nanoHUB-U Fundamentals of Nanotransistors/Lecture 2.4: Flatband Voltage
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[Slide 1] This is lecture four in unit two of our course and what I want to do in this lecture is to go back and talk about something that we left out in the earlier pieces. Just a minor little detail we get it straight we can correct the results that we been discussing and then we'll be ready to proceed.
[Slide 2] So the issue is we have been talking about something, a hypothetical ideal MOS capacitor. And what made it ideal was the properties of this metal that we were assuming. So we were assuming that we had a particular metal. I know you'll recall that, there will be some Fermi level in the metal and it will have some value before we apply any voltage to the gate and then when you apply a positive voltage to the gate you'll lower the Fermi level in the metal. So this is a situation where we been assuming that there is no voltage applied to the gate and that the Fermi level just happens to line up with the Fermi level in the semiconductor. You know, well, what are the chances of that? You know, not much, but it made everything conceptually easy to explain. Now we have to fix that. So in this particular case we would say that the flatband voltage is zero when we apply zero volts to the gate everything lines up, all of the bands are flat. There are no band bending. Another way to say this is that there is no difference between the work functions of the metal in the semiconductor. All right.
[Slide 3] Okay. So let's look at this in a little more detail. If we just look at a metal, a metal has some work function capital phi sub M, that is the energy it takes to pull an electron at the Fermi level out and into the vacuum level, into free space. We'll call that energy E0. So the work function is some well-known property. Each metal has a different work function. So let's use the vacuum level as our common reference. So we'll have an insulating layer, oxide in our structure and there what's critical is the electron affinity. The electron affinity tells us what the energy it would take to move an electron from the conduction band of the insulator. Of course there aren't very many there into the vacuum. So that's a property of the material as well. And then we have a semiconductor and a semiconductor the bands look like this. The semiconductor has an electron affinity. That's the energy it takes to move an electron from the conduction band of the semiconductor into the vacuum, and this is a typical value. This is a value for silicon. And it also has a bandgap and it has a Fermi level that can be somewhere in the bandgap depending on how we have doped the semiconductor. So this is what makes it difficult. It's hard to change the work function of a metal. You basically just have to get another metal, but it's easy to change the work function of a semiconductor. You just change its doping. Now you can see when we put everything together, the Fermi level has to line up and be constant in one position. So what's going to be critical is this energy difference. We'll call that q times the built in voltage, because a positive voltage is going to have to develop on this metal, if we assume the semiconductor is grounded at V equals zero. A positive voltage is going to have to develop here that is just enough to pull the metal down, line the Fermi levels up, and everything is in equilibrium then. That positive voltage is the built in voltage of this structure. And you can see, well, it's determined by the location of the two Fermi levels. The location of the Fermi level in the metal is determined by its work function, and the location of the Fermi level in the semiconductor is controlled by its work function. The distance between the vacuum level and the Fermi level. So this potential difference that has to develop is just minus the difference between the work functions, minus phi. Okay. So the work function difference between the metal and the semiconductor is what's going to be critical.
[Slide 4] Now, we put these two together. We set a positive built in voltage has to develop on the metal. That's this Vbi here. If there's a positive voltage on the metal, before I've applied any positive gate voltage then that positive voltage is going to cause the bands to bend down, and we will have some positive surface potential before we've applied any gate voltage. So that is a situation in a realistic structure with work functions like those that we've shown. So this built-in potential is just minus the work function difference divided by q. We'll usually write that as minus small phi sub MS. So that's a property of the metal and a property of the semiconductor and how we have built the semiconductor. So the flatband voltage, this is going to be a critical voltage , you know. The flatband voltage V equals 0. When I-- if I apply zero volts on the gate. I am not at flatband conditions. I am at this condition. There's a significant amount of band bending going on. In order to make the bands flat I'm going to have to apply a negative voltage equal in magnitude to this built-in voltage and pull everything up until they are flat. I'm going to have to apply a voltage on the gate, an external voltage that's equal and opposite to that built-in voltage, meaning the voltage that I'm going to apply is phi MS, the metal semiconductor work function difference in volts. All right.
[Slide 5] So let's do an example of these work function differences. Here's the semiconductor we've been considering. Here's our relation between Fermi level and valence band edge in the semiconductor. We can use this to solve for the location of the Fermi level in the bandgap if we know the value of the effective density estates, and here it is in silicon. That's a well-known value numerically at room temperature. So we can find out that the Fermi level is located just a little bit above the top of the valence band, .08 electron volts. When I dope this silicon 10 to the 18th per cubic centimeter. Alright so it is very close to the valence band. Okay. Now if I use aluminum its work function is 4.08 electron volts. I need to figure out the work function of the semiconductor. So first I go down to the conduction band, that's the electron affinity of the semiconductor. Then I go down to the valence band. That's the bandgap of the semiconductor, but I've gone too far because the Fermi level is .08 EV above the valence band. So I've got a subtract off that difference. I know numerical values for the electron affinity of silicon and for the bandgap of silicon. We can plug in numbers and the work function of the of the semiconductor in this example is 5.09 electron volts. The work function difference in volts is minus 1.01 volts. So it means the flatband voltage, in this particular example, is about minus 1 volt. You have to apply a negative 1 volt to the gate of this MOS capacitor in order for the bands inside to be flat. So that's an example of how a calculation like this would go.
[Slide 6] Okay. So we've been talking about this very simple relation between ideal gate voltage and charge in the semiconductor and surface potential. We can easily fix this now. The actual gate voltage is the ideal gate voltage, which had a zero flatband voltage. Now we just have to add in what the actual flatband voltage is. We just add this term in. So we add the real flatband voltage and now we have expressions for the gate voltage in terms of the charge, and potential in the surface and the flatband voltage as determined by the difference in these work functions. Okay. And that's presumably something that's known. If we know the properties of the metal, and if we know the properties of the semiconductor.
[Slide 7] So let's go back to this threshold voltage example that we discussed in the last lecture. So we work through some numbers in our ideal case with the ideal metal and we deduced a threshold voltage that was 1.28 volts. And I told you that's too high for this doping, this oxide thickness, that's not a reasonable value to get. Well, the reason it isn't reasonable is we had not accounted for the minus 1 volt threshold-- flatband voltage. If we account for that flatband voltage, we find that the real threshold voltage of this MOS capacitor is about .27 volts. That's a very reasonable number. So this flatband voltage correction is not something that we can neglect. It's there, its big enough to be important. We always have to account for it.
[Slide 8] Now, there's a second factor that it used to be very important to account for. Now it's relatively minor, okay. But now as silicon dioxide is being replaced with other material it's becoming a little more important again. And that's the fact that there might be a little bit of charge at the oxide silicon interface and we'll only consider the type of charge that it's just fixed. It doesn't change with gate bias or surface potential. It's just a sheet of charge. There's a bunch of dangling bonds there when we terminate the silicon lattice. So it's not unreasonable to believe that there might be charge at that interface. What effect would that have on the flatband voltage? So we're going to assume that there is some charge per square centimeter in a sheet. We'll write it as q times the number of charges per square centimeter, that's N sub F, and we ask what effect does that charge have?
[Slide 9] Well, let me go back and just remind you of how we did this derivation earlier without that charge there and then we'll add the charge. Okay. So we began with Kirchhoff's law, which said that the gate voltage was the sum of the volt drop across the oxide plus the volt drop across the semiconductor. The volt drop across the semiconductor is the surface potential. Volt drop across the oxide is the electric field times the thickness of the oxide. The displacement field in the oxide is equal to the charge in the semiconductor. That allowed us to solve for the electric field and then solve for the volt drop in the oxide, and then we expressed everything in terms of oxide capacitance. And we ended up with this expression that we've been using frequently. Now, what if there's a charge at the oxide silicon interface. What changes?
[Slide 10] Well, it turns out not very much changes. We do the calculation in the same way. We calculate the volt drop across the oxide, that's the electric field across the oxide, but now when we use this relation, the displacement field in the oxide is equal to the charge in the semiconductor. We also have to add the charge sheet at this interface. So we have the charge in the semiconductor that we had before, but we also have this fixed charge, both charges matter. So we put both of them in. Now when we compute the volt drop across the oxide we just have more charge. That's all. Oxide capacitance is just what it was before. Everything is just what it was before. It's just that we have two charges. We have a charge per square centimeter in the silicon, in the semiconductor, whatever it is. And we have a charge per square centimeter at the oxide silicon interface. We just have one more term, just sort of like the metal semiconductor work function difference. We just have to add a term to what we had before.
[Slide 11] So here's our expression. Now, this is our gate voltage versus surface potential expression. We add the correction. Due to the flatband voltage the correction consists of two parts. One part is the metal semiconductor work function difference. We've discussed that. The second part is due to any charge that happens to be at the oxide silicon interface. All right.
[Slide 12] Let's do another example or let's do example. That example we've been looking about. We computed a .27 threshold voltage for that particular example, assuming there was no charge at the oxide silicon interface. Let's now add some charge. Let's say that there are 10 to the 11th charges per square centimeter and that they're positive, which is typically the case. What effect would that have? So that's actually a reasonably large charge these days. If you're very careful with an Si/SiO2 interface you can get it down to 10 to the 10th. For some bad interfaces it might get up to 10 to the 12th. So 10 to the 11th is a moderate charge density. Let's see what happens. We can compute the charge density per-- in coulombs per centimeter squared by multiplying by q. And then we can take q over C ox and that will give us only a .01 a 10 millivolt shift. When we compute the threshold voltage we have to add that to the metal semiconductor work function difference that we had before. So we're just adding a negative .01 volts, so the threshold voltage in this example changes from .27 to .26. Not particularly important. One of the reasons it's not important is that these charges in modern MOS structures are usually well-controlled to some reasonably low value. The other is that as over the years as we scaled oxides thinner and thinner we made this capacitance bigger and bigger, and that's also happen-- works to lower the impact of this charge on the threshold voltage. But you should be aware that it's there in case you're working with some non-ideal system for which that might be important.
[Slide 13] Okay. So just a wrap up. We have a simple relation with a flatband voltage. And the ideal case it's zero and the real case it's nonzero and it consists of a part due to the metal semiconductor work function difference and a part due to charge at the oxide silicon interface. The relation between gate voltage and surface potential is given by this relation. We simply need to add in the flatband voltage, the real flatband voltage, whatever it is and then we can treat real problems. All right. Well, with that background, we are ready to dive into looking at this inversion layer charge. The charge that actually carries the current in a MOSFET and that will be the subject of the next two lectures. So we'll continue this discussion in the next lecture.