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So far, we have assumed that the gate behaves like a perfect conductor.

And it would be close to that if a gate were made out of metal, but so far, we

have assumed that the gate is made out of polysilicon which is a semiconductor.

And because of that, it does not behave like a perfect conductor.

When the oxide is very thin and the electric fields are high the polysilicon

can actually deplete. And the effect of that on the IV

characteristics of the MOS transistors will be discussed in this video.

Now, we have seen how the body gets depleted and inverted.

It's a semiconductor after all, and we expect these effects by now.

we have also said that the gate is composed of polysilicon, which is

polycrystalline silicon, that's also semiconductor.

So we can expect that such effects can happen in the gate, too.

And this is what the this talk is about. So here, I show you a polysilicon gate,

not too heavily doped. It's assumed not to be heavily doped so

that I can demonstrate these effects for you.

So here, we have the depletion region atoms, and the inversion layer electrons.

You can think of this negative charges, as repelling the corresponding negative

electrons on the end and plus poly gate. And if, they leave behind positively

charged bound, donor atoms, shown here with the circle pluses.

So, just like this region is depleted, this region can get depleted.

And this has nothing to do with short and narrow channels, it can occur even in

long channel devices. But it will mostly occur when the field

in the oxide is strong, and this can mostly happen when the oxide is thin.

This is why this effect became worrisome as oxides were made thinner and thinner

in more advanced devices. So we have a poly region that includes a

depletion region consisting of donor atoms positively charged and body region

which is a p region consisting of negatively charge acceptor atoms and

electrons. Just like there is a surface potential cs

across this depletion region there is another potential psi poly across this

depletion region. And of course, there's an oxide

potential. Now, what is the effect of this situation

over here? If you have a bias VGB and around that,

you vary the gate body voltage by a small ac amount, you will see some capacitance

effect. But, the, rather than the charges on the

gate being right next to the oxide, and varying according to this ac voltage that

you have applied. Now, they will be varying by depleting or

covering donor atoms over here. In other words, the region where the

delta cues appear is now farther away from the body, and as an effect, you can

expect to see a decreased capacitance. And sure enough, when you plot the small

signal capacitance Cgb versus VGB, you find that at large gate voltage value is

where these effects happen, the total capacitance is smaller and keeps getting

smaller. Now, another effect that can happen is

that eventually when Vg is very large. The, the this region here becomes so deep

and psi poly becomes so large, that eventually, you invert the surface and

you start seeing here holes right next to the surface.

When this happens, you effectively have a capacitance which is equal to the oxide

capacitance, because you have the delta cues are being developed due to the

presences of halls right above the oxide and the minus delta cues due to the

presence of electrons right below the oxide.

So eventually, this is expected to go up and reach C oxide, but it is shown, this

is not shown here. Now, let's see how we would analyze this

structure. First of all, the potential balance

equation says that VGB shared between psi poly, psi OX, psi s, and psi MS, the work

function potential difference. As we have seen before, this is

practically the same equation we've been using, except for the presence of psi

poly. the total charge balance equation must

include all charges. So it's the charge in the poly, here,

shown because of depletion region charges.

As the interface, effective interface charge Q0 plus inversion layer charge

plus depletion layer charge is equal to 0.

For the oxide capacitance, we have this relation which we've seen again and

again. The charge above the oxide is psi OX

times psi OX. [UNKNOWN] psi OX being the oxide

potential. And this charge happens to be too be q

poly in this case. And just like we had found the charge in

the depletion region in the p body. As a function of psi S, a same type of

equation is valid for these charges here. It looks exactly the same.

Instead of QB you have Q poly, and instead of psi S you have psi poly.

In addition, we have some relation between inversion layer charges and psi

S. We have seen such relations before.

And we have a relation between QB and psi S.

Again, we have same [UNKNOWN] before. So, you see here we have a total of six

equations and six unknowns. So we can in principle, solve.

Now, let me write a few more equations that we will need.

The depth of the depletion region in the poly is governed by an equation of the

same form as the depth of the depletion region in the body.

You just replace DB, we had [UNKNOWN] D sub B here, by D poly, and psi S by psi

poly. And, NA for the substraight by N poly,

the concentration of donor atoms in the poly, and you get the equivalent equation

to here. And if you differentiate this expression

with it [UNKNOWN] to psi poly you find the corresponding capacitance of the

depletion region in the poly, which is given by the same type of equation as we

have seen before. So if you replace, in the corresponding

depletion region, capacitance relation, NA by N poly and psi S by psi poly, you

would get this equation here. Now, let me take weak inversion, which is

very simple to cover in the presence of poly depletion effects.

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First of all, I can neglect QI in weak inversion, and for simplicity, I will

also neglect the effective interface charge.

So effectively, we only have two charges, Q poly and QB, and they must balance each

other out. The, this result and what I'm about to

show you below, that can be derived with very simple algebra working with the

equations I showed you on the previous slide.

I will only show you the result. Very easily, you can get that psi poly,

the potential across the poly depletion region is related to psi s, the potential

across the p-type depletion region by this very simple ratio.

this is consistent, let's consider an extreme case where N poly is extremely

high. If N poly is extremely high, the

depletion region depth will be negligible and it's as if you have a perfect

conductor for a gate and then indeed you find from this here that psi poly goes to

negligible level. So you're back to what we've been

assuming so far, that there is no potential drop vertically across the

gate. You also find that psi poly the

capacitance of the depletion region on the poly gate is related to the body

capacitance by this expression. And you find that the capacitance

combination of the poly capacitance and the oxide capacitance, of course,

equivalent. They are in series for reasons we have

explained when we, back when we discussed the capacitance of the two terminal

structures. So we can write this expression.

And from here, you can derive and it is done in the book.

If you're interested, please take a look. The extra capacitance you get in series

where the oxide capacitance is equivalent to an increase in your oxide thickness.

So you can assume that, the effect on, that this has, is the same as the effect

that you would have if you didn't have poly decreasion but depletion, but

instead the oxide were thinner, excuse me, thicker, by an amount delta t OX,

which is given by this. The OX is the permittivity of the oxide,

ES is the permittivity of the silicon here.

And d poly is the depth Of the polydepletion region.

Finally, in weak inversion if you remember there is an n factor which

appears and affects the slope of the log i versus VGS in weak inversion.

And it's easy to find that there is an effective increase in the weak inversion

slope factor n by this amount. In other words, the slope deteriorates

because n decreases by this amount. And if a poly is very large, then you

have a highly conducting gate. It's as if you don't have a poly

depletion in that case, and that then goes to negligible.

Now, let's briefly talk about strong inversion as well.

There is a very simple way to take into account poly depletion effects in the

case of strong inversion. Assuming we are near the threshold, we

are somewhere above threshold, so we can be in the, in, in strong inversion, but

not much above it. QiIfrom our strong inversion analysis

before can be expressed as minus Cx times VGB minus the threshold.

Now, instead of this threshold, we need an effective threshold.

It has to be different because of the presence of the poly.

The question is how can I find this threshold?

If you go around this way, you find that psi poly, which is positive here with

respect to this point is encountered with a negative sign.

So, it is minus psi poly plus VGB. In other words, what psi poly does is

opposes VGB. It makes it more difficult for inversion

to happen. So you can effectively think psi poly as

the voltage source in series with VGB in the opposite direction.

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That makes the threshold larger by psi poly.

And, we're going to use for psi poly, the expression we found for the weak

inversion region, because we assume that we're not in really heavily strong

inversion, just like strong inversion near the bottom of the strong inversion

region. So, I'm going to assume for psi poly the

same value that we found for weak inversion, which was this one.

And for the surface potential, I will assume the peak value that we know we

have in strong inversion. So this is now an estimate of the new

threshold, this is by how much the threshold increases in strong inversion

as a very rough approximation. There are better ways in doing this, in

fact, you can write a non-region model for including polidant depletion.

You end up with a set of equations that must be solved implicitly, but typically,

approximations are made in order to derive explicit equations.

there is a short summary of how this is done in the book.

Now, as the oxides are made thinner, this problem becomes worse and worse, because

the higher fields in the oxide mean that not only you have a significant poly

depletion region, but the delta TOX, the, the effective increase in the oxide that

takes place there becomes a significant fraction of TOX, the oxide thickness.

So to avoid this, you either need an extremely heavily doped n poly, which is

not easy to do or you can go to a metal gate, which is what some modern processes

are doing. So in this video we talked about the

effect of poly depletion. This is an effect that can even affect

long channel devices. And it is due to the fact that the

negative charges on the body can repel the electrons on the n-type poly gate.

And reveal a depletion region, which helps makes the effective oxide thickness

larger, which of course, is a negative effect.

Because the current, if you remember, it is W over L mu C OX.

And if you make the effective oxide thickness larger, C OX becomes smaller.

So, for the same gate voltage, we have less of a current drive.

In the next video, we will see a short summary of the so-called quantum

mechanical effects.