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When the voltage is supplied to a transistor vary very fast, and the

quasi-static operation assumption is no, not valid.

We have to allow for non-quasi-static operation.

This requires a different form of analysis as you will see.

We will make a transition from quasi static modeling to non-quasi static

modeling by using the multi segment idea I have presented in the previous video.

Remember that in order to help extend the validity of a quasi static model, we

split it into m segments. And the higher the speed were trying to

accommo to model, the larger the number of segments that we must split the device

into. We can take this idea to the limit and

allow m to become infinite, in which case the length of every subdevice here

becomes infinitesimal. And then we are led to non-quasi-static

model. Notice that in the general case, we can

expect that the current will be a function of position in the channel, and

of course the time, since the voltages are changing.

And the same we can say for the inversion layer charge per unit area.

It will depend on position and on time. Key to quasi, non-quasi-static modeling

is an equation called continuity equation, which expresses charge

conservation at the point in the channel. So let me go through a derivation of that

equation. Here we have a part of the channel, this

is the width w of the channel, this is a very small length delta x.

And we're going to consider the current that goes in, the current that goes out,

and the charge accumulation inside this section of the channel.

Now, let's call the current coming out i. The current coming in in general can be

different, let's say by amount delta i, because of the transit effects to the

channel. That means that a larger number of

carriers per unit time go in than come out, and therefore charge will accumulate

here. Let us say that in time delta t, the

amount of charge accumulation here is delta qi.

What is delta qi prime? It will be delta qi divided by the area

of the element as we look at it from above, which is W times delta x.

And what is delta qi? It is the charge that enters in time

delta t minus the charge that leaves in time delta t.

So it will be the current entering times delta t minus the current leaving times

delta t. And in this difference, you can see that

i delta t cancels out, and we end up with this equation.

Now if I rearrange the terms in this equation, I can write it in this form.

And now I can allow delta x to go to zero, in which case this becomes the

partial derivative of i with respect to x.

And I can allow delta t to go to zero, in which case this becomes the partial

derivative of qi prime with respect to t. So we end up with this equation, and this

is the continutiy equation. Again, this equation is a statement of

charge conservation. Now how do we apply this equation in the

non-quasi-static transient modeling of the device?

I will give a strong inversion example. We will write the equations we know from

DC analysis, and we will also write the corresponding equations in

non-quasi-static operation. First of all, the charge per unit area

according to the simplified source reference strong inversion model, is

given by this. You can look at the book.

But when we discuss strong inversion and the simplified model, you will see this

equation there. VFB and V0 plus gamma times this square

root is the local threshold voltage at the point in the channel.

Then the corresponding equation for non-quasi-static operation can be

obtained from this one, by replacing qi prime by q, small q capital I, meaning

the total charge. That is varying with respect to position

and time. Replace VGB by VGB of t, replace VCB by

VCB of x comma t, because now the channel body voltage will vary both with position

and time. And you do the same thing over here.

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Now for the DC analysis we have also written the drift current equation, at

the point in the channel which is mu WQ i prime dVCB dx.

I remind you that is the same derivative as[UNKNOWN] dx where[UNKNOWN] is the

surface potential. We can develop the corresponding equation

for non-quasti-static operation, it will look like this.

So we have replaced VCB by VCB of x comma t.

And because now VCB depends on two variables x and t, the derivative becomes

a partial derivative with respect to x. qi has been replaced by qi of x comma t,

and the transport current has been replaced by current of x comma t.

Now, in this analysis the final fact that was needed was that the current at any

point in the channel, which will called as the transcu transport current, is

equal to the drain source current. Now we can no longer make a corresponding

statement for non-quasi-static operation. Instead, we have to take into account

that the current values from point to point, and take charge conservation into

account, which is expressed by charge by the continuous equation.

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Which we derived on the previous slide, shown here.

So we have these three equations, which I repeat over here.

These are three differential equations, at least two of them are differential

equations. But this is a system of equations in

three unknowns. And the unknowns are the inversely

charged per unit area, the current at any point in the channel, and the channel

body voltage at any point in the channel. All three of these quantities both on

channel position and on time. To solve these equations, we need the

so-called initial and boundary conditions.

If you have studied differential equations, you know what these are, but I

will summarize. Let's say we're solving for the inversion

layer charge per unit area. The initial condition means at t equals

0, what is the value of the charge at each position in the channel?

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Let's see how we can apply what I just described in the case of this simple

example. So here we have a transistor with VDD, a

large voltage. So, when the transistor is turned on, we

can assume that the drain end of the channel is in saturation or it's in

pinch-off. And VG of t that had to be slow in the

past, is a step. It varies infinitely fast.

In the non-quasi-static analysis I just described, it's supposed to give you

valid results. So let's see how this is done.

Here is the device. This is the current, i of x comma t.

This is x equals zero at the channel end next to the source.

And this is x equals L at the channel end next to the drain.

Now qi prime at any position next at time t equals 0 is equal to 0, because the

channel before t equals 0 was empty. The device was off and there was no

inverse layer charge. So you can assume that right at t equals

0, no inversal layer charge has been formed yet.

So you can say the charge is everywhere 0 at t equals 0.

This is the initial condition. Now the boundary condition at the channel

next to the source, can be obtained from the DC equation I showed you on the

previous slide. You just replace the gate source voltage

by V. And the rest is valid.

Why is it we can use a DC equation in this case?

Because the channel end is right next to the source, there is immediate

communication between source and the source end of the channel.

There is no delay involved, so you can assume that the end of the channel next

to the source, responds infinitely fast to what happens at the source, or rather

to what happens between gate and source. Which really what happens is that the

voltage changes from zero to capital V. At the drain end of the channel, you get

another boundary condition. Which is that because we have pinch-off,

remember we're assuming saturation with VDD being large enough.

We always have pinch-off there, and therefore the channel charge at the drain

end of the channel is always 0, like that, and remains 0.

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So now with these conditions, we can solve the equations I showed you on the

previous slide, and I will show you the results.

First of all, if we know that current at any position x, then if we let x become

equal to L, we get the drain current. And as far as the source current is

concerned, it is the current at position x equals to 0.

But with a minus sign, because i s is defined as a current going into the

source whereas i goes in the opposite direction.

So here are the results. This is the circuit we're analyzing, VG

is a step function like this. And here I have plotted qi Prime,the

inverse layer charge per unit area versus position.

And here I have the corresponding drain current versus time.

So let's start from qi Prime. There is a bunch of curves here, each of

them is for a different time. As time increases, you go from one curve

to the next. So a time t equal to t sub a for example,

we see that the inversion layer charge has only reached up to point x sub a.

So you seeing that the charge is travelling from source to drain, and it

has only gotten to the point x equal x sub a at this point.

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Then a little later, a time t equal t sub b, the charge has reached to position x

sub b, then it continues at time t equal t sub d, it first reaches the drain.

And as soon as it reaches the drain, at time t equal t sub d, the drain current

starts flowing as you can see on the plot, on the right.

Then after a certain time, the distribution of the charge changes and

eventually a time t equal infinity in principle, it obtains it's distribution

as expected from this considerations. And by then the drain current has reached

asymptotically its DC value captial I sub d.

So now you can see that this behavior is consistent with a measurement result I

showed you earlier. Because you're allowing the device to

operate non-quasi-statically in your model you can predict all of these

effects. In particular, this wave, if you like, of

charge that gradually travels towards the drain, is very well predicted by a

non-quasi-static analysis like this. the current here, although asymptotically

goes twice at D, it has practically reached the value of I sub D for all

practical purposes. Where at some value tau 0, which is the

transit time of the device, and there is a delay, t sub d, that is about 40% of

the transit time. In analogue and RF circuits, the

non-quasi-static modeling capability is very important.