nanoHUB U Fundamentals of Nanoelectronics: Basic Concepts/Lecture 4.3: Heat Current
========================================
 >> 

[Slide 1] Welcome back to Unit 4 of our course, and this is the third lecture. 

[Slide 2] Now, in the last lecture, we talked about the Seebeck effect. In this lecture, we'll talk about what is called the Peltier effect. Now, let me explain. You see, so far, whenever we have discussed this current flow, and we use this idea of elastic resistor, we say that no heat is dissipated in the channel, but where the electron actually loses energy and generates heat is in the two contacts. So the way you think is, the electron comes in from the battery at mu1, exits at mu2, and goes back to the battery, so obviously in the process of getting from left to right, it lost that much energy. Every electron as it goes through. So this is-- and how much is that, well that is q times the voltage. That is this much. And where is it dissipated? Well, a little bit in this contact and a little bit in that contact. Now, what I want to point out though is that in general that is the total heat dissipated, but both contacts need not be heated in general. One could be cooled, for example. So let me show you what I mean. Consider these n-type materials, which means where you have states available, not in this window, but let's say above this window. So you've got lots of states up here. Now something like this, if you look back at what we discussed in the last lecture, you'll see that it will give you a strong n-type Seebeck coefficient. Now, but in this context, the point I want to make is that now when you follow an electron, it comes in from the battery right around the chemical potential at the Fermi energy, but then in going through the channel, there's no states here. So what it has to do is climb up, then go through, and then of course, when leaving, it again leaves that mu2, so it will have to give up that much energy. So overall, of course, what happened is an electron came in at mu1, left at mu2, which means it got rid of an amount of energy equal to mu1 minus mu2, exactly the same as here. Except that in this picture, you kind of generate heat at both ends. Here, in this contact, you actually absorb heat from it, which means that if that contact is to be maintained at a fixed temperature, then what would happen is heat would come in from the surroundings, and you could use this to cool something. So you could put your Coke on top of that contact, and basically what the contact will do is take heat from the drink and the drink would get cooled down. So this is something you could use for refrigeration in the sense that one contact is getting heated, that you're not using that, but the one contact is getting cooled every time an electron goes through. And you could have the reverse when you have a p-type material, that is where all your states that are conducting are below the chemical potential, down here. Because now in order for an electron to flow, it first has to give up a lot of energy, cross over and then climb up again and then get up. So now, what would happen is, it is this contact that would get cooled. This is where it will be absorbing energy. So you have these two complimentary things, which in the one case, it is the first contact that gets cooled, and the second case is the second one that gets cooled. So there is p-type and n-type in that sense, affect reverse of sign in some way, just like the Seebeck coefficient. And as we will see when we do this quantitatively, the two coefficients are related in the sense that a material that gives you a strong Seebeck coefficient, also gives you a strong Peltier coefficient. Now one thing I should point out is that often in discussing conductance, I've said that, you know, it's easier to think at zero temperature. And zero temperature means, of course, that the Fermi function is such that all states below it are filled, everything above it is empty. Now if that were the case, then of course it is hard to understand why there should be any flow at all. And that is very true. At zero temperature, you wouldn't have this effect. This effect you only have at non-zero temperatures, because at non-zero temperatures, the Fermi function doesn't just go to zero, there is a tail out here. And because of that, there is electrons out there which can flow, and the same in this picture as well. So yes, at zero temperature you wouldn't have that. 

[Slide 3] Now, let's do this quantitatively. What you want to define now is so far we have always talked about the electron current, which is like the charge current. What we want to now define is a heat current. So the way you can do it is, so this is our usual expression for the electron current. This tells you the rate at which charge is flowing from one site to the other, and if you divided by q, you get the rate at which electrons are flowing. So you then have a q squared there, if you wanted to write the rate at which electrons flow. Now, every time an electron goes from left to right then, you would have this, an energy that is taken from the battery, which is mu1 minus mu2. And so if you wanted to write the energy current from left to right, it would be whatever we had for the electron current times mu1 minus mu2, divided by q. Divided by q in the sense that every electron that goes through carries a charge of this much, and this is the charge current. So if I divide by q it becomes electron current, and then the point is that the electron carries that much energy, this mu1 minus mu2. Now if you want to look at how much energy is actually taken from the source, then the way you can think is as far as the source contact is concerned, electron came in at mu1, left with energy E, so every electron, I guess, is taking out an energy from the source, which is E minus mu1. Because E is leaving, mu1 is coming in, so the amount it is taking from the source is E minus mu1. How much energy is taken from the drain? Again, you look at the basic charge current, and then consider if you look at the drain, electron comes in with E, leaves with mu2 to gain how much was taken from the drain? Well mu2 minus E. And what you can convince yourself is, if you take those three things and add them up, the answer is zero. You see, E minus mu1,  mu2 minus E, mu1 minus mu2. But this is kind of the overall balance of things. This tells you how much energy is flowing per second out of the source, of course, steady state, if the source is being maintained at a constant temperature, it means that energy would also be flowing in, because you are continually taking it out from the surrounding energy will come in and keep it at a constant temperature. This is the energy from the drain, and this is the energy from the battery, because from the battery you are taking electrons out at mu1, and returning them at mu2. 

[Slide 4] Now the way you define the heat current is by saying that just look at the energy that is being taken from the source, and what is the source, and what is being taken from the drain? So one is E minus mu1, the other is mu2 minus E. So this is of course what is from the drain. If you had written how much energy was being dumped into the drain, it would have been E minus mu2. This is what is being taken from mu2 minus E, but if you reverse the sign it will be what is dumped into the drain. So we are now talking of low bias, which means mu1 is almost equal to mu2, so you could say that what electron flow is doing is, is taking an amount of heat from the source, which is E minus mu0 over q, mu0 being this average mu. And it is taking that energy and dumping it into the drain, E minus mu0. So you could say that heat current is flowing, which is exactly this, but with the mu1 replaced by mu0. So this is the amount of heat that is flowing from source to drain. So we have this expression for the heat current, and we can now linearize it. That is, turn this f1 minus f2 using the Taylor series Expansion, and you do it the way you have done before, f1 minus f2 is this del f zero del E times delta mu and then the system with delta E. This is what we did in the last lecture, when we were discussing the Seebeck effect. And we are using exactly the same thing. The difference is, the last lecture we applied it to the charge current. Now we are applying it to the heat current, and the heat current has this extra E minus mu over q. So when you put those together, then you get a heat current that is proportional to the delta V and a term that is proportional to the delta T. So you've got these two terms. And the expressions for GP and GQ, again, look like these averages over energy. For the GP you have this E minus mu over q that comes from here. For GQ, it comes with a square, because there is a E minus mu from here as well. So there is one from here, and another from there, that is how you pick up the square. 

[Slide 5] So what we have then is, that if you start from this expression for the heat current, linearize it, then you get these two transport coefficients, this GP and GQ. And what we have already done before is written the charge, taken the charge current, linearized it and got, again, a part that is proportional to delta V and a part that is proportional to delta T. And we had these expressions for G0 and GS. So what we have here then, these are the basic equations of thermoelectricity. This is often used as a starting point for people who are looking at materials, interpreting experiments, designing devices. That is often the starting point. And of course what we have been discussing so far before was just that I equals G0 delta V. We were interested in only the electrical conductance. What we did in the last lecture was got that one. What we did in this lecture was explain what a heat current is, and then expanded the heat current to get these two terms. And we get all these expressions, and these are all the standard expressions for the thermoelectric coefficients. These are what are normally derived from the Boltzmann equation. And the point to note is that we can get these much more easily, much like we did in the last lecture and this lecture, because we are using this idea of elastic resistor. So we get, but the results themselves though are standard, and they are usually obtained from the Boltzmann equation. These expressions for the transport coefficients. Now, before finishing up this lecture, then, I just want to mention one other thing. I guess a couple of other things. One is that there is something called the Kelvin relation. I mentioned earlier that this Peltier effect Seebeck effect, the one we talked about in this lecture, what we talked about in the last lecture, are relative things, in the sense that if you wound a material with a good Seebeck coefficient, it will also have a good Peltier coefficient. The two are kind of related. And you can see that from these coefficients actually, if you look at it, because one has an extra T0 in it. Otherwise, it is exactly the same expression. So what you can show is that this GP is actually this temperature times GS. So any material that gave you a big GS at a given temperature would also give you a big GP. And of course, you see that at zero temperature, you don't have much of a Peltier effect, because it is T0 times that. And there is a Seebeck coefficient and a Peltier coefficient, which are actually not GP and GS, but it is related to that. And let me explain that point. 

[Slide 6] You see, the equations the way we have obtained it, and when you are doing a theoretical derivation of trying to get these things, this is what comes out naturally. On the other hand, when you are doing this experimentally, it is more convenient to have the voltage on this side, and the current on the right-hand side. That is, think of temperature and current as your independent variables, and voltage and heat current as what you measure. So you could take that first equation and move it around so that delta V is written as a resistance times I minus GS over G0 times delta T. And this is, of course, as I said, the resistance. It is the inverse of the conductance, and this second one is what we call the Seebeck coefficient. Remember, last lecture we said that the Seebeck coefficient is the open circuit voltage for a given temperature difference. So open circuit means I equal to zero. That means this term isn't there. So you can see delta V is equal to S delta T. Now, similarly, if I now use this relation, then I can put it back in here for delta V, and I'll get an IQ is equal to something times current plus something else times delta T. And this is then what you would call the, this would be the Peltier coefficient, and this is this heat conductive conductors. And I will come to that in a minute. But the point you'll note then is that these are the coefficients that are usually experimentally measured and that is why they have a name. That is resistance, Seebeck coefficient, Peltier coefficient, heat conductance. Where as this, these are the things that come usually from a theoretical model. And they have expressions associated with it. Now one thing to note is this heat conductance, we kind of have two heat conductance here. One is this one, and the other is this one. You see, and this one is what we have there, minus something. So what is the physical difference between those two in the sense that what this tells you is how much heat current will flow for a given temperature difference if the voltage is zero. So that is what you call the short-circuit heat conductance. On the other hand, what this tells you is for a given temperature difference, how much heat current you have if the current were zero. So that is like what you would call an open circuit heat conductance. And the interesting thing is that of course if you have a material with a strong Seebeck coefficient, where GP and GS are sizable, then the two heat conductances are different, and their difference kind of tells you how strong the thermoelectric effects are. And before I move on again, let me just mention one little thing, and that is that the way we have defined things, the current and voltage are written as electron voltage and electron current, because of which these coefficients, this one and this one, might have an extra negative sign compared to what you normally see in the literature. It doesn't matter for this coefficient than that one, because here I've reversed the sign of both voltage and current, and so the coefficient has the same sign. It doesn't change. Here, we are talking of heat current and temperature, and have done nothing with those. But it is these ones that kind of I've changed the convention with respect to current, because we are talking electron currents, and here again I've changed the convention with respect to voltage because we are talking electron voltages, and that is why the sign is reversed with respect to what you would normally see in the literature. But of course the way I have defined things, Seebeck coefficient in the literature is negative for n-type materials, and that is exactly what this is also. I have defined it so that is what you will get from here also. 

[Slide 7] Now, what we will do in the next lecture is look at a one-level device. That is, in general, we are talking of a channel with many energy levels, and these coefficients involve an average over all the energies. And it is instructive to think of a very simple device where there is only one energy level of interest. And ask the question what are the thermoelectric coefficients of a single-level device? And that gives you a little bit of an insight into what these various coefficients mean, because we have kind of seen mathematically what it is all about, how you calculate them quantitatively. What we will do in the next lecture is look at this one-level device.