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Large-Signal Dynamic Operation – Evaluation of Charges

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МОП-транзисторы

Колумбийский университет

4.6 (оценок: 126) | Зарегистрировано учащихся: 29K

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.

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4.6 (оценок: 126)

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SS

Apr 12, 2017

I come from electrical engineering background. I found the material in the course very insightful and easy to follow.

RS

Jul 18, 2017

Extremely useful and well planned course! Must for every analog circuit designer...

Из урока

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.

Large-Signal Dynamic Operation – Evaluation of Charges9:23

Large-Signal Dynamic Operation – Transit Time5:22

Large-Signal Dynamic Operation – Transient Response Using QS Modeling13:01

Преподаватели

Yannis Tsividis
Charles Batchelor Professor of Electrical Engineering

Текст видео

We have seen that in order to predict the charging components of the currents of

the terminal currents in the device, we need to know the charges in it.

In this video, we will talk about how to evaluate those charges.

Let us take an example of a strong inversion.

The gate charge is the integral of the gate charge per unit area over the gate

area, as we have already shown. This gate charge per unit area is

typically given in terms of potentials in the channel, not in terms of the position

X in the channel. So, in order to evaluate this integral,

we need to relate potentials in the channel to position.

And this is easily done, because we know that in strong inversion we have a

current which is given by the drift current equation, which we have derived

in the past: mobility times channel width times the negative of the inversally

charged per unit area times the rate of change of the channel body voltage with

respect to position. I remind you, that this rate of change in

strong inversion is the same as the rate of change of the surface potential c s

with respect to position. From this now, if we rewrite it like

this, we have a simple way to relate the x to tvcv.

We plug it in here and we get this equation, and after we move out the

constant one of this, we end up with this.

This then is the equation that allows us to find the gate charge.

If we know the gate charge per unit area and the inversion layer charge per unit

area. And we typically do depending on the

model we are using. Similarly, we find a corresponding

equation for the bulk charge and for the inversion layer charge.

And because, we have split inversion layer charge into a drain charge and the

source charge, we also have the corresponding equations derived in the

same way for those charges. Now, for saturation, we do the same thing

that we did with the current. QI will be given by an expression in

terms of the terminal to terminal voltages, in this case it's VGB, VSB, and

VDB, as long as VDB is less than the pinch-off voltage, so we are in

non-saturation. And once we enter saturation, we simply

replace VDB by its maximum value non saturation being the pinch-off voltage

value V sub p. Let us now take an example, we take the

simplified source reference from inversion model.

I remind you that we have found a very simple way to write the current.

Now, by defining that parameter theta, in this theta was called the degree of non

saturation. It was one, when VDS is zero, and it went

to zero when VDS reaches saturation value VDS prime.

And then, it stays at 0 for VDS, larger than VDS prime, over here.

And we showed back then that we can write the drain source current in both

saturation and non saturation using this simple equation.

So, if you use this either to simplify the algebra, which by the way is quite

long, and it is described in the book, you find for example, that the inversion

layer charge, total inversion layer charge, is given by this expression

depending on the value of VDS. You find the corresponding value of ETA,

and from that you find the corresponding QI using this, and similarly for the

other charges. So, for this device, if you evaluate the

charges in this way and again, the details are in the book you find that the

charges versus VDS look like this. So let's take one of them for example,

let's take QB. QB becomes more and more negative, why is

that? Because as you increase the drain

potential, you increase the reverse bias, the effective reverse bias between the

channel and the body so that the depletion region widens and overall it

contains more negative charges than before.

That's why it becomes more and more negative.

Now, why does QI become less and less negative as you increase the drain

voltage? The reason is, as you increase the drain

voltage the reverse bias in the channel here makes the inversion less heavy here.

You go towards pinch-off, so the overall inversion layer charge seems to have less

and less negative charges here it becomes less and less negative, like this.

And then, you split QI into QD and QS and corresponding components are shown here.

Now, let us take special cases, I will take two special cases.

One, that VDS equals 0 and 1 in saturation, and see what happens there.

If VDS equals 0, we have this situation V sub d is equal to V sub s.

The channel is uniform and underneath it you have a uniform depletion region.

The calculations I showed you, give you these results.

QB is given by this and in fact, this is, can easily be shown to be equivalent to

just taking the total charge in the depletion region, per unit area and

multiplying by the unit, by the channel area w times l.

QI turns out to be given by this, and the parts of QI that have been called QD and

QS are equal because of this symmetry and each of them is half of this.

Now, for VDS larger than VDS prime, we are in saturation and we have this

situation where the channel is strongly inverted near the source and becomes less

and less heavily inverted until you reach pinch-off near the drain.

QI now is found to be less than before and the algebra shows you that there is a

factor of 2 3rd involved. Qd turns out to be this, and QS turns out

to be this. All of these results are derived after a

long algebra which is summarized in the book.

Note the following, Q sub d from here depends on the gate voltage.

Why is this the case? Because when you vary the gate voltage

you vary the inversion layer charge, and the two components of the inversion layer

charge, QS and QD also vary. So you can expect that QD in general will

depend on VG. However, QG does not depend on V sub d.

And why is that? Because once you're in the pinch off

region in saturation for a long channel device varying V of d does not have any

effect in the channel to the left of pinch off.

So, all of these charges in the channel stay the same correspondingly the gate

charge remains the same. So notice then, that the action of VG on

QD, and the action of VD on QG, are not reciprocal.

We would have a chance to discuss this phenomenon later on.

Recall that until further notice the channels of the devices we are

considering are assumed to be long. Now, I would like to mention that

sometimes you see the following statement, the pinch-off region isolates

the inversion layer from the drain in saturation, and therefore q sub d is

zero. This statement is not correct.

What happens is the following, when the gate voltage goes up, the magnitude of

the inversion layer charge must go up. And for this to happen, you have charging

components that in general come from both feed source and the drain.

And therefore, the drain chart, Q sub d, must be such that its rate of change with

respect to time is equal to the charging component of the drain.

And it so happens that Q sub d is non-zero in that case.

In the book, we'll find also a result for weak inversion, depletion, accumulation,

and all-region models. So, in fact, you can calculate the

charges for all-region models as well. The calculation is somewhat more involved

so I will leave it for you to read it in the book.

In this video, then we have shown how we can in principle evaluate the charges in

the device and we saw some specific results in strong inversion.

Now, that we know how to find the charges we will revisit the subject of transit

time and we will evaluate the transit time in strong inversion and weak

inversion.

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