Good morning, all.
Good morning. I hope you guys did
not spend all of last night
celebrating the Red Sox victory,
but there is one more tonight. OK.
Let's see. I trust the quiz went
OK. What I will do today is take
off from where we left
off on Tuesday.
And continue our discussion of the
large signal and small signal
analysis of our amplifier.
Today the focus will be on "Small
Signal Analysis".
So let me start by reviewing some of
the material. And,
as you know, our MOSFET amplifier
looks like this.
One of the things you will notice in
circuits, as I have been mentioning
all along in this course,
is that certain kinds of patterns
keep repeating time and time again.
And this is one such pattern. A
three terminal device like the
MOSFET with an input and the drain
to source port connected to RL and
VS in series in the following manner,
this is a very common pattern.
There are several other common
patterns. The voltage divider is a
common pattern.
We keep running into that again and
again and again.
The Thevenin form,
a voltage source in series with the
resistor is another very common form.
The Norton equivalent form,
which is a current source in
parallel with a resistor is also
very common. And it behooves all of
us to be very familiar with the
analyses of these things.
Voltage dividers in particular are
just so common that you need to be
able to look at it and boom,
be able to write down the expression
for voltage dividers.
I would also encourage you to go
and look at current dividers.
When you have two resistors in
parallel and you have some current
flowing into the resistors to find
out the current in one branch versus
the other very quickly.
The expression is very analogous to
the voltage divider expression.
And some of these very common
patterns are highlighted in the
summary pages in the course notes,
so it is good to keep track of those
and be extremely familiar with those
patterns to the point where if you
see it you should be able to jump up
and shout out the answer just by
looking at it without having to do
any math. So here was an amplifier.
And then we noticed that when the
MOSFET was in saturation it behaved
like a current source.
And this circuit would give us
amplification while the MOSFET was
in saturation.
So we agreed to adhere to the
saturation discipline which simply
said that I was going to use my
circuit in a way that the MOSFET
would always remain in saturation in
building things like amplifiers and
so on. And by doing that throughout
the analysis I could make the
assumption that the MOSFET was in
saturation. I didn't
have to go through --
Analysis became easier.
I didn't have to figure out now,
what region is the MOSFET in? Well,
because of my discipline it is
always going to be in saturation.
But in turn what we had to do was
conduct a large signal analysis.
Again, in follow on courses you will
be given circuits like this.
In fact, this very circuit with a
very high likelihood.
And you will be looking at more
complicated models of the MOSFET.
Or you will be given the MOSFET
like this and,
let's say in that course the
designers do not adhere to the
saturation discipline,
in which case you have to first
figure out is my MOSFET in its
triode region or in the saturation
region? And depending on the region
it is in you have to apply different
equations. So it is one step more
complicated than in 002.
In 002 we simplified our lives by
following a discipline.
And let me tell you that following
a discipline is quite OK.
When it simplifies our lives and we
can do good things with it,
it is quite OK to do that. We are
not wimps or anything like that.
It is quite OK to have a discipline
and agree that we are going to play
in this region of the playground and
build circuits in that manner.
By doing so, we could assume the
MOSFET was in saturation all the
time. And analysis simply used a
current source model.
By the same token,
what becomes important is to figure
out what are the boundaries of valid
operation of the MOSFET in
saturation? To do that we conducted
a large signal analysis.
And it had two components to it.
One of course was to figure out the
output versus input response.
And what this usually does is that
it does a nonlinear analysis
of this circuit.
If it is a linear circuit it is a
linear analysis.
And figures out what the values of
the various voltages and currents
are in the circuit as a function of
the applied inputs and chosen
parameters. And the second step we
said was to figure out valid
operating ranges --
-- for input and corresponding
ranges for the other dependent
parameters such as VO.
You could also find out the
corresponding operating range for
the current IDS and so on.
So by doing this you could first
analyze the circuit,
find out the "bias" parameters,
find out the values of VI and VO and
so on.
And then you could say all right,
provided, as long as VI stays within
these bounds my assumption that this
is in saturation will hold and
everything will be fine.
The reading for this is Chapter 8.
And today we will take the next
step and revisit small signal
analysis. In the demo that I showed
you at the end of last lecture,
I showed you an input triangular
wave.
And the input triangular wave gave
rise to an output.
And we noticed that we did have
amplification,
I had a small input and a much
bigger output.
I did have amplification when the
MOSFET was in saturation but it was
highly nonlinear.
The input was a triangular wave and
the output was some funny,
it kind of looked like a sinusoid
whose extremities had been whacked
down and kind of flattened.
And its upward going peak had been
shrunk. So it was a kind of weird
nonlinear behavior.
I will show that to you again later
on. And so it amplified but it was
nonlinear. And remember our goal of
two weeks ago?
We set out to build a linear
amplifier. So today we will walk
down that path and talk about
building a linear amplifier.
So to very quickly revisit the input
versus output characteristic,
VI versus VO, this is VT and this is
VS, this is what things looked like.
Also to quickly review the valid
ranges, until some point here the
amplifier was in saturation,
the MOSFET was in saturation and
somewhere here I had VO being equal
to VI minus a threshold drop.
At that point the MOSFET went into
its triode region and I no longer
was following the saturation
discipline. So therefore this is my
valid region of operation.
We also know that the output was
given by VS minus K (VI-VT)
all squared RL over 2.
Again assuming the MOSFET is in
saturation. It is very important to
keep stating this because this is
true only when the MOSFET is in
saturation, when I am following the
discipline. Notice that this is a
nonlinear relationship.
So VO depends on some funny square
law dependence on VI.
The key here is how do we go about
building our amplifier?
Take a look at this point here.
At this point here let's say I have
a VI input. Corresponding output is
VO. Focus is this point.
And left to itself this was a
nonlinear curve.
Remember the trick that we used in
our nonlinear Expo Dweeb example?
We used the Zen Method.
Remember the Zen Method?
We said look, this is nonlinear,
but if you can focus your mind on
this little piece of the curve here
this looks more or less linear.
If I look at a small itty-bitty
portion of the curve and I do the
Zen thing, and kind of zoom in on
here. This looked more or less
linear. This means that if I could
work with very small signals and
apply the signal in a way that I
also had a DC offset of some sort.
Then I would be in a region of the
curve, I would be delineating a
small region of the curve which
would be more or less linear.
This was a small signal trick.
And what we will do here is simply
revisit the small signal model.
Most of what I am going to do from
here on will be more or less a
repeat of what you saw for the light
emitting expo dweeb.
Just that here I have a three
terminal device,
with a little bit more complication.
The equation is different. I don't
have to resort to a Taylor series
expansion. I will just do a
complete expansion of this
expression and develop the small
signal values for you.
Recall the small signal model.
It had the following steps.
The first step will operate at some
bias point, VI,
VO, and of course some corresponding
point IDS. This is Page 3.
And then superimpose a small signal
VI on top of the big fat bias.
Remember the "boost"?
So VI is the boost.
Boom. And above VI,
I have small signal VI that I apply.
And our claim is that response of
the amplifier to VI is
approximately linear.
The key trick with this is that for
my small signal model here,
this is Page 3 here, and Page 2.
The key trick here is that with the
small signal model,
I operate my amplifier at some
operating point,
VO, VI. I superimpose a small
signal VI on top of small VI on top
of big VI. And then I claim that
the response to VI is
approximately linear.
And let me just embellish that curve
a little bit more.
Notice that in this situation this
was my VI, which is my bias voltage,
this is VO, which is the output bias,
and of course not shown on this
graph is the output operating
current which is IDS.
One nice way of thinking about this
is to redraw this and think that
your coordinate axes have kind of
shifted in the following manner.
This is VI. This is also on your
Page 3. This is VT.
Remember this was the operating
point, VO and VI.
And notice that we were operating
in this small regime of our
transfer curve here.
And in effect what we are saying is
that I am going to apply small
variations about VI and call those
variations delta VI or small VI.
And the resulting variations are
going to look like delta VO.
Also referred to as small V,
small O. So I will have small
variations here.
And they give rise to corresponding
small variations there.
One way to view this is as if we
are working with a new coordinate
system. Another way to view this is
that so the capital VI and capital
VO correspond to my VI and VO as the
total voltages in my circuit,
but at this bias point I can think
of another coordinate system here
with small VI and VO out there.
And for small changes to VI,
I can figure out the corresponding
small changes to VO.
Just that all the analysis I
perform here is going to be linear.
And I will prove it to you in a
couple of different ways in the next
few seconds. When I am doing small
signal analysis I am operating here
in this regime at some bias point.
You have also seen this before.
How do I get a bias? This is my
amplifier RL and VS.
This is Page 4. VO.
The way I get a bias is I apply
some DC voltage VI and superimpose
on top of that my small signal small
VI. This is my DC bias that has
boosted up the signal to
an interesting value.
And because of that what I can get
is by varying VI as a small signal
with a very small amplitude,
I am going to get a linear response
here. And I can draw
that for you as well.
This is my bias point here.
And if I vary my signal like so then
my output should look like this.
This is point VI, this is point VO,
and this is my small signal VI and
this is my small signal VO and this
is capital VO.
So this small thing here is VI.
I would like to show you a little
demo.
I will start with the same demo I
showed you the last time.
I showed you the amplifier.
In the demo I am going to apply a
triangular wave.
And initially I start with a large
signal. And you will see that the
output looks really corny,
is going to look something like this.
That's large signal response.
And then I will begin playing with
the input making it smaller,
and you can see how it looks
yourselves.
There you go. So this is where I
stopped the last time.
The last lecture I applied this
input, time is going to the right,
and the purple curve in the
background is the output.
It looks much more like a sinusoid
with some flattening of its tips.
Nothing like an interesting
triangular wave.
What I will do next is that let me
make sure I have enough of a boost
here, enough of a DC voltage so that
I am operating at some point here.
I believe I already have that.
Notice that I can shift up the
triangular wave input,
or I can shift it down.
So let me bias it here.
I have chosen a VI that's about,
I forget how many volts per division
it is, but I have chosen
some VI here.
And I biased it such that this is
the input. You get a nonlinear
response. It is amplified.
It is much bigger. What I will do
next is make VI that I apply smaller
and smaller. I have already done
the boosting. Boom,
that's a boost. So I have boosted
up your VI already.
Next is I am going to shrink it,
and hopefully you will see that if
all that I am saying is truthful
here you will see a triangular
response.
Let's go try it out.
Watch the yellow. I am going to
shrink the yellow and make it
smaller and smaller.
There you go. It is great when
nature works like you expect it to.
I have never seen a triangular wave
looks so pretty in my life.
It is awesome. Look at this.
Here is a tiny triangular wave.
And the output is also a triangular
wave but it is much more linear.
Yes. Question? What's that? The
question is that the output here is
only as big as the input used to be
before. That's a good question.
What I have done here is I am
showing you a laboratory experiment.
And let's assume that this input is
the input I am getting from some
sensor in the field.
Assume that this is my input,
not what I had before. Assume that
this is my input to begin with and
this is the amplified output.
What I can also do is I can also
change the bias.
And we will see this at the end of
the lecture, in the last ten minutes
of lecture. How do you select a
bias point? By changing your bias
point you can change the properties
of an amplifier to give you a
preview of upcoming attractions.
Let me ask you,
what do you think should happen if I
change the bias point?
I have not shown you the math yet,
so intuitively what do you think
should happen?
If I increase the bias what do you
think is going to happen?
Yes. Good insight. Higher bias
will be more amplification.
Let's see if our friend is correct.
Let me set a higher bias.
Not necessarily,
I guess. You're actually right,
by the way. I am playing a trick on
everybody here.
As I change my input bias.
Notice that under certain
conditions my output becomes smaller
and gets more distorted.
Under other conditions what is
going to happen to my output is that
it is becoming smaller and is going
to get distorted again.
So there are a bunch of funny
effects happening that reflect on
the bias point,
but for an appropriate choice of
bias point as I increase the bias
the amplification should increase.
And I will show you that in a few
minutes. But it is a complicated
relationship. Yes.
This is finally getting fun.
Here is the question. Professor
Agarwal, we love your song and dance,
but if you really want to get a high
signal at the output and you want to
amplify your big input signal
how do you do it?
So the question is let's say I have
an input that is this big here,
if it is this big, I have shown you
how I can get things that are this
big, but what if my input was this
big? How do I get an output that is
this big? Well,
I will use one of those learned by
questioning methods and have you
tell me the answer.
Someone tell me the answer.
How do I do that? Yes. Use
another amplifier.
So the answer is I will use one
amplifier to go from here to here.
And the suggestion is use another
amplifier to go from here to here.
And, in fact, I believe that you
may have a problem in your problem
set where you will do that.
And so you have only yourselves to
blame. So how do you
make this work?
What you have to do is this VI has
to be much smaller than the bias
point VI on this one.
I have to build a different
amplifier, choose a different set of
parameters such that VI prime,
which is the VI for this guy, is
much less than V capital I prime for
this guy. It's a design question.
You need to design it in a way that
the signals of interest need to be
much smaller than the bias voltage
of this amplifier.
So you may have to use much higher
supply voltages.
My amplifier, I believe,
has a 4 volt supply or 5 volt supply.
You might have to use an amplifier
with a much bigger supply,
different values of RL and so on.
And I know that the course notes
also have some exercises and problem
sets that discuss that in more
detail. Yes. This is even more fun.
The question is, good question.
The question is why do you need this
guy here? Just use this guy,
right? Why do you need this guy?
Big guys rule, right? Who needs
the little guys?
Well, let me use the Socratic
method again. Why don't you give me
the answer? You guys are smart.
Why do you need little guys? Why
do you need the small guy here?
Anybody with the answer?
Yeah. The big guy may not be as
sensitive. I like that.
You know what? He is almost
correct. I will show you why in a
second. Anything else?
Any other reason? Yes.
Bingo. That is another good answer.
So let me address both the answers.
The answer given was that look,
this amplifier is amplifying the
signal by a certain amount,
by a factor of 7. And I have
designed this such that this
amplifies a signal by
a factor of maybe 10.
So in all I am getting an
amplification of 70.
This would be a great design
question for lab next year.
I give you a bunch of components
and ask you to design an amplifier
given the constraints with the
highest amount of amplification.
It turns out that when you design
your amplifier,
in order to meet the saturation
discipline and so on,
you have to choose values of RL and
VS and stuff like that and be within
power constraints so the amplifier
doesn't blow up and stuff.
And by the end of it all you are
going to get a measly 7X gain out of
it. The same way here,
to be able to deal with a very small
signal here and get some
amplification,
another set of values and you get
10X. So they multiply.
It is much harder to build one
amplifier with a much larger gain.
You know what? I just realized
that we will be looking at this in
the last five or seven minutes of
lecture. I am going to show you
what the amplification depends upon.
It depends upon K. It depends upon
RL. It depends upon VI.
Now the question is I have had all
this time to think about how to
stitch in sensitive into this,
and I believe I can. It turns out
that when you have large voltages
and so on and you have practical
devices, it turns out that the more
current you pump through devices
they tend to produce noise of
various kinds.
So very powerful amplifiers are not
very good at dealing with really
tiny signals because they have some
inherent noise capabilities.
And so I guess that is sensitive.
It is sensitive to noise.
Another question? Yes.
Ask me the question again.
I didn't follow.
Let me just explain it.
It turns out that I will not be
able to pass this through the big
amplifier to begin with because it
is just going to give me a gain of
just a factor of 7.
However, if I have a signal that is
this big to begin with then I may
just need this amplifier.
I don't need the smaller guy.
If my signal was this big to begin
with, if I had a strong sensor that
produced a strong signal to begin
with, yeah, I can deal with
just a single stage.
I don't need to two stages.
It is all a matter of design.
And it is actually a fun design
exercise. Given a budget,
dollars, right? You go to your
supply room and look at the parts
that you have and you go to build
what you have to build with the
parts that you have.
And so sometimes you need to build
two amplifiers to get the gain or
build a signal amplifier.
It's all a design thing.
All right. Moving on to Page 7.
That brings us to the small signal
model.
Page 5.
What I showed you up on the little
demo was that provided the signal
input in this example VI was much
smaller than capital VI out there as
I shrank my input,
I was able to get a more or less
linear response at the output.
And so to repeat my notation at the
input, the total input is a sum of
the operating point input plus a
small signal input.
This is called the total variable.
This is called the DC bias. It is
also called the operating point
voltage. And this is called my
small signal input.
It is also variously called
incremental input.
This is more a mathematical term
relating to incremental analysis or
perturbation analysis.
So VI, call it small signal,
call it small perturbation, call it
increment, whatever you want.
Similarly, at the output I have my
total variable at the output a sum
of the output operating voltage and
the small signal voltage.
I do not like using Os in symbols
because big O and small O is simply
a function of how big you write them.
It is not super clear.
And in terms of a graph,
let me plot the input and output for
you. Let's say this is the total
input and that is the total output.
I may have some bias VI.
And corresponding to that I may have
some bias VO. Hold that thought for
a second while I give you a preview
of something that we will be
covering in about three or four
weeks. Notice that as I couple
amplifiers together,
the output operating point voltage
of this amplifier in this connection
becomes the input operating point
voltage of this amplifier, right?
So when they connect this output to
this input, the output operating
point voltage becomes coupled to the
input here so it becomes the input
operating point voltage here.
Now I have a nightmare on my hands.
As I adjust the bias of this guy,
the bias of this guy changes,
too. The two are dependent. It is
a pain in the neck.
And we being engineers find ways to
simplify our lives.
And you will learn another trick in
about three or four weeks.
And that trick will let you decouple
these two stages in a way that you
can design this stage in isolation,
go have a cup of coffee and then
come back to this stage and design
this stage in isolation.
For those of you who want to run
ahead and think about how to do it,
think about it. What trick can you
use to get them in isolation?
Moving on.
What I would like to do next is
address this from a mathematical
point of view.
And much as I did for the light
emitting expo dweeb analyze this
mathematically and show you that if
VI is much smaller than capital VI,
I indeed get a linear response.
This time around I won't use Taylor
series because it turns out that
this expression can
be expanded fully.
So you don't have to buy into Taylor
series and so on.
I am going to list everything down
for you. We know,
to begin with, that VO for the
amplifier is VS-RLK/2 (VI-VT)^2.
What I am going to do for this,
much as I did for the LED,
what I'm going to do is derive for
you the output as a function of the
input when the input
VI is very small.
In other words,
when I substitute for VI,
V capital I squared plus small VI.
Much as I did for the expo dweeb, I
want to substitute for VI a big DC
VI. So VI is much smaller than VI.
And show you for yourselves that
the output response,
V small O is going to be linearly
connected to VI.
Notice that, let me write another
equation here.
This is a total variable.
This simply says that if the input
is VI then the output is going to be
VO, which means that the operating
point input voltage should satisfy
this equation,
correct? In other words,
the operating point output voltage V
capital O should equal
VS-RLK/2 (VI-VT)^2.
This is at VI equals capital VI.
This is very simple but may seem
confusing. All this is saying is
that look, this equation gives me
the relationship between VI and VO.
Therefore, if I apply capital VI as
the input, I'm given that my
corresponding output is capital VO,
so they must satisfy this equation,
right?
Those are bias point values and that
must satisfy this equation.
Simple. I know that. So hold that
thought. Stash it away in the back
of your minds.
Now let me go through a bunch of
grubby math and substitute for VI in
this expression here.
Let me go ahead and do that.
VS-RLK/2((VI+vi)-VT)^2. When I do
something that is other than math I
will wake you up.
I will just keep doing a bunch of
steps that are pure math.
No cheating. No nothing.
Watch my fingers. When I do
anything that is not obvious math I
will wake you up.
Next I am going to simply move VT
over and rewrite this as follows,
RLK/2((VI-VT)+vi)^2. Again, I
haven't done anything interesting so
far. I have just substituted this.
I am just juggling things around
just to pass away some time,
I guess. All right.
Next what I am going to do is simply
expand this out and write it this
way RLK/2, expand that out and treat
this as one unit VS -
RLK/2((VI-VT)^2+2(VI-VT)vi+vi^2).
Nothing fancy here.
This is like the honest board.
Nothing fancy here. Standard stuff.
Only math. I will move to this
blackboard here where I do some fun
EE stuff. Yes.
Good. At least one person
isn't asleep here.
Thank you. So just math here.
Nothing fancy. Plain old simple
math. I have not done any trickery.
I still have all my ten fingers.
Now what I am going to do,
now watch me. I am not using Taylor
series here because this expression
lends itself to this analysis.
Notice VI squared here.
I made the assumption that VI is
much smaller than capital VI,
so what I can do is assuming that VT
is small enough that VI minus VT is
still a big number compared to small
VI, what I can do is ignore this in
comparison to the capital VI terms.
So I have a capital VI term here.
I am going to ignore VI squared.
So, for example,
if capital VI was 5 volts and small
VI was 100 millivolts 0.
, so 0.1 squared is 0.01.
So it is comparing 0.01 to 5.
So I am off by a factor of 500.
So now watch me. Now I begin
playing some fun and games here.
I eliminate this, and because I
eliminate that it now becomes
approximately equal.
What I do in addition is let me
write down the output.
The total variable is the sum of
the DC bias and some variation of
the output. And let me simply
expand that term and write it down
again. VS-RLK/2(VI-VT)^2-RLK/2.
I get a two here.
And I get VI-VT. I won't forget
the VI this time.
Again, from here to there nothing
fancy. This is the one step where I
have used a trick.
I have said small VI is much
smaller than capital VI,
and so I have simply expanded this
out and written it here.
So do you see the obvious next
trick here?
From star look at this guy.
I can cancel this out from star
because I know that at the operating
point these two expressions are
equal, and so therefore I can cancel
out the operating point
voltage and this.
What I am left with is small VO is
simply minus RLK(VI-VT) times vi.
Only one place where I did
something funny.
Other than that it is purely math.
So this is what I get.
Notice that this whole thing is a
constant, minus RLK(VI-VT).
This whole thing is a constant.
And so VO is equal to some constant
times VI. Let me just define some
terms for you that you will use
again and again.
For reasons that will be obvious
next lecture, I am going to call
this term here GM.
I am going to call this term a
constant, K(VI - VT).
It is a constant for a given bias
point voltage.
So I am going to call that GM.
And then I am going to call this
whole thing A.
And of course this is VI.
There you go.
I have my linear amplifier.
A is the gain times small VI.
And the gain has these terms in it.
I just call this GM. You will see
why later. But notice that the gain
relates to RL.
The size of the load resistor RL,
how big it is, 1K, 10K, whatever. K,
this is a MOSFET parameter,
and VI minus VT.
That is a constant for a given bias
point voltage and small VI.
So VO equals small VI.
I won't give you a graphical
interpretation,
but I encourage you to go and look
at Figure 8.9 in the course notes.
And it gives you a graphical
interpretation of that expression.
Move to Page 7.
Another way of looking at this,
another way of mathematically
analyzing it, here I went through a
full blown expansion and pretty much
deriving the small signal response.
What I can also do is take a
shortcut here.
So let me just give you the
shortcut. You might find this handy.
VO=VS-KRL/2(VI-VT)^2.
And my shortcut is as follows.
My small signal response is simply
this relationship.
I find the slope at the point
capital VI and multiply by the
increment. Slope times the
increment gives me the incremental
change in VO as follows.
d/dI (VS-KRL/2(VI-VT)^2) evaluated
at vI=VI times vi.
This is math again.
I want to find out the change in VO
for a small change in VI,
and I do that by taking the first
derivative of this with respect to
VI substituting V capital I and
multiplying by the small change
delta VI or small VI.
So this is simply the slope of the
VO versus VI curve at VI.
And so therefore taking the
derivative here of this.
This is a constant so it vanishes.
But twice 2 to cancel out, so I get
KRL(VI-VT) times small vi
evaluated at capital VI.
So I get twice KRL,
VI evaluated at capital VI,
so it is VI minus VT times small VI.
Same thing. Oh, and I have a minus
sign here.
I get the same expression that I
derived for you up there,
and this is just taking the slope
and going with it.
And this, as I mentioned before,
this is A. The last few minutes let
me kind of pull everything together
and also hit upon something that
many of your questions
are touched upon.
And that all relates to how to
choose the bias point.
So here I have taken an analysis
approach. When teaching we often
teach you are given something,
you analyze it, but as you begin to
master it you can begin to design
things where you can ask a lot of
questions and so on.
And here what we have is an
analysis given a value
of RLK, VI and so on.
How to choose the bias point becomes
more of a design issue.
If you are designing an amplifier,
you asked me the question, how do I
choose two small amplifiers versus
one big amplifier,
that sort of stuff?
It boils down to how do you choose
the bias point?
How do you choose VI?
How do you choose RL and so on?
What I would like to do is touch
upon some of these things.
First of all, gain or the
amplification.
One of the most important design
perimeters for an amplifier is what
is the gain? Let's say you get a
job at Maxim Integrated Technologies,
and they say we would like you to
build a linear power amplifier for
cell phones. You can say I know how
to do that. And then they say the
next stage needs a 100
millivolt input.
While this thing coming from the
antenna is only a few tens or a few
hundreds of a microvolt.
So you sit down and say oh,
my gosh, I need an amplification of
so much, and you go design an
amplifier. So gain tends to be a
key parameter.
And notice that gain is
proportional to RL.
It relates to VI minus VT,
so proportional to VI.
It is also related to RL.
The second point is the gain point
determines where I bias something.
If I choose my bias too high I get
distortion, or if I choose my bias
too low I get distortion.
So depending on how I choose my bias
point, as a signal goes up it may
begin clipping or begin distorting.
And I will show you a demo the next
time on that particular example.
So bias point will determine how
big of a signal you can send without
getting too much distortion.
And the other thing is that,
relates to how big of an input,
what is a valid input range? So
let's say you have a signal.
And you want that signal to have
both positive and negative
excursions of the same value.
Then, depending on where you choose
a bias point, your input range may
become smaller or larger.
And we will go through these in the
context of and amplifier and look at
some design issues in
the next lecture.
