You can follow along with this presentation using printed slides from the Nano Hub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. So we had a question that said when we talk about high mobility channel materials like three-fives or graphene replacing silicon, how would it exactly be better? Is it that we can make transistors with the same current levels at longer gate lengths as using longer gate lengths, it's easier to manufacture and control the channel length? So it's a good question, I can't say a lot about that, and when you make a transistor, transistors people would like to have a lot of current when the device is on, so add a given voltage as low as possible, a volt, if you could do a half a volt it would be even better. You'd like to get a lot of current so that you can charge and discharge all of the capacitances, do the wiring and everything on the chip and run very fast, and you would also like the channel resistance down here to be as small as possible so that the current goes up very rapidly and saturates and it doesn't take along a large drain voltage in order to saturate because power goes as voltage squared so you could operate at lower voltages.  So down here, traditionally we write the current as W mu, they call it effective mobility, we'll just say mu here, figure out what mu it is, gate capacitance, Vg minus Vt times Vds. So this is a lot of the reason that people were really interested in 3.5 resistors or graphene because the mobilities are very, very high. They looked at this, maybe this is 200 centimeters squared per volt second for silicon, maybe you could get 10,000 on some of these 3.5's. That looks like an enormous boost, but it wouldn't quite work that way, that would give you more steepness but you're going to saturate and also you can't get steeper than the ballistic limit. So this mobility really here, one over that mobility is one over the actual mobility which might be enormous for these materials plus one over this ballistic mobility and that's the thing that depends on channel lengths. So that ballistic mobility is thermal velocity times mean free path but in the ballistic limit the mean free path is the channel length, divided by 2, that's diffusion coefficient and then the Einstein relation divided by kT over Q. So as you scale and get transistors denser and denser and denser, this ballistic mobility gets smaller and smaller and smaller and it tends to dominate so you get some boost but you don't get nearly the boost that you would expect just by looking at the differences in the bulk mobilities. But there is this advantage alluded to in the second question, is it that we can get the same current levels at longer channel lengths? Yes, so with relaxed geometry, especially if you're looking at RF devices you can get very high performance, oops this should be W over L. I can go to much higher channel lengths where the ballistic mobility is bigger so it doesn't limit things as much. Now we're limited more by the actual mobility and that's so much higher in three-fives or graphene that in a longer channel length device you could still get very good RF performance or high-speed performance. So if you're looking for RF performance that's an advantage. If you're looking for density in a digital circuit you're going to make these transistors as small as you possibly can and then this ballistic mobility is going to limit you. Now also up here I mean you tend to get more current, you can see that this mobility has a thermal velocity, so three-five devices, it's a little bit harder to think about it in graphene but you have to think about what the effective mass is in graphene but similar things happen. If I have a, frequently I have a high mobility because I have a very light effective mass. That light effective mass is also going to give me a very high thermal velocity. That thermal velocity, remember what the on current was. It was W times C ox times the ballistic injection velocity times VG minus VT. So what's giving me the high mobility, part of it is due to low scattering, part of it is due to light effective mass. The light effective mass is also giving me a high injection velocity. So that's boosting my on current, that's good. I'll have a higher ballistic limit in those materials. But there's always a gotcha somewhere, right? It's always a trade-off. So there's one more issue that we can't, I probably can't do justice to but I will point out in an MOS transistor, so let's say I have a source, I have a drain and we have a gate insulator and we have a metal. When I've been talking about the gate capacitance, I've just written the oxide or insulator capacitance. And that's epsilon of the oxide divided by the thickness of the oxide, which is like a little over one nanometer these days. So this is the capacitance in Farads per square centimeter. But that's not quite true because the semiconductor channel itself also has some capacitance. So the actual capacitance is this insulator, or insulator if you've replaced SiO2 with something else like Intel has, then it's some other insulator capacitance with a different dielectric constant. But it also has in series with it a semiconductor. These people call this the quantum capacitance. These two are in series. So here's my gate terminal. Here's the surface of the semiconductor or the middle of the channel. And the actual gate capacitance is the series combination of two capacitors, the physical capacitor formed by this insulator, parallel plate capacitor, and the capacitance due to the semiconductor. This probably won't come up in your lectures later on, but you can show that this semiconductor capacitance under simple conditions is q squared times the density of states at the Fermi level. The 2D density of states is effective mass over pi times h bar squared. So in three fives, the light effective mass that's giving you the high mobility that might not be so important because your ballistic mobility is going to limit you, but it's also giving you the high thermal velocity, which is important because it gives you a higher drive current, which you want, is also giving you a lower density of states capacitance here. So if I put these two capacitors in series, it's going to lower the overall gate capacitance. So there's a trade-off. The boost that you get from the light effective mass is offset to some degree by the drop in the gate capacitance. Your charge is going faster, but you have less charge. So the net effect is that you get a benefit, but again you don't get quite as much benefit as you might have initially anticipated. So people are looking very closely at this. It looks like there might be a win there. It's not quite the slam dunk that you might think of if you had only looked at mobility and you had only looked at ballistic injection velocity, but it looks like if you're careful, there could be a place there where you could design a device and get some improved performance. Okay. Anything else I can add to that? In the ballistic mobility of the thermal velocity VT, so what's the effective mass to use? Is the effective mass in the metal or effective mass injection? Because is the injected from the source? Yeah, yeah. So, yeah, make sure Siprio agrees with this. What we're doing, electrons are coming in from the source and we have this top of the barrier. We're populating these states with a Fermi level according to the band structure at the top of the barrier, which we're thinking of at the beginning of the channel. So it's the effective mass at the top of the barrier here. We're assuming we know the density of states at the beginning of the channel. Is that right? Then there are questions. If you get into really small devices, is the effective mass in this very short channel device in that region the same as it is in the bulk silicon? To first order, it's pretty good to remarkably small devices. These days, one of the other reasons that 3.5s aren't quite as attractive an alternative as they were initially is that silicon never stays still.  There's been a lot that's been done in recent years on strain engineering by straining silicon effectively to lighten the effective mass and get a lot of these benefits, which makes it even harder for anything to compete against silicon. But as I said, you work the numbers, it looks like there's an advantage there, but it's got to be done in a well-designed device. You forgot the contact resistance of 3.5 and your ballistic. So you forgot the ballistic mobility of 3.5. It's better than silicon, right? Yeah. So the ballistic, for a given channel length, I guess this philosophy will be higher, right? But then we probably get that there's an issue of always straining resistance of getting constructed, which is a real resistance. All of this parasitic source-drain resistance, which, yeah. So some of these practical issues which are very, very challenging because you're thinning everything down, you're making all of these parasitic resistances higher. These resistances are probably scaling in the wrong direction. They want to go up with scaling because you're shrinking all the dimensions. You're shrinking the cross-sectional area of this resistor. You're shrinking the channel length, and this is going down. What's the VT, the potential velocity? Pardon me? What would the velocity difference be between the silicon and the resistances? Oh, this velocity here? So in silicon, it's... So one of the things I really should have done, this really depends on where the Fermi level is in the band. When you're a non-degenerate semiconductor, it's independent of the gate bias and it's 1.2 times 10 to the 7th centimeters per second. When you start pushing the Fermi level up into band, that velocity gets higher. So under typical on-current conditions, I think this is more like 1.4, 1.5, something like that. If you go to a 3.5, that number, as I recall, depending on the effective mass and how light it is, that can be more like between 2 to 3 times 10 to the 7th centimeters per second. So there's a possibility of almost a factor of 2 there. Okay. All right, if there are no more questions, then I'll turn the microphone over to Professor Dada. Thinking where to start, there's a number of different issues that people have raised. So maybe one thing we could do first is take a simple example and see how these coefficients, how these things like density of states, density of modes, et cetera, what they look like. So supposing I have a two-dimensional conductor with W length L, and we'd like to get the density of states, number of modes, et cetera, and usually you assume some kind of an energy momentum relationship here, EP relation. So usually you'd say equals E squared over 2M, that would be the standard one. So if you want the velocity, then it would look like DEDP, that's the general formula, which would be P over M. That's the standard one. Now the other material of great current interest is graphene, for example. Where E looks like some constant times P. So this one, if I were to plot it, it would have looked parabolic. This one, if I plot it, would look like this. And the question is what is the velocity here? Suppose DEDP, so basically it's just V0. Actually I don't need this absolute value around because I can think of P as the magnitude and then there's any angle. Then really V equals V0, that's it. Now one interesting question is what is the mass in this case? Because the way I said it is that in order to use Groot formula or things like that, the mass you should really think of as the ratio of P to V. Just as here of course that's this mass, no problem. But if you're thinking of graphene, then if you're wondering what mass to use, it's really P divided by V. And there you'll notice that you could write P as E over V0. And this V is fixed. So that's like E over V0 square. So the point is as the energy increases, the mass increases proportional to the energy essentially. So if you're trying to use a formula like, let's say you're talking of mobility. Mobility is usually think of it as Q tau over M. And as electrons have higher energy, let's say because you have, when you increase the carrier density, as you know the Fermi energy is going up. And so the average energy of the electrons is higher up. And usually as you go up, if the mobility goes down, you say that maybe the scattering time is going down. But not really. It could be just because the mass is going up. That's it. So most graphene materials, when people actually measure the mobility as a function of carrier density, they find it goes down with energy as it increases the carrier density. Most of it really comes from the mass going up essentially. That's got nothing to do with the mean free time as such. Really. If you're thinking of it this way. Right? Okay. So now... I mentioned that the equation there says E equal to MP squared. Right, right.  E equals MC squared. It looks like that, yeah, yeah. But people do say the electrons behave like a relativistic part of the graphene. Right. This is kind of why they say that. Right, so graphene has lots of very interesting... And this is why I say that it's important to separate out the simple things from the rocket science. And what happens with this usual way of thinking of transport, I feel is a lot of simple things look like rocket science. Which it shouldn't. That's my point. A lot of this is really pretty simple. Nothing much is involved here. Now, in terms of density of states, I say usually that the best thing is to first write the total number of states that you have below a certain momentum. And that is kind of the same no matter what EP relation one talks about. And that is the thing I wrote in the morning. That N of P would be like if it was a two-dimensional conductor, a one-dimensional conductor, you'd have written it something like this. And if it's two-dimensional, then you'd actually write it something like this. And this is... Here the simple argument is that you're just looking at... This is P. X, P, Y. And you're asking how many states do I have whose energy... whose P is less than some maximum value? And so you want to know how close together these states are. And that is where you're bringing in the wave nature of electrons. Everything else, it is all particles. But this is where you're saying, well, the de Broglie wavelength has to fit into the box. That's where the wave nature came in. And that's this de Broglie wavelength fitting into the box type of idea. And that's what then gives you this N. And till this point, there's no quantum mechanics at all in all this discussion. Now, once you have this N, so let me write it here. It's equal to whatever I wrote there. Pi P square. So that's the area of... that's the area of that circle. And then each state occupies a region sort of like this H over L, H over W. That's how you're getting it. Now, from this N, the way you get a density of states then, that's where the EK relation will come in. Because now you're trying to get a density of states. And that is when you say we'll take dN, dB, dE, which is like... you could take dN, dB, and then dP, dE. And so the density of states that you finally get would look different here and here. For graphene, it would look a little different from that. Although, as I said, at this level, it was all exactly the same. Nothing was different. This thing, velocity, depends upon energy and momentum. So this is the graphene case. So energy is linear with momentum. And velocity, velocity, that's this constant. This constant. So that is why, you see, often one of the things, I said that the masses should be used as P over V. And that is what I can show is what you need if you want to use Groot formula. But usually in textbooks, what you learn is mass is actually, I mean, 1 over M is usually dV, dP. That is what is usually defined as mass in the textbooks. And from this point of view, in graphene, the mass is actually infinite. And the argument is very simple. You see, in electron and graphene has the same velocity no matter what you do. So the argument they go over there, when you put an electric field, its momentum keeps increasing and dP dt is the force. And because of that force, supposedly the momentum keeps increasing. But velocity doesn't change. So acceleration essentially is zero. So you say, yeah, mass is infinite. That's why you can't. And of course, if you use mass infinite in the mobility formula, you get nonsense. I mean, that would basically tell you it shouldn't be conducting, which of course isn't the case at all. This is why I'm saying a lot of simple things start looking like rocket science. But you can't understand anything. But all I'm saying is not really. If you look at it the right way, it's actually not that complicated at all. Anyway, so the velocity is constant. So the point I'm making is, although this is what is usually described as mass, in order to use the root formula, you should really be defining it this way. So instead of dV dP, you should really be defining it as dO dP. Then you could think that way and it would give you sensible answers. Because you still have a velocity, but the momentum is zero. Because this is a constant slope. Velocity is the slope of that, which is the same everywhere. So when you are here, it's like you got a lot of velocity without any momentum. The mass is zero. So what happens to the mass here? What happens to it? The mass is zero, so it's got five-fourths. So this will have to be very constant. I mean, this should have a like infinite velocity at the end, right? Yeah. The way you are supposed to use the semi-classical theory is that dP dt is equal to force in its electric field. So you're supposed to use the Newton's laws. In semi-classical theory, when you apply it, the force should be equal to derivative of the momentum, time derivative of the momentum. So the way you should think is if an electron happens to be here, its P will keep increasing, like this. The velocity will stay the same though. It won't. It's just that you're not thinking a lot of other things are built in that we think of this as mv, and then we like to say that the acceleration is Qe over m. And what I'm saying is this is correct in semi-classical dynamic. This isn't. I mean, that should not be the starting point. The starting point should always be this one. When you are applying it to any relations, you should really be starting here. So if you look at this dn dP, that would be proportional to P. You can see. So I guess I could write this. So I won't bother. So you could write this out 2 pi P Lw over h squared. And then I have 1 over velocity. So what I've done here is I wrote dn dP because this, as I said, just comes from fitting waves into a box. It doesn't have anything to do with eK relations. Even if you are talking about phonons and trying to figure out density of states, you could still use that. After that, when you do omega K relation, that's where the rest of it will change. This just comes from fitting wavelengths into a box. So dn dP you'll get that. And then dPD, that's the inverse of the velocity. Now this is where you see when you try to get density of states, you'll get two different answers for this and that. Because here, what will happen is the velocity itself is proportional to P. So it will be P over m. And then the P will cancel out. And that's this well-known result for parabolic bands that the density of states is constant. It doesn't change with energy at all. Because you see that P that cancels out and then what you're left with is just constants. You know, it will be this m over pi h square type of thing. That's the standard answer. On the other hand, when you do that in graphene you'll get something a little different. Because up to here is fine. But then this velocity is not P over m, it's constant. Some number B0. Whatever it is. So what that means now is that density of states is proportional to P and E itself is proportional to P. So basically you'll get a density of states that is proportional to energy. And that's the standard result again in graphene. When you draw the density of states it's like proportional to energy. So the picture you usually often draw for density of states is something like this. That will be it. So these are things I'm sure you have seen before. So the part that is probably a different part that we haven't that you may not have seen as much of is this number of modes. And this is where I guess what we said was that this m over h was equal to this dVz over 2l. And this is what is like d V over 2l and then there is a numerical factor that comes from averaging over the angle. Because in general if you think of the velocity like this is the z direction and you have a distribution of angles here. So if you want the average value of the z component you have to kind of take this cosine theta and average over that. That you can check. But this is dV over 2l. Now the interesting result is that while this is true and if you take the d and put it all back in here you'll find that irrespective of the eK relation finally this number of modes becomes how many wavelengths you can fit in here. Just as we wrote this n by saying how many wavelengths you can fit into that box the corresponding modes is like how many wavelengths you can fit into the width. This is the part that's not obvious but this is the point I was trying to make that is because of this general relationship that dVp is n times the number of dimensions and so dV is like n over p and so just as n tries to fit wavelengths into this two dimensional box m kind of tries to fit wavelengths into this cross-section. This is the part that if you try out a few things you'll see how it works. So this one we can do it. So I guess I erase that. So we had this expression for density of states times 1 over v. That's d. And what we are trying to find is what is m and m is equal to h dV over 2L and it's supposed to be this vz so when you average it I think you get 2 over pi. Not sure if I'm remembering the factor correctly. We can work it out but for the moment let me just assume that. So this is d and what I want is h dV over 2L times that. And note that here I've still not made use of the ek relation at all. Which means whatever I've done so far should work just as well for graphene as for anything else. I haven't done anything so far. So if I take this I need h dV so I can take the 1h out of here and 1v out of here and put it here. So that's h dV. Next what I can do is I have to divide by 2L I'll put a 2L here and I'll put a 2L here and then I have this extra 2 over pi. So what I've done is essentially h dV over 2L times 2 over pi and that basically should be mass. So whatever I have on the left hand side I'm not mass. Basically should be this number of modes. So whatever I have on that side should be the number of modes. And again I have not yet made use of ek relations or anything. So it should work in general. So now I think you'll see a number of things will cancel out 2L pi out and so basically what you get is 2 over h over p which is basically as if you are taking this width and fitting de Broglie wavelengths into it. That's it. So this is the point that we're trying to make that although this definition of m kind of doesn't look like you're fitting wavelengths into anything as it stands. It's like density of states times velocity which makes a lot of sense. Density of states tells you how many electrons you can put there to carry current of course they need to be moving. They need to have a velocity. So it's not enough to have states. So that's why current depends on modes and modes depends on density of states times velocity. That makes good sense. But yes. Momentum. So momentum equal to h bar k. Right? At any energy, right? So in this discussion that's a good point. Let me explain. So whenever we are discussing this the way I think about it is that we are talking about this elastic resistor where electrons are traveling at different energies and at the same time it is moving at a particular energy. It's elastic. It takes to the same energy. So when I think of conductance, number of modes, density of states, anything I write it is at some energy. Now you say, okay, but then for this conductor how do I know what exactly to expect? Well if it is low bias it will basically involve integral dE minus del F0 del E of whatever quantity we are talking about. So when I calculate number of modes it is a function of energy. So here for example the number of modes we got depended on P and that I can translate back into energy because what that would mean is number of modes in a parabolic material would look like square root of E because M is proportional to P and the parabolic material means E is proportional to P squared. So it would look like square root of E. On the other hand if you are talking of graphene then P is proportional to E. So the number of modes when you translate it into energy will look like square root of E for parabolic materials and will look like proportional to E for graphene. In terms of P it is still the same because it just fitted how many it is. And if you are talking of ballistic conductance how much is it? It is basically Q squared over H times this number of modes. Number of modes at what energy will over all energies but then you have to integrate DE minus del F0 del E because all conduction properties tends to be average this way. For low bias this usually works very well. Why doesn't DE actually be a very strong function? Right. Is that it from the whole? Right. So very good. I want to get on to that. Before I move on I hope this basic math I guess is clear. What I am trying to say is that usually I always start from this function up to P how many you have and then you can take its derivative all this stuff. And all this then finally will give you a density of states or a density of modes. The basic concepts as I mentioned where D, F, M and N density of states Fermi function M is number of modes N is number of states up to a certain energy and conductance usually will depend on that quantity averaged this way. This is one of the points I tried to make that this DEL F0 DEL E looks like if this axis is energy looks like something that is peaked right around the chemical potential, wherever that is. So Fermi function when you look at its derivative it has a peak there and the area under the curve is 1. That's the important thing. We are taking an weighted average of this. So the E you have is around the Fermi Except that let me draw this picture that I had originally this density of states and then there is a mu1 mu2 energy and this is where I said the Fermi function would probably look like a peak like this. But the same concept though would work if we are talking of semiconductors what I mean by that is supposing this mu instead of being here we are talking of something down here let's say you know with a small bias so in that case this DF0 DE looks like this and of course all the conduction is happening on the tail of that function and this is the limit the non-degenerate limit that Mark talked about in a lot of the calculations in semiconductor texts are usually in this limit because in this limit you see the Fermi function as you know looks like this and of course down here these two things don't look at all similar but up here those two functions are essentially just 10 times each other and that's the approximation that Mark was using for non-degenerate conductors which is that del F0 del E is equal to F0 over KT that holds up here that's it down here actually 1 minus F0 kind of looks like that but otherwise those two functions are very different but this one I feel this function kind of expresses this basic fact the point I've been trying to make is that conduction depends on things right around that chemical potential for low bias and for high bias I'd say what matters is let's say you put mu1 here mu2 here then what matters is this entire window and in this case though often I'd say neglecting inelastic processes is not a good idea because that can change a lot of things in this part and that I'd like to say a few words about after this please go ahead if there's more questions about what we just talked about is this is this point generally clear you know the basic concepts are again the D, F, N and M and for a given EK relation how to calculate them and as I said often things like modes as a function of p will look the same no matter what material you are in you just take width and divide it because you're just fitting wavelengths into this thing so I guess you're really talking about spin problems taking a spin that's true that is one point I should say because these little factors especially when you're beginning can really take up a lot of time and it can really get you very confused what often happens is when people write modes they don't include spin in it because one thing we all know is that all know in the sense you've heard it many times is that states usually come in pairs if there's an up spin state there's a down spin state what you often do at the end is remember to multiply by 2 now the thing is that usually when people talk of density of states the way I have done this it's like as if there's no spin so when others I'm talking of density of states per spin and this number of modes that's also number of modes per spin it's just that when people talk about this they usually talk of the density of states including spin but then they don't include it in the modes the modes they say well it's like modes per spin and so then instead of a 2 you see a 4 here because their density of states includes spin but their modes doesn't so that's what it's all about and of course the advantage of that is that it connects with most of the literature so it'll look like what you'll see most of the places but in order to keep track of it I like to keep and to me in a way spin is a little bit also like valleys that in some cases you have to include valleys and now I'm saying is spin, valley all these things you can bring in later you know okay I've got two spins and two valleys and these are all very so when you look at the ballistic conductance for example what is the quantum of conductance and I said well it's q square over h right or if you invert it h over q square that's like 25 kilo ohms now if you actually measure the conductance of something of a small piece in gallium arsenide what you get is not 25 kilo ohms so the minimum you get the resistance you get is actually 12 kilo ohms more like 12.5 why well because there's always the two spins in parallel so that's the quantum of conductance again per spin per mode but then you always have two spins in parallel but then when you measure it in carbon nanotubes you get 6 kilo ohms why is that well because there's two spins and there's two valleys that go with it okay so whatever I said here everything I tried to write everything in a way as if there is just no spins no valleys etc and then you have to multiply by spins valleys as needed and sometimes equations may look a little different because people might include the spins and valleys here but not there and things like that because again because of conventions and how they're doing it okay any other questions please how about to the example again if you measure the resistance it's going to be like 6.5 right so the smallest carbon nanotubes ballistic will give 6 kilo ohms well because even the smallest one kind of has these four levels four channels in parallel the two spins and the two valleys 25 case per channel and then channels can come from spin from valley etc okay and yeah were there any other questions related to this valleys okay it's like when you look at this energy momentum relation what happens I said that the energy momentum relation looks like this E versus P actually no they said that one of the and this is this part of our band structure that we have in crystalline solids what the E energy momentum relation looks like there and it is really a fairly complicated function that goes on and what you try to do is look at the minimum because those are the ones that contribute to conduction usually because as I said if your chemical potential is around here so usually there is a valence band and a conduction band and you try to look at the minimum here and the maximum here and what could happen is there could be multiple minimums so you could have one minimum here and another one here for example so you could have something that looks like this and then you got one here and one here and whatever we did would kind of apply to that one we got two valleys and we did all the calculation there but then this is in parallel with that so you they are all in parallel so in carbon nanotubes there is like these two valleys that need to be included and this is true in silicon there is six valleys actually so gallium arsenide is simple that when one valley it keeps it straight that is within single limit cell that is within single limit cell any material there is nothing about that six valleys inside one unit cell of silicon like in case of graphene there are two valleys inside one unit cell of graphene so here are six unit cells inside one unit cell of silicon so you are saying the number of valleys is related to the number of atoms in a unit cell is that right? I say because in carbon nanotubes in graphene there are two or one atoms per unit cell that is true the reason I am hesitating is gallium arsenide and silicon both being zinc blend probably have the same number of atoms per unit cell but values are different because what can happen is this you could have one material looking like this and another material where this happens to be d-value so in this case you would say I have got only one valley in this case you would say I have got two valleys because those are the lowest ones so it is hard to predict the lowest one how many there would be there is probably no simple connection there any other question please on this now the other question I have had from number of people I suppose is again about the role of electric fields in this current flow and there let me make a comment that and this is something I think Mark Lundstrom also touched on that if we are talking about let's say semiconductor this is mu 1 this is mu 2 and you have a density of states like I was drawing say somewhere here and for this discussion let's assume you know this is fixed so I had a electric field like this now one point I made was that supposing you had this fixed like this and goes down when I calculate the current it doesn't seem to depend very much on the electric field in here as you think about it because all that matters is the fact that these two chemical potentials are different and the way we are calculating it you might have you would obtain a current that would look like u over h number of moles lambda over l plus lambda I guess this was the expression that we have been working with with f1 minus f2 now and the one point that Mark made I think is that when you put in that actually if you are looking at the saturation current that is when in this energy range of interest f2 is all zero which means you can drop this because there is nothing coming back from this side when you look at the saturation current actually the current is a lot bigger than what you might think from this because although the length of the devices says some length that based on the low field current you might have thought that it is 50 mean free paths but when you look at the saturation current it looks like a whole lot less looks like it's only 10 mean free paths and the question is why is that part of it and this is where the way I think about it is that this is something that kind of requires you to go beyond a point about what is happening inside the device the spatial distribution of things because if you look at the density of states inside it is sort of like you have a density of states here but then because of the electric field it is going down here like this so because as I said when you have an electric field the electrostatic potential is different and because of the electrostatic potential as I mentioned the density of states should be changing this way so if the electrostatic potential here is lower because you have a positive voltage the entire density of states should be going down so what I am trying to plot here is this axis is energy this axis is z and I am trying to show where the density of states are so at this end it is 0 this is sort of like your band edge there is 0 below this but here it has kind of gone down so now if you think of again you are still talking elastic channel model so let's say electron is just going elastically and you use this formula to figure out the current now you see you have a problem you don't know what modes to use because at this end the number of modes corresponds to this energy at this end the number of modes corresponds to that energy I mean then physically you just say that at this end elastic energy you have got all this extra energy but the point is the number of modes has gone up so it is almost as if this conductor kind of looks like it got wider because here you got 10 modes let's say by the time you come here you got 100 modes so it will be almost as if you have widened out the whole thing and if you widen it out then you can see that if you just when you are calculating it this way it is as if it never widened out you are missing that whole thing and so finally you get a lot more current due to this electric field here so in that sense it is not as if electric fields cannot increase the current or cannot have any influence on the current it is just that in this language I will think of it a little differently but the part I felt can be very misleading overall is always to think of electric field as a driving thing and because when it comes to small devices it is very difficult that gets very confusing because as I said there is electric fields everywhere in a solid and it is hard to tell which ones are then causing your current and which ones aren't and right but so for example if you have two devices one of which has this and the other there is no electric field here it looks like this so let us say you manage to you have a really good gate that really holds it there it only drops off at the end so two different devices same mu1 and mu2 one has an electric field and the other has hardly an electric field except right at the end so in this case though when you draw this picture it would look more like this and this should have a whole lot less current than that one and you would say well you would be tempted to say that well it is because of the electric field I guess in a way you could say that that is true are you still in scattering is it true or not is scattering occurring this argument requires scattering right if you had a ballistic transistor none of this would matter because if you had a ballistic channel that is of course one important point that if you had a ballistic channel then the fact that you have hundred modes here makes no difference finally the conductance is limited by the ten modes absolutely right right so because in a ballistic case you do not have this factor and then indeed the electric field has zero effect on what is going on really in this context really do not do anything in the ballistic context unless it can pull this end down if it can pull this end down then of course it would give it more modes here but if it cannot pull this end down then nothing would happen that is why I am telling you but because if you have scattering then it looks like this then as Mark pointed out in a device like this it looks as if this L is a whole lot less right there is some effective length that you could use to describe this so but otherwise I feel that and I think anyone who actually is applying this diffusion type of equation I think they always start from something like sigma D I guess what I wrote as chemical potential BZ or sigma del mu that is where people always start from in the literature I mean this is where they would always start from and then they would say that well this chemical potential you could divide it up into various pieces and then try to extract a drift term and a diffusion term if they want to or when you have a heterostructure you could have a varying mass of things of that sort and all of that would usually then you try to figure out what terms you get but this is the starting point this is what is usually this is what is really believed and I'd say more correctly it is Df or rather del f I mean actually there is a function that tells you the occupation of all the states in energy of different states in energy at a given Z and you should look at the derivative of that that's really what is important and this could be replaced with this if you assume that somehow it is close to a Fermi function and that is sometimes a good approximation sometimes it's not that's part about whether you can replace this or and this let me make a say a few words about and that is so let's go back to our old picture then so here you have a Fermi function that looks like this now question is what is the occupation of the states inside here now if you assume this again this elastic model then the way Mark wrote it he said that well you had this density of states half the states are filled according to this the other half are filled according to that actually and so if you wrote down the occupation of states here what you'd see is something like this because in this region this is zero that is zero so you get zero in this region this is one that is zero so you get about half I mean actually it is like half the states are well occupied half the states are not but if you looked at average occupation it would have looked like that and then at the other end it's one and this of course doesn't quite look like a Fermi function especially if you put enough voltage on it and this is actually what the distribution would be if there were no inelastic processes inside your device if there are inelastic processes it would tend to try and make this into a Fermi looking function because all scattering processes try to restore this kind of function essentially basically would try to take things from here and put it there that's what it would try to do and what often works is often people use at least even though it's not quite justifiable is to say that well let's assume there's so much inelastic scattering that it makes it look like this so we assume that there is some local chemical potential local Fermi function that's what people assume again just to simplify the whole thing but one interesting point is that if the amount of voltage involved this mu1 and mu2 if that difference is less than KT if it's less than KT then actually it wouldn't look like this it would actually automatically look almost like a Fermi function anyway you can take two Fermi functions which are just a little bit off less than KT and add them up it looks almost like a Fermi function that's this Taylor series idea then even without scattering it's almost a Fermi function which means after that you put in some scattering it won't really do much to it but it will restore a Fermi function if that's already there it won't so one last comment maybe and are there any other questions about this that would like me to answer maybe that would be the question that step distribution so you would potentially see that the device that have that delta mu1 and mu2 then you would have an ID curve that would look like oh this is the occupation no but this is not density of states though this is the occupation this is the occupation we are writing no this is not density of states this is the occupation and I would say that in this case you are better off you'll get a much better description by not assuming a single chemical potential but rather two quasi Fermi levels essentially so what I mean by that is don't think of it this way just think of it as one thing that is like this and another thing that is like this and it's an average of the two and then call this one mu plus all that one mu minus and work with that that's a much better description of whatever is going on than to try to get one mu into the whole story is this even voltage for high voltage like if we do like for okay a high voltage on a ballistic conductor that's the part I'm thinking yeah even at okay let's first do the ballistic case ballistic case the thing is that half the states will have this the other half will have that so the best way to describe it is two chemical potentials mu plus and mu minus right and ballistic that is it right right because if it is what I mean is you have one group of states whichever this Fermi distribution and this is the plus group and then there is the minus group which has something looking like that and this is kind of an equilibrium within itself this is within equilibrium within itself and ballistic means what we are assuming is there's nothing to take somebody from here to there and as long as that is true even out with all the inelastic scattering makes no difference because this is already a Fermi function what else can inelastic scattering do to it now once you have things that take things from here to there then of course what will happen is this will tend to come down and this will tend to fill up a little more and then exactly what it is that requires you know the ballistic limit it's really that clear and inelastic processes won't make any difference with two but with two quasi Fermi levels this is the point I was trying to make that if you write i equals sigma d mu d z and try to solve this then you wouldn't quite get the right answer but if you recognize that there are kind of two equations and then set one boundary condition on this on the left and a boundary condition on that on the right then you would get all the effects the ballistic part everything but that is the important itself so just to make that why we do not depend on u plus n minus we do not depend on the time of the ratio right so usually the way these work is so the way I usually think about it is that in these devices in general you'd have one equation which I've written on top that is like what you might call a transport equation now where did that come from well what is inside is like the current this is sigma d mu d z as I was saying and current must be the same everywhere so d i d z is 0 that's kind of like the top equation then there is the second equation which is what I'd call the Poisson equation kind of looks similar except that what that epsilon is is the dielectric constant and that's the potential and that depends on electron densities now in general all this has to be solved self consistently so usually what happens is the following that so supposing you have a conductor here and you have put some chemical potential mu 1 and the equilibrium chemical potential was let's say here and you have put all these extra electrons in and so you're solving the transport equation and you're trying to figure out how what the electron density is in here now if you do it self consistently with Poisson then what will happen is the bottom of the band and all will automatically adjust so that the carrier density because if you just so let's say this is the equilibrium mu 0 and you have voltage here to raise it and you pump all these extra electrons in here that would of course make this very negatively charged and if it is negatively charged when you solve it with Poisson equation this will then automatically go up somewhat etc so all these would be at high bias you'd have to this point that we discussed that saturation current of a transistor can get higher when you put a voltage because of this requirements of space charge that is how to keep the electron density constant so the picture I have in mind usually is that there's two things one is mu the other is u and the u has to do with this how the band edge moves etc and mu tells you how far it is filled and the difference is proportional to n electron density because you have all these states to fill up and what tends to happen is Poisson equation likes to keep this constant because it really likes to get rid of extra charges in general and so often in any solid if the mu varies in a certain way u also tries to vary much in the same way so as to make this charge density constant except that it may not be able to follow it exactly and so there will be local charges which you have to simulate and calculate so there could be local charges but the picture I have always known this is the mu that's the u and the electron density is proportional to that and the point I tried to make is that current is really determined by the slope of this one which you can break up into the slope of this and the slope of that one is drift, one is diffusion etc but to me that's a derived secondary step that under certain conditions may be useful but to me always the starting point is current is proportional to derivative of the quasi formula yes please I am using a course range picture because we are using classical in all kinds of equations so in some sense we are not looking at atomic states so when does this portion what is the problem of scale at least this portion not be able to use algorithms to actually find out the potential okay that's the part I am everything we are doing here is to call this semi classical transport semi classical means there is a density of states which comes from somewhere which actually could come from some quantum input that's why it's semi classical not fully classical since the density of states has some quantum input in it but then after that it's like cars on a highway I mean there is no more quantum physics in this that's the picture here now so that is as far as the transport equation goes now the second equation that's this Poisson equation that has to do with this electron-electron interaction which in general is a very complicated thing I would say in solid state understanding the theory properties of solid state that is the one unsolved problem no one quite knows how to do it accurately and the Poisson equation is an approximation to that and there people usually correct for it what they know is that the actual potential an electron feels due to other electron is less than what Poisson gives because Poisson is what you usually call the Hartree approximation and everything else usually will subtract from it and so people often during this they will use something less than that there will be the exchange correlation correction to all this but that has to do with the Poisson equation the second one now transport part of it I'd say that you are by doing this you are missing things like tunneling for example that is one reason why you might want to go to the quantum formalism like tomorrow morning what I'll talk about the other thing is usually interference effects and that is where I guess tomorrow I'll try to show an example that when you use a quantum formalism you actually see interference effects you get there but often at room temperature when you measure it you never see that so in that sense it's almost like the semi-classical one is more correct correct in the sense if correct means matching experiment of course so in that sense and why is that well that's because in real solids when you do the quantum picture usually you're assuming that an electron is going through and see some nice static potential and then if you have multiple paths you'll have interference but at room temperature it's not a static potential this whole thing is just jiggling it's like a very turbulent medium and so most of these interference effects coming from quantum mechanics are not observed at room temperature although these days there are experiments where people have seen a lot of these effects at low temperatures in the 1990s there were lots of experiments like that which show that so interference effects can be observed but if your objective is to model a conductor at room temperature then there are most cases you are better off using a semi-classical picture and I'd say that almost the only reason to go to the quantum is if tunneling effects are important there are all kinds of barriers where like Mark also showed how in the quantum meter MOSFET there were all these tunneling things that needed to be accounted for yes Is that possible that the bias is too large and the current is too high so the charge inside the contact does not change fast enough to read the so you're saying that if the contact is not connected well enough okay that is too long the charge inside the contact does not work that is again so in this whole discussion usually we are assuming that the contacts are always maintained in equilibrium right and this is the one point I always I want to stress is that current flows because you know electron goes from here to there and then all kinds of external forces talk about kind of relax it and then take it out this taking it out and you see if you never took it out then you see we wouldn't have a resistor you really have a capacitor because then what would happen is you'd have lots of electrons piling up here maybe and then after sometime this would float up some this would come down what you would have is a capacitor so when you take a device that's the point I make about electric fields being the driving thing I mean you could take the same device and put it inside two capacitor plates you would have the same electric field but it would just be a capacitor that's it there would be no current current flows because something not in your Hamiltonian something we don't even talk about is continually taking electrons out from here and putting them back in there if that didn't happen and this is the part where we are kind of including it in our model by saying that contacts are always maintained in local equilibrium this f1 and f2 so this very complicated process that goes on how electrons relax and how they get out and all that we are avoiding that whole thing as a boundary condition f1 and f2 the point you are making that's true that in nanoscale devices one can argue whether your source and drain regions are they acting as good contacts in the sense are they really being able to maintain f1 and f2 or can they get depleted and that's a conceptually quite possible in which case you should think of them also as part of the device and the real contact being somewhere else I mean where you actually have the solder I mean they actually have the blob outside that's the real contact and the source and drain are also kind of part of the device that can go somewhat out of equilibrium you are effectively saying that the channel is very inelastic but the context the channel is very inelastic but the contact are very inelastic you are actually saying that one has some coherence in the other one totally not coherent so the channel is small enough because this is a small region and we are saying the limit where everything is quite clear about how to think about it is this elastic resistor which describes a lot of small devices very well that is an electron goes from here to there and then loses all its energy here so this energy dissipation that is because you see this is where I say the transport theory is so complicated because it mixes these two things the dynamics and the thermodynamics the next part is reversible in the sense that an electron going from here to there is just as likely as an electron going from here to there if you could picture an electron going from here to there and ran it backwards it would not look funny at all it would look exactly quite reasonable on the other hand an electron that loses energy and comes out that is an irreversible process if you saw the reverse something from here for no good reason I am very surprised how do you delineate the context from the channel I mean is it so clean that one is so well elastic so so the basic difference is just in the contact has many degrees of freedom so to me the real thing is that you have this interstate with limited number of lanes which suddenly broadens out into something with many many lanes so that is what really to me is the difference between the two that you have limited channels and then lots of them right so for example when we this question came up earlier that when you draw these f's you know I drew it like this and then the question was but should this be separated or should I join them right and the argument is the what is the argument for joining it well it's sort of like this that the current is q over h times m times mu plus minus mu minus now inside I have got only 10 lanes and so let us say you have a certain difference here that is needed to carry the current here because you have 1 million lanes and you don't need any big difference to carry that same current but current of course is the same everywhere so the thing about the contact is it can carry all that current without having any significant difference in chemical potentials between the plus and the minus but I really wanted to stress that the elastic resistor is just a model that we are saying that let's assume this is elastic and real channels will have elastic processes in the channel and under certain conditions it can change the current and one of the problems I think I will talk about on Wednesday I will try to show an example where the conductance cannot really be written as f1 minus f2 at all the basic equation we had there gf1 minus f2 I will try to give examples of even a small conductor where this would give you totally wrong answers this wouldn't be it at all so I do not mean to imply that this covers everything but all I said was it helps you understand a lot of things a lot of things that are we used to call it to describe bipolar generation transition yeah that's something I guess we have been working on but actually the bipolar is a very good example of where you need something a little different and the reason is something like this actually this example that I give of where ignoring inelastic scattering wouldn't work at all is something like this supposing you have a density of states here that is connected to one contact this is contact one and you have a density of states here which is connected to two let's say so one group of states connected this way one group of states connected that way now if you had an elastic resistor model you know only elastic processes go on inside no current can flow the electrons come in here they can't go anywhere as soon as you turn in inelastic scattering of course there is communication here and current can flow so this is one example of something that could be quite small and the elastic resistor model would be just completely wrong because here one channel that is connected to the left one channel that is connected to the left now the p-n junction is a lot like that because those who are familiar with band diagrams when you draw a p-n junction as you know this is the picture we usually draw this is the p-type p-side with its quasi-formal level somewhere here this is the n-side and if you ignore these things which are not very important here that's a little bit like what I drew there that these this density of states is filled from this contact this density of states is filled from that contact this side and the current flows because of recombination processes that take you from one to the other and that's well known that the IV characteristics of a p-n junction are strongly influenced by the nature of the recombination processes that go on in the depletion region so obviously if you do not include recombination generation processes you wouldn't get it right okay I think yeah thank you for your attention and see you tomorrow


Good evening, everyone. I am Dr. Amar Prakash. I am here to give paper presentation on prevalence of interstitial lung abnormalities in HRCT chest. My guide is Dr. BK Soni, Professor, HOD of the Department of Radio Diagnosis, INH Ashwini. I am finally a resident in Institute of Naval Medicine, INH Ashwini, Colaba, Mumbai. So, in introduction, we will, since we will see interstitial lung abnormalities are radiologic abnormality. And on chest CT, it is thought to be an early and mild form of pulmonary fibrosis. There is a growing awareness of the clinical significance of interstitial lung abnormalities incidentally found in the chest CT. So, a possible association between interstitial abnormalities and idiomatic pulmonary fibrosis has been found in the literature. And there are reports that interstitial lung abnormalities can progress to clinically significant interstitial lung disease. So, with keeping this in mind, we have to have a how problems are these interstitial lung abnormalities. So, with the recent development of endic fibrosis therapy for the patient with interstitial pulmonary fibrosis, early identification of these interstitial lung abnormalities has become important. And for their management, we have to start early treatment. So, and however, there are very few specific study on this topic, which is why I am here. So, my aim is to found the prevalence and to know the significance of interstitial lung abnormality which can progress to idiomatic pulmonary fibrosis. So, my study design is observational study at Ashwini Colaba. And my duration was two years with ethical committee approval. And my sampling was 114 and data collection was obtained after HRCT chest of the patient who give their consent to the study. And data which included name of the patient age, basic details and the chest finding and interstitial lung abnormalities finding and data was collected and compiled. The data was divided into classified into radiological abnormalities like reticulations, mosaics, ground glass, intraction, bronchitis and honey combing. So, what I have found is that there is a overall incidence is 4% which is quite high and we have to detect it early and we have to recommend according to the situation of the patient. And to the progression of the disease so that it can be, that progression of this interstitial lung abnormality to the fibrosis can be stopped at that point of time. So, what I have found is reticulation found in 66% of the patient who have found out to be a interstitial lung abnormalities out of 115-4 patient. I have found out in 45 patients which is 4% and out of that total 45 patients, patient were having reticulation was 66% ground glassing in 57% traction bronchitis in 37% mosaics in 8.8% and honey combing in 8.8%. So, there is a possible correlation between idiopathic pulmonary fibrosis and interstitial lung abnormalities. So, the my emphasis is the incidence of interstitial lung abnormalities to the patient who are undergoing HRCT chest is quite high that is 4%. So, we have to have in mind where if the interstitial lung abnormalities have been found we have to accordingly guide the treating physician or the guide to the patients. So, there is a high probability of interstitial lung disease that radiologist would recommend a general approach to the interstitial lung disease if there is a high probability of interstitial lung disease in interstitial lung abnormalities patients. So, basically the radiologist should be aware of the relevant and irrelevant radiological finding of interstitial lung abnormality and review them with consideration of clinical symptoms and pulmonary function. So, my emphasis is on to do not ignore the interstitial lung abnormalities and these are quite high with respect to the patient who are undergoing HRCT chest. So, there is a high probability to get converted into idiopathic pulmonary fibrosis and so that radiology should be aware and recommend a general approach to interstitial lung disease. These are my references. Thank you very much.