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.
Complete All-Region Model
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來自 Columbia University 的課程
MOS 晶体管
117 個評分
從本節課中
The Long-Channel MOS Transistor – Part 1
This is the first module we are dealing with the complete transistor; much of this material, as well as next module's material, will form the core of what we will be covering in the rest of the course. We hope you will spend extra time studying to make sure you understand this material in depth. The total viewing time in each of these two modules is shorter, which should help.
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Yannis TsividisCharles Batchelor Professor of Electrical Engineering
Fu Foundation School of Engineering and Applied Science
In this video, we will derive models that are valid in all regions of inversion.
We will call them all region models. So let me show you again that chart that
describes our development in the next several lectures.
we will, we're going to start talking today about all region models and of
these in particular we will discuss the so called complete all region model which
later we will simplify. First let me review some of the basic
facts that we will need. From the two terminal MOS structure
discussion, you remember that there is a potential balance equation that simply
says that the externally applied gate body voltage is distributed among the
oxide potential, the surface potential, psi S, and the work function difference
potential, phi MS. We have also seen there is charge balance
equation that says that the gate charge plus the effective interface charge Q0,
which we assume is fixed. Plus the inversion layer charge Qi, plus
the depletion region charge Qb is equal to 0.
And we know that for the oxide potential and the gait charge we have a linear
relation that basically describes a 2 terminal
capacitor with capacitance C ox. I remind you that all[SOUND] quantities
that are shown as primed are per units of quantities but for simplicity when I talk
about them, I don't mention this fact. Finally, we have derived this relation
for the flat band voltage. It is phi MS minus Q0 over C ox.
Now, if you work with these equations and eliminate some of the unknowns, we very
easily get this equation for the inverse[UNKNOWN] charge.
And I suggest that you do that to get some practice with these equations.
And Qb. We had derived an expression for it, and
we had shown it is proportional to the square root of the surface potential cs.
Now, if you plug in this expression into Qb and collect terms, you find that you
have this. So this will be the form of the versa
layer charts that we will be using in this develop.
in addition, you will remember that when we had a 3 terminal structure, we had an
external terminal c, to which we applied bias VCB.
And that. The modified the inversion level.
And our development there, lead to the following equation.
Which relates the surface potential. To the externally applied gate body
voltage. And also to the externally applied
voltage between the third parallel C. And the body B.
The horizontal electric field is much smaller than the vertical one.
That means that the field is practically vertical in this device throughout the
channel. This assumption may be violated near the
source and near the drain, but we said those regions where it is violated are
very small compared to the channel length and for now we will neglect this fact.
We will assume that the electric field is everywhere practically vertical.
The variation of the horizontal component with x is much smaller than the variation
of the vertical component with y. These assumptions are referred to as
gradual channel approximation. And finally, I already covered this one,
channel length must be much larger than the depletion region widths around the
source and the drain. Now with this..
Said, we're ready to start the detailed derivation of the old region model.
We're going to concentrate on a very small chunk of the inversion layer, shown
here, between position x and position x plus delta x.
So let me now expand and show you only this part of the structure, only that
chunk that we're talking about. Looks like this.
Psi S of x is the surface potential, on the left, psi s of x plus delta x is the
surface potential on the right. And the difference between CS of x plus
delta x and CS of x is delta CS which is the surface potential difference applied
across this chunk. The width of the chunk is the same as the
width of the channel. And the electrons flow in this direction.
They're negatively charged. Which is equivalent therefore to a
positive current going this way. And the current at position x will be
called i of x. In the following, we will follow delta x
and delta psi x to approach 0. Now you remember that current can in
general be due to two causes, drift. And diffusion.
We will start by assuming that the current can have both of those components
at the same time. And then we will see that in particular
regions of inversion, one of the two dominates.
For now, let's allow both of them to be present.
We had derived equations for the drift current and the diffusion current
components back then. In the, in our background review.
And if you remember, or you can go back and check, this was the expression we had
derived for the drift component. This was the expression we derived for
diffusion. The drift depends on the gradient of the
potential, and the diffusion component depends on the gradient of the inversion
layer charge. Now, we want to find a way to find the
current, so I'm going to multiply both sides of this equation by dx and
integrate. Now this current i of x is the same
throughout because we are in this situation, we are assuming steady state,
so the current is the same throughout the channel.
So I'm going to replace i of x by IDS. And integrate from the source to the
drain. So now we have, because the variable of
integration here is x, we go from 0 to l, from the beginning of the channel to the
end of the channel. Because this one is, because the variable
of integration here is cs, the surface potential, we go from the surface
potential at the source, psi s 0, to the surface potential at the drain, psi s L.
And because in the last integral the variable of integration is Qi, we go from
Qi at the source to Qi at the drain. The symbols are self-explanatory here.
So now, to do this integration, notice that in the first integral, IDS is a
constant so I can move it outside the integral.
And then the integral of dx is x and you evaluate it from 0 to L.
So it becomes L itself. So the right hand side is I DS times L.
I'm going to divide both sides by L and I finally get this.
This now shows you how to find the drain source current, the current between drain
and source. From 2 integrals.
1 is due to drift, and the other is due to diffusion.
And a well known geometrical ratio, w over l, appears in front of everything.
Now, notice, that, at this integral, you have mu inside the integrals in both
cases. This mu, if we assume it is constant, we
can move it outside the integral and make things simpler.
So for now, I'm going to assume that mu is a constant.
I move it outside the integrals, and I finally have this form.
In the second one here, I have already done the integration.
It was integral of, of d dqi, which becomes just qi evaluated between drain
and source, so you take the difference like this.
This is, I remind you, the component we got due to drift and this is the
component that we have due to diffusion I would call the the first id as 1 and the
second id as 2. Now for Qi we can use the expression I
showed you when I reviewed our material from the two terminal structure
discussion a couple of slides ago. So if you now plug in Qi from this
expression into these integrals. And you will do the integration.
You find this result for IDS1 and this result for IDS2.
Now, notice the following thing. For the component due to drift, which is
IDS1. We have a term, CSL minus CSO.
Then we also have the difference of the squares of these quantities and also the
difference of the 3 half pies of these quantities.
Notice that it, is it to trays where these comes from, when you plug-in this
expressed on interval integral, integrated shears, we ended up with the
squares. And when you have the square root, which
is power to the 1 half, when you integrated, you ended up with the 3 half
powers. It will turn out later on that we can
simplify things if we avoid things like that, but for now let us keep them there
to have the full, complete origin model. Similarly, when you go to the IDS2, the
component due to diffusion, you have the difference between 1 half powers, so for
psi sl and psi so, again, this come because when you integrate it, the
equation shown in the previous slide, you end it up with a difference between two
QI's and these QI's from this equation. Had 1 half powers in them and this is
where we got this 1 later on again we will simply find matters.
Now, this is our all region model but something is missing because we don't
know yet how we haven't at least[UNKNOWN] explicitly how to evaluate CSO and CSL.
We'll do that in a minute. Note the symmetry in this.
If you reverse CSL and CSO, you're going to get the same magnitude of currents,
only they would be int he opposite directions so that means that if Ritter
change their own of source and drain, you're going to get the same current
flowing about in the opposite directions. Same in magnitude, opposite direction.
This surface potential-based, model is the basis in certain, of certain CAD
models. For example, HiSIM.
And then after we simplify this equation, it forms the basis of the PSP model.
We'll have more to say about these models later.
Okay, here is the device again, biased in the same way as before.
We have shown that the drain source current, it consists of 2 components.
One due to drift and one due to diffusion and each of them depends on surface
potential of the source and of the drain. So if this is the surface potential CSO.
And this is the surface potential CSL. In order the evaluate the current we need
to know what these are, unfortunately we don't know what CSO is, we know what vsb
is. We have to take the complete expression
we developed back then, when we discussed the three terminal structure.
And instead of what we called VCB back then.
For the source, we're going to VSP. And that will give us the surface
potential at the source. But notice that CSO is everywhere here.
And there's no way to solve explicitly for this.
Similary, at the drain, you get the same equation.
Only, now, you have VDB over there. And you have CSL, the surface potential
of the drain. Again.
This equation cannot be solved explicitly.
It is the set of this three equations together that forms our first general
all-region model. So here is now a comparison between the
Model we just developed. Which are the lines at two full
semiconductor equation solution. Which makes no, no approximations of the
kind I showed you. And you can see that the exact is the,
the matching is practially exact. The only difference between these 2 sets
of curves is that in one case, we have VSB equals 0, and here, we have VSB equal
2 volts. And you can see that the curves kind of
shrink. And the reason for that, of course, is
the body effect. Again, this is the equation we have
developed, 1 due to drift, another component due to diffusion.
So if you plot these things now, what do we get?
IDS1, the component due to drift, is shown here.
And IDS2, the component due to diffusion is shown there, and when you want them
together, you get IDS. Now, the horizontal axis is VGB.
The gate body voltage, the vertical axis is the current axis, it's logarithmic to
show several orders of magnitude along that axis, and what do we see?
We observe that IDS, the total current practically coincides with IDS1 in strong
inversion. So, that is because drift is the dominant
mechanism in strong inversion. On the other hand, in weak inversion,
IDS1 deviates drastically from IDS. And instead, IDS2 practically coincides
there with IDS and the reason is that in weak inversion as we will see, diffusion
happens to be dominant. Now, unfortunately, in the model
inversion region, which as I said before. a lot of circuits these days operate.
Both IDS1 and IDS2 are important. You cannot neglect either of them and
that makes modeling of the mother conversion difficult.
So we have seen then a simple way to derive a very general all region model.
In the next video we will see how we can simplify the model we have just
presented.
