nanoHUB-U Fundamentals of Nanotransistors/Lecture 3.4: The Ballistic MOSFET
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[Slide 1] Okay, we are ready now to apply our understanding of the Landauer approach to the ballistic MOSFET. 

[Slide 2] So let's dive in. Remember we've been talking about this nanodevice, this device that has these very special contacts, we call Landauer contacts that maintain thermal dynamic equilibrium and it consists of electrons moving across the device in independent energy channels. And we've developed an expression for the current through a device like this and the number of electrons or electron density in a device like this. And now we'd like to apply this to a MOSFET 

[Slide 3] and it looks a little different from a MOSFET, so let me talk about how we think about the MOSFET as a nanodevice. Here's an energy band diagram and see there's a drain bias, there's a gate bias that has pushed the barrier way down. And as we've seen, you know, most of the action of this device takes place at the top of the barrier. So the top of the barrier we can move up and down as we change the potential in the semiconductor with the gate voltage and we've seen how if we properly design the MOSFET MOS electrostatics will obey classic 1D electrostatics at this top of the barrier or virtual source. So we're going to think about a very small part of this device near the top of the barrier as being our active nanodevice. That's where we're going to apply the Landauer approach and we will think about the region to the left as just a contact that lets electrons in an out and the region to the right as a contact that lets electrons in and out and we'll focus on our little nanodevice here. 

[Slide 4] Okay, so we'd like to compute the IV characteristics of the ballistic MOSFET. Let's begin by doing that with low drain to source voltage in the linear regime. There the Fermi levels 

[Slide 5] of the two contacts are approximately the same. We've considered that case when discussing the Landauer approach and we developed the case and expression for the conductance in this linear regime. I'll call it the channel conductance now because we're talking about a MOSFET. Okay, I'm going to go through the derivation here fairly quickly. If you want to see all of the steps or read about them I'll refer you to the lecture notes, but we'll discuss how the derivation goes here. So we've seen that we can derive an expression for the number of channels at any energy, so this assumes a 2D channel electrons move in the xy plane. We're assuming the ballistic case or the transmission is one, we know what the Fermi function of the contacts are, but we're going to assume that we can use Maxwell  Boltzmann statistics just to keep things simple. And if we do that, then we can simplify the Fermi function into an exponential. We have also developed an expression for the electron density and again, we'll simplify that so instead of having a Fermi direct integral, we'll have an exponential. The constants out front are the 2D effective density of states and involve fundamental constants and parameters 

[Slide 6] like effective mass and valley degeneracy. Okay, so let's evaluate this integral. We know everything in the terms, we can take the derivative, evaluate the Fermi window function. Since it's an exponential it easy to take the derivative. We can then plug everything in the expression for the integral assuming the transmission is one, we can work the integral out. If you do that this is the expression that you will get, this is the final result. Now I've lumped together a set of constants and parameters like temperature and I've called it v sub t because they have the units of velocity. So v sub T is defined as the square root of 2 kT over pi m and this is a velocity that will appear over and over again in our discussions, so we'll talk about it just a little bit. Okay, so our final result then for the channel conductance is it's the negative sign because the electron charge is negative, width thermal velocity divided by 2 kT over q. So that's what we set out to do, that's our channel conductance in the linear regime. 

[Slide 7] Let's talk a little bit now about this velocity that appeared in the discussion. So in thermal equilibrium if the electrons are non-degenerate, then they're bouncing around in random thermal motion and their distribution of velocities obeys a Maxwellian distribution and if I plot it, it looks something like this. Now we can see where that comes from. If we begin with the Fermi distribution which tells me the probability that a state is occupied. If we simplify it for Maxwell Boltzmann or non-degenerate carrier statistics, then we're just dealing with an exponential. If we assume a parabolic energy band, then we can write the energy of electrons as the energy at the bottom of the conduction band plus 1/2 m*v squared where m is the effective mass. And then if I pull out the constant terms, the Fermi level in the bottom of the conduction band I have some constants that multiply e to the minus 1/2 m, magnitude of velocity squared. That's my Maxwellian velocity distribution, it's plotted here versus velocity in the x direction at some specific y velocity and you can see from the symmetry of this distribution that in equilibrium the average velocity in any direction is zero. Because everywhere I have a plus x velocity I have an equal probability of having a minus x velocity and it all nets to zero. 

[Slide 8] Okay, now what do we mean by v sub t. So the average velocity of this distribution is zero, but if I just look at the positive half the average velocity of the positive half has some finite value. It's equal and opposite to the average velocity of the negative half, but the average value of that positive half is what we call v sub t, the unidirectional thermal velocity. Now we've seen if we want to compute this, this is how we would compute it. At any particular energy we've seen that when we average over angles that the average velocity in the x direction at some particular energy is 2 over pi times the magnitude of the velocity at that energy. Now if I want to average over all of the energy states in the conduction band that are occupied, I should do it this way. I should take that angle average velocity, I should weight by the number of states that are present at that energy, I should weight by the probability that the states are occupied. And then do that integral from the bottom of the conduction band and account for all of the states. And then I divide by this integral to normalize the probability. Okay, if we do that and if we take the non-degenerate case we'll find that that average velocity is one that appeared in our derivation, square root of 2 kT over pi m, that's the average velocity of the positive half in the non-degenerate limit. And that is an important velocity that we'll see over 

[Slide 9] and over and over again. Okay, so we've done the linear regime, here is our result. 

[Slide 10] Let's now look at the saturation regime and find out what the on current of this ballistic MOSFET is. Again, we'll begin with our Landauer expression, we will assume that we have a large drain bias, so that the Fermi function and the source is much larger than the Fermi function of the drain. So we can ignore the f sub d 

[Slide 11] and we have another integral to evaluate. All right, so this is the integral we need to work out. Again, we know everything inside the integral, we know the number of channels as a function of energy, it's a ballistic device, the transmission is one, the Fermi function is being approximated with an exponential because we're assuming non-degenerate statistics. The thermal velocity will pop out again and the electron density is simply related exponentially to the location of the Fermi level with respect to the bottom of the conduction band. Now there's a factor of 2 here and let's look a little more carefully. Why do we have a factor of 2 here? Where in the linear case we didn't have that factor of 2? 

[Slide 12] So we can see if we can look at -- remember n sub s refers to the electron density per square centimeter at the top of the barrier, at the virtual source. Now if I look at say an Ek band diagram there in the small device and assume that it's the same Ek that I would get in a bulk piece of semiconductor and that's not an unreasonable assumption. Then if I draw my Fermi level of the source in, the states at the top of the barrier will be occupied by electrons entering from the source. Only the positive velocity states because those are the states that can be filled when electrons enter from the source. And the probability that they will be filled will be determined by the Fermi function of the source. Okay, so we can write an expression for the electron density at the top of the barrier due to those electrons that came in from the source and half of the states have a positive velocity, so we have a factor of two there. Same thing happens for the drain, electrons with negative velocities come in from the drain, these negative velocity states can be occupied by the drain. We will get an expression just like this for the drain except we would have the drain Fermi level. Okay, now if I've applied a large bias, which is what I'm doing under on current conditions, the drain Fermi level has been lowered a lot. There is an insignificant probability that these negative velocity states will be occupied, I only have positive velocity states occupied. Therefore, the total electron density at the virtual source is equal to the electron density with positive velocities that entered from the source and that total electron density is determined by MOS electrostatics in a well-designed transistor, meaning that that's fixed by the gate voltage and by the threshold voltage and by the oxide capacitance, it never changes. So even though only half of the states are being occupied in this case, in the linear regime both of the states, both negative and positive velocity states are being occupied which is why I didn't have the factor of 2. Now only half of the states are being occupied, but I don't have half the number of electrons, MOS electrostatics demands that I have the same number. So what happens is that the Poisson equation self-consistently readjusts itself in order to be sure that all of the charge and the gate balances at an equal amount of charge in the semiconductor. That happens by pushing this barrier down a little bit under high drain bias, so that more positive velocity electrons come in and we're left with the same number that we had under small drain bias. 

[Slide 13] So that's what the factor of two in this expression is all about. Now we can just go ahead and do the math, work out the integral. Transmission is 1, this is the number of channels, Fermi function is going to be an exponential, put everything together, work it all out this is what we get. The current is width times charge at the top of the barrier, virtual source times this thermal velocity again. So we have a very simple expression all right q times n sub s, that's our mobile charge, we understand how to compute that as a function of gate to source and drain to source voltage. That's what we discussed in unit 2, that's an MOS electrostatics problem. The velocity is the thermal injection velocity. So we've derived the on current 

[Slide 14] and we've derived the linear regime current, 

[Slide 15] what about in between. In the virtual source model we connected those two empirically, but in the ballistic case we can calculate the entire IV relation. We simply have to evaluate this complete integral. This first integral is what we evaluated when we did the on current. There's one just like it for the drain current and we would evaluate it the same way and we would just subtract the second term from it. So our net current we can write as two integrals, each of those integrals works out just like the on current integral that we just went through. So the result will be for ID1 when we work out the first integral, we will just have charge due to positive velocity electrons. For ID2, we will just have charge due to negative velocity electrons. If we want the net current we have to subtract those two components. And I'll simply factor out a charge due to positive velocity electrons and write the result that way. 

[Slide 16] Okay, so here's where we stand. We have the net current expressed in terms of the charges of these two halves of the density estates, the half with positive velocities and the half with negative velocities. Now this is some physics that's going on inside the device, we want to relate everything to the terminal voltages. So in order to do that we have to work on these charges a little bit. So remember the total charge, now this we know. The gate to source voltage determines it, the drain to source voltage through DIBL affects it a little bit, but we know how to calculate the total charge. It consists of these two components with positive and negative velocities, so I'll just write it this way, factor out the Qn plus. Okay, now I can solve for Qn plus in terms of the charge that I know and this ratio and then I can take that expression and I can put it in my expression for drain current and I'll have this complicated looking expression, but we're were almost there. Because the only reason that the negative velocity charge and the positive velocity charge is different is because the Fermi levels in the source and the drain are in different places. And under Boltzmann statistics, the probability of these states being occupied so the amount of charge that's there goes exponentially with the location of the Fermi energy from the bottom of the conduction band. The Fermi energy in the drain is lowered by the drain voltage, so this ratio is very simple, it's simply e to the minus q V drain the source over kT. Now we're done. Here is our full IV characteristic of the ballistic MOSFET and this should look familiar to you because this is the same expression that we developed in lecture 1 when we went through the argument in terms of thermionic emission arguments. The difference is we've done it in the context of this Landauer approach, so it will be easy to generalize this later in unit 4 

[Slide 17] to include effects such as scattering. So here is our complete model. Here is our model that's valid for any drain to source or gate to source voltage. Here is how it simplifies for small drain to source voltage. Here is how it simplifies for large drain to source voltage. So we have our full range IV characteristic of the ballistic MOSFET. 

[Slide 18] Now in practice when we're quantitatively comparing to experimental data, we will often times or sometimes need to consider Fermi Dirac statistics, exponentials become Fermi Dirac integrals, things get a little more complicated looking, but the basic ideas are the same. This is what the final answer looks like and for those of you that would like to see this derivation again, I'll refer you to the lecture notes where it's all worked out. These equations we could use to calculate an IV characteristic for a ballistic MOSFET and we would do it this way. We would begin with a charge expression because we know how to calculate the charge due to MOS electrostatics, so the left-hand side is known for a specific gate and drain voltage. And then we would solve the right-hand side to find the location of the Fermi level in the source. And then in the drain it's just lower by an amount equal to the drain voltage. Once we know the location of the Fermi level we can calculate this injection velocity. Under non-degenerate conditions all Fermi direct integrals reduced to exponentials in this term is one and this ballistic injection velocity is just the unidirectional thermal velocity that we've seen before, but in general it depends on the location of the Fermi levels. Well now that we know the ballistic injection velocity and the Fermi levels, we can compute the current at this specific gait and drain voltage and that way we could map out the entire IV characteristic of a ballistic MOSFET with Fermi Dirac statistics. 

[Slide 19] And if we do that for a typical device, this particular device is an extremely thin SOI MOSFET model, you know, after that using typical series resistances and gate oxide capacitances and things like that. I refer you to the lecture notes for details of the device structure. But the point I just want to make here is that if you compare these two, you will see a significantly higher on current when Fermi direct statistics are considered than when Maxwell Boltzmann statistics are considered. And that occurs because of the injection velocity which changes in the presence of Fermi direct statistics. 

[Slide 20] So this ballistic injection velocity is really a critical parameter to the control of the on current of a MOSFET and it's something that I would like to talk about a little more and will make that the subject of the next lecture. Thank you and I'll see you at the next lecture.