nanoHUB-U Fundamentals of Nanotransistors/Fundamentals of Nanotransistors Unit 2: MOS Electrostatics Lecture 2.1: Introduction Mark Lundstrom
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[Slide 1] Okay. Well, we're ready to begin unit two of this course. So unit one was a review of some MOS fundamentals for some of you. For some of you it was an introduction. But we just-- we have the basic concepts that we need now to dive into things, and that's what we'll do in unit two. So unit two is material that is traditionally taught in MOS transistor courses. For those of you that have had advanced courses in this area will find most of it familiar. Those of you that have had only elementary courses will get the basic concepts that are needed for the remainder of the course, and if you only know basic semiconductor physics you'll still be fine. 

[Slide 2] So there are a number of topics that we're going to be discussing in lecture two. They all have to do with this general topic of MOS electrostatics. So let me say a little bit about what I mean by MOS electrostatics. 

[Slide 3] We talked in lecture five of unit one about this energy band view, about how the essence of a MOSFET is we control the flow of charge out of the source and across the channel and out to the drain by manipulating energy barriers for the voltages that we apply to the terminals. And we drew energy band diagrams like this to explain the operation of the transistor. All those energy band diagrams are determined by electrostatics. Now, another way we can look at this is to write the current equation for a MOSFET. We're hoping to understand the current of a MOSFET and current is a product of charge times velocity. So two important factors. In unit two we're going to be focusing on charge and that's all about electrostatics. The charge comes from the source when we push the barrier down and it flows into the channel. We're not going to say much about that detailed process here. We're just going to focus on can we establish the charge in the channel as a function of the voltage on the gate and the voltage on the drain. And that's our goal in unit two. Then we will be ready for the really novel part of this course, which is how the electrons flow across the channel. And that's the subject of units three and four. 

[Slide 4] Okay. So a word about energy band diagrams. We talked a little bit about this in unit one in lecture five, and in that case we drew an energy band diagram along the top surface in the X direction here. But the device is really a two-dimensional device or even a three-dimensional device. There's another dimension Z that comes out of the page. So we really should be thinking about two or three-dimensional energy band diagrams. Well, we won't need three-dimensional energy band diagrams, but we will need to talk about two-dimensional energy band diagrams. So in principle you remember that an electrostatic potential, a positive potential pulls the electron energy down. So in order to understand these two-dimensional energy band diagrams we need to understand the two-dimensional electrostatic potential inside the MOSFET. That's this quantity psi of (x,y). That's our goal. 

[SLide 5] So to find that it's very clear how to do that. We go to the Poisson equation. So the Poisson equation you'll recall can be written as divergence of the displacement field is equal to the charge density. You'll remember that the displacement field is dielectric constant times the electric field and the dielectric constant of a semiconductor is its relative dielectric constant times the permittivity of free space epsilon knot. You'll remember that the electric field is minus the gradient of the electrostatic potential and we need the charge density, which can be quite complex. It's the positive charge density due to the mobile holes minus the negative charge density due to the mobile electrons plus the positive charge density due to the donor dopants that are fixed in the lattice minus the negatively charged density due to the acceptor atoms that are fixed in the lattice. So if we put that all together we could rewrite the Poisson equation as del squared psi is equal to minus Rho over epsilon. Either one of those two equations would be a solution to the Poisson equation and would give us what we need to know. Now, in general it requires numerical techniques to solve this equation. It turns out that in a couple of cases we can derive simple analytical solutions that are very useful. And in the next lecture we'll see that if we make something called the depletion approximation we can easily solve this equation and get some very useful insight. 

[Slide 6] So we need to understand the electrostatic potential in order to draw the energy band diagram, but, you know you can always draw an energy band diagram qualitatively relatively simply, even if you don't have precise numerical values. And when you're drawing an energy band diagram like that, what you're really doing is a qualitative solution to the Poisson equation. That's why it's important to have an ability to draw these energy band diagrams. 

[Slide 7] Now, here is our cross-section of the MOSFET flipped on its side that we're going to be interested in most of the lectures in this unit in drawing energy band diagrams normal to the surface. Normal to the channel, into the depths of the semiconductor we're to call that the Y direction. We're going to ignore the presence of the source and the drain and just assume they're so far away that they don't really affect things. They do in practical devices because these channel lengths are very short. And that brings up some very important factors called two-dimensional electrostatics that we'll talk about towards the end of this unit. 

[Slide 8] Now, let's review just briefly energy band diagrams. We've talked about this before. If we have a semiconductor with the bottom of the conduction band and a bottom-- and a top of the valence band. I'm writing Ec0 here to indicate that this is where the band is when the electrostatic potential is zero. So we have some electrons in the conduction band. The number, the density of electrons is determined by the location of the Fermi energy with respect to the bottom of the conduction band. The higher the Fermi energy, the more electrons. We have holes in the valence band and the lower the Fermi energy is the more holes we have in the valence band. And remember that if we are in equilibrium and we're going to stay in equilibrium in this unit then np is equal to ni squared. And ni squared is 10 to the  10th squared for silicon and it has different quantities for different semiconductors. Okay. Now, some of the key assumptions that we're making here. We're in equilibrium. It turns out that's a pretty good assumption. Even when we apply a voltage to the gate, there is an insulator below it that blocks current flow so the semiconductor stays in equilibrium. Assuming Boltzmann carrier statistics you can see that, because there's exponentials here in these relations for the carrier densities, not Fermi direct integrals. And I'm assuming in this particular picture a uniform electrostatic potential, so the bands are flat, they don't bend. 

[Slide 9] So we're interested in asking ourselves what happens when we establish a non-uniform electrostatic potential, how do we draw the energy band diagrams? So let's think about this situation. We have a P type semiconductor with the bottom grounded. We have a metal plate, that's going to be our gate. We apply a positive voltage greater than zero that effectively puts some positive charge on the gate. That positive charge repels the mobile holes in the semiconductor and pushes them back. It pushes them back to some depth W sub D. We call that the depletion layer thickness, because that's the layer over which the mobile holes have been pushed out. It's depleted of mobile holes. But what's left behind are the P type dopants that were there and gave us the, the mobile holes to begin with. Each one of those P type dopants from column three, which had three valence electrons acquired an additional fourth electron when it built the semiconductor and created a hole in the valence band. So each one of those acceptor atoms is negatively charged, so when we pushed the mobile holes back we expose these negative charges. If I plot the electrostatic potential as a function of depth into the semiconductor, well it's positive at the gate electrode, it's going to be positive at the surface, and it's going to be zero at the bottom of the silicon, because we grounded the bottom. So my electrostatic potential is going to increase from zero at the right side of the semiconductor, and then increase to some value that will play an important role in our discussions in this unit. Psi sub S is known as the surface potential. It is the electrostatic potential at the surface of the semiconductor, assuming that the potential is grounded and is zero in the bulk. 

[Slide 10] So the energy bands are going to bend now because we have a non-uniform electrostatic potential in the semiconductor. Wherever this potential is positive, the bands will bend down. 

[Slide 11] So then we'll sketch our energy band diagram easily, as electrostatic potential increases as we go towards the surface, both the conduction band and the valence band bend down. So we have band bending like this. The slope of the energy band is q times the electric field, so we can read the slope off easily and see whether it's positive or negative, large or small. Okay. Now, what we would like to do is to talk about looking at different varieties of band bending. And under different conditions the bands can bend up, the bands can bend down, the bands could be flat. 

[Slide 12] So this is a nice reference place to start. So let's assume that we have a P type semiconductor uniformly doped. Everything is uniform, the potential is flat and constant and zero. Let's assume that we put an insulating layer, silicon dioxide on the top of this silicon, between the silicon and a metal plate. And that insulating layer is just going to tie up dangling bonds and make sure that there are no charges there and make sure that the surface is nice and passivated. And then let's put a metal gate on top of the-- of this structure. And this is a very special metal gate. This is a metal gate with a work function, such that the Fermi level in the metal just happens to line up perfectly with the Fermi level in the semiconductor. When that happens there's no charge transfer. There's no band bending, everybody is happy. You put the materials together and nothing changes. We call that the ideal flat band condition and it's a nice reference point for us. Okay. Now, let's apply a voltage. You know, when we apply a voltage the voltage is going to lower the Fermi energy in the metal and pull it down. So we'll pull it down by an amount qVG prime. So we will then pull that down and we'll see next what happens when we do that. Now in this particular case there's zero votes on this gate electrode and we say that this is the flat band voltage. The flat band voltage is zero. I apply no voltage, everything is flat. This only occurs under a very special condition, which will never happen in practice, but it's useful for us pedagogically. That is it only occurs when we have this magic metal whose Fermi level just happens to line up with the Fermi level in the semiconductor. That's what the prime is. When I write a VG prime, it means I'm assuming I have this hypothetical magic metal where the Fermi level just lines up with the semiconductor, when I apply no voltage to the metal and everything is flat at VG equals zero. We'll see in a couple of lectures how we fix that up for more realistic conditions, but for now it's an easy place to start. Okay. Now let's apply a voltage. 

[Slide 13] We apply a voltage on the gate, this hypothetical gate. A positive voltage pulls the Fermi level down. So if we lower the Fermi level, now we have a Fermi level in the metal that's below the Fermi level in the semiconductor. You'll remember that in equilibrium the Fermi level has to be constant. It's not constant anymore. But it's constant in the semiconductor, it's constant in the metal, and the insulator separates the semiconductor in the metal so we have two separate systems, each of which is in equilibrium. That simplifies the analysis quite a lot. Now we've grounded the back of the wafers, so the potential is zero there. That's just an arbitrary reference. The bands are going to bend, because as we-- the electrostatic potential increases as we move towards the positive voltage on the gate, the bands are going to bend down. We can measure the electrostatic potential at any location. It's just two times the electrostatic potential is how much the bands have lowered due to that positive electrostatic potential. And at the surface we have a potential that we call the surface potential. And you can see that the bands are bending in the insulator too. They're being pulled down as well. And so you can also see from the slope I have an electric field in the insulator. I have an electric field in the semiconductor. Their slopes are different, means those two electric fields are in the same direction, same sign, but they're different magnitudes. Okay. So the surface potential is a very important quantity. When I apply a positive gate voltage I get a positive potential at the surface of the semiconductor. 

[Slide 14] What if I apply a negative gate voltage? Negative gate voltage pulls the bands up. Okay. If we pull the bands up, it means we're inducing a negative potential in the semiconductor. The surface potential is now negative. If I pull the bands up you can see I'm pulling the valence band towards the Fermi level. That means that as the bands bend up, due this negative surface potential, the hole density is increasing exponentially, because the top of the valence band is getting closer and closer to the Fermi level. So if I look at my expression for the hole density in terms of the top of the conduction band in the Fermi level, you can see that it's going to increase near the surface. If I look at my net charge, you can see that now I'm -going to pile up many more holes at the surface, so I'm going to get a net positive charge at the surface. That's being attracted by the net negative charge on the metal surface of the metal, and if I integrate that charge throughout the depth of the semiconductor I'll get some sheet charge and coulombs per metered squared or semiconductor people like to quote things in coulombs per centimeter squared. So this is called the accumulation charge. In a P type semiconductor the majority carrier holes accumulates or pile up near the surface. 

[Slide 15] All right, let's go back to a positive surface potential again. We apply a positive gate voltage. We bend the bands down, that means we have a positive surface potential in the semiconductor. We now have a charge density in the semiconductor, because that the positive charges on the metal gate have pushed back the holes from the semiconductor. Another way that you can see that is at the valence band has been pulled way below the Fermi energy, so it means that there's less and less probability that there are any holes in that region. So that region, that X-- Y equal  equals zero to the depths WD, that's our depletion region. It's depleted of these mobile holes. Okay. So if we look at that a little more carefully, our hole density versus position is given by the conventional relation between Fermi level and top of the valence band. The electron density we're pulling the bands down, so we're increasing the electron density as well, but the electron, the conduction band is still well above the Fermi level, so we can assume that there are negligibly small number of electrons there. If I look at the charge density, really the only charge I have in this region near the surface between Y equals zero and Y equals W sub D, the depth of the depletion layer is the charge due to the ionized acceptors, due to the, you know, NA minus. And if I go beyond that the semiconductor is neutral, because I have an equal number of positively charged holes and negatively charged receptors. So that's the situation when we apply a positive gate voltage. We call that by a situation depletion of the structure. 

[Slide 16] Now, what if we keep increasing the positive bias and we pull the Fermi level in the metal down more and more. We pull the bands down more and more. We deplete more and more and we push the depletion layer depth deeper and deeper into the semiconductor, but we also bring the Fermi level closer and closer or bring the conduction band closer and closer to the Fermi level, which isn't moving. That means our electron density is getting higher and higher. When we reach a critical potential and we'll call that critical surface potential 2 times psi sub B then the density of electrons at the surface is equal to the density of holes at the-- in the bulk of the semiconductor and we could no longer ignore those holes near the-- or those electrons near the service. So remember, in the bulk we had a small number of electrons. At the surface we are pulling the conduction band down towards the Fermi energy, so the electron density is increasing exponentially, as we go towards the surface. It, in fact, we can write it as the density in the bulk, that small number times e to the q surface potential over kT. So it increases very substantially. Now, when we bend the bands enough such that this electron density at the service is equal to the hole density P naught in the bulk. When we make those two equal we have reached a critical band bending. And that critical band bending we define as 2psi B. If you work through the algebra here you can see that psi B is directly related to the doping density of the semiconductor. kT over q log doping density  divided by intrinsic carrier density. So psi B plays an important role in our discussions on MOS capacitors. 

[Slide 17] So if we look a little more carefully at the inversion condition, we have pushed the depletion layer deep. In fact, we now have  bent the bands by 2psi B, so our depletion layer thickness we're going to call W sub T. It's a sort of the terminal depth that we can push the depletion region. And we just evaluate it with the-- a formula that we're going to derive in the next lecture, and we just evaluate it at a surface potential of 2psi B. We call that depletion charge. We have a certain charge due to the ionized acceptors that are there. But we also have these electrons in the conduction band and beyond inversion they're beginning to build up very substantially. So when I apply a voltage that's greater than this critical voltage, I'm-- I would say I mean inversion and there are lots of mobile electrons near the surface they're going to give me the current through my MOSFET. 

[Slide 18] Okay. So just to summarize in MOS electrostatics we can visualize it in terms of these energy band diagrams. When we apply zero volts on this hypothetical gate metal we have zero potential everywhere in the semiconductor. We're at flat band conditions. When we apply a negative voltage to the gate we bend all the bands up. We have a negative surface potential. Holes accumulate near the service. We call that accumulation. When we apply a positive voltage to the gate we bend the bands down. We have a positive surface potential. We deplete the semiconductor of majority carrier holes, but if our voltage is large enough we also invert the semiconductor and make the surface end type if the bulk was P type. So those are the energy band views of what band bending looks like. 

[Slide 19] We could also sketch the charge in the semiconductor. The total charge integrated through the depth of the charge in coulombs per square centimeter as a function of surface potential. If I apply a negative surface potential, all of those positive holes pile up near the surface, that's accumulation. So we get a very large positive charge density that increases exponentially with the band bending. If we apply a positive gate voltage, positive surface potential we deplete the semiconductor and we'll see later that the charge increases slowly as the square root of the surface potential. But if we bend the bands enough beyond this critical point 2 Psi B, we'll find that the inversion layer electrons begin to build up very rapidly exponentially with service potential as well. 

[Slide 20] Okay. Now, if you want to test your understanding of these concepts, a good way to do that is to repeat what we have just done, but assume an n-type type semiconductor. So that's a semiconductor we would use to make a P-channel MOSFET. And draw the band diagrams for accumulation flat band depletion and inversion for an n-type type semiconductor bulk. That's a good way to test your understanding. 

[Slide 21] Our overall goal is to understand MOS electrostatics. Our first goal is to solve the Poisson equation. In general, we need to do it numerically, but under certain conditions under this depletion condition things simplify we can get simple analytical solutions and that's the subject of the next lecture in this unit. I'll see you there.