0:56

And I will differentiate it with respect to V G S as required by the Phoenician of

the Gate trans-conductors. in order to find a value for that

trans-conductance. Which we have called G M Gm is a partial

derivative of ids, with respect to vgs. And for brevity here, I do not show the

fact that you have to keep constant vds and vsp, but it is implied.

And from this differentiation, you get this expression for non-saturation.

9 saturation you can use the same expression but evaluate it at the top

point of the upper limit of non saturation.

It plays vidius by vidius prime as we've done for the karen also.

And expect, obtain this expression for gm in saturation.

Now, instead of vds prime, I can use the value derived for ds prime back when we

discussed that model. It was vds minus vt over alpha, and I had

mentioned there that alpha very often Is assumed to be 1 in very approximate work.

So now we have 1 expression for gm in nonsaturation and another expression for

gm in saturation observe that in nonsaturation gm depends only on VDS.

2:30

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.

7:51

And because that VGS equals VT. VGS minus VT becomes zero, this gives you

Zero for the trans-conductance. For equation number three on the previous

slide, which was this one, I see that if place IDS by the [UNKNOWN] that in I, I

find this expression for gm. And for on equation four on the previous

slide, which was gm equals 2 IDS over VGS minus VT.

I has a finite value but VGS is equal to VT, so the denominator goes to zero and

this expression gives me an infinite GM. So I have used what I claim to be three

[UNKNOWN] expressions for GM and the first gave me zero transconductance, the

second gave me. A non-0 transconductance but finite and

the third expression gave me infinite transconductance.

So, here's my puzzle to you. How can three equivalent expressions give

three different values for the transconductance?

In fact, not just different, but hugely different.

9:03

[BLANK_AUDIO] Now, if we include velocity saturation, we have shown that drain

source current is given approximately by this expression.

If you differentiate the expression move VGS, you'll find your threads conductance

in the presence of a loss of saturation. Mu is E, back when we discussed velocity

saturation, we found it was the maximum drift velocity, and therefore the thrust

conductance is given WC ox prime VD max, and does not depend on L.

When you have heavy velocity saturation. Similarly, for other short channel

effects, you write the corresponding expressions, you differentiate them, and

you find the transconductance. Sometimes, this is a very long procedure,

but I think we have already illustrated it with enough examples so far.

Let's now go to weak inversion. In weak inversion, we know that the drain

source current. Depends exponentially on VGS, in fact the

exponent is VGS over N phi T, where n is the so-called weak inverse and slope

factor, it's of the order of one point one, one point two, and so on.

And the, whatever was in front of the exponential, I will represent by a single

quantity which I will call I sub K, because it doesn't matter what this one

is for my purposes now. So now I want to find the

trans-conductance. I need to differentiate the current with

respect to VGS. And of course we get 1 over n phi t times

the rest of the expression like this. Now if you look at this part which is

equal to IDS from here and there gm is nothing but one over n IDS over phi t in

the weak inversion region which is very different from what we had seen in strong

inversion, strong inversion. If you knew the current, you didn't know

the transconductance. But here, you do.

For a given temperature phi T, is fixed, is known, if you know the current you

know the transconductance. Now, let's compare this to a bipolar

transistor. Which has a collector current in the

forward active regent, which is a constant times an exponential of the base

and neither volt is divided by the thermal voltage.

If we find the transconductance for this one, it would be the derivative of the

collector current with respect to vbe. It is given by this and this one here is

the same as i c. And therefore the transconductor is i c

over f t. So now notice that the bipolar transistor

and the my's transistor weak inversion behave in a similar manner.

But because n is always larger than 1, somewhat larger than 1.

For a given current, the transconductance you get for a, weakly invective MOS fit

is less than the corresponding one for the bipolar device.

11:59

So, this sometimes is called the Boltzmann limit, and in weak inversion,

we approach the Boltzmann limit, but we do not reach it.

For a gain, the same current, the two devices, the MOS transconductance is

less. Dante bipolar transistor transconductors.

Now I'm going to compare these two more generally, not only weak inversion but

also in moderate and strong inversion in a minute.

Just keep in mind that in weak inversion, if you know the current, you know the

transconductance and there's no dependence on W over L.

So here, now, I have plotted the log of GM versus the log of I.

This corresponds to the bipolar transistor, this is the [UNKNOWN] limit

that we already mentioned. And the broken line is the somewhat

smaller trans-conductors that you get for the [UNKNOWN] [UNKNOWN] inversor /g.

Now let's talk from very low currents. At very low currents you increase the

current, you get a higher trans-conductant, but eventually you

deviate from this. Broken line because the current becomes

so large that you are no longer in weak inversion, now you are in moderate

inversion and eventually you go to strong inversion.

13:15

So this particular curve is for W over L equal to 0.1.

If instead I do it for W over L equal to 1.

Then it will take a little larger current before I get out of weak inversion but

eventually I'm in moderate than in strong inversion.

So now the important facts that this graph summarizes are the following.

First of all in weak inversion the most fit has a transconductance that behaves

qualitatively like that of the bipolar transistor but it is somewhat smaller.

The w variable doesn't matter in weak inversion in terms of the

trans-conductance because all of the curves merge into one.

And in strong inversion, not only does trans-conductance very slower with a

current, but it also depends on W/L. This summarizes rather neatly the

different degrees of freedom that an analytic circuit designer.

Has when he or she's designing circuits using using bipolar devices, or using

[UNKNOWN] devices. [SOUND] Now, we finished with our

examples of the gate transconductance, we are now going to talk about body

transconductance. Gmb.

You may recall that gmb was the partial derivative of the drain source current

with respect to the body source voltage, keeping the other voltage fixed.

if you go through the algebra, you find that it is convenient to express the

result in terms of the gate trans-conductance times some point of of

b sub g. For example strong inversion b sub g.

Turns out to be this. I will leave the algebra to you, and to

the book. If you have very low VDS or VGS, we show

in the book that the subject, which is the ratio of the body trans-conductors to

the gate trans-conductance, is approximately this.

15:10

And from the body effect equation, you can show that this is nothing but the

rate of change of threshold with respect to the source body voltage.

So if you have the body effect low, the threshold voltage vs VSB is given by this

curve here. And this one, shows that b sub g is

nothing but the slope of the body-effect curve.

Now, if you play with this equation, a little, as it is done in the book, you

can also show that this approximately can be written like this, where t ox is the

thickness of the oxide. And dBm is the depth of the depletion

region. Es is the permittivity of the silicon,

and Eox is the permittivity of the oxide. So this equation intuitively makes sense,

because gmb over gm represent the fighting between body and gate to control

what happens in the channel. So the larger the g n b, the stronger is,

the, the stronger the effect of the body voltage on the channel current.

The stronger g n is the effect is the larger of the effect of the gate-voltage

on the channel current. So the ration of the two tells you

something about the relative strength of these two.

And this says that if Gmb over Gm is small, then the gate transconductance is

much larger than the body transconductance.

And so, it makes sense that this depends on tox because if tox is small, the gate

is very close to the channel, and it can keep very good control of it.

On the other hand if DBM is small then the edge of the depletion region is close

to the channel and the body can keep control of the channel's current to a

significant extent. And of course you have to weight this by

the 2 permittivities which effect the fields.

the effective fields. For example, the E ox will affect the

field from the gate to the channel, and so on.

By the way, in weak conversion, b sub g turns out to be.

Approximately n minus 1 and approximately this given by the same expression as this

one. Now the strong and weak inversion

expressions fail in moderate inversion, and this is shown by this plot.

So, here we plot VGS, and here we plot two things.

First of all Gm normalized to the maximum volume, so the peak value here is one.

17:45

And also we plot gm over i, normalized to the peak value of the same quantity.

So the peak value here is one. Now, in weak-inversion, gm over i is

approximately constant, as predicted by the expressions I showed you before.

It's not exactly constant. This is obtained by an all region model

so it does a more accurate job than what I showed you.

It's approximately constant as you see by the fact that this curve is flat.

But G M over I, it peaks in weak inversion and then it goes down.

In the strong inversion it is much smaller.

On the other hand, G M as a whole, is very small in weak inversion and it peaks

in strong inversion. So roughly constant G M, is a

characteristic of strong inversion. And constant G M over I is a

characteristic of weak inversion. Clearly, in moderate inversion, neither

of the two is valid. You don't have constant gm, and you don't

have constant gm over I. So, none of the expression I showed you

represents the device adequately in moderate inversion.

For that, you have to go to an all-region model and then, of course, the

expressions for GM and GNB become very complicated.

You can see the book for details. So in this video, we have discussed the

gate transconductance and the body transconductance.

We gave some examples of how these behave, depending on the bias.

The voltages in the device and the bias currents.

In the next video we will talk instead about the source drain small signal

conductance, and the combined conductance that one sees when one looks into the

drain of the transistor. That conductance will be called the small

signal output conductance.