So, let's now develop the equations for this current.

I will start with a Body-Referenced Model.

If we take on All-Region Model which has two components, a drift and a diffusion

component, and you neglect the drift component, you end up with this simple

equation. We have shown this back when we discussed

the All -Region Model. QI0 is the value of QI next to the

source. QIL is the value of QI next to the drain.

And from our three-terminal MOS structure discussion, we have this rather long

equation. The inversion layer charge depends on psi

sa, which is the depletion, the depletion region surface potential.

[COUGH] You can see it here and there and it depends on VGB.

And it also depends on what used to be the bias between our third terminal and

the body. For us, in this case, VCB will be VSB

next to the source, and VCB will be VDB next to the drain.

So, I'm replacing these values for VCB and the result in the top equation.

We find this equation for the current where I hat of VGB is this expression.

Notice that I hat depends only on VGB. And then, you have two factors.

One depending on, excuse me, you have two terms, one depending on VSP and one

depending on VTB. In fact, this form is reminiscent of the

corresponding Ebers Moll equations and bipolar transistors.

And it is not surprising that this happens, because in bipolar transitions,

we also have currents due to diffusion. So now, we have a Body-Referred Model.

The gate voltage with, with respect to the body is here, the source voltage with

respect to the body is here, and the drain voltage with respect to the body is

here. And very simply now from that model, we

can also derive the Source-Referenced Model.

So again, this is the diffusion component of the complete on region model.

[COUGH] I can write it in this form by pulling out QI0, the inversion layer

charge per unit area next to the source, pull this out as a common factor.

[COUGH] And from the previous slide, we know that each of these two charges can

be written as that long expression times an exponent next to the, for QI next to

the drain, the exponent involves VDB, and for the inversion layer charge next to

the source, it involves VSB. Now, if you plug in these two things in

here, you get to the ratio of two exponentials which, of course, gives rise

to an exponential of the difference, VDB minus VSB divided by phi t.

But VDB minus VSP is simply VDS, the drain source voltage, so we have this.

So then, this equation here reduces to this one.

Now, we need an expression for QI0. Of course, being next to the source, the

inversion layer charge there will depend on VSB.

So, if you look at our material on the three terminal MOS structure, you find

how to represent QI0, and I will show you the final result by passing a lot of

algebra. it will involve VSP I will approximate

the surface potential there as we do for the three terminal MOS structure and the

resulting equation, as I've said after several steps, is this.

Now let's collect some terms together. All of this I will call IM Prime, you

will see the reason in a minute. So, we have that the drain source current

is W over L times IM prime times an exponential that involves VGS and VM, and

VM turns out to be the upper limit of Weak Inversion, in terms of VGS, you'll

see it in a minute graphically. And here, we have a term that includes

only VDS in terms of terminal voltages, nothing else.

So, let's write this expression again. This is what we have obtained.

[COUGH] Now, the interesting thing about this equation is that it neatly separates

the factor that gives rise to saturation, which is this one.

As VDS values, this factor values , we'll see how in a minute, and the rest of it

is something that depends only on VGS, not on VDS.

This is how things look like. Take a given VGS, for example,

corresponding to this curve. The first factor, this one times W over L

gives you this value for the current. And this factor here, initially is 0.

So, when VDS is equal to 0, this term is 1, 1 minus 1 is 0 so we're done at the

origin. And as you increase VDS, the exponent

becomes more and more negative and this becomes negligible and eventually all of

this factor reduces to just one. So eventually, it takes you to the

maximum saturation current. And the transition from the origin to the

saturation is about 3 phi t simply because once[UNKNOWN] is equal to 3 phi

t, then we have E to the minus 3 phi t over phi t, which is equal to minus 3.

It's a rather small number, so we have this behavior.

Now, if you choose a different value for VGS, the only thing that will change is

that the saturation current will be different.

For example, let's make VGS smaller, we are here.

But the way you approach saturation is still the same.

At above VGS equal to 3 phi t, you have reached saturation.

So, the saturation point, so although they're not clearly defined, can be

assumed to be about 3 phi t and you can see that one is above the other.

This is different from what we have seen in the Strong Inversion case.

[COUGH] Now, a few other things. VM, as I mentioned before is the value of

VGS that then shoot to the upper limit of Weak Inversion.

Above it, above VGS minus VS, which is represented, excuse me, above VGS equal

to VM represented by this curve, you are in moderate inversion.

So, VGS, VGS equal to VM is the maximum value of VGS for Weak Inversion and when

VGS goes below that, then VGS minus VM becomes negative and the exponent becomes

smaller. Now, when you make VGS equal to VM, this

entire thing here reduces to just IM prime times W over L and this is why this

top curve is IM prime W over L. So, what is IM prime?

It is the current at the top of Weak Inversion per W over L if you like.

So, let us know, plot our Weak Inversion region equation, and compare it to an

All-Region Model. The solid line here is the All-Region

Model, to which we have added leakage currents is that our currents that flow

because of reverse biased p-n junctions, but they are tiny.

Notice this is a logarithmic axis so it represents orders of magnitude of, of

current change. So, this current here, the leakage

current, is much, much lower than the currents over here.