I remind you that before we discussed this effect, we had this expression

already, but it was this part. Now the new thing is here.

As we did for the long channel transistor now, we're going to integrate both sides

from source to drain. So this will be become an integral in

VCB, and so will be this one. This will become an integral of DX, so

you multiply through by DX and then you integrate both sides from source to drain

and when you do that this is the result. Now VDP Minus Vsp is the ds, the drain

source voltage. So I will replace this by Vds and then I

will divide both sides by this. And I get this expression.

What I've done here also is I divided by l and multiplied by l so the denominator

becomes like this. I did that in order to reveal w over l

which is A quantity we want to have. We had it for the long channel device.

In fact, this here is exactly the expression for the long channel device

that I showed you a couple of slides ago. The new thing is here.

So you see then that when you include the velocity saturation, the drain source

current is what you would have without counting velocity saturation divided by

this. Quantity here.

This is what takes care of velocity saturation effects.

So for example, let's take the simplified, source reference strong

inversion model. It will give you this expression.

And again, I remind you, W over L mu C OX times this bracket, is exactly what we

had in this model, and this factor in the denominator, is what Takes care of

velocity saturation. So lets plug this.

If we don't have velocity saturation we see the top curve shown in broken lines.

And when we do have a velocity saturation.

We get the bottom curve. You can see two effects.

First of all. The current is smaller because you're

dividing by the quantity in the denominator.

And also it gets to be horizontal faster. As you see here, so Vds prime becomes

smaller. So now how do we find the new value of

Vds prime We do it the same was as we did for the long channel transistor.

You take the derivative of this, of the equation that gives you this curve over

here, and you set it equal to 0, and you solve for V D S, and that is the value of

V D S prime. When you do that for the equation I

showed you on the previous slide you get this expression.

And, this one can be recognized at V D S prime for the log channel device.

And this is the new thing that you have to multiply with, in order to get, the

new V D S prime. This here is less than 1.

You can easily check this is less than 1, it makes V D S prime to have this value

Rather then the long channel value which would be here.

Now how do we find the value of the current in saturation.

We simply take the non saturation expression and evaluate it at Vds prime.

We find this expression. So this expression now gives you the

saturation current. It is clear.

That it is the same as the saturation current for long channel devices shown

here, by the denominator times W over L mu C OX, divided by something that takes

care of, velocity saturation. Here is a set of curves, for two devices

having the same W over L, but one is a long channel device.

So, we don't include velocity saturation. And this is a short channel device, where

we include velocity saturation. I should emphasize here, that the plots

include other short channel effects, which we have not yet discussed.

In particular, this is the reason that the, the curves, in saturation,

increasing with V D S. Or you might say, they are both the same

length, but here we forget to take into account the velocity saturation, and here

we do. You can see the big difference.

Not only for the same V G S is the current smaller, for example, this one

becomes this one Not only the points where you roughly reach saturation become

smaller. They correspond to smaller v d s voltages

because v d s prime became smaller but also.

The fact that the quantity in the denominator has the effect of compressing

the curves in such a way that the spacing of them is nearly equal.

Whereas here the spacing went up corresponding roughly to square long

modified by the effective mobility effect.

Here you nearly have a spacing that is independent of VDS.