PLEASE NOTE: This version of the course has been formed from an earlier version, which was actively run by the instructor and his teaching assistants. Some of what is mentioned in the video lectures and the accompanying material regarding logistics, book availability and method of grading may no longer be relevant to the present version. Neither the instructor nor the original teaching assistants are running this version of the course. There will be no certificate offered for this course. Learn how MOS transistors work, and how to model them. The understanding provided in this course is essential not only for device modelers, but also for designers of high-performance circuits.
Small-Dimension Effects – Channel Length Modulation
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來自 Columbia University 的課程
MOS 晶体管
120 個評分
從本節課中
Small-Dimension Effects 1
In this module we are dealing with phenomena that occur when the transistor dimensions are made small; this is important knowledge to have when one is dealing with modern devices.
與講師見面
Yannis TsividisCharles Batchelor Professor of Electrical Engineering
Fu Foundation School of Engineering and Applied Science
We are now ready to discuss another short channel effect called Channel Length
Modulation. In fact, this effect is present in long
channel devices also and it has been known for a very long time.
Perhaps it is the most ancient of the short channel effects.
finally we will be in a position to discuss.
In more detail than what have done in the past, what happens in the so-called
pinch-off region of the device. And how that affects the current in the
saturation region. Let us begin by looking the transistor at
the verge of such a region so we assume that Vds is equal to Vds prime.
And according to the very simplistic model we have used for storing inversion
up to now, the channel is just pinched off next to the drain, as you see here.
Now, we know that, as we go from source to drain, the, effective reverse bios
between the inversion layer and the body, if you want to look at inversion layer as
as an n type region. Increases, it's Vcb and it goes from Vsb
to Vdb in this pictures. And that the inversion layer becomes
lighter and lighter as you go from source to drain.
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.
