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.
Large-Signal Dynamic Operation – Terminal Currents in QS Operation
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Из курса от партнера Columbia University
MOS Transistors
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Large-Signal Dynamic Operation
In this module there is a significant departure from what we have done up to this point: we will allow the terminal voltages of the transistor to vary with time. We will determine the resulting terminal currents, which will now include a “charging” component. We will do this both for moderate and for very high “speeds”. The transient response of circuits involves similar calculations, only done by computer for an ensemble of transistors and other devices.
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Yannis TsividisCharles Batchelor Professor of Electrical Engineering
Fu Foundation School of Engineering and Applied Science
In the previous video, we talked about what quasistatic operation means.
We will now assume that we are in such an operation and
we will discuss how to predict the terminal currents in it.
Here is the situation we have been discussing.
A long channel device with slowly varying terminal voltages so
that we can assume quasi-static operation.
I will consider one special case as an example I will assume that the source,
body and drain voltages are fixed and
only the gate voltage is varying, it is fixed then it slowly goes up.
Then it is fixed at the larger value, then it goes down to its previous value again
and then it goes up even more slowly to the same value as before.
Now as the gate voltage increases it will demand a larger
inversal layer chart's magnitude.
So you can expect that the magnitude of q i, the total inversal layer charts,
will go up.
And then when the gate voltage goes down, magnitude of q i will go down again.
Then this goes up slowly, q i will go up slowly.
Now as q i goes up, because of the charge
balance equation you can expect that q b will change and q g will change.
Since now q g is changing the rate of change of this charge will be
felt outside as a gate current.
So we have i g of t is d q g d t.
Likewise, because there's a change in q b.
The rate of change of this charge will be felt as a change in the body current.
I remind you that this, variation of qB is not due to
immobile ions moving but rather to holes that either cover or
deplete more acceptor ions.
What about the drain and source currents?
In order to reason about those two, think of it this way.
What is the drain current?
The drain current is a quantity that is proportional to the rate
at which electrons leave the channel through the drain.
And what is the source current?
The source current because it is defined in opposite direction for
iD is a negative quantity so I will instead talk about the absolute value.
The absolute value of i s is proportional to the rate at which electrons enter
the channel through the source.
So now if these two rates are the same, then
at any given time, as many electrons would be entering through the source
as living through the drain and if that were the case you could never change Q,
I, but here we want Q, I, to up which means that momentarily.
More electrons should be entering through the source than leaving through the drain.
And in fact, when you plot, i s and i d, this is what you observe.
This is the magnitude of i s, momentarily becomes larger, then
i d and the difference between the two is what allows q i to build up.
Now i sub t here is the transport current.
This is the current you will have obtained if you were using just
the transport card equation which was established for this here operation.
So, clearly, the two currents will be both different from i sub t.
Then eventually the currents merge together because here we reach,
this is steady state.
And when v g goes down, and magnitude of q i must go down momentarily,
i g becomes larger than the magnitude of is which means.
Momentarily, more electrons exit per unit time through the drain that come
into the channel through the source so that the channel can partially empty and
the magnitude of qI can decrease like that.
And then when VG goes up very, very slowly and qI goes up very, very slowly,
the same things will happen but now, because this change lasts for
a longer time, i sub d and the are closer to each other,
because the longer they are different, over a longer time, is enough to build
the extra q i required here.
Now as I mention already, let's take this part, the mitrodavis and
the mitrodavi different from the transport carbon I sub T,
which would have been obtained using DC type equations.
So this difference I will call I sub DA.
Here is the transport current, here is the actual drain current, and
the difference I will call I sub DA.
I can write that the total drain current is what I would have obtained from DC.
Derived equations plus a new component which is
what we would call the charging component of the drain.
And I can do the same for the source,
I can say that the source current is what it would have been from
these equations plus a charging component which we call i s a.
So we can see that in quasi-static operation
we are not implying that the drain and source currents are the same.
All we're saying when we say a device operates quasi-statically is that
the charges obey DC like laws in terns of the terminal voltages,
but the currents have to be different to allow for QI to change.
These equations will be used repeatedly in what follows.
Now, if you take the sum of the charging component at the drain and
the charging component at the source.
The sum of the two is the rate at which the inversion layer charge is changing.
So let me now take this equation, we need to satisfy this equation by the two
charging currents, I'll repeat the equation over here and
what I will do is I will define two fictitious charges, qd and
qs such that the rate of change of qd is equal to iDA and
the rate of change of qs is equal to iSA.
If you plug in these two into this equation at the top, you find this result.
So, I'm defining two fictitious charges, qd and qs, such
that their derivatives with respect to time, add up to dqi, dt.
The rate of change of the universal air charge.
Now if you define a function only through its derivative it is not defined uniquely,
of course.
But we can agree that we take the simplest possible functions for
q d and q s in such a way that their sum is equal to q i.
Okay.
So this will be called the train charge and
this will be called the source charge and they adapt to the inversal layer charge.
Notice that I will never use qd and qs by themselves, but,
rather, I would use them only to calculate the charging components of the drain
current and the source card, like this.
Now, to decide what QD and QS is is not a simple matter.
It is discussed in all the appendices of the book and
here I will only show you the solution.
Let us go back to DC operation.
Capital Q sub D is obtained by
integrating the inverse layer charge over the gate area, WDX, but
also waiting Qi prime, there saw that the closer we get to the drag,
the larger x becomes and the larger the waiting factor here becomes.
Go through this integration, you find some function f sub d
of the terminal voltages the same happens for Qs.
Only now the weighting factor becomes larger as you go towards the source,
where x goes towards 0.
And more and more of Qi prime is associated that with the source.
Going through such an integration gives you another
function F sub S over the terminal voltages.
Note that if you add this and
this, as you can see from here It adds up to the integral of q i prime.
D x times w, which is the same as the inversal layer charts so
we satisfy this equation.
Now for quasi-static operation the same.
Their functions will be used only.
I will replace VD, VG,
VB and VS by time varying quantities but
I will assume that otherwise, QD is given by the same function fD.
As before.
And the same for qs.
This is consistent with our assumption of quasi-static operation.
Now the way I showed you as to how to evaluate qs and
qd which of course I didn't go through and I if you're interested,
I would refer you to one of the appendices in the book.
This way is physically and
mathematically sound as you can find from the proof in the book.
It also agrees with current measurements under quasistatic conditions.
It agrees with results from perturbation techniques which is another way to
solve such problems and it agrees with non-quasi-static
models if the speed is reduced to such an extent that
the non-quasi-static operation in the limit becomes a quasi-static operation.
There are other approaches, one of which is to assume that all of the inverse
solar charges associated with the source and none with the drain.
And another approach is to assume rather arbitrarily that half of
the inverse solar charges associated with the source and half with the drain.
We will not be using these approaches, rather we will be using the what I
call the physically and mathematically sound approach that I just described.
So we have seen how under quasi-static operation we can evaluate the terminal
currents, from the rate of change of the charges in the device.
In the next video, we will concentrate on the charging components of those currents.
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