nanoHUB U Fundamentals of Nanoelectronics: Basic Concepts/Lecture 2.9: The Nanotransistor
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[Slide 1] Welcome back to Unit 2 of our course. This is Lecture 9. 

[Slide 2] Now in the last lecture, we talked about this transistor. You know for the first time I guess we introduced this third terminal into it, a gate terminal. And we talked about how the conductance is changed through the gate voltage. Now what we want to talk about in this lecture is the full current versus voltage characteristic. That is, usually we talk about the conductance right around there, close to the origin. In a small voltage, how much is the current? And in that region, it's like a straight line. So you just look at the slope. You call that the conductance. What we're now interested in is what does the entire current voltage characteristic look like? So you'd say, well, that shouldn't be much of a problem because the expression we had for conductance, we actually started from this expression if you remember back in Unit 1. This was the expression for the full current. And what we had done was we then used this Taylor series approximation and got this df/dE, and that's how we got the expression for the conductors. So you say if you want the entire current voltage characteristic, just use that. Now if you did that, then you would get a curve looking like this, and which actually isn't quite right. There's a few things here. And that's kind of what I want to tell you about. Now first thing to notice, if I just use that expression, why do I get a current that saturates so perfectly, saturates meaning that once I increase the voltage beyond a certain point, the current doesn't increase anymore. The reason is that you see when you apply a voltage, the way you are thinking is the source is our reference. That's what we call zero, and then the drain has a certain voltage on it. So as you put a bigger and bigger positive voltage on the drain, the chemical potential in the drain goes down. And current flows in the window between mu 1 and mu 2. So in the beginning when you put a little voltage, of course, more current flows because there are more channels available for conduction. But once mu 2 has dropped below the band, you see there are no new states out here. So it really doesn't matter whether the mu 2 is here, or is it here, or is it here? It doesn't make any difference. You still get the current flowing through just this part. And that is why you see you have this very nice saturation within this model. Now of course if you pulled it down further so you actually get the other band, then again current would start flowing. But as I said, usually in this discussion, the voltage ranges are such that you don't worry about that at all. So you are really talking about just this range. And in that case, the current would saturate perfectly. Now one point I should point-- make right away is that actually what would happen is you'd get somewhat more than what we have calculated here if you take into account this interaction energy that we talked about, namely you see if you take into account this factor that we have talked about, that once the number of electrons changes, then if it increases, then it causes a negative potential. If it decreases, it causes a positive potential. That's something we talked about in the last lecture. And just for that reason, though, you'd expect this to go down somewhat. Why is that? Well, here we had a band that was full of electrons, and now that you have applied a voltage, you have much fewer electrons in here. Why? Because as long as it was in equilibrium, all these states were filled. Now that it's out of equilibrium, we have one contact which wants to fill it up, another contact which wants to empty it. And so by and large, these states are kind of half full. So whatever number of electrons you had at equilibrium, let's say we had one thousand of them, then out of equilibrium, it would become like five hundred. Why? Because these states used to be full all the time, now they're only filled half the time because as soon as one electron comes in, the other one pulls it out. So on the average, you have like half the electrons. And so as a result of that we have fewer electrons, so positive potential, which would make the band go down. See? And that would then give you more current. So what you'd expect is because of this effect, you should get more current than what you have calculated when you neglected it. Mathematically, how do you do this? Well, exactly what we have done before, namely define this channel potential U and write it in the form of something that depends on the gate voltage like we did before. This term doesn't matter here because we are assuming a fixed gate voltage. And then there is this term which is discharging energy. This U zero times the change in the of electrons, and this is the quantity that when you apply a bias, what happens is the number of electrons goes down because these states used to be completely full. Now they are half filled. So this goes down, and so this becomes negative and as a result pulls it down. So mathematically, that's how it would enter. And this U of course would have to be reflected in our current equation. So you'd have -- instead of G of E, you would have this G of E minus U. And what about the number of electrons? How do you calculate that? Well, number of electrons is usually calculated by integrating the density of states times the Fermi function. Now this is out of equilibrium. So we've got two Fermi functions, one in the left contact, one in the right contact. And what you assume is so the occupation of the states would be like the average of the two, f1 plus f2 divided by 2. So if you take this set of equations and solve it self-consistently, you see here we have a U that depends on N and the N that depends on U. So you'd have to solve is self-consistently, and there are numerical techniques for doing that, which I am not going into here. But if you solve it self-consistently, you would actually get a much bigger current than what you calculated. So this is the -- what you calculate when you put U zero equal to zero. That just ignores this lowering completely. And then if you put a large U zero, then you'd get a much bigger current because a large U zero means because this is large, this tends to be small. So it means as if this band floats down enough so that the number of electrons, let's say, used to be one thousand, was trying to go back to five hundred. But instead this goes down enough so that it's almost one thousand. So under those conditions again, you get more current like I've shown here. 

[Slide 3] Now there is another factor, though, that is missing in this discussion, and that is what you see any expert, if they looked at this curve, of course, what would bother them is this perfect saturation. Actually, see, that's a very good thing. If you had a transistor with such perfect saturation, that would be great because when you are trying to design circuits, it helps if your transistor current in the on state is independent of this voltage. So what you really want for a good transistor is that the current is controlled entirely by your gate and not at all affected by the actual voltage between the source and the drain. That's what you'd really like. But the point is this model seems to predict that this just comes easily. You don't even have to worry about it. But that's not true at all. And the reason it's getting such great results is because it's ignoring a very important effect, and that is that the potential in the channel that I've written doesn't just depend on the gate voltage here but is also affected by the voltage that you put on the drain. So you have to include that, you see? So just as we put in a beta factor times VG, there should be an alpha factor times this V, which is the voltage on the drain. Usually people write that with a V sub D to indicate the drain. But since most of this course we haven't been using two terminals and I just used V, I just kept U, calling it V. But once you include that, then you'll see that you find that the current doesn't saturate so nicely, but keeps increasing. And you see physically why, because as you increase V, what happens is the channel potential wants to go down more and more, which means this band wants to sink down more and more. And as it goes down, the current keeps increasing. And indeed, the whole art of designing a good nanotransistor is how to get a small value of alpha so that the current will saturate well. That is, as long as devices were long, usually people got very nice saturation. And you can see why, because you see you have a gate voltage here, which can close the potential through the beta. And you have a drain potential which changes it through the alpha. Now over the years as the devices got smaller and smaller, the point is the effect of the V got bigger and bigger because now this is a very small device. So how effective the gate potential is depends on the fitness of the insulator. How effective the drain is depends kind of on the length. And as the devices got smaller, the drain got more and more effective in terms of controlling the channel potential. I mean sometimes people call this DIBL, that is drain-induced barrier lowering. And as I mentioned before, these days devices are small enough that this active region is like a few hundred atoms across. Now that, of course, is amazing. But the thing is in order to make sure that the gate really controls the potential better than the drain, which means you want to make sure that this beta is a whole lot bigger than the alpha. In order to ensure that, you see you have to make the insulator off of the ten. So if this is like a few hundred atoms, the insulator needs to be a few atoms. It needs to be more like ten atoms, five atoms. And that, of course, is extremely hard because once it's ten, it also means there will be leakage currents, more and more of leakage currents as electrons tunnel through that insulator. Not to mention you know it's very hard to control something through ten atoms across billions of devices. So that really is the major engineering challenge you know in terms of how to make billions of these nanotransistors work. It is how you can make the insulator thin enough that the beta-- that the gate controls the channel potential much better than the drain controls it. See that's kind of what determines a well-designed transistor. So all this you can kind of see from this simple model, and if we're trying to get qualitative understanding of how a smal -- how a transistor works, this is a very good model. On the other hand if you want to be quantitative, 

[Slide 4] then this isn't adequate. You have to go much deeper because there will be other issues that you have ignored. And the most important one is, of course, that in this discussion I put this channel potential U as a single number. In practice, the potential will vary across the channel. In another words, you will have one value at this end, something else at the other end. It will change continuously. What that means is when you look at the density of states of the bands, this conductance function, it will keep going down, because as you know, when the U changes, the bottom of the band kind of tracks that, goes down, okay? So this U is not just one number that you can calculate from this equation. What you need is a differential equation to calculate it, and that is what is usually called this Poisson equation. So if you're doing a serious device simulation, what you would use is the Poisson equation, which would tell you this potential as a function of the electron density. And what we have done is kind of a poor man's version of that. We took a -- kind of put a single number U, put an algebraic expression to it, just to get a rough idea what the factors are. And actually, just these few perimeters you can often understand the real observed characteristics fairly well. You can see the physical features quite well. But if you really want to understand transistors, that's really not the objective of this course. You should take a-- you should take a course that's focused on transistors from an expert. Now one thing that it misses here also is that remember our entire model here was based on this elastic cross transport, which means the assumption was that an electron coming in from here, as it goes through the channel, does not change energies. And if it's a ballistic device, that's usually a good approximation. That's fine. But on the other hand, if you have diffusive processes going on so that electrons lose momentum in here, then you see there would be a lot of situations where these electrons, instead of being confined to this energy channel, they would want to go down here. And once they go down, they are able to carry on to the drain. So the point I am trying to make is if you have a model which allows electrons only to travel to the channel they started in, in energy, then you tend to calculate a much lower current than what you actually get because in practice, in elastic processes that allow an electron to relax will actually increase the current because an electron from here will go down in energy. And once it's down here, it's kind of hard to get back to the source. So instead we'll proceed onto the drain. So you get a lot more current. All those effects, of course, you can kind of qualitatively describe. But if you want to do it in a quantitative way and a convincing way, then what you really need to do is what I had indicated in Unit 1, that when you have a long device, think of it as lots of little ones in series. See? And you could use then this-- what's usually used for device modeling is this drift diffusion equation, which is written as current is the-- proportional to the slope of the electrochemical potential. So usually we talk about in electrochemical potential at this end at this end. And we never talked about what it is in the in between. But if you have-- if you assume then that in between also the electrochemical potential is given by this mu and changes along the way, you could write the current in terms of that. And that's this drift diffusion equation that is widely used for device simulation. It has to be coupled with something called a continuity equation. That is, in one dimension, the current cannot change with length. So if these are the equations that are generally used, and these are, of course, not based on this new perspective. These are old equations. So what did our new perspective add to this whole thing? Well, what it added of course was a different way of understanding the sigma. As I mentioned, the old perspective of sigma is this q square n tau over m, the Drude formula, whereas we had a way of thinking about it which I believe is more direct and gives you a different way of looking at it. But more importantly, it also gives you this extra thing called a ballistic resistance. 

[Slide 5] What I mean by that is you see the expression we get-- we got in the last, you know, few lectures and I guess the last unit, is that the conductance is given by conductance times area divided by L plus lambda, whereas what you would learn in freshman physics is usually sigma A over L. That's this usually Ohm's law. What we have seen is even when the length tends to zero, the conductance doesn't go to infinity, but rather goes to a fixed number. Now you could turn that around and write the resistance. That's the inverse of this. And you could write that as the ballistic resistance times L over lambda. This is the part that follow's Ohm's law, and plus a constant. And of course, in the early days, the major controversy was about what this constant is. And what we did during this course is we gave you a way of looking at this so you can understand where this comes from, how to understand this, and so on. But one thing we didn't quite address, though, is what exactly is this resistance. What is it associated with since it's independent of length? And the point is modern understanding is that that resistance is actually an interface resistance that appears at the two ends, whereas the middle part has a resistance that's proportional to L. So the main thing that this new perspective adds to our old understanding, the most important thing is this idea of a interface resistance. Of course, everyone knows that when you go to-- when you measure the resistance of any conductor, you always have some contact resistance. That's well known. But it's just that usually you think of this contact resistance as kind of a undesired thing, which you can always get rid with enough engineering. What is not recognized is that there is actually a fundamental limit, that no matter how good your contacting technology is, finally you will always have this ballistic resistance appearing at the interfaces. Now this is something we haven't talked about so far. You see we didn't quite talk about where this RB comes from. And that's what we'll do in the next unit of this course. That's what we'll get into as to why this appears at the interfaces. And so when you are using a drift diffusion equation, what we'll show is you can use that as long as you are careful to include these interface resistances. But that's for the next unit really. 

[Slide 6] So that brings us, I guess, to the end of what we wanted to talk about in this unit. And it's time to sum things up in the next lecture.