nanoHUB-U Fundamentals of Nanotransistors/Lecture 2.3: Gate Voltage and Surface Potential
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[Slide 1] Welcome back. We're now about 1/3 of the way through Unit 2, talking about MOS electrostatics. And we have discussed how the surface potential is what determines what happens to the semiconductor. Negative surface potential means accumulation for a p-type semiconductor, positive means depletion or inversion. But in an actual device or experiment, we're able to control or apply voltages to the gate not directly to the surface of the semiconductor. So we can only affect that surface potential indirectly by the voltage that we apply to the gate. What we need to talk about in this lecture is how that gate voltage is related to the surface potential. 

[Slide 2] Okay. So the question is, we were given a semiconductor situation, we have some surface potential. What gate voltage did we apply to produce this surface potential? 

[Slide 3] All right. So let's look at this. The gate voltage is the sum of two voltages. The voltage is 0. We've grounded the semiconductor. The volt drop across the semiconductor is the surface potential. That's how much the voltage changes across the semiconductor. But there is also a volt drop across the oxide. You can see that because the conduction band at the top surface of the oxide is lower than at the bottom surface down by the semiconductor. So there is a volt drop across the oxide as well. So, Kirchhoff's voltage law says we add those two voltage drops up and that gives us the gate voltage. Volt drop across the semiconductor is the surface potential. The volt drop across the oxide is just the electric field times the thickness of the oxide. Ok. Electric field has units of volts per centimeter, thickness times centimeters, we get volts. Now, we're going to use this relation. Remember, I told you that there is this very useful relation between the displacement field and the charge in the semiconductor. We're going to use that relation now. The displacement field in the oxide is equal to minus the charge in the semiconductor. That's going to allow us to deduce the electric field in the oxide. I'm assuming here that there's no charge at the interface of the oxide in the semiconductor. We'll come back later and fix that up. But for right now, we have a simple relation between a normal D-field and a charge in the semiconductor. We can then write-- We can solve this equation for the electric field. We can multiply by the thickness of the oxide. We now know the volt drop across the oxide. It's related to the charge in the semiconductor and the thickness and the dielectric constant of the oxide. Now, we'll define something. If I just made a parallel plate capacitor and put the oxide in between, I would get some capacitance dependent on the area of the plates. If I ask what is the capacitance per unit area of that capacitor, it's dielectric constant of the oxide divided by thickness of the oxide. So I can write this quantity as 1 over the capacitance per square centimeter of the oxide. And we get the final-- very simple, but very useful and very important relation. The gate voltage is the volt drop across the oxide which is simply minus the charge in the semiconductor divided by the oxide capacitance per square centimeter plus the band bending or surface potential. Notice that in this condition, I have a negative charge in the semiconductor, so the minus sign means, here. My surface potential is positive, that gives me a negative charge in the semiconductor, but I have a negative sign out here, so both terms in this equation are positive. They simply add up two different contributions to the gate voltage. 

[Slide 4] OK. So that allows us-- We now have a sort of a qualitative understanding of how the charge in the semiconductor varies with surface potential and we can now use that understanding to deduce, you know, how do we relate what's going on inside the semiconductor to the voltage that we apply on the terminal of the gate. The way this usually works is sort of backwards. The way you would like to do it is you'd like to apply a gate voltage and figure out what the surface potential is. It's a lot easier because this equation is a little bit complicated and nonlinear. It's a lot easier to work backwards. If I give you a surface potential, say a half a volt, you can deduce what the charge in the semiconductor is. We know how to do that if we're in depletion, if we're not depletion, we'll have to discuss that in the next couple of lectures. But we can, in principle, always deduce what the charge in the semiconductor is, and then we can use this expression to find what gate voltage produced that surface potential. 

[Slide 5] So that allows us to very easily, for example, calculate, make a plot of surface potential versus gate voltage. So if I increase the gate voltage, I expect the increase of the surface potential and the curve will do something like this if I just go through that procedure. Initially, if I apply a positive gate voltage, I get a positive surface potential, they're almost in proportion, the slope isn't quite 1. We'll talk about why that's the case later. But if I continue to apply a more and more positive gate potentials, we'll find that the surface potential increases and increases until it gets close to this critical value and then it starts to slow down. And beyond that point, it actually gets very difficult to bend the bands anymore. We say that the surface potential is almost pinned at 2 psi B. It's very difficult no matter how large I make the gate voltage to bend the bands anymore and to get a significantly higher surface potential. We'll talk about that a little more, sort of intuitively, you can see why it makes some sense because once we've inverted the surface and a lot of electrons pile up at the surface, it's almost like the surface of the semiconductor is a metal and you may remember that a metal is an equal potential. It's very hard to change the potential of a metal. Ok. So there-- A semiconductor is not an ideal metal. So it turns out there is a little bit of increase in the surface potential, but above this critical value 

[Slide 6] of 2 psi B, that's very small. Now, we can use this information to calculate the threshold voltage of a transistor. How much voltage does it take to create that inversion layer? So we'll begin with our relation between the gate voltage and the surface potential. When we have a surface potential of 2 times psi B, we're at the critical value and an inversion layer is beginning to form, so we'll simply evaluate the charge in the semiconductor at 2 psi B, the surface potential is 2 psi B. Now, the charge in the semiconductor, we have charge due to the depletion layer, we also have charge due to the mobile electrons. If we are just at the onset of inversion, we're at this critical point where the inversion layer is just beginning to form. We can say that, well, at this point, the inversion layer charge is negligible, we will just assume that the total charge is in the depletion layer. We have an expression for that. It's easy to evaluate that and we can get a simple expression for the charge in the semiconductor. And that means we have a simple expression for the threshold voltage. We can plug in numbers. We've established before what the depletion layer depth is in terms of surface potential. We use that expression. We have an equation now. If we're given the doping density, oxide thickness, other quantities of interests, we can calculate how much voltage does it take on the gate to create an inversion layer. And remember, if I have a thicker oxide, I'm going to have a lower capacitance. A lower capacitance is going to get-- then give me a higher threshold voltage. If I have a higher doping density, I'm going to get a, you know, more charge in the semiconductor and I'm going to get a higher threshold voltage. So we can tune the threshold voltage to what we need it to be by playing with the doping in the semiconductor and the thickness of the oxide. 

[Slide 7] Now, I want to say just a little bit about capacitance because we're going to need this from time to time. You remember if we have two parallel plates with cross-sectional areas of A, then their capacitance is dielectric constant of the insulator between the plates divided by the separation of the plates times the area of the plates. So that's the capacitance in farads. Now, usually, we're going to ask for the capacitance per unit area because we can always change the area of the gate. So we'll be more interested just in dielectric constant divided by the thickness of the plates. 

[Slide 8] Now, what if I have two different dielectrics between these two plates? Well, you'll see why we do it later. If you do, this is an interesting simple problem in electrostatics that we often use for homework. I'll just give you the answer. This is as though you took two separate capacitors with two different insulators in different thicknesses and put them in series. And then the rule for adding two capacitors in series is given by adding the inverses. The first one is just the capacitance, assuming I have one insulator, the second one is the capacitance assuming I have the second insulator. All right. So I haven't derived that, but it's sort of-- should look intuitive to you. 

[Slide 9] Now, let's ask ourselves, what is the capacitance of this structure in depletion? That's something we could easily measure and it would tell us something useful about this structure. Well, if I look at this, I've got a metal plate here for the gate. The undepleted semiconductor down here, where I haven't pushed things away from the surface, I have lots of holes, that's like another metal. So I have two metal plates and I have two different things inside between those two metal plates. I have an insulator with a certain dielectric constant and I have a depleted semiconductor, which is like an insulator because there are no mobile carriers, and it has a different dielectric constant. This is exactly the situation that we were considering before. So if I look at this situation, my two capacitors are the oxide capacitance and the capacitance of this semiconductor layer which the oxide capacitance per square centimeter is dielectric constant of the oxide divided by thickness of the oxide. The semiconductor capacitance or depletion capacitance is dielectric constant of the semiconductor divided by thickness of the depletion layer. 

[Slide 10] Okay. So that's the capacitance that we would measure if we did that measurement in this structure. Okay. Now, if I go to inversion, you can see what would happen. I would pull the bands down, I would create a inversion layer with lots of electrons if I bend the bands beyond 2 psi B. Now it's almost like I have a metallic layer there, right at the bottom of the semiconductor, right at the bottom of the oxide. So now, I have a metal plate on both sides of the insulator. You would guess that the total capacitance that I would measure would just be related to the thickness of the oxide, just the oxide capacitance. And that's about the right answer, you know. But it turns out that, you know, the semiconductor is not a perfect metal. You know, it's not an infinite supply of carriers, so this isn't quite right. We'll talk later about why the capacitance is actually a little bit less than the oxide capacitance, 

[Slide 11] but that tells us what to expect. Now, another topic that I want to mention here is this relation between gate voltage and surface potential. This is the correct relation. If we can compute this charge correctly, then we can solve it and we can relate the gate voltage to the surface potential. There is another solution, an approximate solution that is frequently used that I want to mention here now. And the way we think of this is we have a gate electrode and we have two capacitors in series just as we were discussing. We have an oxide capacitance and we have a semiconductor capacitance. The only difficulty with the semiconductor capacitance is it's a nonlinear capacitance because the depth of the depletion layer, which is the separation between the two plates of that capacitor, goes as the square root of the surface potential. So it's a voltage-dependent nonlinear capacitor. It makes the problem a little more complicated. But if I were to ignore that and just say I'll just use some average depletion layer capacitance, then the problem would actually be relatively simple. Because I could say that, well, if I'm in depletion, my semiconductor capacitance is dielectric constant of the semiconductor divided by depth of the depletion layer. I could then do a simple circuit theory. Voltage division on this simple circuit would say if I'm interested in a surface potential, it's the second capacitor over the sum of the two capacitors. So that's this ratio here. And the first capacitor is the oxide capacitor, the semiconductor capacitance is the average depletion layer capacitance. So I could write my surface potential in terms of my semiconductor and oxide capacitances in that way. And I could then define a parameter, we'll call m, which will occur over and over and over again in our subsequent discussions, and I can simply say that the surface potential is the gate voltage divided by this numerical parameter, m. So m is always greater than 1. It's just 1 plus the ratio of these two capacitors. So the bottom line is that if I apply a gate voltage, some fraction of that gate voltage gets into the semiconductor and appears as the surface potential. This is approximate because I'm assuming that this complicated nonlinear capacitor is actually a constant average capacitor to do this, but it is a useful, widely-used approximation. 

[Slide 12] All right. So let me do an example here of how we would do this. Here's the example that we've been talking about and we want to see what the parameter, m, is. The parameter, m, is just a ratio of these two capacitors. Well, we know what the depletion capacitance is, we know what the oxide capacitance is, so we get an expression for this numerical factor, m. We have the numbers. We simply plug them in and we'll find that m is about 1.2, a little bit bigger than 1. That's a typical number, a little bit bigger than 1. You can see that it can never be less than 1, not unless you're really, really clever. Okay. So what that means is that 83% of the gate voltage gets into the semiconductor and appears as the surface potential of the semiconductor. 

[Slide 13] So if we go back to this plot, in this initial part of the plot, the slope of this is 83, 0.83. So we apply a gate voltage, 83% of it gets into the surface potential. So we've been able to discuss this part of the curve where we knew that the semiconductor capacitance was the oxide capacitance. 

[Slide 14] Okay. Now, you know, we also did an example earlier where we computed some numbers in depletion for an MOS capacitor. Assuming that we had a half a volt surface potential, and I think we computed the depth of the depletion layer and the surface electric fields and things like that. These are the numbers that we got. We also could have gotten the depletion layer charge and we have these numbers here. So let's continue with this example and let's ask ourselves what gate voltage produced this half-volt surface potential. 

[Slide 15] Well, we simply go to our relation between gate voltage and surface potential, we assume that the charge in the semiconductor is a depletion charge. We were given that psi sub S was a half a volt, we were given the oxide thickness. We compute all of the numbers, plugged them into this expression. And what we find is that it took three quarters of a volt in this example to establish a half a volt band bending in the semiconductor. All right. So that gives you a feel for how these numbers play out in a real example. 

[Slide 16] Okay. Now, let's go to inversion. Let's bend the bands to inversion and let's ask ourselves what is the threshold voltage of this MOS capacitor we've been discussing. Well, we have to compute this critical value, psi B, but we know the doping density in ni. psi B , we already evaluated, it's 0.48 volts, 2 psi B is 0.96 volts. So we're asking ourselves, what gate voltage does it take to produce 0.96 volts surface potential? That's the threshold voltage of this MOS capacitor. 

[Slide 17] Okay. So if we go ahead and, you know, compute some numbers, we can first of all assume that we put 0.96 volts in and then bend the bands, we can then reevaluate our depletion layer thickness. It's now a little deeper, it's 35 nanometers now. We can reevaluate the charge in the depletion layer. We can reevaluate the electric field. It's now higher because the surface potential is higher. So we know those numbers, and then we can find 

[Slide 18] out what the threshold voltage is, what gate voltage produced those numbers. Well, we know 2 psi B, we know W, we know depletion layer charge, we know oxide capacitance. Basically, we know everything to insert in this expression, compute the threshold voltage, plug in numbers, do the calculation. It takes 1.28 volts to create an inversion layer in this particular semiconductor. Now, that's a little too high. That's not a number that you would expect for these values of parameters. The reason is remember this prime-- this is the threshold voltage prime. What does the prime mean? The prime means I'm dealing with a hypothetical metal that just happens to have the same work function as the semiconductor, so when I apply 0 gate volts, everything is at flat band. We're going to need to look at a real metal in the next lecture and when we do, we'll come back and find out that this threshold voltage is actually significantly lower. But that's how we do the calculation and we'll do the correction measure. 

[Slide 19] Okay. So just to wrap up this lecture, we've covered a bit of ground. These are the main takeaway messages. The gate voltage induces the surface potential that controls the band bending in the semiconductor. There is a simple and exact relation between the gate voltage and the surface potential. Here it is. Now, there is an approximate relation between the gate voltage and the surface potential that works well in depletion. One way we could do that is just use a depletion charge here. We know how to compute that, it goes as the square root of the surface potential. That would give us a quadratic equation that we could solve and find the surface potential for a given gate voltage. Even that quadratic equation is a little too complicated sometimes. So there's an even more approximate solution that is even more useful which says that the surface potential is just some fraction of the gate voltage that we apply given by this parameter, m, which will appear over and over again in our discussions. All right. And we've been doing these discussions for a hypothetical metal that had a work function that was equal to the semiconductor. In the next lecture, we're going to need to make this more realistic and you'll see it's very easy to make a correction and just add a correction to everything that we've done and then we've got realistic structures. So we'll continue with that discussion in the next lecture.