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 – Capacitance Evaluation and Properties
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MOS Transistors
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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).
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Yannis TsividisCharles Batchelor Professor of Electrical Engineering
Fu Foundation School of Engineering and Applied Science
In the previous video, we saw how we define small signal capacitance
parameters. In this video, we will discuss how to
evaluate those parameters. We can now evaluate the capacitances by
using the definitions we have. Shown on the previous, in the previous
video. Those involve differentiations of charge
expressions with respect to voltages. I will skip the algebra here, I will just
show you some results and I will refer you for the details to the book.
For example, Cgs. After differentiating the gate charts
with respect to the source voltage and taking minus DQGDVS you end up with this
result coming from the approximate strong inversion model where Cox is the total
capacitance of the oxide. It's the area of the gate times the oxide
capacitance per unit area multiplied by this.
I remind you that Eta is the, degree on long saturation.
Is 1 when V D S is 0. And linearly reduces to 0, a at the
saturation point. And then stays there in the saturation
region. Similarly C B S, the body source
capacitance, turns out to be given by a very similar expression.
Only here you have Cbc, where Cbc is the average body channel capacitance.
It is the area of the channel multiplied by the average channel, body channel
capacitance per unit area and so on. Again we are skipping a lot of derivation
and results which are in the book. This is what capacitances versus VDS look
like, this is the nonsaturation region, this is saturation region.
We see that Cgs goes up and stays there. Cgd goes down, you will see y.
CBS and CBD do something similar to this too, only they're in general smaller.
As CGB increases from zero to some non-zero value in the saturation region.
And these, the broken lines and solid lines correspond to the.
The approximate model that we're discussing here and a very accurate
surface based, based model, which actually shows that the results from this
simple model are not too bad in the strong inversion.
Now I'm going to do, try to, make sense of these results by considering.
Two special cases. One is VDS equals zero.
In other words, we'll discuss the value of the capacitances here.
And another case is in saturation. So, let us first take VD equal VS, this
gives us this symmetric situation where we're applying equal VS and VD here.
Therefore. You have no potential drop across the
channel, the channel is uniform and so is the depletion region underneath it.
The total capacitance between the channel and the gate is C ox, the total oxide
capacitance. So, we remember that we had one
capacitance connected between source and drain in our small signal equivalent
model, which we have called CGS. And other capacitance between drain and
gate, which we have called CGD. And the sum of these two is suppose to
give you C ox. So, it's not surprising that by symmetry,
CGS is equal to CGD and each of them is C ox over 2.
You can do the same with the body capacitances, body source capacitance and
body drain capacitances, capacitance. Each of them is equal to half of the
total body channel capacitance. Finally, the simple model predicts the
gdb is 0 for this special case, and intuitively this can be seen as follows,
if you remember, cgb was defined as minus dqg, dvb.
In other words we were varying the voltage V b and we were looking at what
is the effect on the gate-charge. Now, here you have a strong inversion
layer throughout extending all the way from source to drain, right, they.
Everywhere you have strong inversion here because you are deep in non-saturation.
So this conducting layer forms, forms a shield between the body and the gate.
You might say, think of it momentarily as a sheet of metal, totally shielding the
gate from whatever happens in the body. So as you change vb, you will change the
charge qb and you may change the charge qi as well.
But this shield prevents the gate from seeing what happens below the immersion
layer. And therefore, qg is not expected to
change in this simple picture. Therefore, minus dqg dvb zero, which
results in Cgb equal to zero. Let's now go to the saturation region.
In the saturation region, we have a situation that looks like this.
V sub D is assumed to be much larger than Vs so that we reach saturation at this
point. We assume that the channel has been
pinched off here. This is an approximate picture that we
typically use in strong inversion. And the detailed derivations so that Cgs
is equal two thirds C ox. Now does it makes sense that Cgs is less
than Cox, lets see if we had inversion layer throughout, if we had.
The source, connected to a metal plate, going all the way up to here, but
isolated from the drain, because we have the pinch-off region.
Then, you, between the gate and that metal plate, you would have the entire C
ox. But, now, we don't have such a metal
plate, we have a conductive plate, over here, next to the source, but it becomes
less and less. Conducting as you go towards the drain.
And eventually you don't even have strong inversion here.
And that is an indication that the capacitance between the versal layer and
the gate is not all of see C ox but you expect it to be somewhat less.
And it turns out from the algebra. It is two thirds of C ox.
Now if CGD is assumed to be zero according to the algebra of the strong
inversion model, the algebra we applied to the strong inversion model to find
CGD, why is that? Because we have pinch of.
Attained at this drain end of the channels so a pinch of, if you increase V
sub d further, you don't affect the channel.
We're assuming this is a long channel device so, the length of the pinch of
region doesn't matter, right? So, you're already at pinch off, you
increase V sub d, you still remain at the pinch of the situation and the channel
doesn't change, QI stays the same, QB stays the same, and therefore QG also,
stays the same. So, CGD, who was, which was minus DQG DVD
since QG stays the same reasonably can be expected to be zero.
C bs turns out to be, two thirds of C bc, the bodied channeled capacitance for the
same reason as I explained before. It's expected to be less than C bc, and
the algebra gives you two thirds of C bc in fact.
And Cbd is expected to be 0 again for the same reason, because we have pinch-off
that made Cgd equal to 0. Finally, Cgb turns out to be this.
It's one third times the serious combination of C ox Cbc.
So, it is not zero anymore. In saturation, CGB is not zero.
Remember that in non-saturation, we had an inversal layer throughout that helped
us isolate the body from the gate, so changes in the body voltage.
Did not affect the gate charge but here we don't have a very conducting plate to
isolate the body from the gate anymore. Okay, it is very conducting over here but
over there you have hardly strongly inverted channel right.
So, the gate and the body concedes rather through direct field lines.
So, if you change the body voltage you can effect the gate charge directly so
it's not surprising that there is a gate body capacitance.
Now I would like to come to an important concept and that is the so called
Intrinsic Transition Frequency or Captive Frequency.
Let us take, a transistor. I'm only considering the intrinsic part
of it. I emphasize this.
I bias it with fixed voltages between source and body and drain source voltage.
And the fixed VGS DC voltage here on top of which I superimpose a tiny small
signal voltage. Epsilon sine omega T.
So this sine, sine is [UNKNOWN] varying, varying voltage with a [INAUDIBLE] radian
frequency of omega and the small amplitude epsilon.
Now we can develop the small signal equivalent of this situation.
We place the transistor by a small signal equivalent circuit.
We'll have already seen such a circuit. And we replace DC voltages by shorts.
Again, these results of small signal analysis are explained in electronics
circuits books. And we end up with this picture.
First of all, all of the DC voltage are replaced by shorts as I already
mentioned. I see that between gate and shorts I have
a CGS, between gate and body I have CGB. But because the bad is shortcut to source
for small signals the 2 capacitances are in parallel.
Capacitances that don't have current through them have been skipped here.
I already have the 2 capacitances that matter.
In part of the small signal equivalent circuit is that transconductance source
gmVgs which we have seen. And now we have an input small signal
voltage which will have which will cause some current to flow because we see the
capacitor here and there will be some current flowing in the drain terminal
also. The current flowing here will be GM VGS
but VGS is equal to Epsilon sine omega t, so this current is gm epsilon sin omega
t. Now what is the capacitive current
entering the gate terminal? We know that the current in a capacitor
is C dv dt where v is the voltage across the capacitor.
And the voltage across the capacitor is this.
So therefore ig is cgs plus cgb, the total capacitance connected between gate
and source, times the, time derivative of epsilon, sine omega T.
Which gives us this. So now we see the following.
The fact that we're applying the sine total voltage between gate and source
means that there will be a drain current of amplitude Gm epsilon, and there will
be a gate current of amplitude omega Cgs plus Cgb times epsilon.
This is the input current, and this is the output current.
So, in a sense you can think of the current gain between the two.
It will be the ratio of. The output current amplitude to the input
current amplitude. This is A sub i, the pick value of the
train current and divide it by the pick value of the gate current, and it is, as
I said, the ratio gm epsilon to omega cgs plus cgb epsilon, epsilon cancels out and
we get this expression for our current gain.
More precisely, it is the magnitude of the current gain.
and there is also a phase in volts, which we're not considering here.
Now, as the frequency of the input voltage is increased, the current gain
decreases. And if the frequency is made high enough.
The current gain becomes one. What is the frequency at which this
happens? You can set this equal to one, and solve
for the corresponding omega. I will denote this by omega ti, and it
turns out it is simply from here, Gm over CGS plus CGB.
This is called the Intrinsic Transition Frequency.
Again the intrinsic transition frequency is the frequency at which the magnitude
of the current gained becomes 1. It's an indication of how high you can go
in frequency while still expecting that you get more current here, more amplitude
of the current here. Than you get at the input.
For strong inversion, if you go through the expressions we have derived for gm
and for the capaccitances, you find for example that omega ti is approximately gm
over cgs because this is much smaller than cgs and it is.
Three halves of omega 0, for omega 0 turns out to be this quantity here.
Now again this involves only the intrinsic elements, we have not yet
included the extrinsic elements. When you have extrinsic elements you get
something similar but you have to also include the external, the extrinsic
capacitance of this expression. Now in weak inversion, we have a simpler
situation, you have very very few electrons and for that reason it's like
you don't have the bottom plate of the capacitance that we have been assuming.
So CGD, and CGS, and CBD and CBS. In other words the, all of these are
related to channel to gate or channel to body capacitances can be assumed to be
approximately zero. In fact.
You get larger capacitances because of the extrinsic part of the device, which
we have not yet discussed. So these are can be neglected.
On the other hand, the gate and the body see each other directly now because there
is no conductive plate to isolate one from the other.
And Cgb turns out to be non-zero. And in fact from what we had presented
for the two terminal Mos structure, you can find this result.
Cgb depends on VGB in this manner. So the gate sees the body directly and
the corresponding capacitance Cgb is non-zero.
The intrinsic cut off frequency omega Ti for the weak inversion region turns out
to be given by this, where IM is the peak current.
In weak inversion. Deep in depletion, you have approximately
the same situation as before because there you really don't have any inversion
layer charges to speak with. So all of this assumptions become, in
fact, more exact. In accumulation.
In deep accumulation, you have many holes here which help you form the bottom plate
of a capacitance. The top being the gate here.
The bottom being the accumulation layer. So you can expect that between them, you
have a capacitance corresponding to the entire oxide capacitance.
So the Cgb is approximately C ox. And that's the only intrinsic capacitance
we need to take into account. And if you take an accurate surface-based
potential model and you plug CGB versus VGB, then it goes like this.
So you can see that asymptotically this becomes approximately C ox deep in
accumulation. And asymptotically the exact Cgb approach
is the expression I showed you before shown here.
So if you are away from Vfb in other words deep in depletion over here, this
expression is pretty accurate. Now all-region models, the all region
models which we will not discuss because they are very, very involved
analytically, but you can actually derive this results either through analytic
expression or through numerical differentiation of charges.
Here we show capacitance versus vgs, so as you can increase vgs, you go from weak
inversion, over here, to moderate inversion, to eventually strong
inversion. Notice that, for a given drain source
voltage, we are initially in saturation, but if we raise vgs to a large value, to
large values, then the corresponding saturation voltage which is approximately
vgs minus vt, becomes so large that the vds value used to derive this plot is no
longer enough to have the device in saturation.
So, you enter non-saturation. So, you start from saturation.
Then, you go into non-saturation. And so in weak inversions Cgs, Cbs, Cgd
and Cbd are all zero, then they rise in moderate inversion and then they attain
the strong inversion values I showed you before.
Cgb has a considerable value in weak inversion and goes gradually down as we
have already explained. And now, if you have a short channel
device, not on the same process, and not with the same dimensions, this comes from
an entirely different device. So, please don't compare directly the
values, only the general shapes. For a short channel device, short channel
effects matter, charge-sharing matters, everything matters.
And things become different. You cannot really use the long channel
expressions to find the capacitances of a short channel device.
But I'm showing this to you to show you that the general shape, the general
behavior is still proximately the same as for the long channel device.
Now, if you include the extrinsic part, we have already discussed the extrinsic
part when we discussed the large signal dynamic behavior of the transistor, we
saw that we have resistance [UNKNOWN] with a source, the drain, the gate, and
the body. And we have extrinsic capacitances, the
gate source overlap infringing capacitance.
The gate drain capacitance, the body source capacitance, the body drain
capacitance, and the gate body extrinsic capacitance, the sub script E everywhere
indicates an extrinsic capacitance. So you add all of this to the small
signal equivalent circuit with the extrinsic part we have already seen.
And now you get the complete model for the device, that you can use up to rather
high frequencies for analysis and design purposes.
Now, the capacitances for example, inside here there is a capacitance between gate
and source which is, was called cgs. And, you can see now that in parallel
with it you have cgse, the extrinsic gate.
Source capacitance. So you can add one to the other and get
the total gate source capacitance. Similarly for the other capacitances
here. So the total capacitances are shown here.
this is the intrinsic plus extrinsic gate source capacitance.
This is the intrinsic plus extrinsic. Body source capacitance and so on and
they behave like this versus VGS. So we start with a weak inversion, now
you see that no capacitance is zero because you still have the extrinsic
capacitances. Even if you don't have a channel, you
have the capacitances between physical structures in the extrinsic part of the
transistor. So this capacitances are not zero.
And then once the intrinsic capacitances become significant, they add to it and
you end up with something like this. In this video, we discussed capacitances,
specifically how to evaluate capacitance values from the capacitance definitions
by differentiating charges with respect to voltages.
We also showed how we can add the extrinsic capacitors to the intrinsic
ones, and derive combined equivalent circuits.
And we also introduced the concept of the intrinsic cut off frequency.
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