 You can follow along with this presentation by going to nanohub.org and downloading the corresponding slides. Enjoy the show. Okay, so the objective today, this all morning, is to dive in and try to make use of that theory and mathematical formulation that we introduced yesterday to see what we can do with it. So the first lecture is about resistance and you might think that, you know, what can you say about a resistor, but actually there are a few interesting things to talk about. So we'll dive into it. And just by way of reminder, you know, the traditional way we think about resistance, if you've got a long wire in 1D, the resistance should be proportional to the length of the resistor. Make it twice as long, it'll have twice as much resistance. And the constant of proportionality would be the resistivity in 1D, and you could figure out what the units are. If it's a 2D plane, then the resistance is sheet resistance times length divided by width. And if it's a bulk 3D conductor, then the resistance is the resistivity times the length divided by the cross-sectional area, just like you have a lot of resistors in parallel when you make the area bigger. So that's the traditional diffusive way of thinking about large resistors, and we want to get the right answer for large resistors, but we also want to get the right answer for small resistors in anything in between. So we use this model, and this is this Landauer model that we talked about yesterday. So this is our little ideal device. We have ideal contacts that are in thermodynamic equilibrium. The resistor can be 1D, 2D, or 3D, it doesn't matter. We can compute what it is, and current is positive when the current flows into contact, too. And our basic relation, if we apply a small voltage, is that current is conductance times voltage, and we have this formula that we're hoping that you'll get familiar with and remember because it's very important. The conductance is 2q squared over h, and then it's an integral over all the energy channels. Transmission, just a number between 0 and 1. It says what's the probability if an electron comes in from contact 2 that it goes out contact 1. If it's ballistic, the probability is 1. If it's diffusive, the probability is very small. M is the number of channels there are for conduction at the energy E. And the DfDe will talk a little bit more physically about what that means. That sort of picks out which are the important channels that contribute to current flow, and then we integrate over all of energy. So there are people talk about the driving forces for current flow. So there are two driving forces, and one of them is not really a force. But if we take our expression, and this is really the fundamental expression up there on the upper left for current. This is before I made the assumption that F1 was approximately equal to F2, and we did the Taylor series expansion for near equilibrium. That was the starting point. That's really a very good starting point to remember. If you're thinking about a transport problem, that's a good place to begin. I showed you yesterday how you can derive a drift diffusion equation from that if you want to, but it also works when you can't use a drift diffusion equation. So the key point is that when there are differences in occupation, then current flows. When F1 is different than F2, then current flows. When the Fermi levels and the two contacts are different, then current flows. But that's what we talked about yesterday. We did the little Taylor series expansion. But we assumed that the temperature of the two contacts was the same. Another way to make F1 different than F2 is to make the temperatures of the two contacts different, because temperature is in the Fermi function. So that will also produce a force. Now this is a little different kind of force. Temperature doesn't exert a physical force on an electron and move it. It's a statistical force. But people call it a force. So there are two driving forces for current flow. A change in Fermi level, and we change the Fermi level by applying a voltage on the contact. Positive voltage moves the Fermi level down, or a change in temperature. So I'll be focusing in this lecture on the first driving force. After the break, we'll talk about the second driving force. And then we'll have the whole story. Okay, and I'll remind you, we have these different transport regimes. And it all depends on whether the transmission is closer to zero or one. If it's close to one, we say we're dealing with ballistic transport. If it's close to zero, very small, we say we're dealing with diffusive transport. If it's somewhere in between, we say we're dealing with quasi-ballistic transport. Which is more and more the case in modern devices. We're somewhere in between these two limits. And we sort of just asserted that this was a, I showed you that, I developed an, we know that T is one for the ballistic case. And we know that T is mean free path over length in a diffusive case. We got that just by making sure we got the right answer in the long resistor when fixed law applied. And then I sort of asserted that this equation will cover both limits. And actually, I mentioned it's much better than that. And lecture six will derive it. It showed that it really does work very well. Okay, so for the most part, we'll talk about 2D. And then in the discussion section, we'll talk about some other things. So let's do ballistic first because ballistic is easiest. Okay, so here's our resistor. We're looking down on the top of it. The electrons are moving in a plane. Might be a thin semiconductor film on an insulating substrate. Might be an inversion layer of a MOSFET. Two contacts, the resistor has some length and width. In this case, PL is the lattice temperature. They're the same in both contacts. And we start with our expression for conductance. And we'll see where it leads us. Okay, so this is our expression for G up there on the upper left. So inside that integral, I'll just put T equals 1 if we want to see what happens in the ballistic limit. Okay, then we discussed M. M is the number of channels. So we talked about that yesterday. Let's talk about that derivative of the Fermi function. That came from the Taylor series expansion, saying that F1 minus F2 is very small. We did the Taylor series expansion. Well, here's a plot of our Fermi function. And you can see the only place it has a derivative is right around the Fermi level, where it makes a transition from 1 to 0. And the derivative there is negative. And the width of that region is a few kT, right around the Fermi energy. So if I take that derivative, I just get a sharply piqued function around the Fermi level. As I go to lower and lower temperatures, that looks like a delta function. In fact, you can easily integrate that and show that the area under it is 1. So it behaves like a delta function. So let's think of it. We're doing it at low temperature, low enough that we can assume that it's a delta function. Then it makes that integral easy to do. Because it's just a delta function at the Fermi energy. And our conductance is easy to obtain. It's just 2q squared over h times the number of channels at the Fermi energy. All right, very simple, very nice result. That's the conductance of a ballistic resistor. You put numbers in. If I want to think about resistance, then it's 1 over the conductance. So it's 1 over the number of modes and h over 2q squared. And that works out to be 12.8 kilo ohms divided by the number of channels. That's a fairly high resistance. If you're going to make a low resistance, you have to have a lot of channels. Put them in, it's a lot of current flow in parallel. And you can already see that if you have a small structure where you can actually count the number of modes that the conductance is quantized. M will be 1, 2, 3, 10. The resistance can't be in between those numbers. It has to come in those discrete units. Okay, so this is a well-established experimental fact now. This is a beautiful experiment that was done in 1988. This has been done many times or many different types of conductors. People have even put probes on a hydrogen molecule, an H2 molecule and measured it. And this is the resistance that you get. So this particular structure, let me try to explain it, comes from the paper. This is a 2D sheet. This is called a modulation doped film. So if you want an n-type semiconductor and you dope it, then the dopants are scattering centers and it lowers the mobility. So this is a very clever idea that people discovered as a way that you can get mobilities of a million or 10 million in samples. And the trick that you use, if you look at the cross-section there, you can see that the film itself, where the electrons are, is a gallium arsenide film. Mobility, typically if you dope gallium arsenide moderately, I think your mobility may be around 7,000 or so. But the trick here is that you put a wide band gap aluminum gallium arsenide layer on top of it and you dope the wide band gap layer. And the electrons spill out of the wide band gap layer and they drop down into the gallium arsenide. Now you've got electrons in undoped, pure gallium arsenide for which the mobility is 100,000 at room temperature. And then if you cool it, you can get mobilities of a million, long mean free paths and ballistic transport. So that's a technique that in the 80s people pioneered in order to study electron transport in the ballistic regime. So the film there, you can see there's a contact on the bottom, that's like contact one for us. There's a contact on the top, that's like our contact two. There's those two fat regions are like our thermal reservoirs. You can see that they've made it smooth. People worried about how do we make these contacts to be ideal so that there aren't any reflections from them. They kind of sometimes they would intentionally gradually make them smooth so that the electrons could get in without reflecting. And then there's a width of that channel. But what's really not shown here is that this is not an etched channel. There are actually two shot key barriers on top of it and they're depleting the gallium arsenide. And if you reverse bias the shot key barriers, you can deplete it more or forward bias it to deplete it less. So you can electrically control that width with a gate voltage. And then classically the conductance should be proportional to the width. But that's not what they measured. They measured these discrete steps. So here the gate voltage is electrically controlling the width. And you'll see if you apply a large negative gate voltage, you really pinch the channel down to a very small thickness. There's only one channel now. Only one half wavelength that that energy will fit in at the Fermi energy. And now you reduce the reverse bias. And you can see the steps going, these are in units of two q squared over h. So you can see them going from one quantum of conductance to two, to three, to four, to five, beautiful experiment. So there's no doubt that this exists. It's a well-established experimental fact. You get the bigger structures. It's harder to see the steps because there are so many modes. You go to room temperature. When that Fermi function, its distribution is spread out a little bit. And now you have, you know, modes are fractionally occupied. You always have a few of them occupied. So it's harder to see the steps. This was done, I think, at four Kelvin. But it's there. There's no doubt that it's there. Okay. And, you know, the other interesting thing is, you know, classically those expressions that I put on, like the slide number three, you say resistance is proportional to length. You make the length of the resistor twice as much. You get twice as much resistance. As the length gets shorter and shorter, you think the resistance goes to zero. No, the resistance doesn't go to zero ever. The resistance goes to this one over this quantum of conductance. That's as small as it can be. Okay. So that's the basic concept now. You know, if you're an engineer working on transistors or electronic devices or something, you're working at room temperature and you're working with wider structures and you have many more, many more half wavelengths will fit into the width here. The discreteness is hard to see. Although, you know, people are starting to worry. You know, I was asked to calculate this here recently. When you get down to a very small width MOSFET, how many channels are there? You might get to, you know, I think we might be getting to the point where you're starting to see the discreteness where, you know, you always write a MOSFET IV characteristic as the current is proportional to width divided by channel length. So we learn two things. You know, if you get very small, it's not proportional to width. You have to count the number of modes. And if you get very short in channel length, you'll reach a fundamental limit of conductance. So we talked about this, the number of channels in 2D. Yesterday afternoon in the last lecture, we developed this expression. We saw that physically what it corresponds to is just imposing boundary conditions on the wave function so that it goes to zero at both edges of that width. And then we just count the number of modes that are in there. So that gave us, you know, if we're at TE equals zero, then only that derivative of the Fermi function is very sharply peaked at the Fermi energy. The only thing that matters is the energy at the Fermi energy. So we can write the number of modes in terms of KF, the wave vector at the Fermi energy. Now we can also do this in terms of the carrier density if we like to. And this one, you know, in 2D, the area occupied in K space is pi times the Fermi wave vector squared. And you might know that the amount of space that a state takes up in K space is 2pi over L in 1D, 2pi over area in 2D. So 2pi squared over area in 2D, 2pi cubed over volume in 3D. So I have to multiply by two for spin. So the point is I can write a very simple relation between the Fermi wave vector and the carrier density. And the result is that I can express the number of modes. If I know the carrier density, I know the number of modes. And the interesting thing is this, there's no band structure in this. This is valid for any band structure. If I know the carrier density, I know the number of modes. It has to be isotropic because of this pi F squared. So, but it's valid only at TE equals zero, right? But it gives you a close. So if you would like to estimate in a MOSFET, say you have 10 to the 13th carriers per square centimeter and you want to know how many modes are occupied, this expression will tell you. Find out how many. It'll tell you at TE equals zero. And I'll talk about TE greater than zero in a minute. All right, here's TE greater than zero. TE greater than zero, things just get a little messy and unpleasant. TE equals zero, everything is beautiful, right? We have this delta function. The math is simple. Most problems that engineers will deal with are at room temperature. And you have to be a little more careful. And things just get a little messier. So let me go through this because I don't want to, I don't want to spend much time in these lectures on it, but this is what you need to do when you want quantitative numbers, right? So we have this integral. And let's see if we can work this out when we can't approximate minus DfDe by a delta function. We know we just work it out. Okay, so we know everything inside the integral. We know how the modes are related to energy. We know how the Fermi function is related to energy. So you'll notice this DfDe, when I take the derivative of that Fermi function, I've got an E minus EF on the bottom. So if I took, I'll get the same answer just with a different sign if I take a Df with respect to the Fermi energy, right? Now that's kind of nice because then I can just pull the derivative outside the integral, okay? So I get an integral like that. And I'll take the derivative with respect to the Fermi energy later. Okay, so now I have a different integral to do. And I just integrate that Fermi function. Again, I know everything that's in there. I'll make some definitions here. I'll define a normalized energy eta. It's just energy minus the bottom of the conduction band in units of kT. And oops, the second eta there should be an eta sub F. So that's just the Fermi energy minus the bottom of the conduction band in units of kT. And then I can write the Fermi function in terms of these normalized units. And I can write the number of modes in terms of normalized energy. And I can do a change of variables from integrating over energy to integrating over this normalized energy. And this derivative becomes kT times the derivative with respect to Fermi energy becomes kT times the derivative with respect to this normalized Fermi energy. Just math, right? But you have to. So you get an integral like that. Now we have an integral we can work out. Unfortunately, there's no analytical solution for that integral. But you can do it numerically. But it's a well-known integral that pops up all the time. It's called the Fermi-Dirac integral. You can find lots of approximations for it. Its definition is here. So F to the one-half is the Fermi-Dirac integral of order one-half. That's what we've got. So inside the integral, we have eta to the one-half d eta. And then that's the normalized Fermi function underneath. There's a factor out front that's two over square root of pi. That's the definition of the Fermi-Dirac integral of order one-half. Now I'll talk a little bit more about Fermi-Dirac integrals in a minute. But one of its very nice properties is if you differentiate a Fermi-Dirac integral with respect to eta F, you just kick it down one order. So if I do that derivative of the Fermi-Dirac integral, then instead of getting an F to the one-half, I'll have an F to the minus one-half. So when I do that derivative, that's my final answer. Now that's not quite as easy and pleasant as t equals zero. But if we're going to apply these ideas at room temperature, you'll continually run into Fermi-Dirac integrals and you'll have to get comfortable in working out Fermi-Dirac integrals of different kinds. If we do this in one d, we'll get a different type of Fermi-Dirac integral. If we do it in 3d, we'll get a different type of Fermi-Dirac integral. So let me write the final answer then so that it looks similar to the t equals zero answer. Two q squared over h w times, I don't know what I call this, the effective number of modes. At t equals zero is a number of channels at the Fermi energy. When t is greater than zero, it's the number of channels around the Fermi energy. And the number of channels around the Fermi energy is related to this Fermi-Dirac integral. You work it out. Now, so it means that in order to compute that, if I want to find out what is the ballistic conductance of my MOSFET, I have to know where the Fermi energy is. And the way you get the Fermi energy typically is that it's related to the carrier density. If you can figure out a way to get the carrier density, then you can deduce what the Fermi energy is. So in order to find out what the ballistic channel conductance is, I would have to know, say I'm dealing with a MOSFET with 10 to the 13th carriers per square centimeter in the channel. Then I could integrate the density of states over energy times weighted by a Fermi function. I could find out how the 2D carrier density is related to the Fermi energy. That's this expression, n sub 2D. It's a different Fermi-Dirac integral of order zero. And you work that out. So if I measure n sub s, or if I'm given that, I could solve that bottom equation with a Fermi energy a to f, plug it into the equation on top and I could find the effective number of channels, plug it into the equation on the top and I can find out what the conductance is. That's what we have to do. So the point is, this is probably the last thing I'm gonna say about, well almost the last thing I'm gonna say about Fermi-Dirac intervals. The point is you need to get comfortable if you're working on problems like this with Fermi-Dirac intervals. The definition for arbitrary order is given on this top equation. So f sub j is the Fermi-Dirac integral of order j and it's defined like that. So it has a eta to the j's power inside the integral. It also has this normalizing factor. Remember there was a two over the square root of pi? So it has this factor out front which is determined by a gamma function. And remember a gamma function for an integer is gamma of n is n minus one factorial but it's also defined for half integer values. So with these relations, you can work out pretty much whatever you need. In the dotted box, one of the nice properties of these integrals is if you're non-degenerate and a to f is much smaller than zero actually, you will, your Fermi energy will be negative, your Fermi energy will be way below the conduction band and these things all reduce to exponentials. That's Maxwell-Boltzmann statistics. And they have this property, any Fermi-Dirac integral of any order, if you differentiate it with respect to the normalized Fermi energy, you just kick the order down one. Now be very careful and when you're reading papers, people define Fermi-Dirac integrals differently. This is I think is the best way to do it because it has all these nice properties and when you differentiate it, it has this very nice property. You will frequently see a Fermi-Dirac integral and usually people are careful. The script f refers to the one I'm talking about. If it's a Roman f, it doesn't have the normalizing factor out front. And that means when you differentiate it, it doesn't have this nice property that just kicks it down once, you bring out some factors out front. So sometimes when you read papers, students will come to me and they'll say, I've got a factor of square root of two over pi error here, I can't get the same answer as this paper. It's because you're using different Fermi-Dirac integrals. So just be careful about that. We have my student, Rasong Kim, who just defended her thesis last week, wrote a nice set of notes on the Natal Hub. That's sort of everything you need to know about Fermi-Dirac integrals in order to solve semiconductor problems. And you don't need to know a whole lot, but you need to get a little familiar with it. And then I have another student, Zufeng Wang. Is he here? He wrote, just last week, he wrote me this beautiful little app for my iPhone. You can download it, search Fermi-Dirac at the app store and you can compute Fermi-Dirac integrals on your phone. So that's very nice. He saw what, you know, trouble. I was always asking him, you know, I need to know the Fermi-Dirac integral of order three halves for argument 1.5. And I'd have him go and run a MATLAB script and do the numerical integration. And he got tired of doing that. So he said, here, I wrote you an app, you know, figure it out yourself. Okay, so let's get back to physics here. So, you know, one of the things, you know, we always, you know, again, getting back to traditional diffusive transport, we always think that, we like to think that well, conductance is proportional to the number of carriers, right? The more carriers that are carrying the current, the less it is. And the proportionality is n, q, mu, width over length. All right? So is that true, you know? So one of the points we're trying to make is that you get into a lot of trouble if you start here and try to apply these to nanostructures because this is not really a fundamental starting point. But we know what the fundamental starting point is. Now let's hear. So let's do this at t equals zero. So at t equals zero, the number of channels that participate is just the number of channels at the Fermi energy. And we saw that there's a simple relation between that and the carrier density that you can just work out relatively easily. So at t equals zero, the conductance is not proportional to the number of carriers. It's proportional to the square root of the number of carriers. You know, because most of the conductance is occurring only around the Fermi energy. Most of those electrons aren't contributing to current flow. Now, if you work this out, you know, we worked that out, it had a Fermi Dirac integral of order minus one half. If you work that out at finite temperatures, if you assume that it's non-degenerate, the Fermi Dirac integral reduces to an exponential, you can work that out and do a little bit of algebra. It's a nice exercise. What you find in the non-degenerate case is that it is indeed proportional to the number of carriers. So it may or may not be proportional. The conductance may or may not be proportional to the number of carriers. If you're ballistic at t equals zero, it isn't if you're non-degenerate at a finite temperature, it is. Okay, so let's look at a MOSFET. So this is a MOSFET, it's a few years old now, right? You know, Intel, Samsung, probably have better ones, but it's not bad, right? It's reasonable, let's take a look at it. This was a 60 nanometer n-channel MOSFET, unstrained silicon, so that's the IV characteristic. We're talking about near equilibrium transport, so we're talking about small voltages between the source and the drain. So the region that we're relevant to our theory is that red circle near the origin. And we could take a look at that. These are the parameters if you characterize this device, and these are some numbers that my student Chang-Wook was sitting in the back row extracted from this device. It has a mobility of 260 centimeter squared per volt second. When you apply 1.2 volts to the gate, it has a sheet carrier density of about 6.7 times 10 to the 12th. And if you subtract out the parasitic source drain resistance, it has a channel resistance of 250 ohm micrometers. You know, what do those units mean? It means if it's one micrometer wide, its resistance is 250 ohm. If it's two micrometers wide, you have to divide by two. The channel resistance will be half. So since everything scales with width, if you're wide enough that the number of modes scale with width and you don't count them, then people usually quote resistances that way. Okay, so the question is, how close is this to the ballistic limit? All right, can we figure that out? Okay, I'm going to do it because I didn't have my iPhone app when I was making this lecture out. I'm going to do it with Maxwell Boltzmann statistics, just to keep it simple, and it's relatively close. So this is the expression, when we worked it out at finite temperatures for non-degenerate, the expression there on the top left is what you get for the ballistic conductance. So on the top of the numerator is the thermal velocity divided by 2KT over Q. Okay, so we can put some numbers in. The thermal velocity for electrons and silicon, I just put in the appropriate effective mass. It's about 1.2 times 10 to the seventh centimeters per second, and then I can plug numbers in, and if I take one over the ballistic conductance, I get the ballistic resistance. It's about 40 ohm micrometers. So the measured channel resistance is about five ballistic resistors. So we're somewhere, we're definitely not ballistic, but you'd like to see 10 or more to really be comfortably in the diffusive. So you're beginning to get in this regime where we would call this quasi-ballistic transport. Now today, channel lengths are probably half of this maybe. So, and if the mobilities are maintained with strain, you might be even closer to the ballistic limit. So if you'd like some practice, download that iPhone app and redo the calculation with primary direct statistics and you'll get reasonably close. It'll be a little bit different. Well you could also estimate the number of conduction channels and see what a minimum width MOSFET would do. That's an interesting calculation. Now, what if you do this for three, five MOSFETs? I know we have some people from MIT here and we've been looking at Heises-Dalamos 35 hemp. So these are high electron mobility transistors. So it's a complicated epitaxially grown structure, but the main point here is that the channel is indium gallium arsenide, which has a very high mobility. This is a 40, this particular device we looked at was a 40 nanometer channel length, three, five transistor, made out of a material that has very high mobility. So the mobility in ingas is about 10,000 centimeters squared per volt second. Remember in the silicon channel we had about 260 that we measured. So this is very high mobility. Now I gave you this simple relation that I think I'm going to derive in one of the lectures later. There's a simple relation between the mean-free path and the diffusion coefficient. Thermal velocity times mean-free path divided by two. So a simple way to estimate the mean-free path is to take the mobility, use the Einstein relation to get a diffusion coefficient, and then use this simple relation and get the mean-free path. So the thermal velocity for ingas is higher than it is for silicon because the effect of mass is lighter. So it's about 2.7 times 10 to the seventh centimeters per second. If you estimate the mean-free path, it's about 200 nanometers. The channel length here is about 40 nanometers. So this device is almost ballistic. There's very little chance that anything is going to scatter across it. When we compare our ballistic transistor simulations with the measured IV characteristics, it's quite close. The measured current is within about 90% of the ballistic current. So the point is that if you're dealing with high mobility transistors today, you're essentially at the ballistic limit. If you want to make a first cut at analyzing it, you would not take textbook theory that has mobilities in it. You would be better off starting with a ballistic theory. It'll get you closer to what's going on. Now, if you look at the channel resistance of this device, it was basically, it was all the parasitic source drain resistance because that dominated everything. Okay, so those are ballistic resistors. Now we need to put in scattering. Scattering is still very important for silicon and there is some scattering in 3.5 resistors as well. Okay, so this turns out to be easy to do. So we go back to our same resistor. We go back to our same expression. It's just that instead of putting T equal one, we will take our normal transmission, lambda over lambda plus L. Everything works out the same. If I do this at low temperatures, it's easy. And I simply have to do that integral and my conductance is two Q squared over H times the transmission at the Fermi energy, which now may be less than one, times the number of channels. Same result we got before, we just multiply by the transmission. So it's easy to do. So we can go all the way from the ballistic to diffusive. Let's see how we do this. So the conductance is transmission, which is mean free path over mean free path plus length. If I take one over that, then the resistance will be one plus length over mean free path times the ballistic resistance. Well, we just get a higher resistance, that's all. If the channel length is many mean free paths long, we just get a higher resistance. And this sort of also shows you very clearly. As you scale the channel length down towards zero, the resistance isn't going to zero. The resistance is going to its ballistic limit. And there are even people that are beginning to put this in spice models and things. So that your normal spice transistor model as a W over L doesn't behave very well as L goes to zero. And these companies are very good at scaling L so that's getting pretty close to zero. People start to worry about that first term there. Okay, so if we go back to this sample, and we just say the measured resistance is 250 ohm micrometers. And that's about five times the ballistic resistance. If I just put it into this little expression, I can estimate the mean free path. The mean free path is about 15 nanometers. So that was a 60 nanometer channel length device. The mean free path is about 15 nanometers. So the channel is about four mean free paths long. Though it's in this regime, I'd say maybe more diffusive than ballistic, but not clearly in either regime. Okay, so those are the main points. Now there are just a few things I want to discuss to wrap up here. So talk about mobility. You notice I haven't said anything about mobility yet. I'll talk about some different ways to read the right conductivity because you'll read papers and you'll see it expressed in all kinds of different ways. We'll do a 1D example and then we'll talk about a couple of other things too. So what about mobility? We have this expression and this is what we're trying to say. The expression on the upper left is what you should start with when you're thinking about how do I compute the conductance of a structure of it. The expression on the right is what you more often would see in introductory textbooks and valid for bulk samples. What is the connection between these two? Okay. Well, you shouldn't say mobility is Q tau over M because that gets you into all kinds of trouble. What's tau for a ballistic device? It's supposed to be the scattering time. What's M for some of these like graphene? So really the way to do that is just to equate these two. Right? That gives you a definition. That's what the mobility is. Looks a little bit complicated. All we did is equate the expression on the right to the expression on the left solved for the mobility. That's the definition of mobility. Now you see people do this sometimes. I think it's referred to as the Kubo Greenwood formula. You can drive it from the Boltzmann equation. You do some, you do basic things. This in the end is the definition of mobility. Now you see I have a subscript there. I'm calling this the apparent mobility. So I'm going to say a little bit about what I mean by that in a minute. All right, so Q tau over M is nice. I sometimes when I encounter a problem and I'm trying to understand what it means, I sometimes think Q tau over M and that helps me understand it. But you can also get in trouble in thinking Q tau over M, but you won't get in trouble with the one above it. All right, so let's take a look. In the diffusive limit, first of all, then the transmission is just mean free path divided by length. So in that case, this is what people would call the Kubo Greenwood formula for how you compute the mobility of a bulk material. The formula on the bottom is what we would call the Jeword formula. So as Professor Dottas said, if you're reading something like Ashcroft and Merman, you'll see the second formula in the first or second chapter and you might see the later one near the end of the book, I don't remember or not, but it usually comes in more advanced sort of treatments. Now, what if, you know, that was only in the ballistic limit. What if you're somewhere, what if you're ballistic? No, I'm sorry, the previous slide was just for the diffusive limit. All right, what if you're ballistic or what if you're somewhere in between? You know, there's no problem with us in just using the transmission as it is and defining something that we will call an apparent mobility, you know, and we'll apply it in the ballistic case too. So my transmission, if I, what's inside that integral is transmission times length. So T is lambda times L over lambda plus L. If I just divide by lambda times L, what I'll find is that the transmission is related to one over an apparent mean-free path, which is one over the actual mean-free path plus one over the channel length. Whichever is smaller is what dominates. So I can just go ahead and define an apparent mobility, although I don't know what it means, but I'll get some answer even in the ballistic limit. And what this would allow me to do if I wanted to is I could say that the ballistic resistance, the ballistic conductance is NQ times the apparent mobility and I would get the right answer. Even though you might say, well, I really shouldn't be talking about a mobility in the ballistic limit. So let's see how that works out. So I will do this just at T equals zero, just because I don't want to do fair major act integrals. So this is our definition and it's valid ballistic, diffusive anywhere in between. So at T equals zero, that the FDE is a delta function. So the expression simplifies. We get the second expression. Now, we also know what the transmission is. We know what the number of modes is. It's H over four times the average velocity in the transport direction times the density of states. At T equals zero, the velocity is the Fermi velocity. It's right at the Fermi energy. The two over pi was this averaging over angle, so we got the average velocity in the x direction. We can compute the carrier density just by integrating the 2D density of states. The states are occupied below the Fermi energy and they're unoccupied above, so we get a simple expression for the carrier density. We just plug all of that in and find out what we get. And when you do that algebra and separate it out, what you'll find is that one over the apparent mobility is one over the actual mobility plus one over something else. And I'll call that something else of ballistic mobility. Just algebra. What does it mean? So the point is, if I use that apparent mobility, then I will always, and I use the traditional diffusive expression, g is n, q, mu, w over l, I'll always get the right answer. Now some people tell me I shouldn't do this, but I didn't invent this idea. They say, ballistically, you shouldn't be talking about a mobility. What does a ballistic mobility mean anyway? This is nonsense. But people find it useful, and I actually find it useful from time to time too, and I actually think you can attribute some physical significance to it. So I'll explain this to you in a minute. So when you do that algebra, this is what you get. It's like a Matthiason's rule. If you know how, when you have mobilities due to different scattering mechanisms, the way you find the total mobility is you add their inverses. Whichever one is lowest is the one that matters the most. In this case, the apparent mobility, whichever one, it's either the actual mobility or the ballistic mobility. Whichever one is lowest is the one that's important. The actual mobility is given by d over some energy. This is like an Einstein relation. If I had done this for Maxwell-Boltzmann statistics, it would have been d over kt over q. Since I'm doing it at t equals zero for Fermi Dirac statistics, it involves the Fermi energy. d is just related to the average velocity in the direction of transport times a mean free path divided by two. That's the real mobility. Makes physical sense, has a mean free path in it. What are those other terms when I did that algebra? Well, that's a ballistic diffusion coefficient divided by the energy. Well, what is a ballistic diffusion coefficient? Diffusion is random thermal motion. In the ballistic case, there isn't any random thermal motion. Well, it's just some label I've given to those terms. But you can see that the ballistic diffusion coefficient is just the actual diffusion coefficient with the channel length replacing the mean free path. So this was first done by Michael Shore. People started worrying about ballistic effects in gallium arsenide transistors some time ago. So what's the physical interpretation? If you have a channel like this, the electrons scatter frequently in the source. That's what maintains thermal equilibrium. Then they go across the channel. If they're ballistic, it won't scatter at all in the channel. When it gets in the drain, it scatters immediately. Very quickly thermalizes. So if we have a long channel, the electron will scatter many times and the mean free path will be much shorter than the length and the apparent mobility will be given by the real mobility, just what you think. Now, if you're ballistic, then you would ask yourself, the mean free path is the average distance between scattering events. If you're ballistic, there is no scattering in the channel. But you scattered in the source, then you went across the channel and you immediately scattered at the other end. The average distance between scattering events is the physical length of the channel. In that sense, I could give it some physical interpretation. So it's possible to... I guess a little bit later in the discussion, I'm gonna go back and compute the ballistic mobility for that transistor and we'll see how that works out. So sometimes it can be a useful way to think about problems and sometimes you'll read papers and people will talk about the ballistic mobility and that's what they're talking about. Okay, another thing that I wanna mention is different ways that we can write the conductivity because when you read papers, you'll see it written in all kinds of different ways. We've written it... Now, let's talk about a diffusive conductor. So we've written it as two Q squared over H, mean free path times number of modes times widths over length. So our sheet conductance is just given by the second equation. But you'll see this written in many different equivalent ways. Okay, so if we start with the first equation, we know that we can express the number of channels for conduction in terms of the density of states and velocity. And we know that we can express the mean free path. We'll see, you might think that the mean free path is the distance between scattering events. Velocity times the time between scattering events. Now we'll discuss in chapter six, there's actually a statistical factor there, we have to be careful about. In 2D, it's pi over two times that. In 1D, it's two times that. In 3D, I think, is it four thirds that? So we'll see how that works out in lecture six. And I can define a diffusion coefficient, V squared tau over two. And that's consistent with my previous definitions. So if you use those and insert them in the first equation, we can write the sheet conductance as Q squared times the density of states at the Fermi energy times V squared tau over two. So you'll frequently see that in papers. It's the same thing we've been talking about, that's the point. You'll also see that V squared tau over two written as a diffusion coefficient. So you'll see it written in the bottom as Q squared density of states at the Fermi energy times diffusion coefficient. All different ways of writing exactly the same thing. So you have to, when you read papers, people prefer to do it different ways and you just have to be able to convert between them. There's one more way that we could write this that is kind of interesting. So the equation on the top left, this is the way I've been doing it. Now we saw that you could also write this as Q squared density of states times V squared tau over two. Now, one half mV squared, that's the kinetic energy. So V at the Fermi energy squared, that's equal to EF minus EC, that's the kinetic energy for a parabolic band. And the density of states times that energy range is the carrier density at T equals zero. So I could write that equation in another form. No approximations, except I'm doing this for a parabolic band, but I could do it for another. I could write that as NQ mu, the simple diffusive druid expression, where mu is Q tau at the Fermi energy over m. So in the end, you know that there's nothing wrong with Q tau over m. If you're dealing with a parabolic structure, it's the right answer. It's just that the expression on the top is more general, allows us to think about quasi-ballistic transport more easily, allows us to apply it to different band structures more easily. But under the right conditions, we can write it as Q tau over m, if we want to. Now, when we get to thermoelectrics, this is, I'm going to write things a little differently. That kernel, if I look at that kernel, lambda m times df de, that's something that has the units of one over energy. Because when I integrate it over energy, it has the units of conductance divided by energy. When I integrate it over energy, I get conductance. So I could write that sheet conductance as Q squared over h times the integral of the kernel. The kernel is something that people call the differential conductivity. And you'll see that frequently expressed. And that's going to be useful for us in thermoelectrics, and I'll talk about that in the next lecture. So let me get back to this nanoscale MOSFET example here for just a minute. If I were looking at this, one of the things I could do is I could say, okay, this transistor has a measured mobility of 260. If I wanted to think about how close is this to the ballistic limit, one thing I could do is I could compute the ballistic mobility. And the ballistic mobility for this particular device is 1400. Notice that it's channeling dependent. And since the, I should call this the apparent mobility, is a smaller of the two, you can see that the actual mobility is quite a bit smaller than the ballistic mobility. So this device is going to behave more like a traditional ballistic device than like a diffusive one. If I were to do this for the 3.5 hemp, what you would find is the actual mobility is 10,000. The ballistic mobility, I think for that particular device is something like 2,000. In that case, the answer is just the opposite. It's the ballistic mobility that controls the performance, not the diffusive mobility. All right, so I'm still in the discussion section and we're just finishing up two or three things. One of the other questions is where is this power being dissipated? We know that we have a current. We've applied a voltage. So the power must be V squared over R or conductance times V squared. But the resistor is ballistic or if the resistor has scattering, we've assumed that the scattering is only elastic. It doesn't dissipate any energy. So where is the power dissipated? The answer is it must be in the two contacts. So how does that work? So here I've exaggerated. And remember, we're talking about near equilibrium transport. So we have a very small separation between the Fermi levels of the two contacts. I've just blown it up and exaggerated it here. So the electrons are flowing between the two Fermi levels or they're flowing around the region where DfDE is peaked. So the electrons come in and they can flow between these two Fermi levels. So they flow in a range of energies like this. On average the energy is somewhere in the middle and when they leave that contact, they leave behind an empty state. This would be a hot hole. It's got kinetic energy if you wanna think about holes. They flow across and they don't dissipate any energy and on average they come out and enter a state that's a little bit above the Fermi energy in the second contact. So now there's a hot electron in the second contact. They quickly thermalize, scatter, dissipate that energy in the contact and they drop down to the Fermi energy. An electron goes out, comes around the other side, enters because current flows near the Fermi energy where DfDE is peaked, flows near the Fermi energy, comes in, undergoes some inelastic scattering, drops down and fills up that hole. So you can see half of the power is dissipated in the first contact. Half of the power is dissipated in the second contact. In a lot of these very small devices and these things like experiments and carbon nanotubes, the carbon nanotube has very small volume. If you just calculate the power dissipated in these experiments, the power density is quite large if it were all being dissipated in the nanotube. It is easy to vaporize those nanotubes but they would all vaporize if a lot of the power didn't dissipate in the contact. Now another question that people ask is, where does the voltage drop? And there's a really good and somewhat extended discussion of this in Suprius' book. But let me just give you one intuitive piece of that that you can see where the voltage drops and why this quantized conductance, why people call it a quantum contact resistance. Where does the resistance come from? Well, the channel itself has no scattering so you think it has no resistance. Now, let's see. So I have one Fermi energy in the contact and that Fermi energy populates some of the states in the device. I have a second Fermi energy, the blue line, in the second contact and that populates some states in the device. Now the question is, if you put a voltmeter across, what do you measure? Voltmeter measures the difference in Fermi levels. So you always have this problem when we teach our first semiconductor device course, you have a PN junction and there is actually a voltage difference between the P and the N regions in equilibrium. But if you put a voltmeter across the P and the N regions, you won't measure anything. The voltmeter measures the difference in the Fermi levels and it's zero in equilibrium even though there's a one volt potential drop across the junction. So it's really, if we want to ask where the voltage is dropping, we should look at where the Fermi energy is changing. In the contact on the left, we have one Fermi energy, EF1, no question. In the contact on the right, we have one Fermi energy, EF2. In the device in the middle, we have two Fermi energies. Let's just take the average of them and say that the position dependent called that a quasi Fermi level behaves like that. So you can see that half of the drop occurs at the first contact and half of the drop occurs at the second contact. So we would say that's where the voltage is dropping, that's where this quantized conductance is coming from or resistance. Half of it is coming from the drop on the first contact and half of it is coming from the drop on the second contact. Okay, so that's why we call it the quantum contact resistance. All right, now very quickly because I don't want to go through the details, I've talked about 2D resistors but this could be anything. And all I have to do is to work out the integral and I just need to use the right number of channels versus energy for 1D, 2D or 3D. So I should mention, this is probably worth discussing just a little bit. If I have a 2D sheet, I've been talking about how the number of channels depends on the width of this sheet. But this sheet has some finite thickness. Carriers are confined, it's like carriers are in a particle in a box. So it has some thickness in the vertical direction that I haven't been talking about. The fact that it's very thin in the vertical direction means that the electron states are confined like particle in a box states, that's what this Issa-Ban is. This is your first thing you do in a quantum mechanics course. So this is the thickness, this should be zero to T on the bottom, not zero to A. And that gives discrete states. Now I have been assuming up till now that my Fermi energy was somewhere above the first one of those states due to quantum confinement in the vertical direction. And then I have a set of quantum confined states due to the lateral dimension because half wavelengths have to fit in there. But the lateral dimension is very wide. So frequently those are very closely spaced and we don't resolve them. We have an expression, we just say it's proportional to W. But what if my Fermi energy was even higher? It's above the second quantum confined state due to the vertical confinement. Now I've got another set of channels due to the quantum confinement of those states. I just need to add them all up. So when I do my number of channels I'll take the expression that we had before but I have to sum up the number of subbands due to the quantum confinement in the vertical direction. So frequently in a MOSFET under strong inversion you only have the bottom one occupied. There's just one. But under less strong inversion you can have more than one occupied. If you're in 1D then 1D means W is very small. The thickness is very small. We have particle in a box states due to confinement in both directions and you just need to count up how many states are there. If you're in 3D then the width is large, the thickness is large, the cross-sectional area is large, the quantum confinement is very weak. And then if you work out what the number of channels is they're all spaced so what the number of modes is or channels they're all spaced so closely that we can't resolve them. So then we have this continuous expression. This is just H over four times the average velocity in the transport direction times the 3D density of states. So we have these expressions in 1D, 2D and 3D. So the point is if we want to work out problems in 1D, 2D or 3D we just take the appropriate expressions for the number of channels, plug them into the integral and work out a solution. And the solution will always look something like this. It'll be 2Q squared over H times some average transmission times some effective number of channels that are participating. The effective number of channels that are participating is given by the expression we saw before. If this is T equals zero, that DfDE is a delta function and the number is just the number of channels at the Fermi energy. If we're not at T equals zero, we have to work out a Fermi Dirac integral for the particular M. And the average transmission, so this mean free path might depend on energy. And that gets things a little more complicated. You just have to work out another integral that has the appropriate energy dependent mean free path to find out what your average mean free path is. If the mean free path is constant and independent of energy, then the transmission is just mean free path divided by mean free path plus lambda. But we might have to average that over the energy states. All right, so I'll just show you. You can work it all out in 1D. Your effective number of channels involves a Fermi Dirac integral of order minus one. So, you know, you can simplify it for Maxwell Boltzmann and everything. I'll just include this in the handout for reference. You can work it out in 2D. We thought that the effective number of channels depends on a Fermi Dirac integral of order minus one-half. And you can look at how that simplifies at zero Kelvin. You can look at how it simplifies for Maxwell Boltzmann statistics. You can work it out in 3D. The effective number of channels involves a Fermi Dirac integral of order zero. You can simplify that for zero K or for Maxwell Boltzmann statistics. If you do it for graphene, you have a different band structure. You have a different expression for M. In all of these, I've assumed an energy independent mean free path. If your mean free path has some energy dependence, you'll end up getting Fermi Dirac integrals of different orders. So all of these are just things you have to deal with when you want to carefully analyze data, but there are no fundamental new concepts there. Let's take a look at a 1D example. So this is another famous experiment that was done several years ago now. This is a measured IV characteristic of a metallic carbon nanotube. So you can see that there's a region near the origin where the current is proportional to voltage. This is the linear transport region that we're talking about. You can also see that if you go too high in voltage, it becomes nonlinear. And the electrons get a lot of energy. They start emitting optical phonons. You get velocity saturation. Those are problems that we're not talking about in this short course. But we could take a look near the origin and we could ask ourselves, what's the mean free path of electrons in this carbon nanotube? Well, the ballistic conductance would be 2q squared over h times the number of channels. So here we have to know a little bit about the band structure of a nanotube. And it turns out that the band structure of a nanotube, it has a valley degeneracy of 2. Just like you might remember in silicon, there are six equivalent conduction bands. In a carbon nanotube, there are two equivalent valleys. So we have two channels times 2q squared over h. We can compute the ballistic conductance easy. Now, if you look at t equals 0, if you look at what the actual conductance is, it's going to be the ballistic conductance times the transmission. And the transmission is just the average mean free path over the average mean free path plus lambda. You just read the actual conductance off of that. It's 22 microsiemens. And you just solve that bottom expression there for the mean free path and you get 167 nanometers. So it's easy to estimate what the mean free path is. This particular carbon nanotube is one micron long. So that's a lot less than the length of the channel. So this is a diffusive carbon nanotube. OK, so that's pretty much it. So the main points are that conductors display a finite resistance, even when they're ballistic. And this resistance, or we call the quantum contact resistance, and it's quantized in units of these fundamental constants, h over 2q squared. That sets a lower limit to the resistance of any device. And in some devices these days, especially in 3, 5 transistors, we're hitting this limit. Making the mobility even higher won't affect the channel resistance of a transistor like that because we're already at the ballistic limit. And it's very easy. As Professor Dada mentioned, traditionally people think of this area as analyzing ballistic transport and analyzing diffusive transport as two separate areas. You use two separate techniques. But it's very easy to treat it all in this formalism. You just have a transmission that goes from 0 to 1. And the key message we're trying to get across is that this is the place we want to start. Not with nq mu or q tau over m, because that oftentimes can be confusing. All right, so I'll stop there and see if you have any questions. Yes? You talked about many equations about pressure to 0 and pressure equals to 0 on the pressure. Well, you have to speak up a little bit. My wife tells me I'm hard of hearing, and I think she's right. You talked about many equations around the pressure equals to 0 on the pressure. All of these models working for this very close to experimental results. Are these models good for experimental results? Yeah. Yes, yeah, yeah. Yeah, and so the question is, I did a lot, because I wanted to convey what this is all about. I did a lot of talking about t equals 0. And then I did a lot of just assuming Maxwell-Boltzmann statistics, because we've got simple expressions and I sort of estimated numbers for measured data. Yes, when we've, I think we probably have a paper or two where we've done this. If you actually want to know what the mean free path is in this MOSFET, you need to assume a finite temperature and work out the appropriate expressions in terms of Fermi-Dirac integrals. And that's what you need to do when you analyze real data. When we look at that 3.5 transistor, for example, the way we'll treat that is to do an NEGF simulation of the transistor. Do a simulation of a ballistic device. Add the measured series resistance of that device and then compare it to the measured results. And what you find is that it's very close. The, assuming that the intrinsic device is, at room temperature, is ballistic and adding the parasitic resistance of the source in the drain, gives you a current that is about 10% higher than what they measure. So that might tell you that there's a little bit of scattering in the actual device. So things make sense at room temperature with real data. Now, we do this for graphene as well. Yes? Can we go to slide 41? Slide 41. This one here? That's the point S resistance. That was 2q squared by h. But that was what? That was 2q squared by h. Sorry, that's h over 2q. Here, if you were showing each contact having one point in here, can we be in here? No, what I'm trying to show here is so we're on slide 43. And the question is, we know we have this measured resistance of h over 2q squared. And that's current divided by voltage. So the question is, where is the voltage dropping? That's what will give rise to this resistance. So here we have a total voltage drop of v. And the point of this is that half of v drops at the first contact. Half of v drops at the second contact. So half of the resistance must be associated with the first contact. The other half must be associated with the second one. So if you're considering a single ideal as one point in resistance, is the first one having half the minimum resistance? Say that again. So if one point of resistance is h over 2q squared, does the first one have h over 4q squared and h over 4q squared adding 2q squared? Yeah, I mean, you can only do a two probe measurement and measure it. And then the question is, where is this coming from? I don't know that you could ever measure the quantum contact resistance of one contact, because you'd need two probes in order to do that. But yeah, so I think the point we're trying to make is it can't be coming from the channel, because the channel is ballistic and there's no drop in potential across the channel. The only place the drop is coming is from the two contacts. So this must be equally shared between the two. Yeah, I'm trying to think about how they could. If you had unequal contacts, then you would have non-ideal contacts. You might have some finite transmission at the contacts. We would be below the ballistic limit. We would have larger resistances. There are cases where you'll make a contact and you'll have a tunneling barrier. And your T will be smaller now, because your contact is not ideal. But then you're not at this fundamental limit anymore. Skip. Everything you show so far seems to be one of the voltage between the two contacts. But the thing of a washer, you have an agate. And the ability is to function agate at one point, primarily at the surface where you do the base. So conflicting, conflicting, because it seems like, particularly high-gain voltage, you can't get to the base. So it seems like you're always going to be limited, because the zoom scattering is so strong. Yeah. So, yeah, I guess maybe your question really is, can you ever expect to achieve ballistic transport in a silicon MOSFET? So when we analyze that data, I always did it for the highest gate voltage, because I'm thinking that under on-current conditions, when you're applying the maximum voltage and you have the maximum carriers in the channel, what is the mean free path? How close to the ballistic limit are you? At the highest gate voltage, the electrons are very close to the surface. They're very strongly influenced by the surface roughness scattering. The mobility is quite low. Now, one of the interesting things is in silicon MOSFETs, we've been doing these estimates of how close to the ballistic limit is a device for, I don't know, 10, 12 years or so. As channel lengths have gone from 250 nanometers to about 30 or 40 now. And one of the things we find is that you're getting no closer to the ballistic limit in silicon. And it's because as you scale the device down and try to maintain its performance, you make the channel thinner. You might dope it more heavily to try to control electrostatics. Everything you do to make a better transistor gives you more scattering. So even though your channel length is getting shorter, your mean free path is getting shorter. And in the end, when you finally end up with a well-behaved transistor, you're operating at about 50% for the on current, you're operating at about 50% of the ballistic limit. Now, for a 3.5 device, it's much different. Because in a 3.5 device, these inner traces can be almost atomically smooth. The surface roughness scattering is very small. And that's why you can maintain very high mobilities in a 3.5 transistor. And that's why they tend to operate near the ballistic limit. It's hard to see how silicon is going to achieve that. So all of this analysis we did is based on the difference in the form in the source and the current. And we do the analysis based on that. So how do you, like, if the device is becoming smaller and smaller and smaller, the electron interaction also becomes important, so how do you accommodate it? So the question is, how do we accommodate the electron-electron interaction? And it's not explicitly put in here in any case. In the context, there are going to be electron-electron interactions that are going to help maintain thermodynamic equilibrium. Now, you might be concerned about electron-electron interactions in the channel. And what would they do? It's not completely clear. If it's electron-electron scattering, it conserves momentum. So the first order, if the electrons are scattering off of each other, it doesn't affect the current flow. We haven't seen strong evidence in conventional devices of anything unusual happening due to electron-electron scattering. But that's always a very challenging area to treat computationally or theoretically. Obviously, you can't treat it in a simple model like this very well. Even when we do more sophisticated things like non-equilibrium greens function calculations, treating electron-electron scattering is enormously difficult. I don't yet see anything in the devices that I'm looking at that tells me that we have to, that anything is going to fundamentally change. Yes? In traditional device simulation, we use a good diffusion of equations. We put a lot of work into the mobility model and to get really accurate scattering for the defensive on that basis. Now, when we go to these really small devices that we just going to have to throw out and all the drift equation go to NTF, or do you see a way that we could do some effective mobility that would be perhaps ninth dependent with our standard type of simulator actually? Yeah. So this is a good question. It might be a research question a little more. So the question is, you have this enormous infrastructure built up and these traditional CAD tools that are based on diffusive transport work very well. I think you're probably doing a little bit of tweaking of saturated velocities and things in order to get things to work. But you're guided by more sophisticated simulations. And the concern is, if you get down to these really small regimes, do you have to worry? You have to be careful. For one of the things I've seen, I don't know if hydrodynamic sort of balance equation models are widely used now. They were at one time. One of the things I noticed is that models like that don't comprehend ballistic limits. And I used to go to conferences and see people showing results that were clearly above the ballistic limit. You don't get it. That's why it's safer to use drift diffusion because you have a saturated velocity that always clamps you, and at least you never go above the ballistic limit. Now your question is, can you somehow do things make the diffusive model better so that it comprehends ballistic transport or quasi-ballistic transport? I've seen some people who develop spice models that are trying to do it in spice so that instead of W over L, you use something like a ballistic mobility so that things go gracefully into the ballistic regime. We wrote a couple of papers a several years ago about things like ballistic drift diffusion equations. It might be possible to make them work using ideas like this ballistic mobility. I don't think you'll have to ever worry about it too much in silicon because the scattering is so strong even as you go down. But if you're looking for three, five devices, the intrinsic device is really operating very close to the ballistic limit. And then I think you have to worry if you're using a drift diffusion model. It might be masked because the channel resistance is so small compared to the parasitic resistances that you might not notice it in the overall device performance. But in that case, you could probably just go with the really high mobility, and it's really dominated by the external resistance. Yeah, yeah, yeah. Any other? Let's see, we'll start, yeah. Go ahead. I've got the question about this picture, and the question is, where is the voltage drop? And your answer is the half of its drops on the right, and the half on the left. How did we know that this is, that the things are like that? Any experience, because we say that, you see it's like the function and the physics, we don't have a problem. Yeah, yeah. So one way you can answer this question, and we've done this, you do an NEGF simulation. So Professor Dada likes to think of this as a simple way to think about problems. The rigorous version of this approach is NEGF simulation. So you remove, what we're doing here is just sort of giving you a plausibility argument as to why this is reasonable. You can run an NEGF simulation on this device, do it self-consistently, go in and look at where the voltage is dropping, and this is what you'll see. I've seen several papers where the, for example, drop is just on one hand, on the other, not on the other. And there are, I think, all sorts of papers where the drop is linear. You know, if you put scattering in the channel, you'll see a linear drop. But in a ballistic device, you won't see any drop. Now sometimes, and this confuses, I get this question frequently, if I do a short channel MOSFET, say 40 nanometers or so, and you plot out the conduction band profile, and you look in the channel, you will see, if it's a ballistic channel, you will see a slope. But what you're seeing there is not a resistive volt drop, it's a 2D effect. You've got a positive drain voltage at one end, you're basically seeing a two-dimensional potential profile that's altering it. You're not seeing a resistive volt drop. If you do a small device in the ballistic case under low applied bias, you'll get a flat potential inside of the gate. So if you have some of those papers, point me to them, I'd like to see what they're doing, but we've looked at these in our own simulations and this is what we see happening. It's a consistent simulation. Yeah, yeah. It wasn't a shooting device. Poisson and NEGF, yeah, yeah. So you said that the 3.5 device is better because it has less surface scattering. Mm-hmm. Can you please tell me a little bit more about that? Because also sometimes the problem comes because of the oxide and 3.5 semi-contrary interface. Yeah. So the question is, why are the mobility so high in 3.5? And it's really in specially designed 3.5s, right? It's in these epitaxial structures where your top surfaces cannot oxidize. It's grown epitaxially with a lattice match semiconductor and in order to get the dopants in the channel, you would dope the wide band gap layer and let them spill down into an undoped layer so there are no charged carriers left to scatter them and people have been able to get enormous mobilities. I've seen mobilities of 10 million at very low temperatures in structures like this. Now if you just take a bare 3.5 and let it oxidize or try to put a dielectric on it, the mobility will be very bad. In silicon it seems that the oxidation process just inherently creates a rough surface and it's very hard to get around that. It's the dominant scattering mechanism in these very short channel devices. There is no interface between the 3.5 and the oxide and how do you get the channel? Well in those particular, in these hemp structures, it's a, let's see if I can find that 3.5 device. We might have somebody in here who's working on these transistors. Do you work on these? So that gate there is really a shot key barrier to the wide band gap in the aluminum arsenide layer. Now sometimes if MOSFET people like to think of that wide band gap, indium aluminum arsenide layer is sort of like an oxide but it hasn't got an especially high band gap but it's a little bit higher than the indium aluminum arsenide. So the interface, the indium gallium arsenide multi-quantum well, that's the channel. And the interface between that channel and this layer that's acting like the oxide is an epitaxial lattice matched interface and it's a much smoother interface than you get from a silicon MOSFET. So when you talk about 3.5 device, it's a hemp structure. Right, right. Yeah, there's a lot of interest right now in trying to build 3.5 MOSFETs to see whether they would address some of the performance challenges of this silicon technology. And one of the biggest challenges is in getting high mobility and it's that interface. Deposit the dielectric on it, it's very hard to maintain a high mobility. These structures give you that. Skip. How do 3.5s go to density states? Yeah, yeah. There's always a trade-off somewhere. So the 3.5, so Skip is saying that there are some challenges. The 3.5s have high mobilities but they also have low density of states and that makes it hard to get a lot of carriers in there. Right. Yes. For the transmission, for evasion, it has a dependence over the channel lens. So it's a dependence very high on the ETL simulation. You're talking about that formula lambda over lambda plus L. You know, I'll give a simple derivation of that in lecture six. You know, the thing that's been surprising to me is it works so well. Like, so we use it in a lot of things like when we do Monte Carlo simulations and you have all kinds of complex scattering mechanisms going on at the same time and you want to ask yourself what is the average mean free path? You know, what we do is we do simulations with different lengths and we just count the number of electrons we shoot in at the beginning, the number that come out at the end, that's the transmission. If we just plot that transmission versus length, it just falls beautifully on this expression even though I don't have any rigorous justification for it. In practice, it seems to work enormous, it seems to work very well. Yeah. Any other questions? Okay, so we'll take a 30 minute break and then we're going to continue this but we'll talk about adding temperature and we're making progress. Thank you. Thank you.