Now, if you work with these equations and eliminate some of the unknowns, we very

easily get this equation for the inverse[UNKNOWN] charge.

And I suggest that you do that to get some practice with these equations.

And Qb. We had derived an expression for it, and

we had shown it is proportional to the square root of the surface potential cs.

Now, if you plug in this expression into Qb and collect terms, you find that you

have this. So this will be the form of the versa

layer charts that we will be using in this develop.

in addition, you will remember that when we had a 3 terminal structure, we had an

external terminal c, to which we applied bias VCB.

And that. The modified the inversion level.

And our development there, lead to the following equation.

Which relates the surface potential. To the externally applied gate body

voltage. And also to the externally applied

voltage between the third parallel C. And the body B.

The horizontal electric field is much smaller than the vertical one.

That means that the field is practically vertical in this device throughout the

channel. This assumption may be violated near the

source and near the drain, but we said those regions where it is violated are

very small compared to the channel length and for now we will neglect this fact.

We will assume that the electric field is everywhere practically vertical.

The variation of the horizontal component with x is much smaller than the variation

of the vertical component with y. These assumptions are referred to as

gradual channel approximation. And finally, I already covered this one,

channel length must be much larger than the depletion region widths around the

source and the drain. Now with this..

Said, we're ready to start the detailed derivation of the old region model.

We're going to concentrate on a very small chunk of the inversion layer, shown

here, between position x and position x plus delta x.

So let me now expand and show you only this part of the structure, only that

chunk that we're talking about. Looks like this.

Psi S of x is the surface potential, on the left, psi s of x plus delta x is the

surface potential on the right. And the difference between CS of x plus

delta x and CS of x is delta CS which is the surface potential difference applied

across this chunk. The width of the chunk is the same as the

width of the channel. And the electrons flow in this direction.

They're negatively charged. Which is equivalent therefore to a

positive current going this way. And the current at position x will be

called i of x. In the following, we will follow delta x

and delta psi x to approach 0. Now you remember that current can in

general be due to two causes, drift. And diffusion.

We will start by assuming that the current can have both of those components

at the same time. And then we will see that in particular

regions of inversion, one of the two dominates.

For now, let's allow both of them to be present.

We had derived equations for the drift current and the diffusion current

components back then. In the, in our background review.

And if you remember, or you can go back and check, this was the expression we had

derived for the drift component. This was the expression we derived for

diffusion. The drift depends on the gradient of the

potential, and the diffusion component depends on the gradient of the inversion

layer charge. Now, we want to find a way to find the

current, so I'm going to multiply both sides of this equation by dx and

integrate. Now this current i of x is the same

throughout because we are in this situation, we are assuming steady state,

so the current is the same throughout the channel.

So I'm going to replace i of x by IDS. And integrate from the source to the

drain. So now we have, because the variable of

integration here is x, we go from 0 to l, from the beginning of the channel to the

end of the channel. Because this one is, because the variable

of integration here is cs, the surface potential, we go from the surface

potential at the source, psi s 0, to the surface potential at the drain, psi s L.

And because in the last integral the variable of integration is Qi, we go from

Qi at the source to Qi at the drain. The symbols are self-explanatory here.

So now, to do this integration, notice that in the first integral, IDS is a

constant so I can move it outside the integral.

And then the integral of dx is x and you evaluate it from 0 to L.

So it becomes L itself. So the right hand side is I DS times L.

I'm going to divide both sides by L and I finally get this.

This now shows you how to find the drain source current, the current between drain

and source. From 2 integrals.

1 is due to drift, and the other is due to diffusion.

And a well known geometrical ratio, w over l, appears in front of everything.

Now, notice, that, at this integral, you have mu inside the integrals in both

cases. This mu, if we assume it is constant, we

can move it outside the integral and make things simpler.

So for now, I'm going to assume that mu is a constant.

I move it outside the integrals, and I finally have this form.

In the second one here, I have already done the integration.

It was integral of, of d dqi, which becomes just qi evaluated between drain

and source, so you take the difference like this.

This is, I remind you, the component we got due to drift and this is the

component that we have due to diffusion I would call the the first id as 1 and the

second id as 2. Now for Qi we can use the expression I

showed you when I reviewed our material from the two terminal structure

discussion a couple of slides ago. So if you now plug in Qi from this

expression into these integrals. And you will do the integration.

You find this result for IDS1 and this result for IDS2.

Now, notice the following thing. For the component due to drift, which is

IDS1. We have a term, CSL minus CSO.

Then we also have the difference of the squares of these quantities and also the

difference of the 3 half pies of these quantities.

Notice that it, is it to trays where these comes from, when you plug-in this

expressed on interval integral, integrated shears, we ended up with the

squares. And when you have the square root, which

is power to the 1 half, when you integrated, you ended up with the 3 half

powers. It will turn out later on that we can

simplify things if we avoid things like that, but for now let us keep them there

to have the full, complete origin model. Similarly, when you go to the IDS2, the

component due to diffusion, you have the difference between 1 half powers, so for

psi sl and psi so, again, this come because when you integrate it, the

equation shown in the previous slide, you end it up with a difference between two

QI's and these QI's from this equation. Had 1 half powers in them and this is

where we got this 1 later on again we will simply find matters.

Now, this is our all region model but something is missing because we don't

know yet how we haven't at least[UNKNOWN] explicitly how to evaluate CSO and CSL.

We'll do that in a minute. Note the symmetry in this.

If you reverse CSL and CSO, you're going to get the same magnitude of currents,

only they would be int he opposite directions so that means that if Ritter

change their own of source and drain, you're going to get the same current

flowing about in the opposite directions. Same in magnitude, opposite direction.

This surface potential-based, model is the basis in certain, of certain CAD

models. For example, HiSIM.

And then after we simplify this equation, it forms the basis of the PSP model.

We'll have more to say about these models later.

Okay, here is the device again, biased in the same way as before.