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The first effect I will discuss and there would be several such effects is called

Velocity Saturation. We will see that has the fields become

lots, velocity reaches the limit and will discuss how this affects the current in

an MOS transistor. Let us start with a review of some long

channel transistor basics. This is our long channel transistor.

We have been working with this for the past several lectures.

We have assumed that we can neglect the influence of drain and source over

practically all of the channel. We will assume, for now, strong inversion

and non saturation. So the channel is strongly inverted

throughout, from source to drain. And, we know that the surface potential

is p to the value V 0 plus V C B. Where V C B is V, V S B at the source end

of the channel. It is V D B at the drain end of the

channel, and it varies between V S B and V D B along the channel.

We have used the gradual channel approximation to analyze this device.

The field has been assumed to be practically vertical, and so we were able

to use the 3-terminal MOS Structural relations.

We used the 1-dimensional analysis and we found the drift current, which is

dominantly strong inversion, and we know that the driving force for the drift

current is the small horizontal electric field component, e sub x at position x.

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We can perform a change of variables from cs to vcb using using this relation here.

So d cs becomes dvcb, cso and csl become vsb and vdb.

So this equation gives us this strong inversion nonsaturation current for long

channels. And we will see how this gets multiplied

for short channels. Now before we start with small channel

modeling, I would like to make several comments that apply to the next several

lectures. First of all when you have a sharp

channel you are in a situation that shown here.

Where the source and the drain have approached each other.

And the depletion regions of the two extent over most of the channel.

So you cannot just neglect. Any more the presence of source and drain

over most of the channel. It is clear that this is a two

dimensional situation. You can no longer assume that the field

is vertical. It may be vertical in the other center of

the device, but it certainly will have a different direction away from that

center. So you need two dimensional analysis, if

the length is small. And likewise, if the W, if the width is

small, and the length is large, you're going to need a two dimensional analysis

too. Finally, if both L and W are small, as

you can expect, you need three dimensional analysis.

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So we start with velocity saturation. Because this is associated with drift

current, we will conentrate in, on strong inversion.

This is the drift velocity of electrons, versus electric field.

And as we have already discussed in our review of basics Initially the volocity

is proportional to the field, then it deviates, and eventually it saturates.

The inital low field behavior was described by an equation of this form.

Drift velocity's a constant times the horizontal component of the field.

And the constant of proportionality was the mobility.

The maximum velocity obtained, V D max, is shown here, and, when you extend this,

and you find the intersection with this line, this one is called the sub C, the

critical field. So we roughly assume That above the

critical field we have strong velocity saturation effects.

If you plug in E sub c in here you find the value for V sub d maximal and it is

mu sub e sub c. Now we assume nonsaturation.

So the entire channel is strongly inverted.

And, we know that the electric field is the derivative, with respect to pos,

horizontal position of the surface potential, or equivalently, of V C B, D V

C B D X. Now this equation, this behavior here Can

be modeled by a simple equation. I will show you the simplest equation

that is used. There are others that are more

complicated than this, but this will suffice to demonstrate the principles.

So V sub D is the maximum drift velocity times this fraction.

Now you can see that when the fields are very small, the whole fraction, because

of the numerator, becomes very small. And, V sub D Becomes approximately the

sum V d max over e c times e x. But v d max over e c is mu from this

equation. So the whole thing reduces to this

equation here, mu e x At large fields, though, this term is much larger than

one. You can neglect the one.

This cancels this and V sub d becomes Vd max.

And in between it does a decent job of modeling this curve.

Now if you plug in for E sub x, this here, dVcb dx, we find this.

So this is now our drift velocity. At any position x in the channel.

And in order to predict the current, we cannot use proportionality or velocity to

feel. Of course we have to use the entire

velocity eh, this velocity expression here.

So we're going to use the general expression which we derive back in our

review of basics. That the current.

Is the width of the channel times the negative of the versal layer charge per

unit area times the drift velocity. So now you can plug in this expression

here into v sub d of x. And you multiply both sides by the

denominator and you end up with this expression.

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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.

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So these are the main effects, that you see in devices that show velocity

saturation effects. Now let's take this equation, and let us

take an extreme case. So, I'm going to assume that I have a

very small L If I have a very small l this here would be dominant.

And I can neglect one in comparison with it.

So then I have this expression for very small l.

Now I notice that vds prime and vds prime here cancel out.

And so is the case with L, L cancels out. So now we see that we are left with this.

How did I get this? It is simply this expression.

Now, you can recognize UEc from what we said before.

As the saturated velocity, V D max. So this says that the drain current,

under strong velocity saturation effects, is W minus Q I times V D max.

This is equivalent to assuming, that you have velocity saturation throughout the

channel. That's exactly what you would have gotten

from the beginning, without going through all this algebra, by starting with the

equation I equals W times minus Q I times Times the velocity the drift velocity.

Notice that the velocity is fixed throughout the channel, under these

assumptions, and therefore, Q I prime, the inversion layer charge per unit area,

must be fixed to independent of Horizontal position x.

So that the current which is the product of Qi prime and velocity saturation, or

rather is proportional to this product, can be the same throughout the channel.

Right. So we get this result, but it is that in

this result. The current is independent of L.

It just cancelled out here. So the question is, do we leave it that?

Okay, is this just a mathematical coincidence, that the length cancelled?

Or is there something physical we can say about it?

So how is it possible that we have a current in a device That depends on W,

but doesn't depend on L. W is clear enough.

You double the width of the device, you have twice the current.

But why doesn't L appear there, physically, why does it?

Let's assume that we have maximum velocity throughout.

So throughout the channel, all carriers Operated travel at the maximum possible

drift velocity. We know that the current is the magnitude

of the charge, divided by the transit time.

We discussed this back in our review basis.

Now of course, the inversal layer, is proportional to the length And the

transit time is also proptional to the length.

So, you double the device you get twice the charge and also it takes twice for

the carriers to get off. So, of course they depend on[UNKNOWN]

cell sort that's the reason we don't have a dependence of L in the final current

acquisition. And you can actually see this in an even

simpler way. Think of a faucet and let's say I tell

you that the water flows at constant speed.

This corresponds to our maximum drift velocity.

If you know the speed that the water is coming out at.

Can you tell me how much per unit time flows out of the faucet?

And of course you can, right? And you know that to determine that you

didn't have to know the length of the pipe, right?

So just like you don't need to know the length of a pipe here, you don't need to

know the length of a transistor to find its current provided you know The speed

at which the carriers are traveling. So, this was our brief summary of

velocity saturation. In the next video, we will discuss

another short channel effect called channel length modulation.

And we will have a chance to look at the pinch off region in more detail.