 You can follow along with this presentation by going to nanohub.org and downloading the corresponding slides. Enjoy the show. Okay, so good morning again and welcome back to lecture seven. So we're going to be talking about the Boltzmann transport equation in lecture seven. So let's dive into it. So here are the topics that we're going to be covering in the next hour or so. So I'll show you what the BTE will formulate the Boltzmann transport equation. And then we'll talk about how you solve it under some simple conditions and it'll just be steady state spatially uniform. Now we can solve the Boltzmann equation and we can get all of the transport parameters that we've been discussing. We can get the conductivity, the PELTA coefficient, the CBEK coefficient, the electronic thermal conductivity. I'll just do one, just a conductivity to show you how that works. Now one of the advantages of the Boltzmann equation is that there are some problems that you can do very simply with the Boltzmann equation that it would be more difficult to do with the land hour approach that I've been discussing. One of them is magnetic fields. It's easy to put a magnetic field in and see how a magnetic field affects conductivity. So we'll do that as an example. And that's a commonly used technique when you're characterizing semiconductor devices, so it's an important topic. Okay, so here's the idea. Let's say we have some two-dimensional phase space. Well, actually this is a one-dimensional problem. So electrons can move in the x direction only, but they have some momentum, or h bar kx of momentum. So that's in the vertical axis. So we have this two-dimensional plot, momentum space, position space. And we have some location there at some position and at some momentum, at some particular time. We want to know the probability f that that particular state is occupied. It's a number between 0 and 1. And we know what the answer is in equilibrium. It's the Fermi function. But out of equilibrium, it'll be something else. Now, it's interesting, if you reflect back on the Landauer approach, we knew the Fermi function in the two contacts. We never asked what the probability was of the states being occupied inside. There was no Fermi function inside. We're only in equilibrium in the contacts. In the Boltzmann approach, we're looking to try to formulate, find out what the distribution function is everywhere. So we don't know what it is if we apply a bias. That's the whole goal, so we need an equation. So our goal is to find an equation for that thing that's analogous to the Fermi function but out of equilibrium. And then we have to learn how to solve it. And we'll just talk about solving it near equilibrium, because that's what all the lectures are about. And then I want to show you that the answer that we get is the same answer that we've been getting for the last few days. It should be, we're solving the same physical problem. And then we'll add a B field and show you what happens when you add a B field because that's going to be very important for the next lecture when I talk about measurements and how you do Hall Effect measurements and what you can learn from a Hall Effect measurement. Okay, so the equation that we need to solve for F is this Boltzmann transport equation. And we can talk about where this comes from. So this was formulated by Boltzmann, back around the turn of the century before he knew about quantum mechanics. But we still use it. And the idea is that we will treat electrons as semi-classical particles. So they obey equations that are like Newton's laws, but just a little bit different. So the first equation is like the PDT, that's the derivative of momentum, that should be force. Force is minus gradient of the potential energy. So it's minus gradient of the conduction band edge. The force on an electron, if I don't have a magnetic field, the force is just minus q times the electric field. So that's just like Newton's laws. And it works in a semiconductor crystal as well. Now the interesting part about that is there's no band structure. It just tells how you move in k space. So where does band structure come in? Well, if I want to find out how it moves in position, I need to know its velocity. And it's, well, before we go there, I could track out its position in momentum space as a function of time just by integrating that equation. Now if I want to know where it is in position, I need to know its velocity, okay? But if I know where it is in k space, I can look up the slope of the energy band at that particular point in k space. And 1 over h bar times the slope is the velocity, okay? So I know how fast it's moving in position space. And then I can just integrate from its starting point and find out where it's located. So these are the equations of motion. This is what we call semi-classical transport. They're analogous to Newton's laws. We're making some assumptions here. One of the assumptions is that the electric field varies slowly on the scale of the electron's wavelength. So that there aren't quantum mechanical reflections. So we don't have to worry about quantum transport. When you can't do that, then you resort to techniques like non-equilibrium Green's function, which is basically a quantum mechanical version of the Boltzmann equation. Notice that there's no effective mass here. If we have a band structure, we can just trace out these trajectories. So we don't need an effective mass. We just, if we're given a dispersion, and if it's graphene, it's a very unusual E of K, we just take its slope and we can solve these equations and we can figure out what's going on. Okay, so the idea is we have these equations of motion in phase space. Phase space is momentum in position space. So for a one-dimensional problem, phase space is two-dimensional. For a 3D problem, phase space is six-dimensional. That's one of the problems. It gets very complicated to solve these six-dimensional problems. But let's say we have an electron that starts out. We can just integrate with time and trace out its trajectory in phase space. So this is some trajectory. Okay, now what we want to find is if we look at some point that's filled red dot there, what we're asking is at some particular position, at some particular momentum, at some particular time, what's the probability that that state is occupied? It's some number between 0 and 1. That's what we're trying to compute. Well, all I have to do is to look upstream in that trajectory. If I know what the probability that the state, a little bit earlier in position, a little bit earlier in time, or a little bit earlier in momentum, a little bit earlier in time. If I know what the probability that that open state was, then that probability just flowed along the trajectory. And nothing changes along that trajectory. Electron just follows that, so the probability that the state is occupied is just constant along that trajectory. So that gives me a really simple little expression. The probability that the filled state is occupied is just equal to the probability that the corresponding state upstream was occupied at an earlier time, at an earlier position. So what that is really saying, if I subtract those two and divide by the delta t, that's just the total differential of f. It says df dt is 0. There's no change in probability as long as I follow the electron along that trajectory. So that's our equation. This is really the Boltzmann equation. It's actually the collisionless Boltzmann equation, because when I start putting scattering, we'll see things can change. We can get knocked out. But this is the equation that I would solve for ballistic transport. Now, that's a total change with respect to time. f is a function of x, momentum px and t. So I could just expand that derivative out. It's partial f, partial t. And then I could use the chain rule. It's partial f partial x times dx dt plus partial f partial px times dp dt. That's just expanding out the total derivative with the chain rule. And then I can recognize that dx dt is just the velocity of the electron, moving along that. And dp dt is just the force on the electron. So if I want to generalize that to three dimensions, instead of having vx times df dx, I would have velocity dot gradient of f. Instead of having fx times df dpx, I would have force times gradient in momentum space of f. And gradient in momentum space is just the derivative with respect to the p's instead of x. So in the blue box down in the lower right, that's the way Boltzmann wrote down the Boltzmann equation. And we would consider that. We haven't considered any scattering events that could knock an electron off of that trajectory, so it's a ballistic Boltzmann equation. If I solve that, I would get the ballistic conductance. Okay, so this is our result. Now there are other reasons that things might change along the trajectory. I might have a solar cell. I might be shining light on the semiconductor. The photons might be increasing the probability that that state is occupied. So I could add a generation term instead of having that side zero. Or that electron might be going along and that might recombine with a hole and disappear. So there are other ways that the probability could change along that trajectory. And I would have to add those explicit terms in. So those would be problem dependent terms. If I knew what the optical generation was, I could put it in. If I knew what the recombination processes were, I could put them in. We're not going to be treating those today. So, okay, so we have our collisionless Boltzmann equation. And just to remind you the assumptions, we're using these semi-classical equations of motion. We've assumed that we have some crystal with a band structure, E of K. So the total energy is the bottom of the conduction band. And then E of K tells me how the energy goes above the bottom of the conduction band. The velocity is just given by the gradient of that energy band structure. And I'm thinking of this as a particle. So I'm asking what is its momentum and what is its position? And quantum mechanically, I can't specify the position and momentum simultaneously. There's this Heisenberg uncertainty principle. So there's an assumption here that that uncertainty is small enough. My device is big enough that that uncertainty in the location is small enough that I can think of this electron as a particle. And it's not spread out. If that isn't valid, then I go to energy F. Now we're going to neglect these recombination generation processes. That's not really a fundamental assumption. I can put them in if I want to. That's straightforward. Now one of the other things, we've actually made a huge simplification that I've glossed over. This, we're interpreting this equation. This is the probability of finding an electron at some particular position, at some particular momentum, at some particular time. It's really much more complicated than that, because there are all kinds of other electrons. And those other electrons are exerting their Coulomb forces on this one. The probability that this state will be occupied, depends on how close those other electrons are. I can't really write, in principle, I can't really write an equation just for this one electron. I have to write one in this multi-electron space, because they're all correlated. Now the assumption that we usually do is we neglect these correlations. And we say that the effect of all of those other electrons is that we just solve the Poisson equation, and we get some self-consistent potential or electric field. And we include all of those effects in that average charge density. That neglects these detailed particle-particle interactions. Sometimes they can be important. Now one of the things some of you may know about Monte Carlo simulation. Monte Carlo simulation is a statistical technique to solve the Boltzmann equation. But it also does a little bit more, because it can treat these particle-particle interactions. Then you track particles through a semiconductor, choosing random numbers to select scattering processes. And if you put 10,000 particles in, you can compute the Coulomb forces between all of the particles, and you can get more than the Boltzmann equation. If you treat those electron-electron interactions carefully. But most solutions of the Boltzmann equation that you see write this single particle equation and assume that the effect of all of the other electrons can be taken into account in this average sense. But the big approximation that we've made is we've neglected scattering. That's what the zero is about. So let's think about that. The electron we said, well, it's probability of being occupied is just the same as it was upstream. But that's not necessarily true in the presence of scattering. Because I could have electrons scattering into that state from other states. And the electrons in that state could be scattering out into other states. And that's going to change the probability that the state is occupied. Now notice, I'm assuming that this all happens at a fixed position. So that's another assumption that we make. Some people have looked carefully at it and tried to go a little bit beyond it. The assumption here is that the scattering events occur very, very quickly. Electron will come in, it'll get knocked in another direction almost instantaneously, it hasn't got time to move anywhere during that scattering event. Now, there are people that worry about this under very high electric fields. You might be pushing the electron very fast and it might move a little bit during the collision process. It might change its position. That's called an intra-collisional field effect. But it's one of these things that doesn't seem to be important under the vast majority of conditions that we consider for semiconductor devices. So this idea that they occur instantly in time is a pretty good one. So what we have to do is to say that, well, there is some time rate of change of F due to collisions. And I won't talk about how you write that in detail, I'll just write that as C hat. It's some collision operator that describes this in scattering and out scattering process. So when Boltzmann did this, he had a collision term that was just, he had two classical particles colliding. He didn't know about quantum mechanics. When we do this for semiconductor devices now, we treat this process with Fermi's golden rule and we calculate these transition rates. And that's another part of the semi-classical approach. So when electrons are moving along the trajectory, it's like they're obeying Newton's laws, or something that looked very much like Newton's law. And then when they get knocked somewhere else because of the scattering events, we treat that quantum mechanically by Fermi's golden rule. Though it's kind of a hybrid approach. And as I think Professor, yeah, okay. All right, so it's the scattering that makes this equation so difficult to solve. So, and that's why if you use numerical techniques like Monte Carlo simulation, you can treat these scattering processes in great detail and the computer will do it for you. If you want to get analytical solutions and get some insight into what's happening, then you have to simplify the scattering operator drastically. This is a common simplification. So commonly it's said that the rate of change of F due to collisions is minus the deviation of F from its equilibrium value over some characteristic time. Turns out that the characteristic time is the momentum relaxation time. That's not completely obvious, but that's what it turns out to be. So C sub F is minus, that should be a delta sub F. The difference from equilibrium divided by a characteristic time. Now you can look at that, you can see that if you look at the top line, that the first term is an in scattering process. Because that's positive, that's increasing F. The second term is an out scattering process. The second term actually makes a lot of sense, because it says, well, if I have some probability that that state is occupied, then 1 over tau is the probability per unit time that it will scatter out. So the second one is very solid. The first one, it's harder to justify why it should have that form. Now it turns out that you actually can justify this under some conditions. And it takes a little bit of effort, so I'll just, you can find it in my book or in other books. This approximation can be justified. First of all, you need to be near equilibrium. So this delta F has to be small, but that isn't enough. Even near equilibrium, it doesn't always work. It works when the scattering is isotropic, or it works if it's elastic. So for silicon, germanium, the common scattering mechanisms are ionized impurity scattering, which is elastic, so you're fine. Or acoustic phonon scattering, which is isotropic, so you're fine. For materials like gallium arsenide, 3.5 materials, the dominant scattering mechanism is a polar mechanism due to optical phonons, polar optical phonon scattering. It's inelastic, and it's anisotropic. So this created lots of problems for people because you can easily solve the Boltzmann equation for silicon and germanium, but it's much more difficult to solve it for gallium arsenide and 3.5s. Because you can't make this assumption. Now, a lot of times people go ahead and do it anyway. And you get answers that are reasonable, but you can't rigorously justify it. Okay, what does this mean? So let's take this, this is our Boltzmann equation. Let's see, let me, instead of setting it equal to zero, let me set it equal to the deviation from equilibrium over tau. Let me assume spatial uniformity, so there'll be no gradient with respect to r, and let me assume that there's no electric field. So I'm just going to inject some excess electrons into a semiconductor and see what happens. So if you solve that equation, what you'll find, the equation you'll get is this little differential equation that has a familiar solution. It says if I perturb it at time t equals zero, the perturbation will decay as e to the minus t over this characteristic time. So it will just exponentially decay away. So physically what this equation says is that if I perturb the system, the system will react with a negative sign. It'll try to react in the opposite direction and bring me back to equilibrium. And you'll get back to equilibrium in this characteristic time. So it's a nice plausible thing to do. You make assumptions like this in many problems. If you don't know the details of what's going on, it's reasonable to say if I perturb my system away from equilibrium, the system will decay exponentially back towards equilibrium with a characteristic time. So that's what the RTA means. So now we have our Boltzmann equation in a form that we might be able to solve. So I'm going to write it steady state, so nothing is changing with time. So the time derivative is gone. I'm going to get rid of the df dx because I'm going to assume spatial uniformity because all the problems I've been talking about until now were for uniform resistors. I wasn't resolving things in space. It just gets a little more complicated. I have to apply boundary conditions in space. Actually it gets a lot more complicated. And we'll see if we can make the relaxation time approximation. We're going to do that. And we will neglect magnetic fields and just have an electric field. And we'll get an equation that we can solve. And let's talk about solving that. Okay, so this is our near equilibrium steady state Boltzmann equation and the relaxation time approximation. Oh, actually I am going to include. I forgot what I'm doing here. I'm going to let, we'll keep the spatial terms here. So let's do that. Okay, now if you look at this, remember there, f consists, f is just a small, a little bit away from equilibrium. So f is f naught, the equilibrium value, plus a little perturbation away. This is the equation I have to solve. It's reasonable to assume on the left hand side that when I take the gradient, I can just take the gradient to the big part. And when I take the gradient in momentum, I can just take the gradient to the big part. And that should be very close. Now, when I get all done and get a solution, I can plug it back into this and make sure that the gradient of the big part is much bigger than the gradient of the small part, and that is the case. So if you do that, it gets to be a very easy equation to solve because the equilibrium f is the Fermi function. So I can take that gradient, the left hand side is known. I can solve for delta f. That's my perturbation. And then it's relatively straightforward to solve for the Boltzmann transport equation. And this is the solution. Okay, so now a little bit of math. So just, you know, afterwards you can go through the details if you want to. But let's work on this just a little bit to see if we can simplify it. So I'm going to write my Fermi function as 1 over 1 plus e to a bunch of stuff. So theta is e minus quasi Fermi level over kT. And remember, e is a total energy. So it's e sub c, the bottom of the conduction band, plus the kinetic energy, which is e of k, or h bar k is momentum p, then minus the quasi Fermi energy. Okay, now I have to take the spatial gradient. That's the first thing I have to do. So let me use the chain rule. So I'll take the derivative of f with respect to theta, because that's an easy derivative to do. And then I'll take the spatial gradient of theta. I also have to take the gradient of f in momentum space. So I'll take the gradient of f with respect to theta again. But I'll take the gradient of theta with respect to momentum. And you can see here that the df d theta is just like df de with a kT out front. Okay, so if I do it that way, and if I can evaluate those two gradients, then it turns out I'm going to be able to simplify this a bit. So let's do that. So here's my solution now. Remember, theta is just e minus quasi Fermi level over kT. And I'll take the spatial gradient. So there's a 1 over kT out front. When I take the spatial gradient, I can take the spatial gradient of the bottom of the conduction band, e sub c. That's the force on the electron. And I have to take the spatial gradient of the quasi Fermi level because that can be position dependent. But the band structure doesn't change with position. It's one piece of silicon. It's only a function of k or momentum. So there's no gradient there. Okay, but the temperature might also be changing. So I have to use the chain rule and I have to take that stuff in brackets times the gradient of 1 over kT. So I'll just have to work all of that out. I also need to take a gradient in momentum space of theta. But when I do that, the only thing that depends on momentum is the e of p or the band structure. And the slope of the band structure gives me the velocity. So I just get velocity divided by kT. So I can plug those two gradients back into the top equation. And I'll get this stuff in the squiggle brackets. Now, something interesting has happened here. Let me track this down. You can see on the top, there's a minus qe times gradient of theta p. So that gives me a minus qe dot velocity over kT when I take that gradient. Now you can see in the bottom expression, the electric field has disappeared. The reason the electric field has disappeared is because gradient of the conduction band edge in that spatial gradient of theta term, the force on the electron is minus the gradient of ec. Plus the gradient of ec is minus the force. That's plus q times the electric field. That cancels that other one. So all I'm left with then is instead of an electric field, I have a gradient of a quasi-firmly level. And the electric field has disappeared. And then I have a gradient of the inverse temperature. Okay, so I can lump all of those things together. And my solution now starts to look simple. Just because I lump some terms together. It's tau minus df de times velocity dot f. F we call a generalized force. Remember we talked earlier about what the two driving forces are for current flow. They were differences in quasi-firmly level and differences in temperature. The definition of the generalized force in the Boltzmann equation is all those things we were left with when we did the gradients. It has a gradient of the quasi-firmly level and it has a gradient of inverse temperature actually. So the two driving forces are the same two driving forces that we're used to seeing. If I had neglected spatial uniformity that generalized and if I had uniform temperature, that generalized force just would have been minus q times the electric field. But this includes diffusion terms and temperature gradients. All right, so we've solved the Boltzmann equation. At least under these conditions, near equilibrium, that's the answer. So what do we do with it? Well, we'd like to compute the conductivity and all of those other parameters that we can. So let's see how that works. So if I want to compute things that I can measure, what I have to do is to take some moments of the distribution function. Compute some sums. So for instance, if I want to find the total number of electrons that are in the semiconductor, I'll take a chunk of semiconductor. And most of you have probably seen how if you establish periodic boundary conditions, then you have discrete values of k that can exist. You have some finite number of k states. It may be very, very large and a very big sample. And then we convert the sum to an integral. But in principle, there's a finite number of k states and a finite volume. You simply go through and you add up the probability that every one of those is occupied, some number between 0 and 1. And then if I multiply or if I divide by the cross sectional, let's see, that first area, that should be the volume. If I divide by the volume, then I'll get the number per unit volume and I'll get the electron density. Now, when I'm doing the electron density, I'm adding up the big part plus the small part. The small part is negligible. So all I have to do is to sum up the big part. So I'm in near equilibrium. If I apply a small voltage across this resistor, I don't change the number of electrons there. I can just get the number from the equilibrium distribution. But for the current, it's different. So the way I get the current is I go into each state and I say, well, each state is going to carry a current that is minus q for the electron charge times the velocity that it's moving, times the probability that it's occupied. This current is always charge times velocity times number that are there. Now, the probability that that state is occupied is f naught plus delta f. Here, I threw away the big part and I only kept the small part. And it's because I know that in equilibrium, if I sum up all of those states, the current is zero. Every positive velocity is balanced by a negative velocity. So only the small part contributes to the current. So that's where I would plug my solution in. If I want the energy current, I would just do a sum. I would take each state. Current is going to be proportional to velocity. And then I would weight it, instead of weighting it by charge, I would weight it by energy, or kinetic energy. That would give me the kinetic energy current. Some books, you see people identify that as the heat current, but it really isn't the proper thermodynamic definition of heat. If you want the heat current, you should weight each state by e minus the quasi-firmly energy. And when you do this in books that are solving the Boltzmann equation, they'll usually invoke thermodynamic arguments as to why the heat is e minus f, and not just e. But we've seen from that physical picture, the electrons hopping from the Fermi level in the contact to an energy in the channel, that this is the heat they have to absorb. So it all makes sense. So if the sample is big, then we have many finely spaced states. And we convert that sum to an integral. So when we do the integral, we have to be sure that we count the states properly. So that introduces a density of states. So N of k, dk is the number of states that are in that small dk volume. And you may have seen this before. If you haven't, I'd refer you to a lecture. These densities of states, now this is the prescription we use for converting the sums to an integral. So in one d, the density of states is l over pi and 2d, it's area over 2 pi squared. In 3d, it's volume over 4 pi cubed. And if you make that conversion, then we can convert those sums to an integral. Okay, so let's see how this works out. Let's compute the current density in 2d. So the current density in 2d, I'm going to take a sum minus q times a velocity times delta f. I'm going to use my solution of the Boltzmann equation and plug it in. And we're going to see if we can work it out. So when I plug that solution in, I get an answer like this. The first V there is because, that comes because current is velocity times f. The V dot f came from the solution to the Boltzmann equation. So if I look at that, V dot f is a scalar. And then I have a vector current and I have a vector V inside. So everything works well. If I just bracketed things differently, I could do it this way. I could take V times a vector V times a vector V dotted with a vector f. What's V times V? How do I interpret that? That must be a tensor, right? So I'm really getting a matrix. So I'll show you how this works out. We just have to be careful about doing that. In general, it's a tensor. And in general, I can get different conductivities and different directions. Okay, but let's do a simple isotropic case. Let's say that the force is just due to gradients of the quasi Fermi level in the x direction. There are no temperature differences. Then V dot f is just Vx times generalized force in the x direction. Then if I want to evaluate my x directed current, I'll use the x directed velocity inside the sum. And I'll insert that expression for delta f and I'll get a Vx out front and another Vx inside the square brackets. So I end up with an expression like this. So you can see I'm getting what I'm supposed to get. I'm getting something in the big brackets times a gradient of the quasi Fermi level. Current is supposed to be conductivity times gradient of the quasi Fermi level. So what's in the big brackets must be the conductivity. So I'm getting the answer that I should get. I look at all of those terms in the big brackets and that's my definition of conductivity from solving the Boltzmann transport equation. It looks a little different from what we've been seeing before, but it must be the same thing because we're solving the same physical problem. So to work that out, I've got to convert that sum to an integral and work out the integral. So the way we do that is we just convert to an integral over k space. We multiply by the density of states, which is a over 2 pi squared. In 2d, when I do that integral over k space, it's kdk d theta. I'll integrate magnitude of k from 0 to infinity. The Brillouin zone has some finite maximum, but I'm going to rely on the Fermi function to take the probability down to 0 before I get there so I can extend the integral all the way to infinity. And then I'll go from 0 to 2 pi. So let's see, if we do that, if we use that prescription to convert the sum to an integral and we put it in the top equation, we'll get an integral like this that we have to work out. Vx is the magnitude of velocity cosine theta. So then I can separate my integral out into a part due to magnitude of k and a part due to over theta. And let's see, so now I've got an integral like this on the top that I need to work out. And if I do the, I can first of all do the integral over theta, over cosine squared theta. And I get this result. I think there I've just done the integral over cosine squared theta. Okay, now we're going to do a conversion. Remember, everything that we did, we're naturally working in k space here. Everything we did in the Landau approach was in energy space. So I can do a change of variables from k space to energy space. And when I do that change in variables, I'll get expressions that are only in terms of energy. Now, just to make the math simple, I'll assume that the scattering time is independent of energy, so I can pull it out front. And I get an integral like that that I have to do. And now, if I look at that integral, and if I normalize the energy and call the energy minus the bottom of the conduction band divided by kt by eta, and if I define this eta f, which is a normalized Fermi energy, then I can take that expression and I can convert it. Let's see, I can insert my definition of my Fermi function. I can use those normalized energies and I can convert it into the bottom expression here. Now, that's starting to look like a familiar integral, right? It looks like a Fermi-Jorak integral. And it has eta to the first power. So it looks like a Fermi-Jorak integral to the first power. Notice I've done a little trick here without mentioning it. I had a df de inside the integral. But the Fermi function has an energy and a Fermi energy that come in similar position to just one has a minus sign. So I can pull that minus df de out and make it a plus df, the Fermi energy, pull it outside the integral. So that's how it got outside the integral. So it looks like I'm going to get a Fermi-Jorak integral of order one when I do that integral. But the property of Fermi-Jorak integrals is when I differentiate them with respect to their argument, it kicks it down one. So the final answer is this. So like I said, it's a lot of straightforward math that you just have to be careful about and if you're careful about everything and check my algebra, this is the answer that you get. So we look at that and we say, what is that? It doesn't really, you look at that, it doesn't look intuitive. What is it? Well, I could stare at that a little bit. And one of the things that might occur to me is that I remember I can evaluate the electron density. I showed you the sum about how you do an integral. That would just be the 2D density of states times the Fermi function. If you do that integral, you get this expression, which has a Fermi-Jorak integral of order zero. So that term there with the Fermi-Jorak integral of order zero is proportional to the electron density. And actually, if I go through and collect up terms and everything, I'll find that if I use that, if I express the results in terms of carrier density, I get sigma is nq times q tau over m. Now that looks familiar. That's what I'm expecting to get. Now if I had left that scattering time inside the integral, it normally has some energy dependence. It's normally described by power loss scattering. So it goes as e minus ec to some characteristic exponent s, like minus 1 half for acoustic phonons, plus 3 halves for ionized impurities. Then when I do the integral, I've got that energy dependent scattering time. In the end, you get the same answer. You just replace tau naught by the appropriate average scattering time. But it still doesn't look like the Landauer approach. We have no transmission and no m's and anything like that. So it should also look like that. So let's see where that comes in. So if I go back, before I converted to energy space, let's go back here and look at it again. I'll just do my change of variables to energy. And when I do my change of variables to energy, I get the top expression. And now I see g sub v, that's a valley degeneracy. That tells me how many equivalent valleys that I have. That depends on the band structure. Remember, there are six equivalent conduction band valleys in silicon and 3D, things like that. M or pi h bar squared, I can recognize when I look at that expression, I can recognize the 2D density of states. And so if I look at this now, and I have a v squared tau, so I'm going to take one of those v's and put it beside the density of states. Because I kind of remember that number of modes had something to do with velocity times density of states. And I'm going to keep one of those v's next to the tau m. Because the mean free path is something like v times tau. And now I remember that the average x directed velocity when I average over angles is 2 over pi times the magnitude of v. And the number of modes is really h over 4 times the average velocity in the transport direction. And the mean free path is pi over 2. So if I go through, if I insert a pi over 2 and then cancel it with a 2 over pi, then I can start to recognize some things that are very, very close. And when I get done, now if I insert a 4 over h and then divide by h over 4, the bottom line is I'm going to get that expression on the bottom, which now is a very familiar expression. So you can see that if I had not defined the mean free path to be pi over 2, I wouldn't have got the results in exactly the form of the Boltzmann equation. Okay, so in the end, we could write it as 2q squared over h times the effective number of modes times the average mean free path. So it has to be equivalent, and it is, you get the same answers. And similarly, we could do all of the other transport coefficients and we would get all of the answers that we got. Okay, so if you want to see a lot more about this, you can look at these books and online resources. If you want to see a more formal connection about these two approaches done without assuming any simple bands or anything for an arbitrary band structure, you can look up this paper by my student, Changwook. And that's also discussed in an online lecture. Okay, so we have these two different approaches and you might ask, why talk about two different ways to do it? Now the advantage of the Landauer approach is it gives you some nice clear physical insight into what's happening with a minimum of mathematics. It works in the ballistic limit. You can solve the Boltzmann equation in the ballistic limit too, but it gets very, very hard to impose the right boundary conditions on it. And it works in the clausibilistic and diffusive regimes where it's easiest to use the Boltzmann equation. Now the Boltzmann equation has some advantages too. It's easy to put a magnetic field in, I'll discuss that in a minute. It's easy to treat anisotropic materials when all of these transport parameters become tensors, you have different conductivities and different directions. You can solve it if you want to resolve things spatially. You can impose spatial boundary conditions. If you want to do it away from equilibrium, you can do that. There are simplifications and approximations that you have to make. As I mentioned, you can do ballistic transport but not nearly as easily as you can with a Landauer. And a lot of times when you're doing it, you get the feeling that I'm just doing a lot of math and a lot of change of variables and finally you get an answer. Now you see a DfDe here, from the Landauer approach, we know what that means physically. Current flows when f1 minus f2 is positive. And DfDe tells me the difference between f1 and f2. It just came in from some chain rule I did here in the Boltzmann equation and it wasn't so physically transparent as to what it was. So you really should know both approaches. Okay, so there's basically only one more thing that I want to discuss because it's widely used, so we should know about it. But, and it's easy to add a B field to the Boltzmann equation. It's not quite as easy to solve the Boltzmann equation with a B field. So I'll show you how it works without all of the details. So here's our Boltzmann equation again, okay? And we could do this in steady state by eliminating the DfDt. Just set that to zero. We could use the relaxation time approximation so that we can get simple analytical answers. But now we could add the force due to the B field, all right? Just go ahead and solve it. So we get an equation that looks like this. I'm assuming, I'm going to assume spatial uniformity so there's no gradient with respect to position of f. But I'll get two terms for the force. The first term is the electric field, the one we dealt with before. The second term is the magnetic field. And the right hand side comes from the relaxation time approximation. Okay, now, the temptation is to solve this the way that we did the previous problem. When I take those gradients on the left hand side, I'll just take the gradients of the big part and then I've solved it, okay? Now actually, that doesn't work and we should discuss why. So take a look at this. Let's say on the left hand side, we just take the gradient of the big part, okay? Now we say we're done. We've got the solution, we just solve that equation for delta f, okay? But look at that term that involves qv cross b dot gradient p of f naught. Okay, the one with the electric field is fine. That's like the one we did before, no problem there. But look at that other one. If I take the gradient of p of f, the way I could do it is to take the f, the e and then take a dp, d derivative energy with respect to momentum. Just use the chain rule. The gradient of the energy in momentum space is the velocity. So that derivative gives me a term that is proportional to the velocity. Now look, if I put the velocity back into that expression, I get v cross b dot v. V cross b is perpendicular to v. When I dot it into v, I get zero. So in that order of approximation, I get no effect for the b field. So that isn't going to work. So there are different tricks that people use to get around this and find a solution. So this is going to be a more difficult, we just added one term to the force. But now it's a much more difficult equation to solve. But here's one way that you can do it. Remember, when we didn't have the b field, the answer had this form. A lot of times the way you solve a differential equation is to guess the form of the solution and see if it works. So why don't we guess the form of the solution with a b field? Why don't we guess that it's going to look like the solution without the b field? But instead of that generalized force, I'm going to have some other vector g that I don't know. And I'll see if I can plug that in, whether that works, and what the expression for g is. So that's some unknown vector. And if I can plug it in and find a vector that works, then I've solved the Boltzmann equation. Now, this is where it takes some tedious algebra, right? Between here and here is, probably would take me about a half an hour of. You wouldn't want to see it. It's in my book, all right? But trust me, if you do that, if you go through, you can find the vector that solves that equation. This is the answer, all right? It's not pretty, but that's what it is. You'll find that in textbooks. So what is this omega sub c? This is called the cyclotron frequency. If you have an electron in a vacuum and you apply a b field, the electrons will rotate around the b field at some frequency. The frequency is omega c, q times the magnetic field divided by m. Now, we have a term in the denominator called omega c times tau. And if the b field is small, that's going to be much less than 1. We'll talk physically about what that means in a minute. But so I could ignore those omega c times tau terms for low magnetic fields. All right, so let me, I'm just going to basically ask you to, I'm going to show you how this would work out in 2D without going through all of the details. So in 2D, we have carriers flowing in a sheet in a two-dimensional plane. So this vector g is going to lie in the xy plane. And I'm going to assume a small magnetic field. So that means that these, I only have a one in the denominator. And you can see that the last term there goes as a square of the magnetic field. So if I assume a small magnetic field, I can ignore that second order term. It goes quadratic in b field. So I'm only going to get the first two terms. So my g is going to simplify considerably. So if I take that solution and if I put it into my expression for the current density in 2D, then you work it out. And I'm not showing you any of the details here. You'll get a current density in the xy plane that will look like this. The first term is what you would get without a b field. Conductivity times electric field, just what we got before. We get an additional term and since the Lorentz force involves a cross product between velocity and b field, it's not surprising that we get a cross product. So we get an additional term and the sheet conduct in sigma sub s is just what it was before. When you collect up all the constants after you've done all of this, you'll get something that has the units of mobility. But it is not quite the mobility. It is the actual mobility that we've been talking about, multiplied by some numerical factor. And the numerical factor that you multiply it by is the average over energies of the scattering time squared over the, of the average over energy of the scattering time and that whole quantity squared. For an energy dependent scattering time, those two things are different, when you average the square of the momentum relaxation time, you'll get a different answer than when you average the momentum relaxation time and then square it. So it ends up being a numerical factor that is often between one and two. But I've seen in some complex band structures like balance bands and things that can be, people say it's four or five. You need, it's something that it's difficult to pin down because you need to know both the band structure and the energy dependence of the scattering events. And then you can work out these averages. People call that the whole mobility. And that'll be important when we talk about measurements. So if I look at that vector expression, and if I just write out its x and y components, I could write it this way. So what I have is diagonal components that were just like they were before. And I have two off diagonal components such that if I scored in an x-directed current, I'm going to get current flowing in the y direction. Things like that. Okay, okay. So when we did the equations for thermoelectric transport in lecture five, we got expressions like this. In all of these quantities, sigma, pi, kappa sub e, s, they were all just numbers because I had an isotropic material. Now when we apply a b field, even though I have an isotropic material, they're no longer diagonal matrices or numbers, they're tensors. They have off diagonal components. So every one of them is going to involve this cross product and every one of them is going to have some similar structure. Their diagonal components will be the numbers we worked out before. Their off diagonal components will be proportional to the b field. All right, so then, just a couple of more things here. All right, so we have two different ways of writing this. I can either write this as a matrix equation or I can write it as a vector equation. They're the same way of writing two different things. Let me give you a physical picture for what's going on because when I talk about measurements, we'll talk about Hall Effect measurements. And you can see physically what's happening. If I force a current through this n type semiconductor, that means electrons are going to be flowing in the minus x direction. Now that'll actually induce an electric field in the positive x direction. So the force on the electrons is minus q times electric field. That's what's pushing the electrons to the left. But if I take minus q v cross b, which is the Lorentz force, those electrons moving in the minus x direction are going to experience a Lorentz force in the minus y direction. Because of the b field that's pointing out of the page in the z direction. Just take v cross b with a minus sign. That means electrons are going to be deflected and pile up on the bottom side and they're going to be pulled away from the top side where they'll expose some positive charge. That means an electric field is going to develop in the orthogonal direction, in the y direction. It'll go from the positive charge to the negative charge. Which means that if I put a volt meter across the sample in the y direction, I'll measure a voltage, which is called the Hall voltage. That Hall voltage is very, very useful for characterizing semiconductors. You go to any semiconductor lab and there will be a magnet and a setup for doing Hall effect measurements. Because as we'll discuss in the next lecture, you can learn a lot about the mobility and properties of the material by doing those measurements. So we get a Hall voltage that's just proportional to the magnetic field. And you can think about it in terms of this current equation too. The e cross b term and you'll get the same answer in terms of the signs of the effect. Okay. All right, so I think that's mainly what I wanted to talk to you about. Just to acquaint you with the Boltzmann equation and how it all works. So the idea in the Boltzmann equation is that we try to determine what the probability that a state at some position, at some momentum, and at some time is occupied. We formulate this Boltzmann equation. It's based on this semi-classical picture, so it's not exact. It works when the semi-classical picture works. It's a single particle equation, so it ignores these correlations. So when electron densities get high, people worry about this a bit. But it is often taken, when you don't have to worry about quantum transport, it's often taken as the fundamental starting point. And then everything else is a simplification of this. Now notice that it's six dimensions. This is what makes it so difficult. When you take a commercial simulation package for semiconductor devices like Centaurus or whatever, you solve a simplified version of this. A drift diffusion equation can be derived from this. Just as we derived the drift diffusion equation from the Landau approach. It's a simplified version of the Boltzmann equation. People say, well, why don't you just solve the Boltzmann equation directly? There it is. Just impose boundary conditions and solve it. The challenge is that, in general, it's six-dimensional. Because there are three dimensions in position and three dimensions in momentum. And it is very difficult to solve six-dimensional problems. They get very big. This is what people call the curse of dimensionality. We solve this in one dimension for nanowires. It's easy to solve, right? But when you go to six dimensions, the size of the problem just explodes. And it gets enormously difficult to do that. Which is why people use Monte Carlo simulation. Which is a statistical way to solve this equation. And requires much less burden. So you can sometimes solve it analytically. Nerequilibrium, if we assume the relaxation time approximation. That's what I've been discussing. So this semi-classical picture assumes a bulk band structure and a slowly varying potential. So there aren't any quantum reflections that we can think of this as electrons as particles and precisely specify their position and momentum. Under near equilibrium conditions, we can use the relaxation time approximation, not always, right? We can only justify it for specific scattering processes. But what you'll find when you read papers and things is that people use it all the time, even if they can't justify it. Because it's really the only way you can get some reasonably simple analytical solution. And from the solutions, you get all of the transport coefficients, the same ones we got before. And you can go beyond that and get anisotropic transport and other things as well. Okay, so let me stop there and see if you have any questions. And again, we have this fellow with the microphone. So now the next lecture will have relatively little math. Because I'm going to talk about how you do measurements. But we'll use all of these concepts, yes? At the beginning, you mentioned when you derived the cosmic transport equations, you said probably I didn't listen to you carefully. Total derivative of function at with time, zero. Is it due to probability or making assumptions is a bit of a barrier? Yeah, so the physical picture is this electron is moving through its trajectory in phase space. So along that trajectory, the probability that the state is occupied is not going to change with time. That position is going to move along that trajectory. So when I looked at any one position, then I could say the probability that this state is occupied is the probability that the state just a little bit upstream was occupied a little bit earlier in time. Everything just moved along in case space. That's the picture that we have. Pardon me? Actually it's one, every way is one. No, I mean actually we extend that, if f is one-half along that line, then f is one-half everywhere along that line. So if I look a little bit earlier on that line and the probability was one-half that that state was occupied, then my equations of motion will tell me that well that case state will just move up to here a little bit later. And nothing has changed, the electron moved with it and its probability is still one-half. Now, question, I guess I'm supposed to be repeating these questions, right? Okay, so I guess this is, so it didn't end up in two equations. So one of the fundamental assumption is that when there is a collision, the, it's incidentally, so that the, again, the electron doesn't, it doesn't know what happened, what happened just before the collision. So I guess the same concept is used here in the sense that the probability is the same just before the collision or any happening. So how is it fundamentally different from the truth here? But how is it fundamentally different from the, which? Drude equation, except I think that we're doing certain other things, especially magnetic theory and all those things. Yeah, so I guess your question is how is this Boltzmann approach different from just the simple Drude theory? Yeah, so what difference does it make to you using the operation number of the position of the electron? The main difference is that we're resolving, in the Drude theory, you normally just think of the, you think of an average electron. How does an average electron move in the electric field and how does it scatter? Now, here what we're doing is we're resolving electrons in momentum space, or energy space. And we're trying to figure out how are they distributed in momentum space. When we get all done, we will, when we get all done, then we can compute the average quantities and we get things that look like the Drude theory. But we know how to compute the averages properly, because we know how they're spread out in momentum. So for instance, these characteristic times, these average, when you do these Hall Effect measurements and you get these numerical factors, these average tau squared divided by average tau quantity squared. That doesn't come out of the Drude theory, because we have to know how the electrons are distributed in energy in order to compute those averages. And the other thing is, when you apply high biases and you really start accelerating carriers and they get very far from equilibrium, then the Boltzmann equation tells you how those high energy states are occupied as well. But here, we take the assumption that you are not far from the, you are still in the same class. Yeah, yeah. And there I would say that the main, near equilibrium, you can get very similar answers with the Drude theory. But the actual averages of these scattering times, you don't get that from the Drude theory. Was there another? You said you mentioned that actually we can trace both momentum and position simultaneously, which is impossible to trace more dimension. So I wonder if there is any form of limit for the use of Boltzmann equation? Yeah, so the question is, we have this semi-classical picture. And can you establish when it works and when it doesn't work? We worry about the sort of quality. One thing I do is you can compute the average de Broglie wavelength of an electron. And I think for silicon it might be something like 60 Angstroms at room temperature, something like that. And then you can look at how rapidly your potential is varying inside the device. And if it's varying rapidly on that scale, then there's possibilities for quantum reflections and tunneling and things. Then you worry that you're pushing it too far. But precise limits are really hard. People frequently push this beyond where you might be completely comfortable in using it and they often get reasonable results anyway. But it's nice to know when you should be worrying about these things. Was there a question down here? Please, can you turn to the slide number 29? Let's take a look. This GV5, this is this. Yeah, did I insert that at the beginning? And one of the things we always, where did that GV come in? When did I first insert that? Yeah, I haven't been careful about. There's always this valley degeneracy problem, right? You do your calculations in one conduction valley minimum say. And you compute the density of states there. But you might have equivalent valleys in the Broglie zone. So when I'm worrying about density of states, here I should. Is it the number of subbands? No, no, no, not necessarily. Like, so the number of subbands comes from, so let's consider that you have a silicon MOSFET. You have carriers confined in the inversion layer. You have a set of subbands due to that confinement. The lowest subband actually has a GV of two. Because it comes from two of the six valleys in silicon have that. So maybe I can do a better job of this. Let me draw a picture here. So you remember in silicon, the conduction band of silicon has these six equivalent ellipsoids. Now normally when I'm doing these calculations, you're doing them about the minimum of one. And you get a density of states there. But then at the end, you multiply by the valley degeneracy. So in 3D, you'd multiply by six. Because you know, there are five other ones out here that are just like it. But now in this particular case, if I had a silicon MOSFET and I have an inversion layer, I'm going to have quantum confinement. Now how does this work? You remember that the particle in a box, the energies here are something like H bar squared, K squared over 2M, effective mass. If I'm confining in this x direction, and that's this x direction, then I'm confining them in this direction. These two valleys here have a heavy effective mass. So their energies are going to be low. These four other valleys, they have a light mass in the confinement direction. So their subband energies are H bar squared, K squared over two times the transverse effective mass. So they're going to have a set of valleys that are higher in energy. Well, two different sets of valleys coming from these two different types of ellipsoids. Now if I'm just looking at this one here, this is the first subband due to confinement in these who have their ellipsoids oriented in the long direction with the heavy effective mass. That's the first subband, but there are two of these valleys. So in this case, I would say that the 2D density of states for those valleys is m star over pi H bar squared. That's the 2D density of states times two, because there are two equivalent valleys in the Brillouin zone. So whenever you do these problems with a material, you always have to know something about the band structure and how many equivalent valleys there are. I can bring it up to slide number 29. And here we see a Vx, X is a V times the isotropic mass, isotropic mass means the energy is H bar squared K squared over two m or something else. So H bar squared K squared over two m and m is the same in any direction. So it means the curvature of the band in any direction is the same. In general, that's not the case, right? What about cos n delta? Pardon me? Cos n delta is cos n one of the. Yeah, so the picture here is that these electrons are moving in the constant, if this is K y, this is K x. Then my constant energy surface is H bar squared K squared over two m. Is equal to energy. So a constant energy surface would be saying that it's isotropic means that the constant energy surface has the same radius in all directions. Now, the velocity is going to be H bar K over m, magnitude of K. That's the magnitude of the velocity, okay? But the velocity in the x direction is what I need. So that's just magnitude of the velocity times cosine theta. And the velocity, if you measure the mobility, it doesn't depend on the energy to shoot, right? So the mobility doesn't depend on the field vehicles or not? No, so the question is, the mobility doesn't depend on electric field. That's kind of the definition of low field transport or near equilibrium transport. If you apply high electric fields and the carriers get more energetic and they start to scatter more and the tow drops. People, if you use a semiconductor simulation program like Centaurus, you use a field dependent mobility all the time. But near small electric fields, it's independent. Now the same thing with B fields. For small B fields, the whole voltage is just proportional to the B field. And the whole mobility that you extract is independent to the B field. Remember, I dropped out this B squared term, which would give you the non-linear effects. Because I assume the magnetic field is not too big. If it gets bigger, then that B squared term starts to become important. So if we assume that we have some contact with an applied electric field, and a part B field, and we measure the mobility, what we get, what we get, how much is the mobility? Okay, so I'm going to talk about this in when we do the measurements in the next lecture. So very frequently, this is always a problem. People who grow semiconductors and then measure them in their labs, they will typically use Hall effect to measure the mobility. Now, frequently they'll just tell you, they'll just say the mobility is. What they always mean is the Hall mobility is. Because that's the only thing you can measure with a Hall effect. And there is always this numerical factor, which is often thought to be somewhere between one and two. But it's kind of hard to pin down. And it's always an uncertainty there. So you measure something in a Hall effect measurement, that's not the mobility that you want when you're computing conductance. It's. So the regular mobility loses its meaning in semiconductors. With the B field? I wouldn't say that, we have this magnetoconductivity tensor. Let's take a look here, there I put the B field in it. We have this magnetoconductivity tensor. The diagonal components of that tensor are Nq mu. The regular mobility we've been talking about all along. The off diagonal components of that involve this Hall mobility. So both of them are in this 2 by 2 tensor. When you measure the Hall effect, you're measuring these off diagonal components and you're really deducing the Hall mobility.