 So, today's lecture, this one, the first one tomorrow morning are a little bit heavy going in terms of the amount of math and things, concepts. After that we get over the hump and we start talking about some other things. This afternoon you'll have a change of pace and we'll talk about molecular dynamics and ab initio. So, let's do the second half of what we're talking about. We did one driving force, voltage differences across the terminal. The second driving force is temperature differences across the terminals. And I'm going to try to do this in a little more of a physical approach today. It might not be completely clear to you this morning how can you make that assumption, but hopefully it'll look plausible to you. And then tomorrow we'll just go through and we'll formally show you how you derive the same results. Okay, so we're talking about thermoelectricity and a thermoelectric device can be used in two ways. It can be used to convert heat into electricity and or it could be used to convert electrical power into cooling power. So the challenge I think as Professor Fischer mentioned is that the efficiencies aren't very high and this is a challenge that people have been struggling with for a long time. But there are very important niche applications and they're also important for characterizing materials and for understanding transport physics. So we saw in lecture two actually that if we took a very big sample we could express the current as conductivity times the gradient of the electrochemical potential. So that's a traditional expression. All of our results reduced to that expression if we take a very long sample for which the contacts don't matter. Device much longer than the mean free path. Now in thermoelectricity we're oftentimes going to turn these equations around and write them in different ways. So in the first equation the idea is that we impose some gradient in the electrochemical potential or some voltage difference and then we find out what current flows. I could turn this around and we could say well what if we have a current source that's the independent variable and we scored a current in what voltage difference would result or what electrochemical gradient would result. That's the second. We could always write it, we could rearrange that. And you'll see that in the thermoelectric equations it's most natural to derive them in one form, it's most natural to use them in another form so people are continually going back and forth between these two forms. Now our question is what changes in these equations when we have a temperature difference between the two contacts? And I'll just give you the answer and then we'll talk about where it comes from. So we'll find that the current is conductivity times gradient of electrochemical potential or quasi-thermal level if you're a double E but then there's an additional term that involves the gradient of the temperature and that brings in the C-BAT coefficient. Now alternatively I could turn that around and I could ask what gradient of the electrochemical potential results if I force a current in and impose a temperature difference. So those are just two different forms of the same equation. Resistivity is one over conductivity. So if we want to describe this, thermoelectricity involves the flow of both charge and heat. So we have an equation for the charge current and we're going to see how it gets modified when we add a temperature gradient but we're also going to need an equation for the heat current. Now you would expect heat flows down a temperature gradient so you would expect the answer to be something like this minus a thermal conductivity times the gradient of the temperature. So kappa zero is a thermal conductivity. Now the question, since we're having both, so first of all I should back up. We're talking here about the heat that is carried by the electrons. So these are particles in random thermal motion they can carry heat. Latest vibrations can also carry heat. In a metal most of the heat would be carried by the electrons but in a semiconductor or insulator most of the heat would be carried by the lattice vibrations. So we're only talking about the part here due to the electrons right now. So this is the amount of heat that the electrons, you know we just write it down phenomenologically four years ago, this is what you would expect. But there's current flowing at the same time, you know. So how much heat is conducted when current flows? So I'll just give you the answer and we'll talk about it later. So the answer is going to be that if there's a gradient in the quasi-firmly level that is also going to carry heat. So we have to add that to the heat flowing down a temperature gradient. We'll have two terms. Just like we got two terms for the electrical current, one due to the gradient in the quasi-firmly level and one due to the gradient in the temperature, the heat current will have two terms. One due to the gradient in the quasi-firmly level, one due to the gradient in the temperature. Alternatively I could turn these equations around or I could express them in terms of if I were to force a current and impose a temperature gradient what heat flux would flow? And the answer then is that the heat flux would be pi times the electrical current where pi is the Peltier coefficient and then minus kappa on the temperature gradient. Now there are two different kappas here so we're going to see where these equations aren't intuitively obvious. We're going to see where they come from. But notice in the first equation on the top, if I have no gradient in the quasi-firmly level and I change quasi-firmly levels by changing voltage, so if I have no voltage difference or I'll tell you in another way, if I'm under short circuit conditions the first term on the right would be zero and the heat that would flow would be kappa zero times the gradient to the temperature. So I would call kappa zero the thermal conductivity under short circuit conditions. No voltage difference. If you look at the second one, we just rearrange the equation to get the second one. If I look at the second one and I say what if I open circuit the device so that no current flows, then the heat that flows would be minus a different kappa times the gradient to the quasi-firmly level. That kappa is I would call the open circuit thermal conductivity. There are two different quantities that are related on the bottom. You have to be careful when you're dealing with these and problems that you know which thermal conductivity you're dealing with. If you measure it under different conditions you'll get different answers. As I pointed out, it's both electrons and lattice vibrations or phonons carry heat. We're only talking about electrons right now. Our goal is to understand the physical origins. Where do these effects come from? What are they all about? How are they related to the properties of the semiconductor? First thing we have to talk about is current flow. How it changes in the presence of a temperature gradient. Then we have to talk about heat flow and then we'll talk about how those two flows are coupled. Let's take this n-type resistor. I'm thinking now of a bulk 3D resistor. We could do this for ballistic transport too. People do that. Let's just talk about the diffusive limit, large bulk sample, big cross-sectional area. If I make the contact on the right, contact 2, hot, then just intuitively I'm going to expect that the electrons will diffuse away from the hot region. They'll have more kinetic energy, they're random thermal motion and scattering will cause them to diffuse away. But if it's open circuited, then a positive voltage will develop on contact 2 to attract the negatively charged electrons and pull them back to prevent current from flowing. That positive voltage, the open circuit voltage that I would measure is the C-beck voltage. The temperature gradient will give rise to an open circuit voltage. This is what we call the C-beck effect, very physical intuitively. We would expect a positive voltage for an n-type semiconductor. What if it's a p-type semiconductor? Here's the case where it would be simpler for me to think about holes rather than electrons flowing through the conduction band. So in this case, if that's a p-type semiconductor and the right contact is hot, the holes would want to diffuse away from the hot contact. And I would have to develop a negative voltage to attract the holes and pull them back and have zero current flow. So the sign of the voltage will be different, depending on whether I'm n-type or p-type. The sign of the conductance is always the same, it's positive. But the sign of the voltage of the C-beck voltage is different. So our fundamental picture is that current flow is always due to a difference in Fermi functions. And we've talked about moving the Fermi energy in the second contact with respect to the first, and that was our first driving force. But our other driving force can be temperature. So the temperature controls the width of that, transition from one to zero. If I have a much hotter contact to, say, but its Fermi level is in exactly the same position, then I will have a much broader Fermi distribution. Now, if I'm dealing with an n-type semiconductor, the states are above the Fermi energy in the conduction band. Above the Fermi energy, in this case, f2 is bigger than f1. If I'm dealing with a p-type semiconductor, then below the Fermi energy, f2 is less than f1. So the sign is opposite. So depending on whether the conduction is occurring through the conduction band or the valence band, I'm going to get a different sign of the C-beck voltage. All right, if the states above the Fermi level are important, we'll call that n-type conduction. If the states below the Fermi level are important, that's p-type conduction. All right, so here's a little idealized energy band diagram. Again, to be specific, I'm thinking about an n-type semiconductor. So the Fermi level is located near the conduction band. I have two ideal metal contacts. There's some barrier height. So there's always a, the barrier height is the difference between the work, comes about because of the difference in the work functions between the semiconductor and the metal. It's simply the difference between the Fermi energy in the metal contact and the conduction band in the semiconductor. Now, you know, in general, there'll be a shot key barrier here. There'll be band bending. There'll be a potential drop and none of that will matter if I have a long sample. If I have a long sample, all of those effects that occur near the contact will settle out very rapidly and I'll be concerned about what happens in the bulk of the material. If I get to a very short structure, thermal electricity in a nanostructure could be affected by details of the contacts and people worry about that. But I'm going to talk about in the diffusive limit the more traditional area that we talk about. So we can get by with an ideal contact because the details of the contact aren't going to matter. Okay, so let's say we have the same temperature at both contacts and we apply a positive voltage to contact 2. That lowers the Fermi level in contact 2. If we have an n-type conductor, it just gives us a potential drop in the n-type conductor. We just have a constant slope. The electrons then, there's a positive voltage on contact 2. It attracts electrons out of contact 1. They flow from contact 1 to contact 2. Now, they don't flow exactly at the bottom of the conduction band. So if this is a 3D semiconductor, its density of states is proportional to the square root of energy. There are no states right at the bottom of the conduction band and then the number starts to increase. So if I look at where the average energy is, on average they flow just a little bit above the bottom of the conduction band. So positive voltage lowers the Fermi energy. The electrons flow at an energy that's just a little bit above the bottom of the conduction band. And I'm going to call that energy delta. It just turns out it's 2KT for a parabolic energy band with a constant scattering time. So what's the Seebeck effect? So the idea here is that contact 2 is hotter than contact 1. That means the electrons want to diffuse to the left. But in order to stop that from happening and make zero current, a positive voltage has to develop on contact 2 so the Fermi level comes down. That's the Seebeck effect. So the hot contact would like electrons to diffuse to the left. The lower voltage pulls the Fermi level down on contact 2 and stops that flow. You end up with open circuit conditions. So under those open circuit conditions, then we're going to have something like this. We have two different temperatures of the contacts. We have a voltage across these. This is the Seebeck voltage which stops the current from flowing. We have electrons that would like to flow a little bit above the bottom of the conduction band. And under open circuit conditions, they won't do that. The voltage has adjusted to accommodate the temperature difference so that everything balances and no current flows. Now how can we make that happen? So one way we could make that happen is we could say, well, if I look here at the left end of the n-type resistor, if the probability that an electron comes in at that average energy that the current is flowing, if that probability is equal to the probability that an electron comes in from contact 2, then on average there won't be any current flowing. They'll cancel each other out. So the probability that that state is occupied by contact 1, if it's the voltage drop on contact 2 will develop to make the probability that that state is occupied by contact 2 equal to the probability that it is occupied by contact 1. So zero current occurs when the probability that state is occupied by the two contacts is equal. And those two Fermi functions have two different Fermi energies. I'm thinking here now about a very short section of sample. And I'll tell you why I'm doing that here later. This argument you can see is valid under ballistic conditions. It's also valid under very short samples for which there isn't much scattering. It's also valid when you're near equilibrium. So you can kind of see that it makes sense. OK, so let's just see what it does. If those two probabilities are equal, and if the Fermi level on contact 2 is a little bit lower than the Fermi level on contact 1 because this voltage has developed, I'll just set those two Fermi functions equal. And if I can solve for that voltage, then I know what the Seebeck voltage is. So I just solved that. The arguments of those exponentials have to be equal. You go ahead and set those two arguments equal and solve for delta V. And the result is that we get an answer that delta V is something times delta T. The something is the Seebeck coefficient. So our final answer is that the voltage difference across that thin little sample is minus s times the temperature difference. The minus s is by convention, we think of Seebeck coefficients as negative for n type. And s is EC at the left where we did our bookkeeping plus delta sub n. That's the energy at which the current wants to flow. And the difference between that energy and the Fermi energy divided by Q times T of the first contact, that's the Seebeck coefficient. So let's summarize. So we've derived the Seebeck coefficient. Those two terms, EC plus delta n, that's average E sub j. It's the average energy at which current flows when it does flow. Now I can rearrange it. It's nice to write energy and divide it by KT and take that capital delta and normalize it by KT so I get the third expression. And when I've done that, I pulled out a factor K over Q. Boltzmann's constant over Q, that's 86 microvolts per Kelvin, which is a useful number to remember. For a 3D semiconductor, it's non-degenerate. This little distance above the bottom of the conduction band that current flows is 2KT. So the normalized value is 2. And now, of course, when it becomes degenerate, the Fermi energy goes way up in the conduction band. And current wants to flow near the Fermi energy because that DfDE is only positive near the Fermi energy. So when I become highly degenerate, that delta goes to EF minus EC. You'll notice that that's equal and opposite to the first term. So you can see that when you become highly degenerate, when you become metallic, the CBAT coefficient is going to approach zero. OK, so here's what we have. So we basically, DfNDQ, that's basically the electric field or the voltage difference across this sample. Before we put in the temperature gradient, we would have said that's just resistivity times current flow. That's what we got in the last lecture. Now that we've put in the temperature gradient, we just have to add a term, S times DTDX. And we've got an expression for S. We developed it for a small little short sample. And we used the temperature of the first contact. If I have a longer sample and the temperature is varying from one end to another, then the CBAT coefficient is temperature dependent. And the whole voltage difference I'll have to find by integrating the temperature dependent CBAT coefficient from one end to the other. And you can see what's going to happen if I plot S versus Fermi level. If the Fermi level is way below the conduction band, I'm going to get a large S, but it'll be negative in sign. When the Fermi level is exactly at the conduction band, then the first term and the second expression is zero. And I just get this little term delta, which is about 2. So I get 2 times 86 microvolts per Kelvin. And when the Fermi level starts going above the conduction band, then the current starts flowing near the Fermi level and the conduction band, it cancels out the first term and the CBAT coefficient gets very low. So if you measure the CBAT coefficient in a metal, it's very small. If you measure the CBAT coefficient in a lightly doped semiconductor, it's very big. Just makes a transition between the two. So here's measured data. And this is the magnitude of the CBAT coefficient in N and P type germanium. And you can see that it behaves in the way that it should. So zero here means that the horizontal axis means that the Fermi energy is at the conduction or valence band edge. Negative quantities means that the Fermi energy is way below the conduction band or way above the valence band in a P type. And positive numbers means it's moving into the band. And you see it has the shape. And a little bit later, we'll compute numbers and we'll see we get almost exactly these values. It's easy to compute. OK. So that's the first half. The first question was, how is current flow affected by the presence of a temperature gradient? We have to add a term that involves a CBAT coefficient. The second question is, how do we understand heat flow? So here we're going to take. We have an N-type semiconductor. And let's say I'm forcing a current through it. If I'm forcing a current in contact two, then if I'm forcing a current in contact two, that means the electrons are moving from contact one to contact two. So the current is negative because of the negative charge on electrons, but the flux of electrons is in the positive x direction. So electrons are flowing from left to right. So electrons, if you do this experiment, what you'll find is that the left end will get cold. It'll absorb heat from the environment and it will get cold. The right end will get hot. Heat will come out of the right contact. What's that all about? Well, what we'll find is that there's a heat flux that's going here and that that heat flux is just pi times the electrical current. And the heat flux will be positive because the electrons are moving in the positive x direction and they're carrying their heat with them. So we get a positive number. It must mean that the Peltier coefficient, pi sub n, is negative times the negative electron current. OK. So the question is, why does the heat current equal pi times the electrical current? I'm assuming zero temperature difference here now because I just want to see what the electrical current does to heat flow. If there's a temperature difference, heat will flow down the temperature gradient and I'll just have to add that. So we want to know why that expression is true and we want to know what determines the Peltier coefficient. So to understand, you heard this famous quote that Herb Krohmer has about semiconductor devices. If you're giving a talk on semiconductor devices, you should give an energy band diagram. Otherwise, no one in the audience will know what you're talking about. If you can't do that, it means you don't know what you're talking about. The first step is always draw an energy band diagram. So here's our energy band diagram. Let's see if we can figure out what's going on with this energy band diagram. Same n-type semiconductor. Now we know that if we keep the temperatures the same, we apply a positive voltage to contact two. We pull the Fermi level down. We tip the bands. Current wants to flow from contact one to contact two and it will flow a little bit above the bottom of the band. 2KT, something on the order of that depending on some details of band structure and scattering and things like that. Now, so we have this question. All right, we're making things simple by keeping the temperatures the same. So take a look at this. This is isothermal. Electrons in the metal are flowing near the Fermi energy. Remember that mathematical expression we had where we had a DFDE? The only place that that was positive was right near the Fermi energy. In a metal, that's where the current is flowing, near EF1. Now look in the semiconductor. The metal is flowing in that, or the current is flowing in that dashed line at a higher energy. Okay. How do the electrons get up there from the metal to the semiconductor? Where do they get the energy to hop up and flow at a higher energy in the semiconductor? They absorb it from the environment. That's why that contact gets cold. They absorb heat from the environment. That gives the electrons enough energy to hop up to where they need to be to flow in the semiconductor. The amount of energy that they absorb, that's thermal energy, the amount of energy they absorb is that energy that they need to get to, which is EC at the bottom of the conduction band at the first contact, plus that little delta, that 2KT above. And they already had the Fermi energy. That's where they came in at. So the energy difference is that energy minus EF1. That's the amount of heat they have to absorb in order to get into the channel and flow. Then they flow across and they just dump that heat out in the second contact. They thermalize and come back down to the Fermi energy because that's where the current flows out to contact. So contact two gets hot, contact one gets cold. It's reversible. You change the sign of the voltage, contact two gets cold, contact one gets hot. I think in some of these seat heaters and automobiles, they have all thermoelectric devices to heat them cool and keep the temperature constant. They do useful things. Now there are also, if this is a semiconductor with some scattering and inelastic scattering processes and there's also some power dissipation inside. So that's an I squared R loss. Now when we talk about linear transport, we're only talking about things that are linear in current or linear in voltage. This is quadratic. So this is an extra effect. But it's important in these. So people, you'd have to add that extra dissipation in if you wanna do things like compute the efficiency. Okay, so there's really a simple physical picture here. The idea is electrons flow from left to right when we apply the voltage. So we have some flux of electrons. We just divide the current by the charge. Each of those electrons has to absorb on average an energy of EC plus delta minus EF. So the flux of heat that flows from left to right is just EC plus delta minus EF times the flux. Those quantities that multiply J and X are just a Peltier coefficient. So we have a simple expression for the Peltier coefficient. There's a negative sign out front so it's negative for an n-type semiconductor and positive for a P-type. Now there's something very interesting here. So we sort of discussed this. These are two different problems. We put a temperature gradient on, we look at the open circuit voltage. We take a temperature gradient off, we just let the current flow and we look at how much heat flows. And so we've got, in the first case, we've got the CBAT coefficient. In the second case, we've got the Peltier coefficient. And we developed expressions for each of them. So on the left we have the Peltier coefficient that we just derived. On the right we have the CBAT coefficient. Notice something very interesting. They're closely related. And this is something that Lord Kelvin discovered in the 1800s. The Peltier coefficient is just temperature times the CBAT coefficient. So these two effects that we sort of treated as two independent things are really intimately coupled. And that's a characteristic of coupled flows that occur in a number of other problems as well. Okay, so let me just mention here very briefly coupled flows. So these are the equations that we've developed. If I have a uniform carrier concentration in the n-type sample, then the gradient of the electrochemical potential is the electric field, the slope of the energy band. If I have non-uniform carrier density, they get diffusion gradients and the connection isn't quite so simple. So instead of writing gradients of electrochemical potential, let me write electric field. Electric field is resistivity times current. That's what we would have gotten from the last lecture. We've had to add to that an extra term due to the C-beck effect. So that's our basic equation. Heat flow, we would normally start by saying that heat flows down a temperature gradient. That's minus kappa n dt dx. But we saw that the heat flow itself is gonna carry heat current from left to right and that's pi times Jn. And we saw that there's an intimate relation between s and pi. And there are four transport coefficients here. Rho is the resistivity, that's one over the conductivity. So the units of rho are ohm centimeters. The units of C-beck coefficient are volts per kelvin. The units of Peltier coefficient are watts per amp and the units of electronic heat conductivity are watts per meter kelvin. So you can just go through dimensionally and verify that for all of those equations. Okay, now if we're talking about what those parameters are, the resistivity or the conductivity is something that we know very well. So we saw that it involved an integral and if I call that integral the differential conductivity I mentioned that we'll find this useful in thermoelectrics. So I bring this in again. So this is our expression for conductivity that we developed in the last lecture. Everything is identical. Now, if we're going to do C-beck coefficient we'll derive this properly tomorrow morning. But I can sort of see that what's happening here is that the C-beck coefficient is, I may have a little bit of a typo here. What I have, I have K over T, EC plus delta, that's the average energy at which current flows. So the C-beck coefficient is really K Boltzmann's constant over minus Q times the average energy at which current flows minus the fermi energy. So if I want to compute the average energy at which current flows I should take the differential conductivity which tells me how much current flows at each energy and I should weight it by that energy difference. So that's an expression that will tell me how far above the bottom of the conduction band is that current flowing. We'll derive that tomorrow but that's, I think that's intuitively that's what it needs to be. That's how you calculate that little quantity. So in the first equation we know everything except delta. The last equation tells us how to calculate delta. And then we have this close connection between the two. If we can compute the C-beck coefficient we get the Paltier coefficient for free. Okay, now notice that that connection, this is a general feature. When you have coupled flows like this what you find is when we have the flow of heat and charge there's always a term. The first one in the first expression that's telling us about the flow of charge that heat is coming into it or temperature gradient. In the second equation where we're talking about the flow of heat the electrical charge is coming into it. Whenever you have two flows like this that are intimately coupled there's always a close connection between the coefficients that couple the two flows. And these are called An-Sager relations. And they're very general. They're usually derived on thermodynamic arguments. They apply to many different types of flows. We've just worked them out and if we had known about An-Sager relations we would have known in advance that this has to happen and we could have just worked out one and deduced what the other one would be. Now there's another relation that Professor Fisher mentioned that we won't be able to talk about rigorously until tomorrow morning but there's a connection between the thermal conductivity and the electrical conductivity. Remember now we're only talking about the heat that flows due to electrons. And let's see. So this is known as the Weidmann-Fran's law. The fact that the thermal conductivity is proportional to the electrical conductivity. And the constant is T times sigma times L and L is known as the Lorentz number. And the Lorentz number depends on details of band structure and dimensionality and all kinds of things. As Professor Fisher pointed out yesterday for a metal with a parabolic band structure the Lorentz number is pi squared over three times K over Q squared. For a non-degenerate semiconductor it's two times K over Q squared. So it doesn't vary a lot. It varies between two and pi squared over three which is between two and three basically. However this is not a law of physics. You expect since electrons are responsible for the electrical conductance and since electrons are the particles carrying the heat, we're not talking about the phonons, there should be some relation between the two. It's not fundamental. It depends on details of band structure and things. You can find for example if you have a delta function of states, you can find that you get a finite conductivity, electrical conductivity and you get zero thermal conductivity. So the number can be zero. So it can change but people would normally carry these numbers around in the back of their head and use them for interpreting experiments but if you're dealing with some structure that has a much different band structure, some quantum engineered structure or a super lattice or something, then you want to be very suspicious about using these. But we'll see tomorrow how you do that. Now there's also this lattice thermal conductivity. So we really need to add, heat can flow by two different ways. The electrons can carry the heat and the lattice vibrations can carry the heat. So we have to add that in. As I mentioned earlier in metal there are lots of electrons to carry the heat flow so the electronic part dominates but in semiconductors in a lot of cases the electronic part is not insignificant but the lattice part is the much more important part. So let's work some examples. These are the thermoelectric, let's look at germanium and let's see if we can just estimate these four parameters. Because if we wanted to do experiments or use these we'd need to know what the resistivity C-beck coefficient, Peltier coefficient and thermal conductivity are. So if I have a moderately doped sample 10 to the 15th doped at 300 Kelvin the mobility for n-type germanium is about 3,200 centimeters squared per volt second. If I'm non-degenerate then there's a simple relation between the carrier density and the fermi level and the effect of density of states I can plug in effective masses for germanium and compute that. So let's compute those numbers. So the resistivity, so the easiest way to get the resistivity is to say it's nq mu. We were given mu, the number of electrons is about equal to the doping density 10 to the 15th so we just plug in numbers and it's about two ohm centimeters, all right. So we have one of the four. C-beck coefficient. So we have numbers, if I assume a non-degenerate semiconductor this little delta is about 2KT above the bottom of the conduction band. The other thing I need for my C-beck coefficient expression is I need to know how far is the fermi level below the conduction band but that I get just from the relation between the carrier density and the fermi level. So that's 9.3 units of KT. We just plug numbers into this expression and it's minus 970 microvolts per Kelvin. I remember that plot I showed you the measured data and that's almost exactly what the measured data showed. Peltier coefficient, all we have to do is to multiply 300K tends to C-beck coefficient of 0.3 and thermal conductivity. So here it's a little bit harder. So let me assume that this Weidman-Fran's law works. It's a non-degenerate semiconductor so the Lorentz number is 2K over Q squared. I'll just work that out, plug numbers in and we end up getting 2.2 times 10 to the minus fourth watts per meter Kelvin. So it's easy to estimate all four. I remember all of these parameters depend on temperature, they depend on where the fermi energy is so depending on what the problem is we have to be careful about doing that. The other point we should also mention is that we computed the electronic thermal conductivity and this is the one under open circuit conditions to be on the order of 10 to the minus four watts per meter Kelvin. The lattice thermal conductivity is 58. So that's, it's really the lattice that's carrying the heat in this material. So I'll mention just briefly then about thermoelectric devices. I don't plan to say very much about them but just so that you're aware that these can be used for useful things, this is the way you would make a thermoelectric refrigerator. So we have an n-type slab and a p-type slab and we hook them electrically in series, they're thermally in parallel. We run a current in the bottom of the n-type slab that means electrons are flowing down on the left slab. They go up through the contact on the top, the current runs down the p-type leg so that means holes are flowing down. So right now it's easiest for me to think about electrons and holes and then holes flow out the other contact. So I would think that well these electrons are carrying their thermal energy from the top to the bottom. So they're gonna remove heat from the top and they're gonna bring it down to the bottom so the top will get cold and the bottom will get hot. That's how you make a thermoelectric refrigerator and the questions you might ask is, how much can we lower the temperature on the top? How much heat could we pump? And what's the efficiency of this device? So these are things that people that design devices worry about. Now we could also operate this in a different way. We could operate this as a power generation device. We could apply heat to the top. If I apply heat to the top, electrons will wanna diffuse away and holes will wanna diffuse away. That means the current will be flowing up in the n-type leg, across the short circuit and down in the p-type leg and then out the contact and through some load, electric motor or light bulb or something that we could do something useful with. So we could convert, in the first case, we can convert electrical power to cooling power. In the second case, we can convert heat into electricity. And we'd have the same kind of questions. What's the efficiency of this process? So I don't wanna go into any kind of detailed analysis here, but let me just give you a flavor for how people do this. Let me just think about one type of those legs. So here we have a cooling device. So we're pumping heat at the bottom, that's Q sub C at the cold junction, and we're moving that heat up to the top. So the top contact is the hot contact. And in between there is some I squared R ohmic losses. And so then we're generating some heat in between. And we'll just assume that half of that flows up the top and half flows down the bottom. Now, if we wanna analyze this device, we can set up a little balance equation. We could say, okay, we've got some heat coming in, that's item one, and everything has to be conserved. So the heat that we're pumping is equal to the sum of the Peltier heat that gets pumped away. But when that heat gets pumped away, the top gets hot and thermal energy starts diffusing down, so we have to subtract that. And while we're doing this, we're generating, current is flowing, we're generating I squared R losses, so heat is being generated in the leg. We'll say half of that comes down. So we can set up this balance equation. The first, the Q sub C is the heat that's being pumped at the cold side. That's the Peltier heat, the first term, minus the back flow from the hot side back down that's subtracting from it, minus half of the I squared R losses, which are also flowing down and subtracting from it. So you can see that there's a maximum. As I crank the current up, we're gonna pump more and more heat by the Peltier effect, and things are gonna get colder and colder on the bottom. But I'll get more and more I squared R losses, so we'll generate heat. If I pumped the current up too high, I'll dissipate too much power in the n-type leg and I'll end up losing that pumping, or it'll turn around and go negative. So I could take a derivative and find where is the current where I can maximize the amount of heat that can be pumped? So I just set the derivative equal to zero. That gives me the maximum current, and it also gives me the maximum heat that can be pumped. And then I could say that, okay, what if I set that maximum to zero? And that would occur when I'm pumping as much as I can, but I've got, the bottom is cold as I can, and the top is cold as I can, such that I've got the biggest temperature difference across it that I can get, and the back flow is exactly equaling the pumping. Everything cancels and I'm at steady state and I've developed a temperature difference across it that is the biggest it can be. How big is that? So it turns out that there's a simple expression. It's one half ZT squared, and Z is something that comes up over and over and over again in thermoelectrics. If you read any paper on thermoelectrics, people will talk about Z. It's a figure of merit. When you calculate the efficiency of a thermoelectric device, and it's not quite so true when you put in all kinds of non-idealities and contact resistances and details, but the first order, the efficiency, whether it's a generator, a heat pump or whatever, is determined by these material parameters in this combination. They always come in this combination. C-back coefficient squared times conductivity divided by lattice thermal or divided by heat conductivity. So that's a very important number. So if we were to calculate the efficiency of this refrigerator, what we would want to do then is to pump the maximum amount of heat and then to find the efficiency to be the maximum amount of heat that's being pumped divided by the heat power that's coming in and that's gonna be my coefficient of performance at maximum cooling power. And if you do that, you get something that's the Carnot efficiency times something that's less than one. And the factors that multiply it depend on the temperature of the hot and the cold side and they also depend on the material parameters in this same combination, Z. And you do power generation and you go through the analysis of the power generation and compute the efficiency. How much electrical power do you get out divided by the heat power that you put in and you'll find that it's also given by an expression that involves Z. So this figure of merit is really important. And an awful lot of thermal electrics, maybe too much because a lot of the practical details depends on contact resistances and device design but an awful lot of the work you'll see is focused on can I find a material that has a good figure of merit? So here's our figure of merit. Usually what people will do, it has units of inverse Kelvin. So usually people will quote a figure of merit times temperature. So you have a figure of merit at a specific temperature. It's a dimensionless number and it is about one for the best thermal electrics and it's been about one for like 50 years. It was developed by this Russian Iofi in the late 40s, 30s I think. It was quickly developed into a practical technology. They found the right materials that had a figure of merit on the order of one and people thought, oh this is terrific. Now if you run through the numbers, if you can get this up to two or three, you can really do a lot. You can generate a lot of electrical power from industrial waste heat economically. You can replace mechanical compressors and refrigerators. It would just have enormous impact and people were very optimistic and there was a lot of work. My PhD thesis advisor measured all kinds of materials to try to find a better thermal electric figure of merit. The bottom line is that for like 40 years nobody could do better than what they had done in the early 50s. More recently in the last few years, there have been some very promising ideas proposed that all have to do with nanostructuring. And people are starting to see real advances. They're starting to see figures of merit and some of you are probably better calibrated than I am. 1.2, 1.4, there's even a report of 2.4. There are arguments about whether you can believe some of them or not. But I think it's clear that some progress is beginning to be made and it all has to do with structuring material on the nanoscale. So it's become a field that is very interesting again and there's always this hope that you can improve the figure of merit. So we should understand a little bit. What is the figure of merit all about? So on the numerator it's about the electrical part. Seabed coefficient, we understand the seabed coefficient. That's mostly determined by where the location of the Fermi level is with respect to the bottom of the band. There's this little delta which tells you how much above the bottom of the band is the energy flow. And you can play with that a little bit but it's hard to do very much about it. Mostly it depends on the bottom of the band and the location of the Fermi energy. And then we have the electrical conductivity that we understand now. And when you're doing electrical conductivity what you wanna do is to get the Fermi level up in a place in the band where there are lots of channels and then you get more conductivity. Now the denominator. So most of the heat conduction is by phonons and we won't talk about that until lecture nine. So let me just talk about the numerator. Actually it turns out that most of the progress that's occurring these days is coming because people are learning how to engineer the lattice thermal conductivity. Now there's this great hope that if you could work on the numerator you could get some more advances. And it's still not obvious that people are being successful on that yet. But so how do you think about this? So we have this problem. This is how the magnitude of the CBAT coefficient varies. If you have a lightly dope semiconductor it's very big. As the Fermi level gets closer to the bottom of the band it gets smaller and smaller. When it's right at the bottom of the band it's about two K over Q. When it gets very deep into the band then the energy flows right near the Fermi level and the CBAT coefficient approaches zero. How about the conductivity? So when the Fermi level is way below the conduction band none of the channels are occupied and you don't have any conductivity. As you push the Fermi level up into the band more and more channels for conduction get occupied and you get higher and higher conductivity. So the figure of merit is the product of S squared times sigma. So it's a product of those two and you can see that it's going to be maximum somewhere when the Fermi level is located somewhere near the bottom of the band. So the power factor or that S squared sigma is going to look something like that. So the strategy is to try to locate the Fermi level at a point where the CBAT coefficient is not too small and the electrical conductivity is big enough. And people are trying to engineer that regime to see if you can get a little bit better performance. Okay. Now one of the other things I have to mention is I told you it's always optional whether you want to think about holes or electrons flowing through the valence band. The mathematical expressions as I mentioned yesterday if you want to compute the conductivity you should just think about electrons flowing through a conduction band or electrons flowing through a valence band. The expressions work in both cases. Now sometimes in problems like this you saw I do this too. Sometimes it's easier to look at a problem and say okay it's easier to think about holes and explain what's going on. But how would you think about this in terms of electrons only? So here's the way we would do it. So think at the top. We have an electron flowing into the n-type semiconductor. That means it's flowing out of the metal on the top contact. It has to absorb some energy to get into the semiconductor. That's why it's cooling up there. It's absorbing energy. It flows down to the bottom contact. It comes, it dissipates that energy when it goes out of the conduction band into the Fermi level of that bottom metal. So the bottom heats. Now how about the holes? So if I, forget about holes. How about electrons on the p-side? So electrons are moving up. So electrons are moving up in the p-type leg. So they move from the bottom contact. They have to fill a hole in the p-type case. That means the electron has to lose energy to drop down in the valence band and fill up a hole in the p-type semiconductor. So the bottom gets heated. The electron flows up and comes out the top contact. When it gets out the top contact, it has to get from the valence band in the p-type semiconductor up to the Fermi energy in that contact. It has to, because that's the only place that there are empty states. So it has to absorb thermal energy to get up there and be, so that's why the top gets cold. So you can always think about any problem in terms of electrons or holes. Sometimes it's easier to think about it one way or the other way. I also want to mention, measuring S, Professor Fischer talked about measuring thermal conductivities and things and some of this gets very difficult. I have a whole lecture where I talk about measurement considerations and mostly it's electrical because some of these thermal electric effects end up being a little more difficult to measure. But I just want to mention one thing here. You could think about measuring the CBEC coefficient this way. This is the top view of a two-dimensional sample. So you have two contacts on it. If I build a little heater here, maybe I just build a little resistor there and run current through it and get that end hot. Then I will make the right side hot, hotter than the left side. I'll have a temperature gradient. I should get a CBEC voltage across those two contacts. And I need to know what the temperature is so then I've got to figure out a way to measure the temperature at those two contacts. Maybe I can stick a thermal couple along each of the two and deduce the temperature. But that isn't quite going to do it because what I want to measure to get the CBEC coefficient of the sample is that voltage difference across between those two contacts. Delta V, CBEC or semiconductor there. But I have a voltmeter and I have one lead of the voltmeter that's attached to the hot contact. So that end of the lead is hot. Then the voltmeter is out there in the ambient. Let's say at 300K. So the voltmeter is sitting at 300K and the other lead that's connected to contact one is at 300K. That means that there is a temperature gradient across that lead. The end that I put on contact two is at a hotter temperature than the end down there at the voltmeter. If there's a voltage, if there's a temperature gradient across that lead, there's going to be a CBEC voltage across that lead. So really what we're measuring here is, turnout, you just add up those voltages and your measured voltage is actually the difference between the CBEC coefficient of the semiconductor and the CBEC coefficient of the metal lead. And when people are doing measurements and you're trying to be very careful, especially if you're measuring a small CBEC coefficient of a semiconductor, the metal has a small CBEC coefficient and people are careful to subtract these effects out. All right, that's about it. So we've discussed, I've tried to give you a physical picture for where these CBEC and Peltier effects come from. We've developed the basic equations. I'm going to go through them. We're going to have one more lecture where we go through this and I do it in a more formal way where we'll start with the general model and we'll just put the driving forces in and we'll see how all of this develops in a more straightforward way. But we have the basic equations and sort of the physical picture for how it works. And we have a little bit of understanding of these four thermoelectric parameters and how you would estimate them. And I've told you just a little bit about how you would build devices that use these effects and what this figure of merit is all about. All right, so that's the end of, what is this, lecture five? Four. Four, okay. All right, are there any questions? Yes. Sir, if you go to slide 13. 13? All right, let's take a look at slide 13. This one? Yeah. So, lots of arguments that we spoke about depending on electron having to take the energy and climb up to the delta n above the. Yeah. What happens in a nano-constituent that is flat and is between the two fermi levels? I'm sorry, the question again was what happens in a. How would this look for a nano transistor where we said the fermi level in the channel is somewhere between the two contours? Yeah. So, your question is, or maybe I'll rephrase. So, your question is what would happen in a nano transistor? Yeah. So, actually, if I were to draw an energy band diagram of a MOSFET, it might look something like this. Here's the fermi level of the source, and the fermi level of the drain is a little bit lower. The electrons in the channel are going to be flowing a little bit above here. So, you would actually expect a MOSFET to be a thermoelectric device, right? You would expect to see a little bit of cooling because the electrons from the source have to get up a barrier and into the channel. There will be a little bit of cooling there. Some people have thought about whether you can put this to useful use. It seems difficult, but when you run simulations, you can see that cooling there, and physically you expect it to be there. Now, let me answer, this is not a, you didn't ask this question, but since we're on slide, what are we on? Slide 13, let me answer it. You know, this physical picture of absorption, what's really going on there? So, we have a fermi level in the contact, EF1. And then, we have a semiconductor with a small slope in here. What's happening? Now, this is really what's happening. If I were to, this is energy. If I were to plot my fermi function this way, so this is F of E. So, here's one. Way below the fermi energy, the probability is one that it's occupied. Near the fermi energy, I have a thermal tail. Oops, I better do this differently. Then I get a thermal tail. Then if I go way above, I get a Boltzmann tail. It starts to decrease exponentially. The electrons that are at this energy, these electrons can flow into the channel. Once they do that, I've now got a highly non-equilibrium distribution in energy. So, people think about this as an analogy to evaporation. This is like a liquid evaporate. The electrons are like a gas. The ones that have high enough energy can evaporate into the semiconductor. But this is a highly non-equilibrium case. Now, what's gonna happen is all of these inelastic scattering processes in here are gonna rearrange themselves and fill in these missing energies and get us back to a thermal equilibrium case. But in order to do this, that's the amount of energy that has to be supplied to the system to get it back into the thermal equilibrium that it started with. So that's a better picture of the process of how electrons get from the contact into the semiconductor. Yes? Can you mention in semiconductor, heat flow is mostly by lattice vibrations. When we have high-dose semiconductor, it's a steel lattice vibration or electrons will play more important moves. You're asking what determines the mobility? No, no, the heat flow is mostly done by lattice vibrations in semiconductor, yes? Yeah. And if you have highly-doped semiconductor, it's still lattice vibration. Yeah, so the question is in heavily-doped semiconductor, it becomes more like a metal. You would expect to see more of the heat conduction coming from electrons. I'm not especially well-calibrated on that, but I seem to remember that you can get numbers on the order of 30% or so, that I'm looking at my students who does these calculations. Is that a reasonable number? Yeah. For a heavily-doped semiconductor like Bismuth Tellurider? One third. One third is reasonable, okay. So there's, so when you do dope these semiconductors just on the borderline of degeneracy, you try to locate the Fermi level right near the bottom of the conduction band, so you still have a little bit of C-beck coefficient, but you've got conductance. And now you have a significant number of electrons that can carry the heat, and Changwook is telling me that if you run the numbers, that could be one third. It's not completely insignificant, so it'll depend on the doping. Yeah. You said that we would play with the delta here to increase the C-beck coefficient. Pardon me? So you said that we can play with delta here to increase the C-beck coefficient. So how would you do that? Yeah, so the question is, the C-beck coefficient is all about the difference in energy between where current is flowing and the Fermi energy. The bigger that is, the bigger your C-beck coefficient. So the easiest thing to do is to make the semiconductor lightly doped and put the Fermi energy way below the conduction band, but that doesn't give you any conductivity. So let's say your Fermi energy is located near the bottom of the conduction band. You'd like to make that average energy of current flow as much above the bottom as you can. And what I said is it's hard to do much about that. It's not easy. One of the things that people point out is that if you're dominated by ionized impurity scattering, and we'll talk about this in lecture six, then the higher the energy the carriers are, the less they're scattered by ionized impurities. So that tends, in that case, you would have an energy dependent mean free path. It would be easier for current to flow at higher energies and at lower energies. That would tend to push this delta up a little bit. Now people also talk about engineering the shape of the band, and they talk about doing things like putting a parabolic band and superimposing a sort of narrow delta function shape band on top of it, getting a different shape of the density of states that kind of promotes more current to flow at a higher energy. I think you can probably do a little bit about it and everything you can do helps. And of course, if you can increase S a little bit, it goes to S squared and that's helpful. But I'm not aware of any ideas of things you can do that dramatically increase that little delta. Yes. What causes the Fermi levels to drop? The conduction can be less and less conductive. Well, I'm sorry. What causes the Fermi levels? Yeah? What causes that to drop below the equilibrium so they'd be less conductive? You mean the Fermi level on the right contact here? Yes. Well, let's see. What am I doing here on slide 13? I'm trying to... Oh. Yeah. Yeah, so on this particular, I mean what caused this to drop is that minus QV. So in this particular example, I'm just saying if I apply a voltage on contact two, I'm going to lower the Fermi energy and the question I was asking is how much heat do I pump under that case? So there I just put a voltage on contact two and that dropped it. If I'm doing the C back coefficient, then I have a temperature difference across and then a voltage develops on that contact to lower the Fermi energy such that it causes the, it stops the current from flowing. It gives me an open circuit case. Yeah. In this case, the temperature was constant because I was just asking, how much heat does the electrical current flow? Yeah. Being dissipated will let the right to go back. Oh. Yeah. Oh, where the heat is, yeah. So while you absorb heat on the left, you drop heat on the right because it has to be dissipated. All of the energy has to come down to the Fermi level. Now, if there is any elastic scattering going on here, that's sort of been out of our picture about what we're talking about. But we know it's present in real devices where we have long legs. So there's an additional I squared R loss that occurs inside that people just add. But remember, linear transport means things go are proportional to I or proportional to V. I squared R is a nonlinear effect, but when you're dealing with thermoelectric devices and trying to compute their efficiency, you have to add it in. Okay, so any more questions? One, yeah, there's one. Speak up, please. My 30, you know, four. 34? All right. Is this the right one? I was following you, first of all, the reason we were going here is because we were looking at what is it being created to do to see the coefficient is very small. And so the end of the part is not that, are not high as we can tell you. So we're assuming that you're all in the bottom and in the center, but close to the bottom. Yeah. So if we were to do like, if we were to go beyond that, here's what we're going to do. How, you know, how, how there'd be a single way you can put a circle in the middle to do a prep and then you see. How we see it like this. Yeah. If you see it like this, it would be complicated if you have to do all the energy. So if I'm understanding your question correctly, it's everything that we've, everything that we're talking about here is near equilibrium. So the electrons are flowing very near the bottom of the conduction band. If you apply a high bias, you get them lots of kinetic energy and they could be flowing at much higher energies. And it's your question, what would happen? So I mean, yeah. You think they look, you see my idea, you would need to like, for example, if I think why then, if we were looking at you see my idea, you would need to think a lot, but not in general, especially for the conduction band, they would look at you. So you see my, you see the elements here, but I'm not, but then in that case, we would have any role of the density of state with other things. So they're easy way to communicate if you see my position, when we're not looking at, when we're looking at something beyond one energy. Oh, so I mean, is your question, what happens if this is a degenerate semiconductor? Yeah, yeah, but if they like to have one energy, for example, how would you be able to see that? Yeah. Well, here they're not in one energy, but they're spread out in a few KT near the bottom of the conduction band. But if we were degenerate, they might be way above the bottom of the conduction band. You know, then I would just go back to, and I haven't given you really the proper mathematical description. You know, we discussed how you would do conductance for degenerate conditions. You get Fermi-Dirac integrals. Okay, when we get the proper expressions tomorrow morning, you'll see that you can do the integrals and you'll get, your C-BEC coefficient will involve Fermi-Dirac integrals of various orders. So you can always do the computation, it just gets a little more painful and messy. Now what I thought you were asking was, you know, what if you go to high bias? You know, do you still have these effects? And you do. But the way I would treat them there is, generally you would go to the Boltzmann equation and you would write moments and you would get these hydrodynamic flow equations. And you get various thermoelectric effects that you can see in these. Under near equilibrium conditions, there are these effects, but they're also present under high biases too. You get additional voltage drops due to electron temperatures when they're hot and things like that. Okay.