Because in the MOS transistor, what I am about to say in this lecture applies to

both the source and the drain. So we call it c.

It's our contact to the inversion layer if you want.

So in this particular situation that I show you here, I have shorted this contact

terminal to the body. And we know from our discussion of pn

junction that there would be a built-in potential, phi bi, between the n and the p

regions, and there will be a corresponding depletion region.

I'm assuming that the n-type region is much more heavily built than the p region.

So the depletion region inside the n-type region is so much narrower that to make

things simpler, I'm not going to be showing it.

I will only show the region on the p side. Now, we know from our p-n junction

discussion that whether this wire between c and b exists or doesn't exist is the

same situation over there. But for now I will have c sorted to b.

The rest of the structure away from the p-n junction and away from its depletion

region is the two-terminal structure we have already studied.

And because we're applying a voltage, V GB equal to the flatband voltage, the bands

are flat in this structure and there is no potential drop as you go from the surface

towards the bulk. So the depletion region is only around the

n region that you see. We will assume that the part of the

structure to the right is long, much longer than the depletion region width to

its left. So, what we're saying applies for

practically all of the structure. We're going to have to follow the energy

band situation for them. I will only show you the conduction band

edge in four cases. So, here is the case a, for this.

I am sketching the conduction band edge, E sub c with respect to horizontal position

x, right next to the surface, as you see at the top where the broken line is.

And you can see that there is a phi bi built-in potential which corresponds to an

energy difference of q phi bi. And because phi bi is positive, the n side

is more positive in potential than the p side which means that the terms of

energies, due to the negative charge of the electrons, you're going the opposite

direction. The energy is lower here in the n type

region, that it is in the p type region. In other words, there exists a, an energy

barrier which makes it difficult for the electrons that are in the n-type region to

cross the barrier and go to the p-type region.

This is why, you don't see electrons in the p-type region in this case.

But you can make them appear, of course, if you increase the gate, body voltage,

sufficiently. So, in the second case I'm showing you

here, I have Increased V GB to some value V GB1.

That is such that it has not only depleted the part below the gate, but it also has

created an inversion layer. Just like we have been doing for the

two-terminal MOS structure. In fact, the situation here, as long as c

is sorted to b, as long as you have this connection, is the same as what you would

have in a two-terminal structure. It is described by the same equation, set

of equations, as long as you stay away, from this part, where the presence of the

n region affects things. So, let us assume that for the given value

of V GB that I'm using, we have a certain surface potential, c1.

So now, the surface is more positive than the bulk, and if you go to the

corresponding energy diagram, you have lowered the corresponding conduction band

at, at the surface. So the, for the case in b, we have the

curve b here. So, effectively, what you've done is

you've lowered the energy barrier that is needed to be crossed for electrons to be

able to go to the channel. And because the energy barrier is low,

indeed, electrons have gotten to the channel and you can see them in the second

figure. Now, I'm going to do something that is key

to the rest of this lecture. I'm going to cut the wire that connects c

to b, and I will introduce a voltage there, which I will call V CB.

And I would like to see how this will affect the concentration of electrons in

the channel. So let's do that.

This is my voltage V CB. First of all, how will this effect the pn

junction? Because the plus is on the n side and the

minus is on the p side, it is a reverse bias.

And therefore for the pn junction the depletion region will widen as you see

here. Now, since the potential on the end side

became more positive than before because of the presence of V CB, the corresponding

conduction band edge E sub c goes down by what amount?

The amount of the potential change V CB times the electron charge Q.

So for this case, we see that, now, we have lowered the energy on the inside by

significant amount QV CB. That intends to increase the barrier for

electrons again. You, you can see that curve c goes up like

this You have lowered this part here, so that the, the barrier tends to increase.

And that is the reason that I'm showing fewer electrons here.

The inversion in the channel is now less heavy than before.

And the question is, what do we have to do in order to restore the inversion in this

case back to its previous value, which you had over here in the second case.

What do we have to do? Well, in the second case, we had the small

energy barrier, as you see on the right for curve b.

So all we have to do is make sure that we restore this small energy barrier to its

previous level, as you see here. And to do that, what we have to do is

lower the energy of electrons in the channel.

Which is equivalent to saying, we need to increase the potential, make it more

positive in the channel. And that, we can do, by simply increasing

the value of V GB. So, that now, you have more positive

potential here, and it is positive enough to allow enough electrons in the channel,

and you have inversion layer as heavy as before.

So from this argument, you can see that just like you changed the energy by QV CB

on the n side, you also have to change by the same amount, the energy at the

inversion layer, at the surface, situation d, looks about the same as situation b.

So that's why the inversion here is about the same as the inversion in b.

So you see then that what counts is no longer, how much the surface potential is.

The surface potential in a sense is competing against V CB.

If you increase V CB, you have to increase the surface potential by the same amount

to restore the inversion layer where it was before.

So therefore, if you follow this argument, and it is done in the book, but there is

no time to do it here. You find something that is not very

surprising. The equation we had for the surface

concentrate, concentration of electrons instead of having Cs, it has CC, Cs minus

V CB. And again, what counts is not what CS is,

but what is the difference of CS with respect to VCB, because the two are

competing. So, you will find that this V CB creeps in

into a variety of equations we have already derived, for example, this is

another equation. Again, here you see psi s minus V CB.

And here, I have lumped V CB together with 2 pi F.