PLEASE NOTE: This version of the course has been formed from an earlier version, which was actively run by the instructor and his teaching assistants. Some of what is mentioned in the video lectures and the accompanying material regarding logistics, book availability and method of grading may no longer be relevant to the present version. Neither the instructor nor the original teaching assistants are running this version of the course. There will be no certificate offered for this course. Learn how MOS transistors work, and how to model them. The understanding provided in this course is essential not only for device modelers, but also for designers of high-performance circuits.
Small-Signal Modeling –Transconductance
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From the course by Columbia University
MOS Transistors
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From the lesson
Small-Signal Modeling I
The material for this module deals with small-signal modeling, and constitutes an important part of the description of MOS transistor behavior. It relates small changes in terminal voltages to the resulting small changes in currents. Small-signal modeling is key to analog circuit design, but not only; it is also used in high-performance digital circuit design (e.g., in designing memory sense amplifiers).
Meet the Instructors
Yannis TsividisCharles Batchelor Professor of Electrical Engineering
Fu Foundation School of Engineering and Applied Science
So far we have defined a number of small-signal conductance parameters.
In this video we will concentrate on what is perhaps the most important among all
of them: the transconductance. The transconductance describes how the
drain-source current changes when you change the gate-source voltage It is a
key parameter in analog applcations, for example, the voltage gain of an inverter
is proportional to a trans conductance. I will begin with deriving expressions
for the trans conductance, in terms of bias quantities in Strong Inversion.
Of course the expressions we will derive depend on the model we're starting with.
I will use the approximate source reference strong inversion model.
Which in non-saturation had given us this expression.
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.
And in saturation G M depends only on V G S.
We can see this dependence graphically as follows.
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.
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.
[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.
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.
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.
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.
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.
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