This is again the definition for G M but the way I have replaced the equal sign by

an approximate sign, this is the way we would measure, GM, by applying a small

change in V G S and measuring the small change in the current and dividing the

two. Now if you take IDS versus VDS and use

VGS as a parameter. you can see that in this plot, I go from

one value of VGS to the next one by adding 0.2 volts to it.

So this is VGS1, this is VGS1 plus 0.2, add another 0.2, you get to this value of

VGS, another 0.2, you get to this value of VGS.

Now, from one curve to the next, you increase VGS by 0.2 volts.

This is not a small signal, okay? So, when we say here, delta VGS, we're

implying a tiny increase in delta VGS. But roughly to, I can make my point by

considering this large VGS [UNKNOWN]. Now, if you take any two of these curves,

the difference between the currents as you go vertically represents delta ids.

And the step of vgs, like you need to add in order to go from one curve to the

next, represents delta vgs. Because this step, delta vgs is always

fixed from one curve to the next. We can say that GM is proprotiona to the

change in the current. So now, let's go to a small value of VDS,

in non saturation. As we go up, we see that as we go from

one curve to the next, the Delta IDS that we get is fixed and it is independant of

the value of VGS If we go to a little to a somewhat larger value of vds, we are

here. Again we see that the delta i we get is

the same, independent of vds. Of course, the delta i is larger, and

this means that the trans-conductance here is larger Compared to the

transconductance here. But, in both cases, we see that as we go

from one VGS to the next, the delta I is not affected.

That is a confirmation of the fact that GM does not depend on VGS.

In non-saturation. Now, if you take a larger VDS and you go

to saturation, now you find that as you go from one curve to the next, the

spacing changes, which is a manifestation of the fact that GM depends on VGS.

But on the other hand it does not depend on VDS because if you go from this value

of VDS to a larger value of VDS, because the curves are flat the delta Is are the

same. So, this delta I is the same as this

delta I for example. So, the trans conductance which is

proportional to the delta I, for reasons we already explained, does not depend on

VDS. In such a race.

Let's now concentrate in the the, on the saturation region.

We have derived this result for the saturation region for the current, and we

can differentiate this, and get this equation For the transconductance.

Let's call this equation one, and this equation two.

Now, from these two equations, I can eliminate VGS minus VT.

For example, I can solve the first equation in terms of VGS minus VT.

Replace the result in the second equation and I get another expression of gm.

This gives me gm in terms of W over L mew Cox over alpha and the current IDS or I

can instead eliminate this quantity, W over L mew Cox.

So, solve the first equation for W over L mew Cox.

Replace what you find in here, and you find yet another expression for gm which

gives you the transconductance in terms of the current and the voltage.

So call these equations three and four. Notice then that we have three equivalent

expressions for the transconductance and strong inversion saturation.

The first one gives you the transconductance if you know.

The geometrical dimensions, W and L of your transistor, and mu C ox, and the

voltage. And of course the, your process will give

you VT and alpha. This one instead will give you the same

Gm if you know W over L mu C ox over alpha and the current.

Third one will give you Gm if you know the current and the voltage.

So you need to know 2 out of 3 quantities that I have just mentioned to find gm

either from this equation or from this or from this.

They are all equivalent. I would now like to, give you a puzzle

involving this 3 equivalent strong inversion expressions that I gave.

Here we have the current ideas versus VGS.

We know that we start with weak inversion, moderate inversion and strong

inversion and somewhere there is the value VGS equal VT.

And for that value we know that of course we do have some current, I will call it

capital I. Now, at VGS equal VT I would like to find

GM, the transconductor. From equation two on the previous slide,

we know the GM is given by this expression.