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Small-Dimension Effects - Charge Sharing
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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.
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Yannis TsividisCharles Batchelor Professor of Electrical Engineering
Fu Foundation School of Engineering and Applied Science
So far in our long-channel discussions, we had assumed that the source and drain
are far away from each other, so they can be, their, their effect can be neglected
over most of the channel. Now, we will allow them to approach each
other and we will discuss how we can take their presence into account in the I-V
equations for the MOS transistor. We will also consider the case where the
channel is not short, but rather it is narrow as opposed to being wide enough
for edge effects on the side to be effected.
So we'll discuss both short channel effects and narrow channel effects.
All of these effects collectively will go by the name of charge sharing.
And you will see the reasons why very shortly.
Let us begin with a reminder. We are talking about short channel
effects. More generally, small-dimension effects,
but in this case, we start with a short channel.
The drain and the source are assumed to be near each other, so we cannot neglect
their presence over most of the channel. I mentioned that you need
two-dimensional, even three-dimensional analysis, if you also have a small width
and these are very, very slow analysis. They have to be done numerically.
We set it up for fast computation. We need approximate analytical models.
And we try to save the long channel expressions by suitably modifying them to
take care of short channel, and in general, small-dimension effects.
I will repeat again, that often, the a h, the steps taken to arrive at such
expressions are not well-justified. empirical approximations are used widely,
and even the parameter values in the empirical expressions need to be
adjusted, again empirically, to make them agree with measurements.
So I will provide the sketch of such approaches for the case of short channel
device and then for narrow channel devices.
Let me first assume that VDS is very small.
A long channel device then would look like this, because VDS is very, very
small, then VSB and VDB are practically the same, the channel is very long.
So what happens around the source and around the drain can be neglected over
practically all of the channel, which is equivalent to saying, instead of this
situation, we have this situation. It, basically, what we did is derive long
channel expressions, assuming this picture over here.
The field is always practically vertical, and we use three terminal MOS structure
expressions, and we went through the derivations we've seen.
Now, if I have a short channel device, then things look like that.
Now, the field is two-dimensional, as we've said again and again.
It may be vertical near the center of the device, but certainly would not be
vertical near the source and near the drain.
Now, in the case of the long channel device over here, all of the depletion
region and all of the inversion layer, are there because of the field coming
from the positive charges from the gate. You can think of vertical field lines,
which terminate on charges in the channel, either depletion region charges
or inversion layer charges. All of the charges here appear, thanks to
the gate and the voltage we apply between the gate and the body or gate and the
source. But here, you have three structures close
to the channel. One is the gate, the other is the source,
and the other is the drain. All of these three structures are close
to the channel, so they all contribute to the charges in it.
The field lines emanting from the gate. And the field lines eminating from the
source, and the field lines eminating from the drain, all help to deplete the
region, and then invert the surface. So the gate has some help from the other
two regions, that are on the other channel.
So now, if you make the approximation, that things should look like that, it
follows you're going to make a big error in your calculations, whereas this here
was reasonable for the long tunnel device.
This approximation is not reasonable for this device.
If you insist on using long tunnel expressions, at least you have to modify
the values of perimeters in it to make this.
Approximate this situation in terms of the current it predicts.
So we have a, the so-called two-dimensional charge sharing, where the
that the job of depleting and inverting the channel is shared by three
structures, gate, source and drain. And, as a result of the help the gate
gets in depleting and inverting the channel, there is an increase in the
surface potential. So that means that even with a smaller
gate voltage than before, than what you had in the long channel case, the channel
can be inverted. Effectively, the surface potential
increases, and of course if you apply a larger drain voltage, now, you are
helping this even more. Just like when you apply a positive
voltage on the gate, you help invert the channel.
When you apply a more positive voltage on the drain, you have invert the channel,
basically for the same reason. So when you plot the I-V characteristics,
if you assume the long channel behavior like this, you would expect this type of
behavior. But instead, for a given VGS, what you
measure is a significantly larger current, like this.
Now, instead of saying that, this curve moves vertically like that.
It's more correct to say that this curve moves horizontally to the left.
So this expected long channel behavior, actually, should be modified and give you
this short channel behavior. You may remember, that with a very small
VDS, if you extrapolate this here, the quantity that you get at this point is
called the threshold voltage v sub t. If you do the same thing for the measured
quantity over here. You get a new quantity which we will
denote by V t hat. And it will be called the effective
threshold. So in such analysis the main task is to
find. The right value for the effective
threshold so that the measurements match your equations.
And I will show you an example of how this is done now.
First of all, if I ignore the effect of source and drain, in which case, we have
this situation as we showed before. We know that the threshold is given by
this expression. This goes back to our discussion of
strong inversion. This is the bulk charge per unit area.
This is the oxide capacitance per unit area.
Now, because I'm neglecting the source and drain here, the charged by unit area
is uniform, I'm assuming negligible, VDS remember, so there is basically no
voltage drop across the inversion layer. The charge is uniform and so I can
multiply by the gate area, both numerator and denominator.
And get this. This is now the total charge in the
depletion region, and this is the total oxide capacitance, no longer per unitary
or quantities. The total charge QB corresponds to the
shaded region over here. Now, QB over COX is equal to QB prime
over COX prime. And we have shown that the whole thing is
equal to this, where gamma is the body effect coefficient, and phi 0 is the pin
surface potential when V is equal to 0. Now, if you plug in this expression into
here, you find this result which we have seen again and again.
It is the extrapolated threshold voltage that is used in strong inversion
equations. So this, all of this is what results from
long channel considerations. Now, let us instead put the source and
drain there. And use this more realistic picture for
the device. As we said before already, several times,
the gate is now helped by the source and the drain and they all have field lines
terminating on the inversion layer charge on the depletion region They all help to
invert the channel. Now, the part of the depletion region
that is inverted, excuse me, is depleted because of the filled lines emanating
from the, from the gate is assumed to be part of the total discretion region under
the gate and it is shown here by this paraphrase, something that looks like a
trapezoid. And the charts inside here will be called
cube b hat. And one is simply there to indicate that
is the first analysis we will do. there will be a two and a three, because
we're going to a total of three analysis. Now, instead of this equation, people
then say, we will have this equation, the effective threshold.
Instead of having QB, we'll have QB hat. And this a very large and rather
arbitrary step that is taken. Do we have the right to take this and
then extend it to this? If you remember, this equation was
derived from this equation, because we said that QB prime is uniformly
distributed along the channels, so I have the right to multiply by the gate area
and get the total QB. But clearly, here now, QB is no longer
uniformly distributed along the channel, so I don't have the right to multiply by
the gate area and get QB. So going from here to there is a large
and arbitrary step. I will just stop at that, but it is a
step that is very often take. Now, QB hat over QB from the discussion I
just had a moment ago will be something less linear and 2B over COX can be
written like this and the extra factor is QB hat over QB.
And this now, the gamma times QB hat over QB is called the effective body effect
coefficient. I will call it gamma hat 1, and from this
development, it is less than gamma, less than what you would have expected, if you
ignore the presence of source and drain, so then when you apply this effective
gamma into the, the threshold expression here you get this.
And because gamma hat is less than gamma, this becomes less than VT.
So, it shows that the affected threshold is less than the long channel threshold.
So as you make then the length of a device smaller and smaller.
The threshold is predicted to decrease from this phenomenon.
There are other phenomenon that may predict the opposite effect.
For now, we're discussing chart sharing between gate, source, and drain.
More precisely, we're discussing how the job of depleting and inverting the
channel is shared between gate, source, and drain.
This chart sharing effect predicts that the effective Will be smaller, than the
long channel threshold. So how do we find Q B hat over Q B?
If you simplify things like this, you assume that around the source you have a
unit from depletion region, going like this.
Around the drain, you have a uniform depletion region going like this and in
the channel you have a uniform depletion region.
You find the points where these depletion region edges intersect here and there.
And you assume that this trapezoid represents the charge that is being
depleted by the gate and the rest of the charge are depleted by the source.
Now, if you didn't have the source and drain, then all of the, this rectangle
would represent a charge in the depletion region which has been depleted because of
the presence of the gate. So QB hat over QB is the ration of the
area of this trapezoid to the area of this rectangle and using simple geometry
you get this result. So now you can see that QB hat over QB,
which is the factor that modifies our gamma, our body effect coefficient
depends on the length of the device, as you expect, it depends on the junction,
depth, d sub j. And, by the way, the junction is assumed
to be a quarter circle over here with radius d sub j, the junction depth, and
it depends on d sub B, the depletion region depth.
And appropriately enough, if L is very large, this whole thing goes to an
[UNKNOWN] amount, and Q B hat over Q B becomes 1.
And then you end up with a long channel expression, but in the general case of
short channel devices, this whole thing is less than one.
So this is how it is done. There are various other approaches, but I
think you get the essence of those approaches with what they just covered.
And here is the plot of the affected threshold versus L.
VT0 stands for the threshold with 0 VSB. You can see that if L is large enough,
you get the long channel threshold, and if L becomes smaller and smaller, the
threshold goes down because of charge sharing effects.
We'll come back to whether modern devices obey this behavior or not shortly.
Now, you have seen that the presence of the depletion regions around source and
drain complicate matters, because they make your threshold small.
If they make it very small, then you cannot even turn the device off.
So if you remember, the depletion region depths depend on substrate concentration,
and the lighter the substrate concentration, the wider the depletion
regions are. So if you want to limit the extent of
depletion regions, you need to make the substrate doping high.
And this is done by using the so-called HALO implants, which help you preserve a
high doping concentration near the source and near the drain shown here by P plus.
But they don't do the same for the rest of the substrate, because if you had
everywhere, highly doped substrate, you'd have a very high body effect coefficient
which is undesirable. So the HALO implants are there to limit
the extent of the depletion regions around the source and drain and limit the
short channel effects. Now, a curious thing happens, as you
make. First of all, although, I show you P plus
being separated from P by a nice line here?
The fact is that you only gradually go from a heavy doping to a low doping, so
there is a distribution of concentration that goes down as we go away from the
source and away from the drain . Now, if you make the length of the device
smaller, this region will approach this region.
And so the high-doped, highly-doped regions kind of merge together and now
you get the substrate that is highly-doped everywhere.
And of course, if it is highly-doped everywhere, the gamma becomes larger, the
body effect coefficient becomes larger, and correspondingly the threshold becomes
larger. So then, as you decrease the length, the
threshold goes up, when you have HALO implants like this.
Now, if the length becomes very small, sometimes, you can also see the charge
sharing effect in which the threshold is supposed to go down as it's shown over
here. But, over the typical range of lengths
used, then, you will, in modern devices, you mostly see the threshold going up, as
the length goes down. Let us now, instead of making L small,
make W small, and talk about narrow-channel devices.
In this discussion, I will assume that the length is long, so that I can isolate
the effect of channel width of the device characteristics.
I will start with a popular shallow trench isolation or STI process, which we
have discussed. Here, I show you a device in
three-dimensions. This is W, the width of the channel, and
this dimension here is the length of the channel.
So the source is assumed to be to the left of the structure, not shown here.
And the drain is over here, again, it is not shown.
This is the depletion region. And you can see that the depletion region
is there, because the gate above it was sufficiently positive charges on it, has
field lines terminating on atoms on the depletion region, but although, the field
is strongest here, where the oxide is thin, there are also some field lines
from the sides from here and from here. So if you look at it from the side it
looks like that, you have field lines from the gate to the depletion region,
then you have extra field lines. These are called fringing field, which
terminate from the gates to the sides of the depletion region.
These extra field lines help deplete a larger portion of the channel than
before. So the fringing field corresponds to an
extra capacitance that help you deplete the channel, which means that it will
deplete more heavily and it will invert sooner.
In other words, for smaller values of the gate voltage.
So you define an effective threshold VT hat 2, because this is the second
analysis we're doing. And instead of QB over COX that we had
before, you have COX 2, which is a larger capacitance than before.
So I, I'm going to bypass the, the various steps here.
I'll show you the results, but they're entirely, the derivation is entirely
analogous to what I showed you for short channel devices.
We end up with a modified or effective body effect coefficient, and it turns out
that this body effect coefficient is the original wide channel body effect
coefficient times COX over COX hat, where COX hat is larger than COX.
And that's why this factor here is less than one and that means that gamma hat is
less than gamma, and therefore, the corresponding threshold is less than the
y channel threshold. It goes without saying that all of this
is not very well-justified. I'm just repeating what is done or has
been done when people discuss charge sharing for narrow channel devices.
COX over COX hat is easy to calculate if you know COX prime, the capacitance per
unit area. You multiply it by the gate area and this
is the numerator. In the denominator, you have the total
capacitance, which is the main gate capacitance plus the side capacitances
from the walls here. So you got to, to this picture here and
you can see that the side wall capacitance is proportional to the
length. So you take the capacitance per unit
length which I call CF double prime here, you multiply by the length.
And because you have two walls, one on the right and one on the left, you
multiply by 2. in practice CF double prime would be,
although, it can be estimated, it is often found from careful measurements.
The result here, of course, is that this is less than 1, which is why gamma hat is
less than gamma, and why VT hat is less than VT.
So as you make the W of the device, the width of the gate smaller and smaller,
the threshold goes down like this. I remind you, all of this is under the
assumption of an STI process. Now, let us quickly also look at what
happens, when you have a LOCOS process that, which looks like this.
We've covered this process. We've said it is not a very model
process, but it is still used for all the processes, and also, it is good for high
voltage processes today. So very briefly, let me recover that too.
Now, you will see that because of the so-called burst burt's beak region, you
not only have a thin oxide here, and suddenly a thick oxide, but you only
gradually go from thin to thick, from thin to thick.
So the depletion region doesn't suddenly stop as it did in the STI case, but as
you go to the sides, you have a smaller depletion region depth here and there.
So if you look at the structure from the left, it looks like that rather than
having only QB at just right below the gate, the extra field lines from the
sites of the gate help you deplete an extra region.
So QB hat, if by QB hat we mean all of the charge corresponding to this region
is larger than QB, which is what you would have assumed if you had, if you
neglected this effect. That will be only the channel, the charge
corresponding to this rectangle here. Going through a similar analysis as
before, we find an effective threshold where instead of QB, you have to QB hat,
and you can write this like this as before, where gamma hat is gamma times QB
hat over QB. From the discussion before you know that
QB hat is larger than QB. And therefore, you get a larger affected
gamma, which will give you a larger affected threshold than for a wide
channel device. How can we find the ratio of QB hat to
QB? If you make the simplification that these
regions here on the sides are quarter circles, it's easy from geometry to find
this result. Where DB is the depth of the depletion
region, and W is the width of the device, and this is of course larger than 1.
And it results in a gamma hat larger than gamma and VT hat larger than VT.
So now, when you plot VT versus W, as you make W smaller and smaller, the threshold
keeps going up. In other words, it goes in the opposite
direction from what you say for the STI process.
So we have seen the chart sharing effect on threshold, and not only on threshold,
but also on the body effect coefficient. We, describe the results that are based
on approximation and some arbitrary steps, nevertheless, these results have
predicted reasonably, the values of gamma and VT for short channel devices and
narrow-channel devices. Now, in practice today devices are so
complicated, that, and also, often, you have a small l and a small w in the same
case, and these analyses are not sufficient to predict the threshold of
those devices adequately. So, what is done is that empirical
expressions are used which depend have gamma and VT depending on l.
For short channel devices, they depend on w for narrow-channel devices and this
they depend on w times l for devices that are both narrow and short.
We will talk about such things more when we discuss models for cat.
In the next video we will talk about yet another short channel effects, the drain
induced barrier lowering effect or DIBL.
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