Having developed a general analysis for the MOS Structure, it is now time to

simplify the results and obtain simple equations that we can use in the specific

case of strong inversion. Let us first, look at an MOS Structure.

And the plot of the charge density, the electric field, and the potential versus

y. I'm assuming that the gate charges, all

pile up in a very, very shallow regions. So, rho is very large, but the width of

this distribution here is very small. The same is true for the interface charge,

here. The electrons also exist in a very shallow

region. Actually called the charge sheet, shown

here, and it's negative. And then, you have a constant density of

charges, due to acceptor, atoms that are ionized, in the substrate like that.

It is easy to show that the area under these things, this is the charge per unit

area of Q'G prime, this is the effective interface charge per unit area of Q'0 zero

prime, this is Q'I prime. The inversion limit charge per unit area

and this is the depletion region charge per unit area, QB prime.

So now, you will recall that we can integrate the charge density, taking into

account the permittivity to produce the electric field, as we reviewed during our

background review last week. So, we have.

We start from a, a neutral point where there's no electric field, then we

integrate this charge so the field goes up, and then, we enter the oxide.

As you go from the gate to the oxide, you go from the material of a certain

permittivity, here the material of a different permittivity.

So, basic electrostatic show that there will be a jump in the field from here to

there due to this fact. And again, you may have to review one of

the appendices in the book if you are not current with such things or if you want,

you can take what I'm saying for granted. Now, in the oxide, there is no charge,

when you integrate, the field does not change.

Then, you reach the interface charge, so it goes up.

The integral of this charge here, shown here, goes up by that.

Then, this is another jump. Why?

Because you go from the oxide, which has a certain permittivity to the semiconductor

that has a different permittivity. So, the ratio of the two permittivities

will determine this jump over here. Then, you integrate the inverse or layer

charge, go down like this, you integrate the depletion region charge like this and

so on. Finally, you take the integral of this

with the minus sign and get the, surf the potential.

So, the potential, since this is positive, goes down here like this linearly because

this is a cons. And then, over here, this one, this linear

variation gives rise to a quadratic variation like this.

So, this actually is the potential versus distance y, and we have seen it before.

And we have said that the sum of oxide potential, surface potential, and contact

potential difference, is equal to VGB, the total gate body applied voltage.

Let's now continue with inversion. First of all, in inversion we're above

depletion, so I assume that Psi S is larger than the [unknown] potential.

And then, the general equation I had shown you for the total charge in the semi

conductors simplifies to this. We saw how it simplifies for the case of

deep depletion. Now, if you do the same thing and you

neglect very small terms in it, it turns out you get this and this is an exercise

that would, I would advice you to go through.

So now, we have this structure, the general charge in the semiconductor is

splitting to 2 charges, inverse layer charge and depletion region charge, QI and

QB. I'm going to assume that all of the

inversion layer charges are piled up in a very, very shallow region.

This is the so-called charge sheet approximation, as I have already

mentioned. And that means that this depletion region

is purely a depletion region without any other charge in it than charges that I

show here, immobile except or atoms. So, we can use the same approximation for

Q'B that we have used for PN junctions in the P type of in the P type of the PN

junction, we have derived this result, we can still use it for this case here.