0:32

Using a strong inversion example. So here we have a transistor with body

shorted to shores the drain value always held constant at the large value.

So as soon as the transistor turns on, it is assumed to be in saturation.

And of varying voltage VG. Notice that only VG varies, the other

voltages are constant. Now you will recall from our earlier

discussion, that the drain current is the transport current that could have been

predicted from DC light. Equations, plus a charging component.

Now this charging component is dqD dt, where qd is the so-called drain charge in

the channel. And that can be written as dqD dvG, dvG

dt, so you can relate it to the rate of change of the gate voltage.

you may recall that there were three other terms in dqD dt.

But here they are 0 because those other terms don't have a dv dt.

The dv dt for them is 0 since these voltages are constant.

The only varying voltage is the gate voltage.

dqD dvG can be evaluated from the drain charge expression which we have shown in

the past. It's this one and when you differentiate

with respect to vG, you find this constant.

Now putting these results together if I assume that vG is a ramp, we get these

results. VG is assumed to be 0 at the time t

equals 0, then it goes up as a ramp, and then at t equal t3 becomes a constant

again. We are making a very simple approximation

that the device is practically off. Before vG reaches the threshold voltage

vt which happens at time t1 here. And then after that, as soon as it turns

on we assume it is in strong inversion. And remains a strong inversion

throughout. And because vDD is large, only saturation

needs to be considered. Now to find the drain current, we need to

know the transport current, and the transport current goes like this.

2:50

It goes like this because once the device turns on at t equal t1, we can apply the

simple square law. Saturation equation we know for strong

inversion. So it goes up until of course vG becomes

constant, and then the transport current becomes constant.

Now, the charging component will be dqD dvG which is a negative constant.

Times dvG dt, the rate of change of the input but that is also constant because

the input goes up as a straight line. So the product of the slope of this and

dqD dvG from here gives you a constant nevative number for iDA.

So, the charging component is 0 then becomes negative.

And a t3, it goes back to 0. Why?

Because now vG is not changing and therefore dvG dt becomes 0.

The total drain current is the transport component plus the charging component so

you add this and this together. And you end up with a wave form that goes

down and then goes up. And then another jump here and goes like

this. Now, how good is this result?

If you compare it to a measured results, obviously for a long time in device, you

find this behavior. So you can see that the quasi-static

model tries to predict this behavior. For example, it does predict that, the

current does not become positive until some time here which is valued over here.

The current starts becoming positive at t equals t2.

The only difference being that the measure of current is 0 before, before

t2. Whereas here, it is predicted to be

negative. So you see problems like this.

You also see another problem here. There is a sudden jump because of the

jump of the charging component as in a real device there is only a gradually

changing current like that. This result cannot be predicted by quasi

static models. One has to do non quasi static modeling

which it will do. But for the time being the question is,

if the speed is slow enough, so we're in quasi-static operation.

How good is the result? And again it depends how close you a,

you, you get to the limit of quasi-static operation.

And empirically it has been found that for, to get reasonable results from a

quasi-static model. You need to have a rise time in your

input wave form, the time it takes for the wave form to go from 0 to it's

maximum value. Which is at least 20 times the transit

time, and I remind you that the transit time is this.

This means that your waveforms have to be slow for quasi-static operation to be

valid. And of course because you have extrinsic

parasitics, which we will discuss shortly in a couple of videos.

You have extra capacity such which necessarily slow your waveforms.

In some cases in practice, speeds are low enough so that for the static model

income do an accurate and adequate job. however once the rise times become short,

was the static models fail. You can save the quasi-static model

statistic stand the reason of validity as follows.

Let's say you have a channel that is too long, so that this condition is violated,

the condition I just showed you on the previous slide.

Notice the L squared dependence. You make L long, eventually.

It would no longer be true that the transit time of a given wave form is

longer than 20 times the transit time. When this happens, you can split the

device into m sections, each of them now has a length L over m.

L over M so, in such a way that, this is smaller then the rise.

You can do it like this. This is the entire length of the channel.

You split it into m segments. And then your model leads segment has a

separate device. Each of these subdevices is assumed to

operate quasi-statically because each length is short enough to allow this to

be true. And then you model this in a very careful

fashion. For example, intermediate points here

correspond to intermediate points here are not associated with the source of the

drain. The source is over here and the drain is

over there. Therefore intermediate points do not

suffer from effects that we discuss when we discuss short channel effects, such as

charge sharing for example. So the only way to really model it is to

really understand what's happening inside the model and be very careful with how

you do it. This is done in some programs where the

number of segments used is typically from 2 to 5.

And it does typically extend the validity of the quasi-static models.

Let me now talk about the drain/source charges.

In saturation we have seen that the drain charges have been given by this

expression. And the source charge is given by this

expression. If you divide one by the other, you find

that the ratio of QD to QS is 40 to 60. So you associate 40% of the inversely

charged with the drain and 60% of it with the source.

There are other partitions that sometimes are used.

One is 50, 50 which is rather arbitrary. And the other is 0 to a 100.

In other words, no charges associated with a drain and no charges associated

with the source. Of course both of these partitions leave

much to be desired but this latter partition 0 to a 100.

Removes the negative part of the drain current plot that we saw a couple of

slides ago and is sometimes used. Sometimes you as a user of a model,

you're asked to choose your partition, so it's good to understand what these

partitions mean and what effect they have.

9:30

Some issues associated with charge modeling.

Sometimes wrong modeling principles are used, for example, time in variance

circuit principles are used. Whereas the actual capacitances of the

transistor, and we will be talking about capacitances in a couple of lectures.

Vary with time because they turn out to be capacitance that the voltage

dependent, the voltage is varied with time, the capacitance is varied with

time. So you can no longer use circuit concepts

from time-invariant circuit. Another problem is that sometimes

transcapacitance terms are omitted. For example the gate current is assumed

to be dqG dvG, dvG dt, for as we know that the correct expression contains

other terms. in addition to dqG dvG, which is the

second term over here we have dqG dvD and dqG dvB and dqG dvS.

These terms are called transcapacitance and they have to be there for correct

modeling. Then sometimes in the software,

implementation of a model, current to evaluate it, and the charges are found by

integrating the currents. Now if the currents have an error, then

when you integrate the error accumulates and that will give you wrong charges.

11:14

Some other effects very briefly. In short channels of course you have

velocity saturation that you need to take into account.

And you have something called the transient transport current.

Because the drain is so close to the channel that it even affects it under DC

conditions, and of course the transport current is also affected.

I will not discuss these effects, there are references to papers in the

literature discussing them if you're interested, and they're listed in the

book. There's one more effect I would like to

discuss. Charge pumping.

Consider a transistor that has been turned on.

It has a gate voltage that is large, and then suddenly the voltage goes down.

Now there is some capacitance between the gate and the inverse layer.

And when you suddenly push the potential in the gate down, because the capacitance

maintains momentarily a fixed voltage across it.

When the potential in the gate goes down, the potential in the invasion layer goes

down, becomes more negative than before. And that can be so negative that it can

turn on the parasitic effective junction if you like between the invasion layer

and the body. This is called charge pumping.

It is as difficult effect to model, and again a discussion of it can be found in

the references in the book. So in this video we talked about how we

can evaluate the transient response of resistors.

Using posistatic modelling . And we saw certain limitations of such

modeling. In the next video, we will talk about non

posistatic modeling. That can take you to much higher speeds.