Good morning, all.
Let's get going. So I guess you had your quiz
review yesterday. I hope you guys didn't beat up
on (name of TA) and who else was it?
(Name of TA) too much. As you know,
the quiz is tomorrow. And unfortunately MIT couldn't
give us one big room so we are broken up into three rooms.
And you will go to your room based on the first letter of
your last name. OK, so today we shall cover a
topic called "Large Signal Analysis".
So in the last couple lectures we looked at one dependent
sources abstractly, and then we looked at an
amplifier built using a practical dependent source
called the MOSFET. Now, the MOSFET had to be
operated in a given region of its operation in order to behave
like a current source. And while it behaved like a
current source you would get amplification or a FET
amplifier. So that was in the past two
lectures. What you are going to do today
is called large signal analysis, and this is a loaded term.
So large signal analysis means something very specific in our
business, and I will describe to what that is.
This analysis involves looking at a circuit containing,
for example, a MOSFET and figuring out how
to get that device to operate in a way that the MOSFET was always
in saturation. So you had to figure out,
based on parameters that you could control,
how to establish those parameters so that the circuit
operated in a way that the MOSFET was always in saturation.
So large signal analysis involves that.
And although the examples we use, use the MOSFET,
the same kind of analysis can apply to any other device.
Remember, your MOSFET is a primitive element that we use as
an example in this course. There are other primitive
elements that you can use. The course notes,
for example, discusses a couple other
devices. One is the "bipolar junction
transistor", the BJT, and works through a complete
example from start to finish involving a circuit containing a
bipolar junction transistor. And you can do a large signal
analysis of that device as well. It turns out that you need to
operate that device in an interesting region of its
operating space, and so you can conduct a large
signal analysis of a circuit containing that device and
figure out how best to operate that circuit.
So that is large signal analysis, and we will do an
example and explain how this is done using an example today.
So to quickly review where we have been so far,
we looked at this little structure here,
our MOSFET amplifier.
Notice that when I write a voltage at a node,
that's a short form for saying I am looking at the voltage
between the ground node and the node at which the voltage is
written down. So VO here and VI applied here.
This is a very, very common circuit that we
use. To emphasize one more point,
in general, in the kind of circuits we look at both in this
course and in real life, there are a few patterns that
we use very commonly that keep repeating all the time.
Very often you don't have to look at every possible
permutation and combination of how things could be connected.
This sort of connecting thing is very, very,
very common. And you will see a lot of this
pattern. And we do the equivalent
circuit for this. In the equivalent circuit we
replace the MOSFET with a dependent source provided this
operated in the saturation region.
So I will just say while operating under saturation the
equivalent circuit would look like this, VO,
VI.
And IDS for the dependent source was given by K/2
(VI-VT)^2. So this was an amplifier.
Here was the equivalent circuit while this device was in
saturation. And to operate in saturation,
I said that certain properties need to be true for the MOSFET.
And there are two properties that need to be true for this to
be operating in saturation. One is that its gate to source
voltage needs to be greater than VT, so VGS for the MOSFET should
be greater than VT. And the second one was that the
output voltage needed to be greater than the input voltage
minus one threshold drop. And this was the same as VDS
for the MOSFET, this was the same as VGS for
the MOSFET. So what are we really saying
here? What we are saying is that
look, we built this circuit using a MOSFET,
and it is up to us as engineers to choose its operating points
in a way that these two properties hold.
For example, to make the first condition
true, I can discipline myself to operate such that VI is always
greater than VT. Similarly, I can choose VS,
RL and VI in a way that this condition is true,
which says that the drain to source voltage across my MOSFET
drain and source should be greater than VI minus VT.
As an example, if VI was 2 volts and VT was,
say, 1 volt, then what I am saying is that
VO should be greater than or equal to 2 minus 1 or 1 volt.
So I need to keep this high, 2, 3, 4, 5, whatever,
a high voltage so that this guy stays in saturation.
The relevant readings for the material that we are going to
cover in the course notes are in 7.5.1 and 7.6.
So that is pretty much a review of where we were.
We said we could build an amplifier.
Its equivalent circuit was shown on the right.
And, provided that, I discipline myself to operate
in the saturation region or to have the MOSFET operating in the
saturation region, then this would work like an
amplifier and all would be good with the world.
So today --
-- we look at large signal analysis of a circuit.
And an example would be this circuit up here containing a
MOSFET. And, again, as I mentioned
earlier, a large signal analysis is a loaded term in 6.002,
or for that matter in circuits. And large signal analysis
involves two steps.
The first step involves writing down the transfer function of
your little circuit. In our case,
VO is the output, VI is the input,
so involves writing down VO versus VI.
Simply write down the transfer function.
In other words, the relationship between the
output and the input for that circuit.
And, in our case, again, we've disciplined
ourselves to adhere to the "saturation discipline".
And the second part of large signal analysis is to find out
the valid input operating range.
Find out for the given circuit parameters, let's say I apply a
VS and I use some value of RL and I use a given MOSFET,
which has a given value of VT. The question then is that what
is a valid set of input voltages that would operate the circuit
in a way that I would be in saturation.
And so find out the valid input range, and this would give me a
corresponding output range --
-- for saturation operation of the MOSFET.
That is what we will dwell on in the lecture today.
So what we are saying here is that if I am careful with how I
apply VI for a given value of RL and VS and for a given choice of
my MOS transistor then I can stay within saturation provided
I select my input voltages carefully.
And the analysis that we will go through today will figure out
what that range of input voltages is.
And, again, I will use this as a motivating example,
the MOSFET amplifier. But in general large signal
analysis would apply to any other circuit as well.
For example, in recitation you may learn
about other circuits containing a MOSFET.
And you can do a large signal analysis of other circuits
containing a MOSFET or you might learn about some other devices
like the bipolar junction transistor, and you could do the
same kind of analysis for that device.
So remember that the MOSFET amplifier here is an example.
I will be using that as a driving example to explain large
signal analysis. So the first step,
as I mentioned earlier, is to get the VO versus VI.
And in general for some circuit that you build the output will
not even be a voltage. There are certain circuits
where the output might be some kind of a current.
Let's say I am building some kind of a circuit where I would
like the output current or the current through some edge of the
circuit to depend on some input. In that case the transfer
function would be the output current versus VI.
And if I had an input current here it would be output current
versus input current, you know, whatever the given
problem tells you. So this is under the saturation
discipline. And I will not rederive this
for you. You can apply a good old
technique like the analytical method.
Or you can use the graphical method to get the appropriate
answer here. I wanted to point out in a
quick aside that why do we care about graphical analysis?
Once you have the analytical method, why do you care about
the graphical method? And a student asked me a
question after lecture last Thursday, and it occurred to me
that it's not obvious why you need the graphical method.
So it turns out that often times you do not have an
equation describing the device. So let's say,
for example, I am a manufacturer.
Let's say I am AMD. As AMD I sit down and my
semiconductor division builds a MOSFET.
And when you build a MOSFET your experiments and your
fabrication division often times doesn't give you an equation
with the MOSFET. They build something and then
you look at it and you experiment with it.
You apply various input voltages and you measure
currents and output voltages and so on.
And so what you end up getting is a graph that describes the
behavior of the MOSFET. And you have seen this in your
lab as well, your 2N7000 or was it 2000?
So your 2N7000,
the MOSFET you use in the lab also gives you a data sheet.
And in that data sheet you see a bunch of curves.
So very often devices come with data sheets.
And when you have a data sheet but no equation then you can
apply the graphical method and solve your circuits.
In this example, assuming I can apply the
analytical method, here was the expression that I
had derived for you in the last lecture.
So VO was related to VI using the square law relationship.
And we can plot and do other fun stuff with this equation.
So here is the input voltage VI.
That is my VT. So notice that VO is VS.
This is true when VI greater than or equal to VT and VO
greater than or equal to VI minus VT.
So these are the constraints of the saturation discipline.
And in our particular situation when VI was less than VT output
would simply be VS. If VI is less than VT the
MOSFET would turn off, switch off, and I would have no
current flowing through RL, and VS would appear at the
output. So until VT,
I have VS, and then following that I get the square law
behavior articulated by this equation.
And this was simply VS-K/2 (VI-VT)(RL^2).
So that's the first part. You have seen this before.
The transfer function shows that I have a square law
dependence between VI and VO. So now I can embark on the
second step of my large signal analysis, and my goal is to find
the valid input operating range. So what does that mean?
What I am looking to do here is, for this little circuit,
is drain, source, gate, VI, VO,
RL and VS. What I am looking to do is that
given the value of the supply VS, RL and a MOSFET,
in our case given a MOSFET implies that it is a given value
of K and a given value of VT for that MOSFET.
So what I am going to do is find out, let's assume VI is my
free variable here. So my goal will be to find out
the range of VI for which this device stays in saturation.
And I will use a couple of methods to do that,
and I will use both a combination of a graphical
method to give you intuition and then apply analytical analysis
to get down to specific answers. So let's start with the
intuitive part. So here is VI,
VO. I will use the transfer curve
VO versus VI to help build intuition here.
So that is what it looks like. So the first step,
looking at this graph, we know that this point here,
that VI needs to be greater than VT to satisfy the first
equation. Let me just write down both
equations here. So VI greater than or equal to
VT is one of them, and VO is greater than VI minus
VT is a second equation. And just remember that this is
the same as VDS and this is the same as VGS.
So VI must be greater than VT for the MOSFET to turn on.
And so therefore the valid operating range starts at this
point and is somewhere up here. So the first part is pretty
easy. Somewhere here --
Somewhere at that point, my output voltage VO.
I'm not quite sure what that point is.
My output voltage VO, as this keeps falling down,
my output voltage VO goes lower than one threshold below VI.
And at that point it goes into its triode region,
and I need to find out what that point is.
So somewhere here I go into my triode region and begin to show
a different behavior than the amplifying square law
relationship there and go into my triode behavior.
So I need to find out what this point is.
Once I find out what that point is then this will be my valid
operating range. So let's figure out what that
point is. At that point the following is
true. Certainly VI is greater than
VT. And at that point the output
goes below one threshold, the input minus one threshold.
So at this point the following is true, VO is equal to VI minus
VT. At that point the output
voltage is equal to the input minus VT.
And if the output goes lower then it will violate this
equation here. It is no longer greater than
that number. So how do we find out what this
point is? The principle intuition.
Let's draw some lines here. Let's assume that VI and VT use
the same scale, say, volts.
So if I draw a straight line at 45 degrees then that is a curve
representing VI equals VO. We all know that.
No big shakes. So the line at 45 degrees here
is the line at which VI equal VO.
And if I take that line now, the VI equals VO line,
and I begin translating it to the right.
So let's take a line here. Let's take a line there.
That line will be simply equal to VO equals VI minus VT.
I have translated that to the right.
And so this line is simply VO equals VI minus VT.
So this line is a locus of points at which VO is equal to
this value. This minus VT shows up as a
translation to the right. So I take my VO equals VI line,
translate that to the right and it becomes VO equals VI minus
VT. Elementary geometry 101 or
whatever. So what do we have here?
Above this we have the condition VO greater than or
equal to VI minus VT, and below that we have VO less
than VI minus VT. If we look at this graph here,
this is the valid input operating range.
Starting at this point greater than VT, and at this point my
output equals VO equals VI minus VT.
This must be the valid operating range for the input
here to here. And correspondingly the outputs
are from here to this point to here like so.
So this is my valid input operating range and this is my
valid output operating range or the corresponding valid output
operating range. So what does this say?
What this is saying is that if I, as the designer of the
circuit, am disciplined enough to apply inputs that are in this
range, VT to some value here, graphically shown here.
Then my MOSFET will remain in saturation.
And correspondingly my outputs will go between VS and some
value here. So hopefully that gives you
some of the intuition behind how we get it.
And let's continue. Let me label this point X.
So continuing with two to get the valid operating range.
I have shown you intuitively where that point is,
but what I will do next is actually compute that for you.
It is a pretty simple computation.
Note that point X is the intersection of two curves VO
equals VI minus VT. And the second curve is VO
equals VS minus K divide by 2, VI minus V2 all squared RL.
So the point X iss at the intersection of these curves,
and I can very easily get that as follows.
What I will do is I will simply substitute for VI minus VT from
this equation here and then solve for it,
so I get VO equals VS minus K divide by 2 VO squared RL.
And so this gives me a quadratic in VO.
And I can solve for it pretty easily.
And I get for a quadratic AX squared plus BX plus C equals 0.
The solution is given by VO is minus B plus/minus square root
of B squared minus 4AC divided by 2A.
And so I am just going to get those numbers here.
So the coefficient of VO, that is B, is minus 1.
Take the positive root because we are up in the positive
voltages here. And square root of B squared,
that is 1, minus 4AC. So I get a plus 4 times K
divide by 2 RL. And 2A is simply 2 times K
divided by 2 times RL. So that is what I get.
That gives me VO. So it tells me that VO,
at the point where the output just equals one threshold drop
below VI is given by the other circuit perimeter such as VS,
RL and so on. Oh, I am missing a VS here.
I just forgot the VS up here. That is my VO.
So what is VI equal to? Remember that at this point VO
equals VI minus VT, so VI is simply VT plus --
I have not taught you anything earth shattering here.
I have just done some grubby math here to solve these two
equations. So this is a straight line at
45 degrees from VT and this is the transfer function.
And I need to find the intersection.
And the intersection is given by this point.
So that point, VI being VT plus something,
is simply the second dot on the X axis.
So therefore I am pretty much done.
My valid input range for VI goes from VT.
So it starts at VT. That is where the transistor
just turns on. And then goes all the way to
this point, VT plus minus 1 plus square root of 1 plus 2 K RL VS
K RL. So this is my valid operating
range. And again remember I won't
dwell on this equation because, in some sense,
you will get a different set of limits for other devices,
for other circuits containing a MOSFET.
Or, for that matter, for other outputs that one may
be focusing on. So what is more important here
is not so much the results that you see but the process that I
have gone through. So what is more important here
is how did I get here? And the way I got here was
looked at the graph and said look, the MOSFET is in
saturation in that regime. And I am finding the bounding
points of the regime of saturation operation.
So now, as an engineer, I can say that hey,
look, if you build a MOSFET circuit like so,
with a given value of RL, a given MOSFET and a given VS,
then if you limit yourself you are operating with input
voltages in this range thou shalt be happy.
If you go beyond that range then you will be violating the
saturation discipline. So the corresponding output
range --
I can write the corresponding output range,
and that goes from VS, when the input is at VT the
output is at VS and goes from VS down to the input minus VT.
Which is simply minus 1 plus --
Let me go back and quickly show you a little MOSFET circuit,
amplified circuit so you can stare at a real transfer curve
yourselves. And indeed convince yourselves
that roughly at the point where proportionately shown in the
curve up there the MOSFET indeed goes into its triode region and
begins heading out of its saturation region.
Notice that here that is the same curve, the transfer
function. And the amplified output is at
VS until input reaches a threshold voltage VT.
And once input goes beyond VT the output begins to drop
precipitously. And at some point here this
begins to go into its triode region.
And what I am going to do is simply increase the input
voltage VI, and you will see that the output them begins to
go into its triode region. It keeps dropping.
And, as you can see, the output begins to go into a
space where the gain is no longer more than 1.
And this is a triode region of MOSFET operation.
So the MOSFET is in saturation, things are going great.
As I increase my VI notice at some point I begin to go out of
my saturation region of the MOSFET.
And somewhere here I go from the saturation region and
transition into the triode region.
And this value shown here gives you the corresponding input
voltage and the output voltage. Other practical devices like
bipolar junction transistors or MOSFETs and other circuits and
so on can be subjected to a similar analysis.
And you can find out the valid operating regions for that
device as well, or for that circuit.
So as a next step what I would like to do --
Out here I began by looking at the transfer function,
the VO versus VI curve, and used that to drive the
intuition behind how we calculated the bounding regions.
You can do the same kind of analysis intuitively looking at
yet another curve, another set of graphs that you
are familiar with, and that is a load line
characteristic. And it is interesting to get
two interpretations. And you can use whichever one
you feel comfortable with. So I will do two alternatively
and show you another set of curves that you can use to get
that.
Here I am going to plot IDS versus VDS, which is the same as
VO. This was the load line graph
that we had seen earlier. And, just for our reference,
remember that VI must be greater than VT for saturation
operation. Similarly VO should be greater
than or equal to VI minus VT for saturation operation.
Those are my limits. The way we got the load line
graph was we superimposed the load line equation over the
device characteristics. And so let me plot the device
characteristics in the saturation region.
Remember that this constraint could be related to the current
as I derived for you in the last lecture as follows.
IDS being less than or equal to K divided by 2 VO squared.
So in terms of my IDS versus VDS relation,
this lateral constraint is equivalent to IDS being less
than K by 2 VO squared. So this is that equation.
So this line is IDS equals K by 2 VO squared.
And in this region I have the valid operating region where IDS
is less than that quality. So here are all my other curves
for various values VGS. So here are my devices curves,
IDS versus VDS. Remember that these curves come
down like this, for the MOSFET,
right? Just that we focus on the
right-hand side because that is where the MOSFET is in
saturation. And on this side the MOSFET is
in its triode region, and we discipline ourselves not
to operate the MOSFET such that it is in its triode region.
So those were the device characteristics.
And then I could plot my load line equation.
My load line equation, if you recall,
was IDS = VS/RL - VO/RL. This was simply obtained by
writing KVL at the loop containing the output node and
the supply VS. Notice there that VO is equal
to VS minus IDS times RL. And that is simply by dividing
by RL on both sides and moving IDS to the left-hand side we get
this equation. And this equation gives rise to
a curve that looks like this. And what is this point here?
This point is where VO is 0. So when VO is 0 my IDS is
simply VS divided by RL. And this point is obtained when
IDS is 0. And under those conditions VS
and VO are equal so this is VS. This is my saturation region
and this is the triode region. This was another interesting
graph. We often times fondly call it
the load line graph. So here is a load line
superimposed on the MOSFET device IDS versus VDS curves for
a variety of values of VGS. So by looking at this curve,
we can also intuitively determine the valid operating
range. So what are the two points
here? I will let you stare at it for
a couple of seconds yourselves to figure out what two points
here bound the valid operating range of the MOSFET,
the valid operating range of the circuit.
I will start. One is this point,
because at this point the output is VS and VGS has just
begun to equal VT. So think about where the second
point is for valid operation. It is here, and,
somewhere along that load line. Remember the load line is a
constraint that must be met by the output VO.
It is the constraint imposed by KVL on the output.
So the output is constrained to operate in this regime for
various values of VGS. So as the output keeps going
from here all the way here, at some point I exit my
saturation region. And that other point is given
by this one. So notice that this is the
curve that bounds. On the left-hand side of this
the MOSFET is no longer in saturation.
It is on the right-hand side, and so therefore this is the
valid operating region.
Here to here. This is good.
This is VS. That is good to know.
And for this point I know that VI, which is VGS,
equals VT. I know VO is equal to VS.
And IDS, at this point, is 0.
So VO and IDS being VS and 0 correspondingly are the output
operating perimeters when VI equals VD.
So that is one point. And let's find out what this
point is. At that point I get my output
just entering the range of the MOSFET triode region operation.
Notice that this point is the intersection of two curves,
this line and this curve. So this curve here is given by
IDS equals K divided by 2 VO squared.
And this is my load line equation.
So that is VS divided by RL minus VO divided by RL.
That's it. So I won't go ahead and solve
that for you. You can go and check it out and
convince yourselves that if you solve these two equations and
find out the VO for this, it should be the same VO that
you obtained using the other graph.
What I have done here, obtaining the valid regions of
operation is no different from what I did here.
The two are alternative approaches to getting to the
same place. Just that over the years what I
have discovered is that there are one class of people that are
output transfer function people, this graph, and another set of
people that are load line people that like to think that way.
I have always been a transfer function person myself,
but some of you may be load line people and so you can use
that to drive your intuition. It is pretty amazing.
As we get into this business and keep going down the path,
it is amazing how some people really kind of get the load line
thing and others feel much more comfortable with the transfer
function. So pick whatever you want.
So what we have so far is we have conducted a large signal
analysis of a MOSFET amplifier. It is an analysis of a circuit,
and we found two things. One is the transfer function
under saturation operation, and we found the valid input
operating ranges and the corresponding valid output
operating ranges for the circuit.
In the last five or six minutes let me talk about a couple of
other issues. And the first issue is what we
have done so far is intuitively and mathematically shown you
what the valid regions are. Now you are thinking that's
fine, but how do I get there? This region is good,
VT through that other point, that's good,
but how do I get there? How do I make my amplifier
operate in that region? The answer is pretty simple,
and let me drive the intuition again using a graph.
So this is a graph. And I showed you that --
That was my valid region here. Take a 45 degree line,
find out where it intersects the transfer function,
then this is the valid region here, VT through that
coordinating function that we developed out there.
If I have an input that looks like so, some input whose
gyrations fall within this range, will constantly keep the
MOSFET in saturation. And the corresponding output
will look like this. If my input is in this range,
my output will be within this range.
And how do I get my input to be here?
Let's say I have a sinusoid that is 1 volt peak to peak or
whatever. How do I get my sinusoid up
there? Well, you have learned the
trick on how to boost things. Remember boost?
All you have to do is boost up your signal by some value
capital VI. And the way you do that is as
follows. VS, RL, VO.
What you do is you apply a DC offset to your input.
You take your sinusoid and boost it up so that all the
gyrations of the input are in the valid range.
This is my input, some VA.
Then I apply some DC offset capital VI given by this value
here. And boost up the interesting
input? My interesting input is the VA.
And I boost it up by capital VI so that this guy is always in
saturation. I would like to show you a
little demo now. I am going to show you an input
that is a triangular wave. And what we will do is I'll
play with a wide variety of offset voltages.
This guy is a triangular wave. And what I am going to do is
apply a triangular wave and we'll look at the output and
convince ourselves that I get amplification when VI is big
enough that the input goes into a valid operating range.
And we will look at a variety of ranges here.
You can put it a little larger.
OK. So the triangular wave is my
input. And this is my output.
This looks nothing like a triangular wave.
And the reason is that I do not have the right offset.
So what I will do is gradually increase the offset on the
MOSFET. So at this point the offset is
very low, a very small near zero offset.
And so therefore my output is a disaster.
My MOSFET is not in saturation all the time.
So what I will do here is apply some sort of offset.
Is this the one? We want to switch.
This is the input. You can see I am applying an
offset by bumping and boosting up the input.
I don't have clipping happening at both ends,
but I get something. And I get amplification.
Now let me apply way too much of an offset.
With this offset I am kind of operating here.
What I will do now is apply an even higher offset so that this
triangular wave begins to move here.
If I apply a very high offset what I am doing is overdriving
the amplifier, boosting it so high that the
MOSFET is going to go into its triode region and you are going
to see that I won't have any gain.
My output is going to shrink noticeably if I overdrive the
input. You will notice the input going
higher and higher. Pull the trigger point down.
There you go. Notice that as I boost up my
input even higher notice that the output is a really small
image of what the right input should be.
The right answer here, of course, is that I apply some
right amount of offset to boost up the input into the right
regime so that the output is seen to be some amplified
version of this input. So I showed you three things.
One is very little offset. That was like so,
as the thing comes down. A very high offset,
it gets killed again. And the right amount of offset.
But notice that we still have a problem, even with the right
offset. The output is not linearly
related to the input. It is nonlinear.
And the answer to get a linear response is good old small
signal stuff. And we will be looking at the
small signal part in the next lecture.