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Good morning,
all.

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Let's get going.
So I guess you had your quiz

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review yesterday.
I hope you guys didn't beat up

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on (name of TA) and who else was
it?

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(Name of TA) too much.
As you know,

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the quiz is tomorrow.
And unfortunately MIT couldn't

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give us one big room so we are
broken up into three rooms.

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And you will go to your room
based on the first letter of

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your last name.
OK, so today we shall cover a

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topic called "Large Signal
Analysis".

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So in the last couple lectures
we looked at one dependent

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sources abstractly,
and then we looked at an

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amplifier built using a
practical dependent source

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called the MOSFET.
Now, the MOSFET had to be

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operated in a given region of
its operation in order to behave

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like a current source.
And while it behaved like a

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current source you would get
amplification or a FET

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amplifier.
So that was in the past two

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lectures.
What you are going to do today

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is called large signal analysis,
and this is a loaded term.

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So large signal analysis means
something very specific in our

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business, and I will describe to
what that is.

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This analysis involves looking
at a circuit containing,

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for example,
a MOSFET and figuring out how

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to get that device to operate in
a way that the MOSFET was always

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in saturation.
So you had to figure out,

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based on parameters that you
could control,

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how to establish those
parameters so that the circuit

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operated in a way that the
MOSFET was always in saturation.

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So large signal analysis
involves that.

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And although the examples we
use, use the MOSFET,

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the same kind of analysis can
apply to any other device.

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Remember, your MOSFET is a
primitive element that we use as

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an example in this course.
There are other primitive

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elements that you can use.
The course notes,

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for example,
discusses a couple other

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devices.
One is the "bipolar junction

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transistor", the BJT,
and works through a complete

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example from start to finish
involving a circuit containing a

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bipolar junction transistor.
And you can do a large signal

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analysis of that device as well.
It turns out that you need to

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operate that device in an
interesting region of its

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operating space,
and so you can conduct a large

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signal analysis of a circuit
containing that device and

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figure out how best to operate
that circuit.

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So that is large signal
analysis, and we will do an

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example and explain how this is
done using an example today.

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So to quickly review where we
have been so far,

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we looked at this little
structure here,

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our MOSFET amplifier.

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Notice that when I write a
voltage at a node,

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that's a short form for saying
I am looking at the voltage

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between the ground node and the
node at which the voltage is

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written down.
So VO here and VI applied here.

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This is a very,
very common circuit that we

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use.
To emphasize one more point,

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in general, in the kind of
circuits we look at both in this

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course and in real life,
there are a few patterns that

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we use very commonly that keep
repeating all the time.

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Very often you don't have to
look at every possible

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permutation and combination of
how things could be connected.

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This sort of connecting thing
is very, very,

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very common.
And you will see a lot of this

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pattern.
And we do the equivalent

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circuit for this.
In the equivalent circuit we

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replace the MOSFET with a
dependent source provided this

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operated in the saturation
region.

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So I will just say while
operating under saturation the

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equivalent circuit would look
like this, VO,

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VI.

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And IDS for the dependent
source was given by K/2

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(VI-VT)^2.
So this was an amplifier.

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Here was the equivalent circuit
while this device was in

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saturation.
And to operate in saturation,

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I said that certain properties
need to be true for the MOSFET.

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And there are two properties
that need to be true for this to

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be operating in saturation.
One is that its gate to source

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voltage needs to be greater than
VT, so VGS for the MOSFET should

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be greater than VT.
And the second one was that the

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output voltage needed to be
greater than the input voltage

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minus one threshold drop.
And this was the same as VDS

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for the MOSFET,
this was the same as VGS for

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the MOSFET.
So what are we really saying

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here?
What we are saying is that

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look, we built this circuit
using a MOSFET,

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and it is up to us as engineers
to choose its operating points

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in a way that these two
properties hold.

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For example,
to make the first condition

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true, I can discipline myself to
operate such that VI is always

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greater than VT.
Similarly, I can choose VS,

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RL and VI in a way that this
condition is true,

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which says that the drain to
source voltage across my MOSFET

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drain and source should be
greater than VI minus VT.

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As an example,
if VI was 2 volts and VT was,

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say, 1 volt,
then what I am saying is that

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VO should be greater than or
equal to 2 minus 1 or 1 volt.

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So I need to keep this high,
2, 3, 4, 5, whatever,

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a high voltage so that this guy
stays in saturation.

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The relevant readings for the
material that we are going to

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cover in the course notes are in
7.5.1 and 7.6.

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So that is pretty much a review
of where we were.

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We said we could build an
amplifier.

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Its equivalent circuit was
shown on the right.

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And, provided that,
I discipline myself to operate

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in the saturation region or to
have the MOSFET operating in the

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saturation region,
then this would work like an

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amplifier and all would be good
with the world.

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So today --

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-- we look at large signal
analysis of a circuit.

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And an example would be this
circuit up here containing a

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MOSFET.
And, again, as I mentioned

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earlier, a large signal analysis
is a loaded term in 6.002,

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or for that matter in circuits.
And large signal analysis

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involves two steps.

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The first step involves writing
down the transfer function of

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your little circuit.
In our case,

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VO is the output,
VI is the input,

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so involves writing down VO
versus VI.

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Simply write down the transfer
function.

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In other words,
the relationship between the

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output and the input for that
circuit.

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And, in our case,
again, we've disciplined

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ourselves to adhere to the
"saturation discipline".

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And the second part of large
signal analysis is to find out

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the valid input operating range.

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Find out for the given circuit
parameters, let's say I apply a

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VS and I use some value of RL
and I use a given MOSFET,

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which has a given value of VT.
The question then is that what

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is a valid set of input voltages
that would operate the circuit

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in a way that I would be in
saturation.

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And so find out the valid input
range, and this would give me a

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corresponding output range --

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-- for saturation operation of
the MOSFET.

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That is what we will dwell on
in the lecture today.

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So what we are saying here is
that if I am careful with how I

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apply VI for a given value of RL
and VS and for a given choice of

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my MOS transistor then I can
stay within saturation provided

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I select my input voltages
carefully.

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And the analysis that we will
go through today will figure out

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what that range of input
voltages is.

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And, again, I will use this as
a motivating example,

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the MOSFET amplifier.
But in general large signal

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analysis would apply to any
other circuit as well.

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For example,
in recitation you may learn

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about other circuits containing
a MOSFET.

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And you can do a large signal
analysis of other circuits

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containing a MOSFET or you might
learn about some other devices

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like the bipolar junction
transistor, and you could do the

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same kind of analysis for that
device.

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So remember that the MOSFET
amplifier here is an example.

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I will be using that as a
driving example to explain large

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signal analysis.
So the first step,

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as I mentioned earlier,
is to get the VO versus VI.

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And in general for some circuit
that you build the output will

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not even be a voltage.
There are certain circuits

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where the output might be some
kind of a current.

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Let's say I am building some
kind of a circuit where I would

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like the output current or the
current through some edge of the

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circuit to depend on some input.
In that case the transfer

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function would be the output
current versus VI.

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And if I had an input current
here it would be output current

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versus input current,
you know, whatever the given

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problem tells you.
So this is under the saturation

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discipline.
And I will not rederive this

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for you.
You can apply a good old

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technique like the analytical
method.

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Or you can use the graphical
method to get the appropriate

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answer here.
I wanted to point out in a

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quick aside that why do we care
about graphical analysis?

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Once you have the analytical
method, why do you care about

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the graphical method?
And a student asked me a

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question after lecture last
Thursday, and it occurred to me

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that it's not obvious why you
need the graphical method.

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So it turns out that often
times you do not have an

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equation describing the device.
So let's say,

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for example,
I am a manufacturer.

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Let's say I am AMD.
As AMD I sit down and my

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semiconductor division builds a
MOSFET.

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And when you build a MOSFET
your experiments and your

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fabrication division often times
doesn't give you an equation

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with the MOSFET.
They build something and then

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you look at it and you
experiment with it.

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You apply various input
voltages and you measure

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currents and output voltages and
so on.

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And so what you end up getting
is a graph that describes the

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behavior of the MOSFET.
And you have seen this in your

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lab as well, your 2N7000 or was
it 2000?

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So your 2N7000,

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the MOSFET you use in the lab
also gives you a data sheet.

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And in that data sheet you see
a bunch of curves.

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So very often devices come with
data sheets.

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And when you have a data sheet
but no equation then you can

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apply the graphical method and
solve your circuits.

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In this example,
assuming I can apply the

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analytical method,
here was the expression that I

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had derived for you in the last
lecture.

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So VO was related to VI using
the square law relationship.

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And we can plot and do other
fun stuff with this equation.

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So here is the input voltage
VI.

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That is my VT.
So notice that VO is VS.

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This is true when VI greater
than or equal to VT and VO

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greater than or equal to VI
minus VT.

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So these are the constraints of
the saturation discipline.

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And in our particular situation
when VI was less than VT output

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would simply be VS.
If VI is less than VT the

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MOSFET would turn off,
switch off, and I would have no

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current flowing through RL,
and VS would appear at the

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output.
So until VT,

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I have VS, and then following
that I get the square law

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behavior articulated by this
equation.

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And this was simply VS-K/2
(VI-VT)(RL^2).

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So that's the first part.
You have seen this before.

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The transfer function shows
that I have a square law

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dependence between VI and VO.
So now I can embark on the

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second step of my large signal
analysis, and my goal is to find

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the valid input operating range.
So what does that mean?

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What I am looking to do here
is, for this little circuit,

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is drain, source,
gate, VI, VO,

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RL and VS.
What I am looking to do is that

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given the value of the supply
VS, RL and a MOSFET,

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in our case given a MOSFET
implies that it is a given value

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of K and a given value of VT for
that MOSFET.

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So what I am going to do is
find out, let's assume VI is my

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free variable here.
So my goal will be to find out

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the range of VI for which this
device stays in saturation.

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And I will use a couple of
methods to do that,

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and I will use both a
combination of a graphical

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method to give you intuition and
then apply analytical analysis

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to get down to specific answers.
So let's start with the

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intuitive part.
So here is VI,

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VO.
I will use the transfer curve

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VO versus VI to help build
intuition here.

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So that is what it looks like.
So the first step,

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looking at this graph,
we know that this point here,

235
00:18:07,000 --> 00:18:12,000
that VI needs to be greater
than VT to satisfy the first

236
00:18:12,000 --> 00:18:16,000
equation.
Let me just write down both

237
00:18:16,000 --> 00:18:20,000
equations here.
So VI greater than or equal to

238
00:18:20,000 --> 00:18:24,000
VT is one of them,
and VO is greater than VI minus

239
00:18:24,000 --> 00:18:31,000
VT is a second equation.
And just remember that this is

240
00:18:31,000 --> 00:18:36,000
the same as VDS and this is the
same as VGS.

241
00:18:36,000 --> 00:18:42,000
So VI must be greater than VT
for the MOSFET to turn on.

242
00:18:42,000 --> 00:18:48,000
And so therefore the valid
operating range starts at this

243
00:18:48,000 --> 00:18:55,000
point and is somewhere up here.
So the first part is pretty

244
00:18:55,000 --> 00:18:59,000
easy.
Somewhere here --

245
00:18:59,000 --> 00:19:02,000
Somewhere at that point,
my output voltage VO.

246
00:19:02,000 --> 00:19:05,000
I'm not quite sure what that
point is.

247
00:19:05,000 --> 00:19:09,000
My output voltage VO,
as this keeps falling down,

248
00:19:09,000 --> 00:19:14,000
my output voltage VO goes lower
than one threshold below VI.

249
00:19:14,000 --> 00:19:18,000
And at that point it goes into
its triode region,

250
00:19:18,000 --> 00:19:21,000
and I need to find out what
that point is.

251
00:19:21,000 --> 00:19:26,000
So somewhere here I go into my
triode region and begin to show

252
00:19:26,000 --> 00:19:30,000
a different behavior than the
amplifying square law

253
00:19:30,000 --> 00:19:36,000
relationship there and go into
my triode behavior.

254
00:19:36,000 --> 00:19:39,000
So I need to find out what this
point is.

255
00:19:39,000 --> 00:19:45,000
Once I find out what that point
is then this will be my valid

256
00:19:45,000 --> 00:19:49,000
operating range.
So let's figure out what that

257
00:19:49,000 --> 00:19:53,000
point is.
At that point the following is

258
00:19:53,000 --> 00:19:56,000
true.
Certainly VI is greater than

259
00:19:56,000 --> 00:19:59,000
VT.
And at that point the output

260
00:19:59,000 --> 00:20:06,000
goes below one threshold,
the input minus one threshold.

261
00:20:06,000 --> 00:20:13,000
So at this point the following
is true, VO is equal to VI minus

262
00:20:13,000 --> 00:20:17,000
VT.
At that point the output

263
00:20:17,000 --> 00:20:21,000
voltage is equal to the input
minus VT.

264
00:20:21,000 --> 00:20:28,000
And if the output goes lower
then it will violate this

265
00:20:28,000 --> 00:20:33,000
equation here.
It is no longer greater than

266
00:20:33,000 --> 00:20:38,000
that number.
So how do we find out what this

267
00:20:38,000 --> 00:20:41,000
point is?
The principle intuition.

268
00:20:41,000 --> 00:20:47,000
Let's draw some lines here.
Let's assume that VI and VT use

269
00:20:47,000 --> 00:20:49,000
the same scale,
say, volts.

270
00:20:49,000 --> 00:20:56,000
So if I draw a straight line at
45 degrees then that is a curve

271
00:20:56,000 --> 00:21:01,000
representing VI equals VO.
We all know that.

272
00:21:01,000 --> 00:21:06,000
No big shakes.
So the line at 45 degrees here

273
00:21:06,000 --> 00:21:10,000
is the line at which VI equal
VO.

274
00:21:10,000 --> 00:21:15,000
And if I take that line now,
the VI equals VO line,

275
00:21:15,000 --> 00:21:20,000
and I begin translating it to
the right.

276
00:21:20,000 --> 00:21:25,000
So let's take a line here.
Let's take a line there.

277
00:21:25,000 --> 00:21:33,000
That line will be simply equal
to VO equals VI minus VT.

278
00:21:33,000 --> 00:21:36,000
I have translated that to the
right.

279
00:21:36,000 --> 00:21:40,000
And so this line is simply VO
equals VI minus VT.

280
00:21:40,000 --> 00:21:46,000
So this line is a locus of
points at which VO is equal to

281
00:21:46,000 --> 00:21:49,000
this value.
This minus VT shows up as a

282
00:21:49,000 --> 00:21:54,000
translation to the right.
So I take my VO equals VI line,

283
00:21:54,000 --> 00:22:00,000
translate that to the right and
it becomes VO equals VI minus

284
00:22:00,000 --> 00:22:05,000
VT.
Elementary geometry 101 or

285
00:22:05,000 --> 00:22:10,000
whatever.
So what do we have here?

286
00:22:10,000 --> 00:22:17,000
Above this we have the
condition VO greater than or

287
00:22:17,000 --> 00:22:24,000
equal to VI minus VT,
and below that we have VO less

288
00:22:24,000 --> 00:22:31,000
than VI minus VT.
If we look at this graph here,

289
00:22:31,000 --> 00:22:35,000
this is the valid input
operating range.

290
00:22:35,000 --> 00:22:41,000
Starting at this point greater
than VT, and at this point my

291
00:22:41,000 --> 00:22:45,000
output equals VO equals VI minus
VT.

292
00:22:45,000 --> 00:22:51,000
This must be the valid
operating range for the input

293
00:22:51,000 --> 00:22:56,000
here to here.
And correspondingly the outputs

294
00:22:56,000 --> 00:23:02,000
are from here to this point to
here like so.

295
00:23:02,000 --> 00:23:06,000
So this is my valid input
operating range and this is my

296
00:23:06,000 --> 00:23:11,000
valid output operating range or
the corresponding valid output

297
00:23:11,000 --> 00:23:14,000
operating range.
So what does this say?

298
00:23:14,000 --> 00:23:19,000
What this is saying is that if
I, as the designer of the

299
00:23:19,000 --> 00:23:24,000
circuit, am disciplined enough
to apply inputs that are in this

300
00:23:24,000 --> 00:23:30,000
range, VT to some value here,
graphically shown here.

301
00:23:30,000 --> 00:23:34,000
Then my MOSFET will remain in
saturation.

302
00:23:34,000 --> 00:23:40,000
And correspondingly my outputs
will go between VS and some

303
00:23:40,000 --> 00:23:45,000
value here.
So hopefully that gives you

304
00:23:45,000 --> 00:23:49,000
some of the intuition behind how
we get it.

305
00:23:49,000 --> 00:23:56,000
And let's continue.
Let me label this point X.

306
00:24:04,000 --> 00:24:09,000
So continuing with two to get
the valid operating range.

307
00:24:09,000 --> 00:24:14,000
I have shown you intuitively
where that point is,

308
00:24:14,000 --> 00:24:20,000
but what I will do next is
actually compute that for you.

309
00:24:20,000 --> 00:24:24,000
It is a pretty simple
computation.

310
00:24:24,000 --> 00:24:29,000
Note that point X is the
intersection of two curves VO

311
00:24:29,000 --> 00:24:36,000
equals VI minus VT.
And the second curve is VO

312
00:24:36,000 --> 00:24:44,000
equals VS minus K divide by 2,
VI minus V2 all squared RL.

313
00:24:44,000 --> 00:24:52,000
So the point X iss at the
intersection of these curves,

314
00:24:52,000 --> 00:24:59,000
and I can very easily get that
as follows.

315
00:24:59,000 --> 00:25:06,000
What I will do is I will simply
substitute for VI minus VT from

316
00:25:06,000 --> 00:25:10,000
this equation here and then
solve for it,

317
00:25:10,000 --> 00:25:17,000
so I get VO equals VS minus K
divide by 2 VO squared RL.

318
00:25:17,000 --> 00:25:21,000
And so this gives me a
quadratic in VO.

319
00:25:21,000 --> 00:25:25,000
And I can solve for it pretty
easily.

320
00:25:25,000 --> 00:25:34,000
And I get for a quadratic AX
squared plus BX plus C equals 0.

321
00:25:34,000 --> 00:25:38,000
The solution is given by VO is
minus B plus/minus square root

322
00:25:38,000 --> 00:25:41,000
of B squared minus 4AC divided
by 2A.

323
00:25:41,000 --> 00:25:45,000
And so I am just going to get
those numbers here.

324
00:25:45,000 --> 00:25:49,000
So the coefficient of VO,
that is B, is minus 1.

325
00:25:49,000 --> 00:25:53,000
Take the positive root because
we are up in the positive

326
00:25:53,000 --> 00:25:57,000
voltages here.
And square root of B squared,

327
00:25:57,000 --> 00:26:06,000
that is 1, minus 4AC.
So I get a plus 4 times K

328
00:26:06,000 --> 00:26:17,000
divide by 2 RL.
And 2A is simply 2 times K

329
00:26:17,000 --> 00:26:30,000
divided by 2 times RL.
So that is what I get.

330
00:26:30,000 --> 00:26:35,000
That gives me VO.
So it tells me that VO,

331
00:26:35,000 --> 00:26:42,000
at the point where the output
just equals one threshold drop

332
00:26:42,000 --> 00:26:50,000
below VI is given by the other
circuit perimeter such as VS,

333
00:26:50,000 --> 00:26:55,000
RL and so on.
Oh, I am missing a VS here.

334
00:26:55,000 --> 00:27:02,000
I just forgot the VS up here.
That is my VO.

335
00:27:02,000 --> 00:27:12,000
So what is VI equal to?
Remember that at this point VO

336
00:27:12,000 --> 00:27:22,000
equals VI minus VT,
so VI is simply VT plus --

337
00:27:40,000 --> 00:27:43,000
I have not taught you anything
earth shattering here.

338
00:27:43,000 --> 00:27:47,000
I have just done some grubby
math here to solve these two

339
00:27:47,000 --> 00:27:49,000
equations.
So this is a straight line at

340
00:27:49,000 --> 00:27:53,000
45 degrees from VT and this is
the transfer function.

341
00:27:53,000 --> 00:27:55,000
And I need to find the
intersection.

342
00:27:55,000 --> 00:27:59,000
And the intersection is given
by this point.

343
00:28:04,000 --> 00:28:10,000
So that point,
VI being VT plus something,

344
00:28:10,000 --> 00:28:15,000
is simply the second dot on the
X axis.

345
00:28:15,000 --> 00:28:20,000
So therefore I am pretty much
done.

346
00:28:20,000 --> 00:28:28,000
My valid input range for VI
goes from VT.

347
00:28:28,000 --> 00:28:32,000
So it starts at VT.
That is where the transistor

348
00:28:32,000 --> 00:28:36,000
just turns on.
And then goes all the way to

349
00:28:36,000 --> 00:28:43,000
this point, VT plus minus 1 plus
square root of 1 plus 2 K RL VS

350
00:28:43,000 --> 00:28:46,000
K RL.
So this is my valid operating

351
00:28:46,000 --> 00:28:49,000
range.
And again remember I won't

352
00:28:49,000 --> 00:28:54,000
dwell on this equation because,
in some sense,

353
00:28:54,000 --> 00:28:59,000
you will get a different set of
limits for other devices,

354
00:28:59,000 --> 00:29:05,000
for other circuits containing a
MOSFET.

355
00:29:05,000 --> 00:29:08,000
Or, for that matter,
for other outputs that one may

356
00:29:08,000 --> 00:29:11,000
be focusing on.
So what is more important here

357
00:29:11,000 --> 00:29:16,000
is not so much the results that
you see but the process that I

358
00:29:16,000 --> 00:29:19,000
have gone through.
So what is more important here

359
00:29:19,000 --> 00:29:22,000
is how did I get here?
And the way I got here was

360
00:29:22,000 --> 00:29:26,000
looked at the graph and said
look, the MOSFET is in

361
00:29:26,000 --> 00:29:31,000
saturation in that regime.
And I am finding the bounding

362
00:29:31,000 --> 00:29:35,000
points of the regime of
saturation operation.

363
00:29:35,000 --> 00:29:39,000
So now, as an engineer,
I can say that hey,

364
00:29:39,000 --> 00:29:43,000
look, if you build a MOSFET
circuit like so,

365
00:29:43,000 --> 00:29:48,000
with a given value of RL,
a given MOSFET and a given VS,

366
00:29:48,000 --> 00:29:52,000
then if you limit yourself you
are operating with input

367
00:29:52,000 --> 00:29:56,000
voltages in this range thou
shalt be happy.

368
00:29:56,000 --> 00:30:01,000
If you go beyond that range
then you will be violating the

369
00:30:01,000 --> 00:30:07,000
saturation discipline.
So the corresponding output

370
00:30:07,000 --> 00:30:08,000
range --

371
00:30:18,000 --> 00:30:22,000
I can write the corresponding
output range,

372
00:30:22,000 --> 00:30:28,000
and that goes from VS,
when the input is at VT the

373
00:30:28,000 --> 00:30:36,000
output is at VS and goes from VS
down to the input minus VT.

374
00:30:36,000 --> 00:30:39,000
Which is simply minus 1 plus --

375
00:30:51,000 --> 00:30:55,000
Let me go back and quickly show
you a little MOSFET circuit,

376
00:30:55,000 --> 00:30:58,000
amplified circuit so you can
stare at a real transfer curve

377
00:30:58,000 --> 00:31:03,000
yourselves.
And indeed convince yourselves

378
00:31:03,000 --> 00:31:10,000
that roughly at the point where
proportionately shown in the

379
00:31:10,000 --> 00:31:17,000
curve up there the MOSFET indeed
goes into its triode region and

380
00:31:17,000 --> 00:31:22,000
begins heading out of its
saturation region.

381
00:31:22,000 --> 00:31:27,000
Notice that here that is the
same curve, the transfer

382
00:31:27,000 --> 00:31:32,000
function.
And the amplified output is at

383
00:31:32,000 --> 00:31:36,000
VS until input reaches a
threshold voltage VT.

384
00:31:36,000 --> 00:31:41,000
And once input goes beyond VT
the output begins to drop

385
00:31:41,000 --> 00:31:45,000
precipitously.
And at some point here this

386
00:31:45,000 --> 00:31:49,000
begins to go into its triode
region.

387
00:31:49,000 --> 00:31:53,000
And what I am going to do is
simply increase the input

388
00:31:53,000 --> 00:31:59,000
voltage VI, and you will see
that the output them begins to

389
00:31:59,000 --> 00:32:05,000
go into its triode region.
It keeps dropping.

390
00:32:05,000 --> 00:32:10,000
And, as you can see,
the output begins to go into a

391
00:32:10,000 --> 00:32:16,000
space where the gain is no
longer more than 1.

392
00:32:16,000 --> 00:32:21,000
And this is a triode region of
MOSFET operation.

393
00:32:21,000 --> 00:32:29,000
So the MOSFET is in saturation,
things are going great.

394
00:32:29,000 --> 00:32:35,000
As I increase my VI notice at
some point I begin to go out of

395
00:32:35,000 --> 00:32:38,000
my saturation region of the
MOSFET.

396
00:32:38,000 --> 00:32:43,000
And somewhere here I go from
the saturation region and

397
00:32:43,000 --> 00:32:47,000
transition into the triode
region.

398
00:32:47,000 --> 00:32:52,000
And this value shown here gives
you the corresponding input

399
00:32:52,000 --> 00:32:59,000
voltage and the output voltage.
Other practical devices like

400
00:32:59,000 --> 00:33:03,000
bipolar junction transistors or
MOSFETs and other circuits and

401
00:33:03,000 --> 00:33:06,000
so on can be subjected to a
similar analysis.

402
00:33:06,000 --> 00:33:10,000
And you can find out the valid
operating regions for that

403
00:33:10,000 --> 00:33:13,000
device as well,
or for that circuit.

404
00:33:13,000 --> 00:33:17,000
So as a next step what I would
like to do --

405
00:33:22,000 --> 00:33:25,000
Out here I began by looking at
the transfer function,

406
00:33:25,000 --> 00:33:29,000
the VO versus VI curve,
and used that to drive the

407
00:33:29,000 --> 00:33:34,000
intuition behind how we
calculated the bounding regions.

408
00:33:34,000 --> 00:33:38,000
You can do the same kind of
analysis intuitively looking at

409
00:33:38,000 --> 00:33:42,000
yet another curve,
another set of graphs that you

410
00:33:42,000 --> 00:33:45,000
are familiar with,
and that is a load line

411
00:33:45,000 --> 00:33:48,000
characteristic.
And it is interesting to get

412
00:33:48,000 --> 00:33:52,000
two interpretations.
And you can use whichever one

413
00:33:52,000 --> 00:33:57,000
you feel comfortable with.
So I will do two alternatively

414
00:33:57,000 --> 00:34:01,000
and show you another set of
curves that you can use to get

415
00:34:01,000 --> 00:34:03,000
that.

416
00:34:10,000 --> 00:34:16,000
Here I am going to plot IDS
versus VDS, which is the same as

417
00:34:16,000 --> 00:34:19,000
VO.
This was the load line graph

418
00:34:19,000 --> 00:34:25,000
that we had seen earlier.
And, just for our reference,

419
00:34:25,000 --> 00:34:31,000
remember that VI must be
greater than VT for saturation

420
00:34:31,000 --> 00:34:35,000
operation.
Similarly VO should be greater

421
00:34:35,000 --> 00:34:40,000
than or equal to VI minus VT for
saturation operation.

422
00:34:40,000 --> 00:34:45,000
Those are my limits.
The way we got the load line

423
00:34:45,000 --> 00:34:50,000
graph was we superimposed the
load line equation over the

424
00:34:50,000 --> 00:34:55,000
device characteristics.
And so let me plot the device

425
00:34:55,000 --> 00:35:00,000
characteristics in the
saturation region.

426
00:35:00,000 --> 00:35:06,000
Remember that this constraint
could be related to the current

427
00:35:06,000 --> 00:35:12,000
as I derived for you in the last
lecture as follows.

428
00:35:12,000 --> 00:35:18,000
IDS being less than or equal to
K divided by 2 VO squared.

429
00:35:18,000 --> 00:35:22,000
So in terms of my IDS versus
VDS relation,

430
00:35:22,000 --> 00:35:28,000
this lateral constraint is
equivalent to IDS being less

431
00:35:28,000 --> 00:35:35,000
than K by 2 VO squared.
So this is that equation.

432
00:35:35,000 --> 00:35:40,000
So this line is IDS equals K by
2 VO squared.

433
00:35:40,000 --> 00:35:48,000
And in this region I have the
valid operating region where IDS

434
00:35:48,000 --> 00:35:56,000
is less than that quality.
So here are all my other curves

435
00:35:56,000 --> 00:36:02,000
for various values VGS.
So here are my devices curves,

436
00:36:02,000 --> 00:36:06,000
IDS versus VDS.
Remember that these curves come

437
00:36:06,000 --> 00:36:08,000
down like this,
for the MOSFET,

438
00:36:08,000 --> 00:36:11,000
right?
Just that we focus on the

439
00:36:11,000 --> 00:36:16,000
right-hand side because that is
where the MOSFET is in

440
00:36:16,000 --> 00:36:19,000
saturation.
And on this side the MOSFET is

441
00:36:19,000 --> 00:36:23,000
in its triode region,
and we discipline ourselves not

442
00:36:23,000 --> 00:36:30,000
to operate the MOSFET such that
it is in its triode region.

443
00:36:30,000 --> 00:36:34,000
So those were the device
characteristics.

444
00:36:34,000 --> 00:36:39,000
And then I could plot my load
line equation.

445
00:36:39,000 --> 00:36:42,000
My load line equation,
if you recall,

446
00:36:42,000 --> 00:36:48,000
was IDS = VS/RL - VO/RL.
This was simply obtained by

447
00:36:48,000 --> 00:36:54,000
writing KVL at the loop
containing the output node and

448
00:36:54,000 --> 00:36:58,000
the supply VS.
Notice there that VO is equal

449
00:36:58,000 --> 00:37:06,000
to VS minus IDS times RL.
And that is simply by dividing

450
00:37:06,000 --> 00:37:12,000
by RL on both sides and moving
IDS to the left-hand side we get

451
00:37:12,000 --> 00:37:16,000
this equation.
And this equation gives rise to

452
00:37:16,000 --> 00:37:22,000
a curve that looks like this.
And what is this point here?

453
00:37:22,000 --> 00:37:27,000
This point is where VO is 0.
So when VO is 0 my IDS is

454
00:37:27,000 --> 00:37:34,000
simply VS divided by RL.
And this point is obtained when

455
00:37:34,000 --> 00:37:39,000
IDS is 0.
And under those conditions VS

456
00:37:39,000 --> 00:37:45,000
and VO are equal so this is VS.
This is my saturation region

457
00:37:45,000 --> 00:37:52,000
and this is the triode region.
This was another interesting

458
00:37:52,000 --> 00:37:56,000
graph.
We often times fondly call it

459
00:37:56,000 --> 00:38:01,000
the load line graph.
So here is a load line

460
00:38:01,000 --> 00:38:06,000
superimposed on the MOSFET
device IDS versus VDS curves for

461
00:38:06,000 --> 00:38:10,000
a variety of values of VGS.
So by looking at this curve,

462
00:38:10,000 --> 00:38:14,000
we can also intuitively
determine the valid operating

463
00:38:14,000 --> 00:38:17,000
range.
So what are the two points

464
00:38:17,000 --> 00:38:20,000
here?
I will let you stare at it for

465
00:38:20,000 --> 00:38:24,000
a couple of seconds yourselves
to figure out what two points

466
00:38:24,000 --> 00:38:28,000
here bound the valid operating
range of the MOSFET,

467
00:38:28,000 --> 00:38:33,000
the valid operating range of
the circuit.

468
00:38:33,000 --> 00:38:36,000
I will start.
One is this point,

469
00:38:36,000 --> 00:38:43,000
because at this point the
output is VS and VGS has just

470
00:38:43,000 --> 00:38:49,000
begun to equal VT.
So think about where the second

471
00:38:49,000 --> 00:38:55,000
point is for valid operation.
It is here, and,

472
00:38:55,000 --> 00:39:02,000
somewhere along that load line.
Remember the load line is a

473
00:39:02,000 --> 00:39:06,000
constraint that must be met by
the output VO.

474
00:39:06,000 --> 00:39:11,000
It is the constraint imposed by
KVL on the output.

475
00:39:11,000 --> 00:39:17,000
So the output is constrained to
operate in this regime for

476
00:39:17,000 --> 00:39:22,000
various values of VGS.
So as the output keeps going

477
00:39:22,000 --> 00:39:27,000
from here all the way here,
at some point I exit my

478
00:39:27,000 --> 00:39:32,000
saturation region.
And that other point is given

479
00:39:32,000 --> 00:39:36,000
by this one.
So notice that this is the

480
00:39:36,000 --> 00:39:40,000
curve that bounds.
On the left-hand side of this

481
00:39:40,000 --> 00:39:43,000
the MOSFET is no longer in
saturation.

482
00:39:43,000 --> 00:39:49,000
It is on the right-hand side,
and so therefore this is the

483
00:39:49,000 --> 00:39:52,000
valid operating region.

484
00:40:00,000 --> 00:40:04,000
Here to here.
This is good.

485
00:40:04,000 --> 00:40:09,000
This is VS.
That is good to know.

486
00:40:09,000 --> 00:40:17,000
And for this point I know that
VI, which is VGS,

487
00:40:17,000 --> 00:40:23,000
equals VT.
I know VO is equal to VS.

488
00:40:23,000 --> 00:40:30,000
And IDS, at this point,
is 0.

489
00:40:30,000 --> 00:40:37,000
So VO and IDS being VS and 0
correspondingly are the output

490
00:40:37,000 --> 00:40:42,000
operating perimeters when VI
equals VD.

491
00:40:42,000 --> 00:40:48,000
So that is one point.
And let's find out what this

492
00:40:48,000 --> 00:40:53,000
point is.
At that point I get my output

493
00:40:53,000 --> 00:41:03,000
just entering the range of the
MOSFET triode region operation.

494
00:41:03,000 --> 00:41:09,000
Notice that this point is the
intersection of two curves,

495
00:41:09,000 --> 00:41:15,000
this line and this curve.
So this curve here is given by

496
00:41:15,000 --> 00:41:19,000
IDS equals K divided by 2 VO
squared.

497
00:41:19,000 --> 00:41:23,000
And this is my load line
equation.

498
00:41:23,000 --> 00:41:30,000
So that is VS divided by RL
minus VO divided by RL.

499
00:41:30,000 --> 00:41:32,000
That's it.
So I won't go ahead and solve

500
00:41:32,000 --> 00:41:35,000
that for you.
You can go and check it out and

501
00:41:35,000 --> 00:41:39,000
convince yourselves that if you
solve these two equations and

502
00:41:39,000 --> 00:41:43,000
find out the VO for this,
it should be the same VO that

503
00:41:43,000 --> 00:41:46,000
you obtained using the other
graph.

504
00:41:57,000 --> 00:42:01,000
What I have done here,
obtaining the valid regions of

505
00:42:01,000 --> 00:42:04,000
operation is no different from
what I did here.

506
00:42:04,000 --> 00:42:08,000
The two are alternative
approaches to getting to the

507
00:42:08,000 --> 00:42:11,000
same place.
Just that over the years what I

508
00:42:11,000 --> 00:42:16,000
have discovered is that there
are one class of people that are

509
00:42:16,000 --> 00:42:21,000
output transfer function people,
this graph, and another set of

510
00:42:21,000 --> 00:42:27,000
people that are load line people
that like to think that way.

511
00:42:27,000 --> 00:42:31,000
I have always been a transfer
function person myself,

512
00:42:31,000 --> 00:42:35,000
but some of you may be load
line people and so you can use

513
00:42:35,000 --> 00:42:39,000
that to drive your intuition.
It is pretty amazing.

514
00:42:39,000 --> 00:42:43,000
As we get into this business
and keep going down the path,

515
00:42:43,000 --> 00:42:48,000
it is amazing how some people
really kind of get the load line

516
00:42:48,000 --> 00:42:53,000
thing and others feel much more
comfortable with the transfer

517
00:42:53,000 --> 00:42:55,000
function.
So pick whatever you want.

518
00:42:55,000 --> 00:43:00,000
So what we have so far is we
have conducted a large signal

519
00:43:00,000 --> 00:43:06,000
analysis of a MOSFET amplifier.
It is an analysis of a circuit,

520
00:43:06,000 --> 00:43:09,000
and we found two things.
One is the transfer function

521
00:43:09,000 --> 00:43:13,000
under saturation operation,
and we found the valid input

522
00:43:13,000 --> 00:43:17,000
operating ranges and the
corresponding valid output

523
00:43:17,000 --> 00:43:19,000
operating ranges for the
circuit.

524
00:43:19,000 --> 00:43:23,000
In the last five or six minutes
let me talk about a couple of

525
00:43:23,000 --> 00:43:26,000
other issues.
And the first issue is what we

526
00:43:26,000 --> 00:43:31,000
have done so far is intuitively
and mathematically shown you

527
00:43:31,000 --> 00:43:36,000
what the valid regions are.
Now you are thinking that's

528
00:43:36,000 --> 00:43:41,000
fine, but how do I get there?
This region is good,

529
00:43:41,000 --> 00:43:44,000
VT through that other point,
that's good,

530
00:43:44,000 --> 00:43:49,000
but how do I get there?
How do I make my amplifier

531
00:43:49,000 --> 00:43:54,000
operate in that region?
The answer is pretty simple,

532
00:43:54,000 --> 00:44:00,000
and let me drive the intuition
again using a graph.

533
00:44:13,000 --> 00:44:15,000
So this is a graph.
And I showed you that --

534
00:44:22,000 --> 00:44:25,000
That was my valid region here.
Take a 45 degree line,

535
00:44:25,000 --> 00:44:28,000
find out where it intersects
the transfer function,

536
00:44:28,000 --> 00:44:32,000
then this is the valid region
here, VT through that

537
00:44:32,000 --> 00:44:36,000
coordinating function that we
developed out there.

538
00:44:36,000 --> 00:44:45,000
If I have an input that looks
like so, some input whose

539
00:44:45,000 --> 00:44:54,000
gyrations fall within this
range, will constantly keep the

540
00:44:54,000 --> 00:45:01,000
MOSFET in saturation.
And the corresponding output

541
00:45:01,000 --> 00:45:06,000
will look like this.
If my input is in this range,

542
00:45:06,000 --> 00:45:09,000
my output will be within this
range.

543
00:45:09,000 --> 00:45:12,000
And how do I get my input to be
here?

544
00:45:12,000 --> 00:45:17,000
Let's say I have a sinusoid
that is 1 volt peak to peak or

545
00:45:17,000 --> 00:45:21,000
whatever.
How do I get my sinusoid up

546
00:45:21,000 --> 00:45:24,000
there?
Well, you have learned the

547
00:45:24,000 --> 00:45:28,000
trick on how to boost things.
Remember boost?

548
00:45:28,000 --> 00:45:33,000
All you have to do is boost up
your signal by some value

549
00:45:33,000 --> 00:45:39,000
capital VI.
And the way you do that is as

550
00:45:39,000 --> 00:45:42,000
follows.
VS, RL, VO.

551
00:45:42,000 --> 00:45:49,000
What you do is you apply a DC
offset to your input.

552
00:45:49,000 --> 00:45:56,000
You take your sinusoid and
boost it up so that all the

553
00:45:56,000 --> 00:46:04,000
gyrations of the input are in
the valid range.

554
00:46:04,000 --> 00:46:06,000
This is my input,
some VA.

555
00:46:06,000 --> 00:46:13,000
Then I apply some DC offset
capital VI given by this value

556
00:46:13,000 --> 00:46:16,000
here.
And boost up the interesting

557
00:46:16,000 --> 00:46:20,000
input?
My interesting input is the VA.

558
00:46:20,000 --> 00:46:27,000
And I boost it up by capital VI
so that this guy is always in

559
00:46:27,000 --> 00:46:32,000
saturation.
I would like to show you a

560
00:46:32,000 --> 00:46:36,000
little demo now.
I am going to show you an input

561
00:46:36,000 --> 00:46:40,000
that is a triangular wave.
And what we will do is I'll

562
00:46:40,000 --> 00:46:44,000
play with a wide variety of
offset voltages.

563
00:46:44,000 --> 00:46:49,000
This guy is a triangular wave.
And what I am going to do is

564
00:46:49,000 --> 00:46:54,000
apply a triangular wave and
we'll look at the output and

565
00:46:54,000 --> 00:46:59,000
convince ourselves that I get
amplification when VI is big

566
00:46:59,000 --> 00:47:05,000
enough that the input goes into
a valid operating range.

567
00:47:05,000 --> 00:47:09,000
And we will look at a variety
of ranges here.

568
00:47:09,000 --> 00:47:13,000
You can put it a little larger.

569
00:47:23,000 --> 00:47:28,000
OK.
So the triangular wave is my

570
00:47:28,000 --> 00:47:31,000
input.
And this is my output.

571
00:47:31,000 --> 00:47:34,000
This looks nothing like a
triangular wave.

572
00:47:34,000 --> 00:47:38,000
And the reason is that I do not
have the right offset.

573
00:47:38,000 --> 00:47:42,000
So what I will do is gradually
increase the offset on the

574
00:47:42,000 --> 00:47:45,000
MOSFET.
So at this point the offset is

575
00:47:45,000 --> 00:47:47,000
very low, a very small near zero
offset.

576
00:47:47,000 --> 00:47:50,000
And so therefore my output is a
disaster.

577
00:47:50,000 --> 00:47:53,000
My MOSFET is not in saturation
all the time.

578
00:47:53,000 --> 00:47:57,000
So what I will do here is apply
some sort of offset.

579
00:47:57,000 --> 00:48:01,000
Is this the one?
We want to switch.

580
00:48:01,000 --> 00:48:06,000
This is the input.
You can see I am applying an

581
00:48:06,000 --> 00:48:12,000
offset by bumping and boosting
up the input.

582
00:48:20,000 --> 00:48:23,000
I don't have clipping happening
at both ends,

583
00:48:23,000 --> 00:48:26,000
but I get something.
And I get amplification.

584
00:48:26,000 --> 00:48:30,000
Now let me apply way too much
of an offset.

585
00:48:30,000 --> 00:48:33,000
With this offset I am kind of
operating here.

586
00:48:33,000 --> 00:48:37,000
What I will do now is apply an
even higher offset so that this

587
00:48:37,000 --> 00:48:40,000
triangular wave begins to move
here.

588
00:48:40,000 --> 00:48:44,000
If I apply a very high offset
what I am doing is overdriving

589
00:48:44,000 --> 00:48:47,000
the amplifier,
boosting it so high that the

590
00:48:47,000 --> 00:48:51,000
MOSFET is going to go into its
triode region and you are going

591
00:48:51,000 --> 00:48:54,000
to see that I won't have any
gain.

592
00:48:54,000 --> 00:48:58,000
My output is going to shrink
noticeably if I overdrive the

593
00:48:58,000 --> 00:49:03,000
input.
You will notice the input going

594
00:49:03,000 --> 00:49:09,000
higher and higher.
Pull the trigger point down.

595
00:49:09,000 --> 00:49:13,000
There you go.
Notice that as I boost up my

596
00:49:13,000 --> 00:49:20,000
input even higher notice that
the output is a really small

597
00:49:20,000 --> 00:49:26,000
image of what the right input
should be.

598
00:49:26,000 --> 00:49:30,000
The right answer here,
of course, is that I apply some

599
00:49:30,000 --> 00:49:34,000
right amount of offset to boost
up the input into the right

600
00:49:34,000 --> 00:49:38,000
regime so that the output is
seen to be some amplified

601
00:49:38,000 --> 00:49:42,000
version of this input.
So I showed you three things.

602
00:49:42,000 --> 00:49:46,000
One is very little offset.
That was like so,

603
00:49:46,000 --> 00:49:49,000
as the thing comes down.
A very high offset,

604
00:49:49,000 --> 00:49:53,000
it gets killed again.
And the right amount of offset.

605
00:49:53,000 --> 00:49:58,000
But notice that we still have a
problem, even with the right

606
00:49:58,000 --> 00:50:02,000
offset.
The output is not linearly

607
00:50:02,000 --> 00:50:04,000
related to the input.
It is nonlinear.

608
00:50:04,000 --> 00:50:09,000
And the answer to get a linear
response is good old small

609
00:50:09,000 --> 00:50:12,000
signal stuff.
And we will be looking at the

610
00:50:12,000 --> 00:50:15,000
small signal part in the next
lecture.