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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.

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And you will see that in the various equations for this structure, the only

difference would be, that instead of 2 pi F, you have 2 pi f plus V CB for the

reason I just described. So I'm not going to go through the

development of these equations, but I will show them to you, and I will indicate V CB

clearly, so that you can see what the difference will be with respect to the

two-terminal MOS structure. So, the general analysis goes along the

same lines as before, but instead of having equations of this type, for

example, at position y, we had n of y that was given by this equation for the

two-terminal structure. Now, it is given by this equation.

And the final result we had derived was, as you see here, by Qc prime, except that

now we have an extra factor with minus V CB in it for the three-terminal structure.

Similarly, we had and this was the, this is the whole contribution, this is the

dopant ion contribution and this is the electron contribution.

Again I believe that this way of writing this equation is new and makes things very

clear. Again, proceeding as for the two-terminal

structure, we obtain this equation from the previous one.

The details I, are again in the book. And, we have a, a relation between the

surface potential and the externally applied gate body potential which is given

by this. And this equation is still the same except

for instead of 2 pi F, we have 2 pi F plus V CB just like it was the case over here.

Other than that, they are the same thing. Now, as we have seen from the two-terminal

structure, certain terms can be neglected in particular regions here, the depletion

inversion and so on, but we will concentrate on inversion.

So inversion, this is the equation for inversion after you delete a number of

terms in the equation on the previous slide.

It looks just like what we had for the two-terminal structure, except again,

instead of 2 pi F, you have 2 pi F V CB. This is the relation between surface

potential and gate body voltage in inversion.

Again, instead of 2 phi F, we have 2 phi F plus VCB.

These are the only things that change, because other than that the potential

balance equation is still the same. The charge balance equation is still the

same. The linear equation between gate charge

and oxide potential is still the same. And the body charge equation is still the

same. It depends on cs in the same way as

before. It just happens that cs now has a

different value if you want a certain the level of inversion, but you use that in

this equation, you get a valid result. And by they way, we have defined parameter

gamma which was the this quantity divided by C OX.

So, instead of this quantity, sometimes we show gamma C OX over here.

Continuing, if you take the equations that I showed you, you can plot various results

versus V GB. And here, I have done this for two cases.

The broken lines correspond to V CB equals 0.

In other words, this source is replaced by a short, and that in fact, is equivalent

to replacing it by on, an open circuit. So, that means that this part of the

structure looks like a two-terminal structure.

So, the equations that we used for the two-terminal structure apply here or you

can take the new equations and apply them with V CB equals 0.

And you'll find the results we've seen before.

As you increase the gate body voltage, Qi goes up and eventually becomes a straight

line in strong inversion. The log of Qi eventually becomes a

straight line in weak inversion. The oxide capacitance goes down and then

up, we explained the reasons why. And, the surface potential tends to

eventually flatten out in strong inversion.

Now, if we apply a value of V CB, which is greater than 0, then everything shifts to

the right because of this and I already explained the reasons for this.

For example if you make V CB larger, you will reduce the electron concentration oh,

in the inversion layer. And in order to restore it to its previous

value, you have to increase the sharpness potential, and to do that, you need to

increase V GB. So to see what you were seeing before, you

have to go to larger values of V GB, for example, strong inversion will encounter,

the encountered at larger V GB values. The weak inversion behavior will shift to

the right. The minimum of the capacitance will also

shift to the right and the flattening of the surface potential will shift to the

right. And instead of being, let's say instead of

having the weak inversion between pi F and 2 pi F, you'll have it between pi F plus V

CB and 2 pi F plus V CB. Other than that, it's the same behavior,

at least qualitatively. So in this video, we have seen how, by

adding a third terminal, which we called c, it's our contact to the inversion

layer, modifies the situation in the channel.

In the next video, we will continue this study, and in particular, we will discuss

the important phenomenon called the body effect.

We will see how the general inversion equations simplify for the cases of strong

inversion and weak inversion.