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 – Transient Response Using QS Modeling
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MOS Transistors
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À partir de la leçon
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.
Rencontrer les enseignants
Yannis TsividisCharles Batchelor Professor of Electrical Engineering
Fu Foundation School of Engineering and Applied Science
At this point, we have discussed quasi-static operation in detail.
And we are ready to apply what we have discussed to the complete transient
simulation of transistors. Provided of course the speed of the
voltage is applied to them is not to high.
So that equals a static operation assumption is valid.
We will discuss the transom response evaluation using quasi static models.
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.
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.
I would say that for speeds that, in which the quasi-static model is valid, 40
to 60 makes sense. or, if that doesn't make sense, it means
that probably the speed is so high that you're no longer in quasi-static
operation and you have to use non-quasi-static operation.
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.
And finally, in some models there is a non-continuity in, in charge voltage
relations. Which results in discontinued capacitors.
We will discuss this effect when we talk about capacitors.
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.
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