nanoHUB-U Fundamentals of Nanotransistors/Fundamentals of Nanotransistors Unit 2: MOS Electrostatics Lecture 2.2: Depletion Approximation Mark Lundstrom
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[Slide 1] Well this is Lecture Two, and in this lecture we're going to solve the Poisson Equation in a depletion approximation. 

[Slide 2] So just to remind you we're talking about this MOS capacitor, just the heart of a MOSFET, and we've been looking at the energy band diagrams. And specifically now we're going to be calculating the electrostatic potential versus position along this line y in the semiconductor. 

[Slide 3] And just to remind you where we are, we're not in accumulation as we haven't applied a negative voltage to bend the bands up-- have majority carriers pile up at the interface. We are in depletion. We've applied a positive voltage. We're bending the bands down, but we haven't applied a large enough voltage to have a significant electron concentration near the surface, so we only have depleted charge. 

[Slide 4] So we're going to be focusing on this depletion region of operation, and there we can solve the Poisson Equation. So our energy band diagram now looks like this. Our surface potential is less than this critical value to create a strong high concentration layer of electrons at the surface. We can see that there's an electric field that's in the positive y direction. We can see that from the slope of the energy band diagram. Positive slope means positive electric field. That positive electric field is pushing the holes back from this region near the surface, and is creating this depletion region. And we'd like to solve for the electrostatic potential drop across that region, and the electric field within that region. It turns out it's really easy to do. But it also turns out that it's really useful in semiconductor work. Well, we're going to have to worry about the charge density in the Poisson Equation. In general we have mobile holes, mobile electrons, and fixed dopants. In this case I'm only considering p-type dopants so they'll be negatively charged. If I'm near this region-- in this region near the surface, y is less than the depletion layer depth W sub D. Then the hole concentration is small; it's depleted. The electron concentration is small; they haven't started to build up yet. So the only charge density is due to these negatively charged acceptors, but we just assume we have constant doping in that regime. That makes it easy to solve the Poisson Equation. Now if I go deeper into the semiconductor beyond y is greater than WD, everything is neutral, okay, so I have no electric field, and no potential drop in that region. 

[Slide 5] So to solve this problem, we're going to solve the Poisson Equation. Divergence D is equal to Rho; this is what it looks like in 1D. Displacement field is dielectric constant times electric field, so I could rewrite it as an equation for the electric field, and the charge density is simply minus q times NA, so I have a simple equation for the slope of the electric field. Now I also know that the electric field is minus the slope of the electrostatic potential, so I can get the electrostatic potential by doing an integration; we're going to get both. 

[Slide 6] So let's look at the electric field first. So if I plot electric field versus y in the semiconductor in this region right here, it's going to look like this. It's going to look like a straight line with a constant negative slope. The reason is because that's my Poisson Equation. It's minus q NA over dielectric constant, and that's a constant. So we have a constant negative electric slope within this region. It must have started at a positive value at the surface- E sub s is the surface electric field. It then decreases linearly; it goes to zero at the edge of the depletion layer, and then it's zero in the bulk. So the electric field versus position in the depletion approximation has this simple form. I can simply write down the equation of a straight line, so this is an expression that goes to zero when y is W sub D, and when I put y equals zero this gives me an expression for the electric field at the surface in terms of the depth of the depletion layer. Now look, if I look at q times NA, that is the volume charged density in coulombs per cubic centimeter. If I multiply it by the width of the region over which we have that volume charged density, then that's going to give me the sheet charged density; the charged density in coulombs per square meter, or per square centimeter, or whatever it is. I'll call that capital Q sub s. The charge in the semiconductor per square centimeter, and it's just simply minus q NA times W sub D. Okay, so for future use we're going to use this frequently. I just point out that if we look at this equation, if we multiply it through by epsilon sub S dielectric constant we can get an expression that says the normal displacement field at the surface is equal to the total charge per square centimeter in the semiconductor. Some people call that Gauss's Law. That relation between the D field and the sheet charge density is going to be very useful for us from time to time as we go throughout the remainder of this course. 

[Slide 7] Okay, so we have the electric field versus position, we have an expression for the electric field at the surface. The only difficulty is we don't have an expression for the depth for the depletion layer thickness W sub D, so we can't evaluate the electric field. However, we remember that electrostatic potential is minus the integral of electric field versus position. If we integrate this characteristic it's simply the area under this triangle. So the total potential drop across this region is the surface potential, because the potential in the bulk with a semiconductor is zero, potential at the surface is psi sub S, the total potential drop is psi sub S; it's just the area under this curve. It's just one half the peak field times the depth of the depletion layer. Well, that allows us then to use this expression for the surface potential. We use this expression for the electric field, put the two together and we get an expression that tells us what the depletion layer thickness is if we know the surface potential of the band bending. That's a really useful relation that we're going to need from time to time throughout the remainder of this course. 

[Slide 8] So we know what the depletion layer depth is as in terms of the surface potential. If we want to know the total charge in the depletion layer- that's what we call Q sub D- we simply multiply the volume charged density times the depth of the depletion layer, so now we have an expression for that total charge in the depletion layer, so we'll call that the depletion charge in a semiconductor. In general there are two types of charge, there's depletion charge and there's charge due to the mobile electrons you know, when we have a positive surface potential. If we have a negative surface potential then we have charge due to holes. But if we talk about a positive surface potential, in general we'll have these two kinds of negative charge. We're assuming that the mobile charge due to electrons in the conduction band is negligible. Okay, so under these conditions we learned that the charge in the semiconductor goes as a square root of the surface potential. 

[Slide 9] Okay, so if we go back to some of these sketches that we were making in the first lecture, we sketched qualitatively how the charge in a semiconductor should vary with surface potential, and now we understand why it varies slowly and as the square root of the surface potential when we're in this region where we have bent the bands down but not down so far that we have a high concentration of electrons. So the semiconductor charge varies as the square root of the surface potential in the depletion region. 

[Slide 10] All right, let's do a quick example just to get calibrated and make sure that we understand some of the numbers involved. So here's a typical doping density if we're a channel of a MOSFET. Let's assume room temperature, and let's assume that somehow we have bent the bands down and established a surface potential of a half a volt. We'd like to find what the electric field is at the surface, and we'd like to find out how deep this depletion layer is; what is W sub D. I'm also going to need to know a couple of other things. I'm going to need to know the intrinsic carrier concentration and kT over q you know, and we'll see why in a minute. Also, we'll need to know what the dielectric constant of silicon is and the relative dielectric constant is 

[Slide 11] almost 12. Now, the first thing that we need to do is to check and see whether we are indeed in depletion, because if we are in depletion, then we know how to solve this problem. If we aren't, we don't know how to solve the problem. So we have to check and be sure that the surface potential is positive but is less than this critical value of 2 psi B. And you'll remember, psi B is kT over q, .026 times log of a doping density 10 to the 18th over the intrinsic density 10 to the 12th-- 10 to the 10th per cubic centimeter. We put numbers in and psi B is .48. Two psi B is .96. We only have a surface potential of a half a volt so we are comfortably in the depletion region. We don't have to worry about inversion layer charge. We can use our depletion layer expressions and compute the quantities that we're after. 

[Slide 12] Okay, so let's do that. So we can go ahead and do the numbers. We have developed an expression for the depletion layer thickness. Now we know we can use that expression. We were given that the surface potential was a half a volt. We plugged numbers in, and we end up with 26-- 25.6 times 10 to the minus 9th meters. Or I could express that differently as 25.6 nanometers. So that gives you a feel for the typical depth of a depletion layer under you know normal conditions- 20-- 25-- 26 nanometers. 

[Slide 13] We can also get the electric field. Remember the electric field and the potential were related by the area under that triangle, so the potential was one half peak field times depth of depletion layer. We were given this surface potential. We've just computed the depletion layer depth. We can solve for the electric field. Plug in numbers and once we do that, we will find that the electric field is 3.9 times 10 to the 7th volts per meter. Semiconductor people like to quote things in volts per centimeter, so 3.9 times 10 to the 5th volts per centimeter, or 390 kilovolts per centimeter. Now that sounds like a big number and you can kind a get a feel that it is indeed a big number, because you know. Think of a centimeter. We know what a centimeter is. Think of two metal plates separated by a centimeter and apply 390 kilovolts across those two plates; that's a pretty large voltage across a pretty small gap. So that's a large electric field, though the typical electric fields that we're dealing with in these semiconductor structures are quite large indeed. 

[Slide 14] All right, so just to wrap up. You know what we've shown is that if we're given a surface potential, and if we are in the depletion region and we always should check to be sure that we are, you know then we can compute depletion layer depths and electric fields at the surface, and quantities that we might be interested in. But the question is where did this half a volt surface potential come from? It came from applying some gate voltage. How much gate voltage? What we need to do next is to elate the gate voltage to the surface potential, because the surface potential is what determines what happens in the semiconductor. That's what we'll talk about in the next lecture,