nanoHUB U Fundamentals of Nanoelectronics: Basic Concepts/Lecture 2.8: Quantum Capacitance
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[Slide 1]  Welcome back to Unit 2 of our course. This is the eighth lecture. 

[Slide 2] As you know, so far we have been talking about this two terminal device. There's two contacts and then there's a channel and this channel we have this density of states and there's electrochemical potential which tells you the level up to which the states are filled. And for this discussion, we are really concerned only with this band at the top like a conduction band. We won't worry about any other bands down here. Now, what we want to talk about in this lecture is how you can control the conduction through a channel, and that's of course the essence of a transistor. But in a transistor there's a third terminal, with a gate voltage V sub G. And this terminal is actually separated from the channel through an insulator, so that ideally no current should be flowing at this terminal. Now, in practice, insulators are often quite thin and there's some leakage current. But that's an undesirable type. Ideally there shouldn't be any current here. So then what does this gate voltage do? Well, what it does is it changes the potential inside the channel. So supposing we'll put a negative voltage here, let's say. So a negative gate voltage means a positive electron energy, because electrons are negative particles. That causes an increase in the energy and so all the energy levels float up. And the electrochemical potential stays where it is, because those are controlled by the contacts. So what that means is here, let's say the channel was conducting well because there's a lot of density of states there. On the other hand now, it's down here and the density of states is small, and so doesn't conduct as well anymore. Now, if you raised it a lot, then you might get to a point where you'd start conducting through this band, but that's usually outside the voltage range of interest and so we won't worry about that. We're just concerned with the top band. Now, how would we model this mathematically? Well, you could say that if you bring in this Fermi function which tells you how states are occupied. At zero temperature everything below mu is filled, everything above mu is empty. But at non-zero temperatures we have this Fermi function which it describes here occupation of the levels, and that's what we discussed back in Unit 1. Now, so this is this Fermi function and that's the mathematical form of the Fermi function. So what's' the number of electrons? Well, it's equal to integral over energy, density of states times the Fermi function. Density of states tells you how many states there are; Fermi function tells you what fraction of them are filled. So D times f is like electron density. Now, what happens if you change the gate potential? Well, it changes the potential energy inside the channel, and the result is that this density of states if U is positive, then it floats up. Now, when something floats up, mathematically you write it as D of E minus U, that is, compared to D of E that function has moved up by U. So this is then the mu electron density. It is given by similar looking integral but with the density of states moved up. Now, what you could do is a little bit of algebra to what we can transformation of variable to turn that integral into this one. You see, what we have done is the E here actually stands for E minus U. So whatever was E minus U here we're replaced with an E and correspondingly, what was E here has become E plus U. So you can transform variables to get from this one to this one. And this form is a little more I guess convenient, which is why we'll be using that for the rest of the discussion. So we have this N given by the expression we just talked about, and for the moment let's think of a situation where the electrochemical potential is kind of below the band, that is not quite into the band like I have shown before but below the band. Now, when it's below the band then you can use what's called this non-degenerative approximation. That is the idea is if it's below the band, then for energy range of interest, E minus mu is a positive number. So by and large this X is a large positive number, in which case e to the part X+1 you can drop the 1. And so the Fermi function becomes exponential of minus X. And this is what's called the Boltzmann approximation. That's often used in semiconductor devices and we talked about it in Unit 1, I believe. So if you use that approximation then, then the Fermi function could be represented, could be replaced with a simple exponential function. And in this form, the algebra gets relatively simple because what you can do then is you have this e to power minus U over kT which you can pull out of this integral because it doesn't depend on energy anymore. So you can pull the e to power minus U over kT out. And whatever's left inside, that's N0. That's kind of what you had when U equals 0. So N is equal to, approximately equal because you have made the non-degenerative approximation, N0 times this exponential. Now what you could do is take the natural logarithm of both sides, so write the natural log of N over N0, you have logarithm of this exponential, and as you know logarithm exponential is just the exponent, so you get that. Now what you could do is the question we want to answer is that when I change the gate voltage, how much does the electron density change inside the channel? And let us assume that when I change the gate voltage, the potential U inside the channel changes by an amount equal to a factor beta times this qVG. That is ideally, what would happen is if I change the gate voltage by say 1 volt, then the potential in the channel would change by one electron volt. So that would correspond to having a beta of one. Now, in practice, the beta is usually less than one. Depends on how good the electrostatic design hits. So this is this non-ideality factor, which is the best it can be is 1 and usually it's something less, .8, .7, .6. So that's this U. So if you use that, we can write the change in this logarithm of N over N0, I'm taking d of this means the change in this is equal to the change in U, which is beta times the change in q times VG, and then divided by kT like before. So next what we can do is express the change in this electron density in terms of the change in the dVG-- that's unless we can turn it around and write it as dVG is equal this times this factor in front. That's a straightforward algebra. Now, you see here we have this kT over q. kT is tweny-five milli-electron volts. kT over q is twenty-five millivolts. So that's this quantity. Now, supposing we want to know how much gate voltage is needed to change the electron density by a factor of ten, let's say. So, we have N over N0 is ten. Now, natural log of ten, that's approximately 2.3. So in that case, the voltage you would need is twenty-five millivolts times 2.3, which is approximately sixty millivolts. So the number that all the experts carry in their head is that ideally what you need is sixty millivolts per decade, decade meaning a change in by a factor of ten. So the change in the gate voltage is the sixty millivolts in order to change things by a factor of ten. Now, if you were say measuring eighty millivolts per decade, people would say well, you need to work on your beta. Beta is probably not 1, probably it's .8, and so that's why you're not getting sixty, you're getting something bigger. On the other hand, if you get up and say I'm measuring forty millivolts per decade, well, you immediately get everyone's attention because either you have made a big mistake, because beta can't be bigger than one, so sixty is about the best you could expect, or you have discovered something important. So the point is that that's the criteria and everyone carries in their head, is sixty millivolts per decade. 

[Slide 4] Now, next what we want to talk about is what happens when the Fermi level is actually inside the conductor, inside the band of states, because what we just discussed actually works well as long as you don't have too many electrons in there, but when you have a lot of electrons, you have to worry about a very important factor. That's what I want to explain. Now, so if you have Fermi energy at this electrochemical potential in the band, then we should not be using the non-degenerative approximation anyway, so instead of this what we'll use is the original form so that when you take the derivative, dN/dU, you'd get something like this, that is the derivative of the Fermi function that comes in here. Now, this is a negative quantity. Why? Because derivative of the Fermi function is always negative. Fermi function goes down as energy goes up. And we have often encountered this before. This is like this thermal broadening function. So this is kind of an average of the density of states. And so we'll denote it as D0, with a minus sign because basically it's negative, because density of states is positive and that quantity is negative, so write it as negative of something that is like the average density of states in the energy range of interest, so that's this. Now, we could write the change in the charge, that is N is the number of electrons, so d of qN. qN is the charge due to those extra electrons, due to a change in the potential, in the channel. The potential energy in the channel is U, but if you divide it by q, minus q, you'll get the change in the potential. So this is like a dq/dv. And so if that quantity is D0, this quantity is like q sqaured dz. And dimensionally it is like capacitance, because it's dq/dv. So that's what is called a quantum capacitance, and I'll try to explain a little further. 00:12:54,376 --> 00:12:59,496 

[Slide 5] Now, you have this quantum capacitance, which is related to D0, the average density of states. Now, what we want to know is then, that how much does the potential energy inside the channel change due to a gate voltage, and you'd say well, we already introduced the beta factor and we know that the ratio of the 2 is beta. Well, that's true as long as you could ignore the charging energy due to the electrons, but when you have a lot of electrons you cannot do that. And so you have to add an extra term in here, and that is this one. That is, the idea is that when the number of electrons goes up, you see, as you lower it the number of electrons is going up, but those electrons will cause an increase in the overall potential energy. That this idea is you see, normally so far we have been treating electrons as if they are uncharged particles but in practice they are negatively charged, which means anytime you have extra electrons, it makes harder for the next electron to come in. So the picture that people use is that you think of electrons as uncharged particles, as they move through but they feel a certain potential due to all the other electrons. Now, as long as the solid was neutral, there was no potential to worry about, but now that you are getting in some extra electrons, this will give rise to a potential energy, which will make the band float up. So in other words, you put a gate voltage, you're hoping it will go down that much, but it won't go down as much. It'll go down less because it will float up due to all those extra electrons. Now, mathematically the way you get it then is you add this term here. So when N was equal to N0, the channel was neutral kind of; that is, all the N0 electrons had a compensating positive charge in all the protons, so it is all basically neutral. And now that you have increased it to N, all these extra electrons give rise to a negative energy, negative potential energy, and this U0 is what you could call the single electron charging energy, that is, it tells you how much the potential would change due to one electron in the channel and an actual change is given by this U0 times the change in the number of electrons. So if you use that then, you get an extra term here. And this quantity here, you could write like this; that is, you have this dN/dq VG, but you could write it as dN/dU, which is D0, as you have discussed before. That's the density of state. And then dU/dq VG. So now you see you have this quantity both on the left and on the right, but what you could do is take this term over to the left and calculate what it is. So it is equal to this beta divided by 1 plus U0 D0. So the point is that in the last slide what I'd argued is that the change in the potential energy due to the change in the gate voltage was given by beta, this non-ideality factor. What you have now shown is that it's actually less than beta. How much less? Well, it depends on how big this quantity is, and what's this quantity? Well, D0 is the density of states, and U0 is this single electron charging energy. Why didn't we worry about it in the last slide? Well, because the electrochemical potential is down here, where the density of states was rather small. Because it was small, you see, this was a small number and you didn't need to worry. But in general, this term should be included. Now, another way to write it is to introduce something we'll call the electrostatic capacitance, to distinguish it from the quantum capacitance; that is, you see D0 is density of states. It's states per unit energy, per electron volt, whereas U0 is energy dimensions, like electron volts. And just as D0 can be written as C over q squared, U0 could be written as q squared over C. But the point is this is all electrostatic in origin. It is basically just the repulsion between electrons, this electron/electron interaction. That is basically the simplest model for taking into account the interaction between electrons. And you could say, well, this in a way is just electrostatic interaction, which is just nineteenth century physics. Although, nowadays people have corrections to it, which is based on quantum theory, but basically it is this electrostatic interaction. By contrast, this one, the reason you call it quantum capacitance is that it depends on density of states, and density of states requires wave nature of electrons, and the starting point is electron in your equation, for example. So that's why the two names, this quantum capacitance and the electrostatic capacitance. And in terms of these two capacitors, you could write that quantity, 1 plus U0 D0 as a ratio of capacitors. This is just straightforward algebra. Once you use these expressions, this becomes that. 

[Slide 6] And you could visualize this in the form of a circuit; that is, you could think of it as if you got an electrostatic capacitor and a quantum capacitor in CDs like this, and what you're doing is you're applying the gate voltage here. At the other end you have a ground, like zero, and at this middle node the potential you get, that's like the potential in the channel. So when you have a very low density of states like we had when you had a nondegenerative case, when this Fermi energy was down in the gap, D0 was small, CQ was small. So this was a small capacitor. So when it's a small capacitor, then the entire voltage appears here, so what you get in the channel is essentially what you apply, because remember, small capacitor is kind of like a small conductance on a high resistance. So if you understand capacitive circuits I think you'll see that if this is small, it means the entire voltage appears there. On the other hand, when the Fermi energy or this electrochemical potential is inside the band, then the density of states is high, quantum capacitance is large. That's like having a low resistance there. And so the potential you actually get here is much less than what you apply. So this is the basic framework in terms of which you can understand how a gate voltage changes the potential energy in the channel and thereby makes this band float up and down and controls the conductance, 

[Slide 7] which of course is the essential physics underlying the operation of a field effect transistor. Now, what we'll do in the next lecture is we'll talk about this full current voltage characteristic. That is, so far everything we have talked about is low bias; that is, you have a I versus v and what you'll do is look right around the origin and look at the slope. That's what we call a low bias conductance. And what we'll now talk about is the shape of this IV curve, which is again a very important factor of course, in the design of a transistor. So that's what we'll do in the next lecture.