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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.

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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.

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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.

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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.