermoelectric coefficients, all four of them. As I mentioned, these are exactly the answers that are ordinarily obtained from the Boltzmann equation. And, they involve this integrals, with this conductance function, which is what we had been talking about from the beginning of the course, about. It's the same conductance function but usually we have only talked of conductance. Now, we have talked about these other coefficients which involve these extra factors. Now, you also talked about this slightly different form for the transport coefficients, this thermoelectric coefficient, where instead of having the voltage on the right-hand side, you have the current on the right-hand side. And, theoretical models usually give you this, but, in terms of our experiment, usually these coefficients are more direct. And so, these are what you are usually measured and compared. And, relating these two, we have these simple expressions, for the Seebeck coefficient, the Peltier coefficient, and the heat conductors. Now, this all for a general case with a continuous distribution of levels. Now, what you want to do in this lecture is, we talk about a special case where you just have one level. And what I want to explain is, that that gives a lot of insight as to what really determines this thermoelectric coefficients. And in principle of course, if you understand one-level, you can always put it together and understand the many levels in parallel. Of course, when things are parallel, you should remember that you can add them up only when they're in this form, and not in this form. Because, all these parallel levels have the same delta V and delta T. But, they don't have necessarily the same I, Now, for this one-level device then, let's first see, mathematically, from our expressions what we get, and then we'll try to understand them, physically. 

[Slide 3] So, mathematically, the point is that if you look at all of these expressions, they all involve integrals over energy, of similar looking functions. Conductance we have seen for a while, this GP and GS, involved this extra quantity, E minus mu over q. Now, so you might say well, what is GP or GS compared to G0? Is it twice, half? And the answer is, I can't tell ordinarily. Why? Because, a different energy's are multiplying by different things. So, after integrating, who knows what you'll get. But, the point is, if the conductance function is a delta function, that it is non-zero only, at one value, then of course, what matters, is just the value of this quantity exactly at equal to epsilon. So, I could put equal to epsilon, and then whatever you have here I could pull it out of the integral so that what remains is G0. So, what I could do is, I could write GP, which is equal to T0 GS, as G0 times whatever this function happens to be, exactly at the equals epsilon. Same with the next one. When we write GQ, its G0 times whatever this happens to be at E equal to epsilon. So, once you have those two, you can find the Seebeck coefficient, it's the ratio of GS to G0, and mathematically this is what you'd get. Similarly you can get the Peltier coefficient, with the ratio, exactly the same with the delta T0, and if we look at the GK, this heat conductance, that's the interesting one. You get GQ minus this, and if you put in all these values, it takes a little algebra, you'll see you get zero. So, in other words, the heat conductance of a one-level device, is zero. So, how can we understand that? 

[Slide 4] Now, what we can argue is, that what is this heat conductance? The point is, if I have a temperature difference across a device but the current is zero, then what is the heat current? And, that ratio of heat current to temperature difference, that's the heat conductance, with zero current. And, what I'm going to argue is zero current means no electrons are flowing, and if no electrons are flowing, there can't be any heat current either, because every time an electron goes through it takes some heat from here and dumps it there. And, I say, well, does that mean heat conductions is always zero? Because, after all, the heat conductance is IQ, with I equal to zero. And, not really, because, the argument I made only applies when you have a single level. Supposing you had two levels. Now you see if the current is zero, does not mean that the current of the individual energies are zero. What it means is just the total is zero. So, actually you'd have a current at-- through this level which will be exact opposite of the current at this level, so that they would cancel out. And, generally, if there are multiple levels some will be going from left to right, some will be going from right to left, but, overall it will all cancel out. So, the current is zero because they all cancelled out, but that doesn't mean the heat current is zero. Why? Because this level carries more heat current than that one. Why? Because this is kind of closer to mu. So, any electron going through this level doesn't carry as much heat as that one. And, so even though the charge currents have cancelled out, it doesn't mean the heat currents will. And, so they'll get heat-current. But, as long as we're talking over one-level device, that is, we just have this level, not the other one, then you see if charge current is zero, it implies, the heat current is also zero. And, that is why the heat conductance of a one-level device is zero. And, the heat conductance that you get for a multi-level device comes from the fact that even though the charge current is zero, there's really currents flowing in all of those levels, and some of them are carrying more heat currents than others, and it all adds up to give you a net heat current. Now, I should qualify this a little bit, because, a one-level device, if you actually had one, and measured it, you wouldn't get zero heat conductance, and the reason is, that in real devices, real materials, a good part of the heat conductance comes from phonons, which has to do with the jiggling of the atoms. Here, we are only talking of electron transport. Now, it is a lot of work actually talking about phonon transport, that is how this jiggling is communicated from one side to another, using this approach, this Landauer, approach, this elastic resistor type of ideas. And, it is discussed in the notes. But, in these lectures I won't go into it at all. The lectures are entirely about the electron transport, and the thing is, in general then, when you measure heat current, you also get the term for the heat conductance due to phonons, and just because this one is zero, doesn't mean that one is. So, whatever we are saying about the one-level device, having the heat conductance zero that applies to this one, not to the phonon part of it. 

[Slide 5] Let's now move on to the Peltier coefficient. See if we can understand that. So, what's the Peltier coefficient? That's this one, there's no temperature difference, both sides have the same temperature, but we have applied a voltage difference so a current flows, and the question is, how much heat current flows for a given charge current? And, the answer is pretty clear, you have all, you have just a delta function device, so there is just this energy. So, when an electron goes through, every time an electron goes from left to right, it carries an amount of charge, q, and it carries an amount of heat, that's epsilon minus mu. Actually it takes epsilon minus mu1 from here, and dumps epsilon minus mu2 over there. And, we are talking low bias, so mu1 is almost equal to mu2. So, the amount of heat it carries is epsilon minus mu1 or mu2. I guess that, equally could use the equilibrium value which is epsilon minus mu zero. So now you understand the Peltier coefficient fairly easily. 

[Slide 6] What about the Seebeck coefficient? Now the Seebeck coefficient how do you measure it? Well, again, current has to be zero, so what is the voltage difference for a given temperature difference? So, the way you think about it is, you have applied, created a temperature difference, so this side is hot, so that side is cold. This is a device where, I guess the level is above mu, so it's like N-type, electrons are flowing from hot to cold. But you see, you don't let the current flow out, because it's open circuited. No currents. So, what that means, is all the electrons going from the hot to the cold side, they pile up on the cold side. And, that's what raises the mu, and creates the voltage difference. So, finally you see, you have a temperature difference, and a voltage difference, which compensate for each other. And that's why there's no current in the long run, after it is all settled. So, how much voltage is generated for a given temperature difference? Well, the basic point is, that if current is zero, and current depends on f1 minus f2, and all that matters now is the energy, E, equal to epsilon, so what must happen is, that the Fermi function in contact one, it's value at epsilon should be equal to Fermi function in contact two, again, at epsilon. Remember, its two different Fermi functions, because the two contacts are different mu, they have different T. So, there's no way they will match up at all energy's. Those are different Fermi functions. But, what must happen for current to be zero is, they must match up exactly at that value. So, how do they match up at that value? Well, the Fermi function depends on E minus mu over kT. So, in order for those two to be equal, epsilon minus mu1 over T1, must be equal to epsilon minus mu2 over T2. And, now, I'm going to use a little algebraic trick, which is that, when a over b, equals c over d, then each is equal to a minus c over b minus d. So, what I have done here is, equated this to T1 minus T2, that's in the denominator, and epsilon minus mu1 minus epsilon minus mu2, which is how you get the numerator. Now, you'll notice what I have here, is actually the Seebeck coefficient. That is you have this temperature difference, what voltage difference does it give rise to? And, this mu1 minus mu2, is basically q times the voltage difference, remember we are using electron voltages. So, what this tells you then, is that the Seebeck coefficient is related to this quantity, it gets divided by q. And, that's exactly what we have here. We're again, talking low bias, so that this is almost equal to that. So, what we use is epsilon minus mu zero over qT zero. So, that's the Seebeck coefficient. So, again, this is the result that came out of the full math. But, you can understand it in relatively simple terms 

[Slide 7] for a one-level device, and this one-level device also gives you some insight about what makes a good thermal electric, for example. So, you're trying to get a good Seebeck coefficient. So, you say, well look at this, the Seebeck coefficient depends on how far the level is from mu. So, what I should do is, take that epsilon, take my-- choose a material in which the levels are way above the chemical potential, way above this Fermi energy. Now, that's really not quite right, and the reason is, if you see, if you think of the conductance's were. Remember, the conductance involves this integral, and as you know, this df/dE looks something like this, so if you have a chemical potential there, it kind of dies out. So, if you have a level out there, then the corresponding conductance will be extremely small. Why? Because this function would have died out exponentially by the time you get there, it's a very small number. So, when you do this integrally you'll get a very low conductance, which means a very high resistance. So, if you are thinking in terms of a circuit, you could represent this material as an open circuit voltage, with a source resistance that's this G0. And, the point is, that by putting this level way up there, you should get a good open circuit voltage, that's what this Seebeck coefficient is. But, it has an extremely low conductance, exponentially low, which means a very high resistance. What that means is, that when you actually try to use this to charge a battery or do something with it, that is, when you connect the load, you really don't get much of a current from it. Because, this resistance is just terribly high. So, in practice, what you want is, a material whose energy levels are, let's say above mu, you get entirely above mu, or entirely below mu, because the two kind of subtract from each other. But, you don't want it to high, you want it, like, within a kT or so. And, so the kind of Seebeck coefficients you expect is, let's say this epsilon minus mu is kT. So, what you expect for Seebeck coefficient then, would be kT divided by qT, which amounts to k over q, the T cancels out. And this quantity k over q, k is both been constant, q is charge of an electron that comes to about 85 microvolts per current. And, that's a number people carry in their head, that's the measure of what makes a good thermoelectric, or a good Seebeck coefficient. Like if it is one micro volt per current, that's not very good. If it is like one millivolt per current, well that's like a giant Seebeck effect. Because this is the number that kind of sets the standard, sets the bar for Seebeck coefficients. And, how do you get that good Seebeck coefficient material? Well, you don't want something like this. Something like this, that's kind of metallic, where you're electric chemical potential is well inside the band. This won't give you much, because whatever is above it would subtract from what is below it. So, metals don't give you a good Seebeck coefficient, and what you want is more like a semiconductor. Something whose levels are above mu, or if you wanted a p-type, whose levels are entirely below mu. But, not too far above, you don't want an insulator. So, within a few kT. So, usually, these heavily doped semiconductors, they make good Seebeck coefficient materials. So, with that then, I think we'll move on, in the sense, as I said, this course we're really are going through details of thermoelectric materials. We have a separate course actually on thermoelectrics, and that is, if you're interested in thermoelectrics, you should learn it from the experts properly. Here, we are introducing it more as a conceptual thing, that how this general viewpoint of elastic resistors gives you a nice clean way of understanding all these diverse phenomena's. 

[Slide 8] And, for the rest of this unit then we want to move in to the more conceptual issues, as I mentioned in the introduction. This idea of this second law, and how it limits the degree to which, the extent to which you can convert, interconvert between heat and electricity. So, will start with that then in the next lecture, with the second law.