0:00

We are now ready to derive expressions of the current in Weak Inversion.

And as we will see, this will be explicit functions of the terminal voltages.

Here is our model chart that will be shown for the last time.

At this point, we have finished with All-Region Models and with Strong

Inversion Models. And in this video, we will talk about

Weak Inversion Models, a Body-Referenced version and a Source-Referenced version.

So, let us look at the transistor in Weak Inversion.

In Weak Inversion, the inversion layer charge is much smaller than the depletion

region charge. In other words, the gate body voltage is

such that the inversion layer has just formed with very few electrons in it.

So, we can say that the magnitude of QI, the inversion layer charge is much

smaller than the magnitude of QB, the depletion region charge.

And that means that the region below the surface looks practically like a

depletion region and I can continue using the expressions I had developed for the

depletion region. One such expression was the surface

potential for a depletion region. We have derived the result you see here.

That goes back to the discussion of the two terminal MOS transistor.

As you can see you have explicitly the surface potential as a function of VGB.

And the so called VCB does not even appear here.

Why doesn't it appear? Because, even if you had a connection to

the channel by using a third terminal, C, for example the, that terminal doesn't

have a hand of what happens in the channel.

It has so few electrons so although by adding a third terminal and inviting the

voltage in it, you can vary drastically the population of electrons.

As long as you are in Weak Inversion, the electrons are so much fewer than the

ionized dopant atoms that you can neglect the charge of those electrons when you

form the relation between surface potential and charge.

So, you basically have just the depletion region governed by this equation.

So, QB, the depletion region charge is given by the equation we have seen

before, this one here. Ad if you replace psi s by psi sa given

by this expression, we have this expression for QB.

And as you can see, because psi s is the same throughout, the depletion region has

the same depth throughout, and QB is independent of x.

Now, psi s, the surface potential, is constant with x, as we can see here.

Psi s does not have x in it. It's a constant, it's the same thing you

would have gotten if you had the pure depletion region.

But psi s being constant with x implies that d psi s dx is equal to 0 and, and d

psi s dx is proportional to the electric field.

This is what would have given rise to a drift current.

If there is not potential gradient There is no drift current.

So, if we do have current in Weak Inversion, what is the only conclusion?

It must all be due to diffusion. So, in Weak Inversion then, we basically

have the fusion, practically no drift. And if we only have diffusion and the

weak conversion current is the same throughout the channel, it means that the

concentration gradient must be constant because the diffusion current is

proportional to concentration gradient or to the, the gradient of QI.

So, QI has to be a straight line. It starts from a value QI0 and goes to a

value QIL at the drain. So, we have a constant inversion layer

charge slope versus position. And that gives rise to diffusion curve.

4:05

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.

12:12

So practically, the solid line is the All-Region Model by itself.

And over here the leakage starts taking effect.

Now, the Weak Inverse region is an exponential as you remember,[UNKNOWN]

VGS. So, on the log current axis, it looks

like a straight line shown by the broken line here.

You can see that it does a very good job in Weak Inversion.

It fails outside the Weak Inversion region here, where we are in moderate

inversion, and it fails in the leakage region.

We have also plotted the Strong Inversion equation shown by the other broken line

here. It has this behavior because as VGS

approaches VT, the Strong Inversion equation predicts that the current

approaches 0. But 0 on the log axis would be at minus

infinity, that's why it goes near the vertical.

But in either case, the Strong Inversion equation we know is only valid in Strong

Inversion. It fails in moderate inversion, just like

the Weak Inversion equation does. Both of them fail in moderate inversion.

So, in the moderate inversion region, you can use All-Region models and in some

cases, you can use interpolation models, you can find some information on those in

the book. But one thing I want to stress is that

the Weak Inversion and the Strong Inversion regions are not adjacent.

There is a whole region in between called moderate inversion, which has to be

modeled correctly. Many circuits today operate in modern

inversion, so a model that does not do a good job in the moderate inversion region

is not acceptable today.