And so we can plot Qi, as shown here. Versus x and the magnitude of Qi goes

down. Now an important question is how come if

Qi goes down the current is maintained the same throughout the channel.

And of course the answer is if you remember.

that the current is proportional to the product of Qi, and the drift velocity.

So if Qi is going down in magnitude, it must be that the velocity is going up, in

such a way that the product stays constant.

So the carriers start here slowly, and as they go towards the drain, they move

faster and faster. As Qi is going down, so that the product

is always fixed, and equal to Ids. The drain source current.

Now let us increase Vds above Vds prime. We know from here, that the inversion

layer cannot support more than Vds prime across it.

So, if you apply a voltage which is larger than Vds , the inversion layer

will keep Vds prime on it. And the extra voltage, Vds minus Vds

prime will be dropped somewhere else in this depletion region, like, region.

The only possibility that this can happen is that the inversion layers.

Shrinks so that it allows some region here over which the extra potential can

be dropped. And in fact if you take the depletion

region charge over here and you apply Poisson equation you can find even the

length of this region that is needed in order to support this extra voltage, Vds

minus Vds prime. So now, of course, this is not a pure

pinch-off region. There are charges here, it's just that

their density is light. Qi goes down in magnitude and when you

reach the pinch off region, Qi is maintained at a very small value.

We assume that here, we have already reached velocity saturation so the

velocity is fixed and very large. And correspondingly, Qi its fixed and

very small so that the product is such that it can support the drain source cut.

The length of the region is i sub p. And of course again, pinch-off region is

an unfortunate name that people have kept historically.

We now know that it is not a pinch-off region, there are charges in it, and they

just travel at very high velocity. So now if Lp's the length of the pinch

off region. Then, the length of the channel itself,

up to the pinch-off point, is L minus lp. So, in a sense, the channel has shrunk.

Now, you remember that the current is proportion to w over l.

So if L has shrunk, the current will go up.

So you do expect That, in the saturation region, as you increase Vds, Vds minus

Vds primewill increase, Lp will widen, L minus Lp will shrink, will become

smaller, and 1 over that, which is what the current is proportional to, will go

up. So this is V D S prime, this is I D S

prime And here I only show you the saturation part of the characteristic.

What I just described to you is the so called channel length modulation.

The effective length of the channel. This distance from here to there is

modulated by Vds. So lets take the value of Vds here called

Vds prime which corresponds to. The left most picture, here.

We know that the current is w over l times some quantity.

Which I will not repeat here. It depends on the model that you use.

Let us now go to a Vds larger than Vds prime.

Then instead of L, as I mentioned already.

We'll have the new shrunk length, l minus l p.

And this quantity remains the same. So now the new current can be found in

terms of the old current. If you eliminate the quantity and the

parenthesis between these two regions. And you get this result.

That ids is the current in saturation at this point, right when you reach

saturation, times l over l minus l p which is larger then one.

And because as we increase Vds, Lp increases, the current will keep

increasing as I have shown here. Now, let's divide through by l here.

We can also write it in this form. Now, normally you expect Lp over L to be

much smaller than one, but if you use this model inside aA, a compact model

that, that, that is used in computer aided design.

While the computer is trying to find the solution to the device equations, it may

try very different values for the voltages from those that it eventually

should end up with. And it may be that momentarily it applies

a large value of Vds such that Lp becomes equal to L And this becomes 1, the

denominator becomes zero and of course then things blow up and you can reach non

convergence. So this is not a numerically stable

equation. There's another way to write it by making

the approximation when a P over L is very small, let's call it epsilon.

One over one minus epsulon is approximately one plus epsulon.

So we can write instead of this expression we can write this one.

Now you can see here that no matter what values of v d s are tried by the

computer. No matter how l p, how large l p becomes.

This never really blows up and we have a numerically advantageous expression.

Let's now sketch how we evaluate LP. You just solve the one dimensional

Poisson's equation. Are all at, across the along rather, the

pinch-off region. And the solution is of this form, where

Vsubd can be expressed in terms of physical constants.

The details, if you would like to read them, are in the book.

However, this is a one dimensional solution.

If you actually take a 2-dimensional solution, which you have to do

numerically, you find the following result.

The inversion layer electrons travel towards the drain and when they enter

what we have called the pinch inversion the pinch off region.

They tend to go below the surface, and reach the drain at various depths.

So this is clearly a 2-dimensional situation.

Also, if you plot the horizontal field, it goes up, and when you enter the pinch

pinch-off region, it goes up, becomes maximum at the bond area between the

channel and the drain and then inside the drain, it goes down.

Now, this 2-dimensional picture cannot be captured by an expression like this.

It is clear that because of what I described happens over here, even the

depth of the drain. makes a difference.

So you need the two dimensional solution, or at least a pseudo two dimensional

solution. Such analysis have been done, and this is

one result. You'll find the reference in the book.

Now, in this expression, L sub A and V sub E Again can be expressed in terms of

physical constants and geometric dimensions.

For example L sub a from this psudo 2-deminsional analysis is given by this.

It suppreses the productivity silicone. It surpresses the productivity of the

oxide. Dx the oxide thickness, and Dj is the

depth of the junction here. Which we already expected should make a

difference, because of what I had, I have described happens in this region.

So I'm not going to go through the derivation of such equations.

I show them to you so that you get a feel for what people are using in such cases.

But the derivation of these things, are long and they're based on a number of

assumptions that cannot always be rigorously justified, but which

eventually leave you lead to an expression that agrees with measurements

provided. You choose the parameters in them

appropriately. So for example, v sub d and v sub e here

Can be allowed to be parameters that you choose so that you can optimize the

matching of these expressions to measurement.

And that is necessary because of the number of assumptions and approximations

we have made On the way towards deriving these expressions.

Now you can take L p from here, and use it inside the expression we just derived

in the previous slide, this one. And you have an expression for the

current in saturation. Now there is an issue with continuity.

Let's go back to the long channel derivations.

We had a nonsaturation expression that was a parabola like this.

Then we realized this is, this is a part that doesn't make any sense.

We stopped it there, and said we extend this.

Horizontally, and you can see that the saturation expression, which is just a

constant, is tangent to this parabola at the zero slop point.

So the slope just continues. It goes towards zero.

And stays at zero. But if you now take a nonsaturation

current expression such as the ones that we derived on the previous slides, and

you plot them you get something like this.

And what is the problem here? Right at VDS equal to VDS prime, the

slope is discontined. It jumps from zero to a non-zero value.

That is a very serious problem, and it has to be avoided at all costs, in

computer aided design models. So it happens often when you start

evaluating the current in one region and the current in another region, then you

stitch the two regions together you may end up with this problem.

the current should be continuous at the boundary between two regions.

The slope should be continuous. Even the second, and higher derivatives

should be continuous. So we will discuss, at a later point, how

we can avoid such issues, of discontinuity.

Of course, single piece models, meaning models that use a single expression for

all regions Are to be preferred. And, increasingly so, such models are

being used in CAD models. Okay, let's look at, a set of iv curves,

in the presence, and the absence of short channel effects.

Excuse me, of a channel length modulation effects.

The broken lines are currents that you get if you neglect channel length

modulation. But you include other types of

short-channel effects for example, velocity saturation and other effects

like that. I have yet to cover in future lectures.

And the solid lines are what you get when you include channel leg modulation.

Lets take for example the curves for 0.8 volts.

It's clear of the channel leg modulation increases the slope.

In the saturation region, compared to the slope, in the absence of channel length

modulation. Okay, let's now go back to the expression

I showed you, that results from the pseudo two dimensional analysis for the

pinch off h, region length. It was this expression.

I'm going to assume that Vds minus Vds prime is much smaller than Ve, which is a

questionable assumption. But I will do it in order to show you how

you can derive an expression, a very, very approximate expression that people

use. Now, I can use, if this is much smaller

than one, call it epsilon. The log of 1 plus epsilon is

approximately equal to epsilon. for a natural log, I mean.

And then doing this here I end up with this result.

So now, I can take this expression for the pinch-off of region length, plug it

into the expression we derived in here and we end up With this expression where

Va is this. Now, this expression predicts that the

current in saturation varies linearly with Vds.

It's a very common approximation especially you see it in circuit books.

Va sometimes is called the early voltage. By extension of the Early voltage

encountered in bipolar transistor discussions.

Where Early is in honor of Jim Early the person who first dealt with this

phenomenon bipolar devices. But the fact that this equation predicts

that the current varies linearly with v d s cannot be taken for granted.

You derive this equation after you go through a whole bunch of approximations.

Which unfortunately are not really valid, so don't expect this equation to be

accurate. It is very crude; it is only good for,

for the crudest hand analysis for circuits work, and sometimes even then.

You need to do better then that. So in this video we discussed channel

length modulation. We saw how increasing the drain source

voltage increases the length of the pinch-off region.

And how the inversion layer effective length shrinks as a result making the

current go up. In the next video we will talk about a

very different effect, charge sharing.