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All right, good morning.
So today, we are going to talk

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about what is both a basic
device in itself,

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the amplifier,
and it also serves as a real

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key example of both nonlinear
analysis and small signal

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analysis.
So, today, dependent sources

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and amplifiers.

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So, let me first spend a few
seconds just pointing out to you

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some of the key points from our
previous lectures.

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I also want to point out that
each chapter in the course notes

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has a summary at the end of it.
And if you take a quick scan of

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the summary at the end of each
chapter, it highlights the major

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takeaway points from each
chapter.

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It stresses what's important,
and if you have to remember a

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few things, what are those
things to remember?

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So, to quickly review,
we talked about a few primitive

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elements: resistors,
voltage sources,

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and so on.
And by now, you should have the

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facility to play around with
these device elements.

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And then we talked about the
Node method, and this is kind of

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the workhorse of 6.002.
When in doubt,

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use the Node method.
OK, and this will work both for

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linear circuits and nonlinear
circuits.

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OK, so if you see a problem,
or if you see a situation in

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real life that requires
analysis, then as a first step,

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you should try to think of
whether you could apply some of

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the key intuitive shortcut
methods, superposition.

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One of my favorites,
the Thevenin method,

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the Norton method,
or the method that involves

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composition, that is very
quickly analyzing circuits that

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have resistors in series and
parallel.

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OK, so if you can apply one of
these quick, intuitive,

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shortcut methods,
go do so.

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If you can't,
then usually you can resort to

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the Node method irrespective of
whether the circuit is linear or

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nonlinear.
So the last week was focused on

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the nonlinear method or
nonlinear circuits,

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and we spent the first lecture
talking about a straightforward

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application of the Node method,
which gave us a bunch of

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nonlinear equations that we had
to solve.

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In the last lecture,
we talked about the small

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signal trick.
What we said is if you look at

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the whole space of nonlinear
circuits, then within that

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space, if we focus on small
variations, small perturbations

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about an operating point,
then even the behavior of

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nonlinear circuits in that small
regime would be linear.

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So small signal method.

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And as an example,
I showed you how I could take a

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highly nonlinear device like the
garage door opener LED,

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and using that,
build a pretty nice transmitter

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that would transmit music.
And as long as we kept the

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signal small,
and operated the device in a

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region where its transfer curve
was relatively smooth,

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and I biased,
or set the operating point

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appropriately,
I would get a linear,

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small signal response.
OK.

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So today, we're going to do a
couple things.

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We're going to look at
dependent sources.

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And the reading for this is
section 2.6 of your course

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notes.
And, the dependent source will

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be a new element in your tool
chest.

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We will also do amplifiers,
and amplifiers are in section

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7.1 and section 7.2 of your
course notes.

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So, before I begin with
dependent sources,

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I'm just a huge believer in
motivating things with real

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world examples.
OK, so let me start by

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motivating: why we need an
amplifier?

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Why do we need to do things
like this?

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Or why do we even bother?
And, spend a few minutes really

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getting you to appreciate that
amplification is fundamental.

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OK, it's as foundational to
life as high fat potato chips

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and stuff like that.
So, let's do some basic

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examples here.
So first, let me talk about,

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why do we need to amplify
signals.

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Why amplify?
Why do we care about building

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an amplifier?
So, an amplifier,

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think of a little box,
and apply some sort of small

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input.
And I get a larger output.

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In this example,
this may be a voltage with a

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swing of 10 mV,
and in this case,

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the output might be another
voltage with a swing of,

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say, 100 mV.
And commonly,

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the amplifier,
in addition to an input and an

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output, input port and output
port, may also contain the power

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port, OK, so that I can apply a
power supply to the amplifier

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because commonly as an amplifier
signal, I'm looking for a power

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gain as well,
an increase in the power

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provided by the output.
So, that's an abstract

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definition of an amplifier,
and let's take a look at an

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example of why we may need this.
So let's say I have a small,

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useful signal,
and let's say the signal has 1

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mV peak to peak.
And, I'm looking to transmit

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the signal over a wire to some
other point.

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But let's say that in this
environment, I get a bunch of

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noise that is in a noisy
environment.

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And in this environment,
let's assume that some noise

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may get superimposed.
And if I have a 1 mV signal,

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and 10 mV of noise,
then what I end up with at the

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output is something that looks
like this.

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And it's really hard to
distinguish my 1 mV signal from

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that large amount of noise.
On the other hand,

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if I do the following,
if I took the signal and passed

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the signal to an amplifier,
and I amplified the signal to

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be a much larger version of the
same signal, let's say in this

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particular situation 100 mV peak
to peak signal.

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OK, so I magnified the signal
by a factor of 100.

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OK, let's say it's a linear
amplifier, I linearly amplified

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signal to be 100 mV,
then in that case,

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if I had a noise on top of
this, it's going to be less

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discernible.
The signal will look like this.

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OK, my 10 mV noise would add on
to it.

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But, this is still pretty
decent.

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I can still recognize the
input.

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And so, this is one application
of amplification.

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If I need to send something
from point A to point B as an

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analog signal,
then an amplified signal is

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less prone to noise attacks than
a small signal.

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Not surprisingly,
a large number of devices that

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are used in everyday life have
amplifiers built into them.

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So, get a little cell phone,
and virtually every single cell

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phone contains an amplifier.
By the way, this is an all

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digital cell phone.
It's a Kyocera,

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I forget the number now.
It's completely digital.

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OK, although they say it's
completely digital,

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it turns out that a significant
fraction of the circuitry is

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analog, in particular,
so digital is sort of a

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marketing term to say that
there's something special about

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this.
But remember,

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there's a bunch of analog
stuff.

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So, here's my little antenna
from the cell phone.

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OK, and typically the first
thing that happens to a signal

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as it comes out of the antenna
in your cell phone is,

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look at cell phone circuits,
or cell phone systems would be

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something that looks like this,
OK, this, and may have a label

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LNA.
If someone were to take a guess

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at what LNA might stand for?
What's that?

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Linear amplifier.
That's pretty good.

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So that's LNA.
Close enough.

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A is correct.
It's amplifier.

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What does L and N stand for?
Low noise.

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OK, so this stands for low
noise amplifier.

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So, I get a really rinky dinky
small signal here,

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and then the low noise
amplifier amplifies a signal.

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And in real cell phones,
and for that matter,

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in your 802.11b,
or 802.11a, or 802.11g wireless

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cards, same thing.
Antenna, low noise amplifier,

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and then you may have a bunch
of processing.

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And commonly,
you have a bunch of analog

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processing.
And then, you convert the

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analog to a digital signal.
OK, I recall last week I asked

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somebody in class here,
how would we transmit the

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signal from point A to point B
without it being impacted way

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too much by noise,
and he said,

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oh, go digital.
Good point.

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OK, so if I go digital,
I can transfer the signal

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without noise being a real
factor.

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But the analog to digital
converters need the signal

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strengths to be a given value
before it can chop it up into

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digital levels.
OK, so an amplifier is very

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fundamental.
OK, and so in this case,

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what may be a signal of a few
tens of microvolts to be

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amplified to some large enough
value that it can be further

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processed.
So, that's application of

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amplification in the analog
domain.

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Let me talk about amplification
in the digital domain.

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So, that's in the analog
domain.

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This amplification is in the
domain that I have both analog

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and digital.
OK, and now let me talk about

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amplification in the digital
domain, OK?

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I'm going to argue that
amplification is absolutely

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foundational to the digital
domain.

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OK, the digital abstraction
would not occur if I did not

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have basic amplification.
OK, and the next minute and 37

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seconds I will prove that to
you, OK?

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So, let's do so.
So, let's suppose I have a very

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simple digital system,
and the system simply contains

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a pair of inverters.
So, if I send a one here,

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it's a zero here and a one
here, which is a very simple,

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trivial, digital system.
And here's the input.

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Here's the output.
And we said that for digital

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systems of this sort to work,
they have to follow a static

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discipline.
OK, our signals and our

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circuits must follow a
discipline for them all to work

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together.
And, the discipline we

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described comprised of signals
adhering to certain voltage

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thresholds so that all the
components in the system could

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agree on what comprised a zero,
and what comprised a one,

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OK?
So the way we did that was we

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said that you would have a
threshold called VIH,

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V input high,
and another threshold called

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VIL, V input low.
OK, and we said that this

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circuit must recognize signals
that are higher than VIH,

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3 V for example as a one,
and simultaneously,

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any signal that has a voltage
level less than VIL,

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say, two volts,
should be recognized as a zero.

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That was the input constraint.
On the output,

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it had a similar set of
constraints, where we had

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tougher constraints on devices,
where we said that the output

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had to satisfy a output low
constraint, output high

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constraint.
What this said is that for this

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circuit to be called a good
digital circuit that satisfies

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the static discipline,
signals that were ones here

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should be recognized as such.
And if I am producing a one as

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an output, then the signal level
should be higher than VOH.

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Similarly, if the signal's a
zero, then it should be less

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than VOL.
So as an example,

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this may be 2 V,
this may be 3 V,

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and this may be 4 V,
and this may be 1 V.

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OK, so input,
I should recognize 2 V and less

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as a zero, but at the output I
have to produce a very,

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very low value,
1 V.

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So, I have some noise margin.
So as an example,

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say if I made a plot of the
input/output,

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so I get my VIL here and VIH
here.

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This is time.
This would comprise a valid

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digital signal:
zero, one, zero,

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one, and so on.
OK, now, I had a tougher set of

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constraints at the output.
I would have VOL,

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VOH.
So, at the output,

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OK, I'm required to stretch the
ones and zeros to be further

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apart from each other so that I
get noise margin,

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and the corresponding signal
for our little circuit there

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would look like so.
Right, if this is a valid

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input, then this would be the
corresponding,

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valid output.
OK, and need I say more?

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OK, you can see that,
intuitively,

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look, there's amplification
happening here,

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and the reason is that VOL is
chosen to be less than VIL,

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and VOH is higher than VIH.
So therefore,

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the signal has to be stretched.
The signal has to be amplified.

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OK, and what's the minimum
amplification needed for the

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system to work?
The minimum amplification is if

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I had a signal that looked like
this.

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OK, that barely skimmed the
VIL, VIH level.

234
00:16:49,000 --> 00:16:53,000
OK, so if signal were this high
peak to peak,

235
00:16:53,000 --> 00:16:57,000
VIH minus VIL,
and what's the absolute minimum

236
00:16:57,000 --> 00:17:02,000
signal at the output?
It would look something like

237
00:17:02,000 --> 00:17:06,000
this.
OK, barely skimming VOL and

238
00:17:06,000 --> 00:17:11,000
VOH, OK, so the corresponding
output level would be VOH minus

239
00:17:11,000 --> 00:17:14,000
VOL.
OK, so this is the absolute

240
00:17:14,000 --> 00:17:19,000
minimum amplification that my
digital circuit has to provide.

241
00:17:19,000 --> 00:17:23,000
OK, and notice,
VOH is larger than VIH.

242
00:17:23,000 --> 00:17:27,000
VOL is smaller than VIL.
Therefore, this quantity needs

243
00:17:27,000 --> 00:17:34,000
to be greater than one.
OK, so I've shown you both a

244
00:17:34,000 --> 00:17:38,000
simple, graphical,
intuitive explanation,

245
00:17:38,000 --> 00:17:45,000
and this is a slightly more
formal proof that even the

246
00:17:45,000 --> 00:17:52,000
digital circuit really requires
to have amplification built into

247
00:17:52,000 --> 00:17:57,000
it, if it is to satisfy valid
static disciplines.

248
00:17:57,000 --> 00:18:00,000
Yes?
Yes.

249
00:18:00,000 --> 00:18:03,000
The question is,
is that the same as gain?

250
00:18:03,000 --> 00:18:08,000
Good question.
Yes, the term amplification has

251
00:18:08,000 --> 00:18:11,000
many, many variants.
You could say gain.

252
00:18:11,000 --> 00:18:16,000
You could say amplification.
You could say increase in

253
00:18:16,000 --> 00:18:20,000
signal strength,
and so on and so forth.

254
00:18:20,000 --> 00:18:24,000
And in fact,
when talking about low noise

255
00:18:24,000 --> 00:18:29,000
amplifiers, people sometimes
talk about having the low noise,

256
00:18:29,000 --> 00:18:35,000
high gain amplifier at the
input stage.

257
00:18:35,000 --> 00:18:39,000
OK, so let me pause there in
terms of motivation.

258
00:18:39,000 --> 00:18:44,000
So, I believe I've motivated
every which way:

259
00:18:44,000 --> 00:18:47,000
pure analog,
analog/digital,

260
00:18:47,000 --> 00:18:50,000
and digital.
OK, so I've covered every

261
00:18:50,000 --> 00:18:55,000
single base here.
And so, we need amplification.

262
00:18:55,000 --> 00:19:00,000
OK, so let's look at how to
build a fundamental,

263
00:19:00,000 --> 00:19:04,000
primitive device called the
amplifier.

264
00:19:04,000 --> 00:19:08,000
Before we do that,
however, let me take a quick

265
00:19:08,000 --> 00:19:13,000
detour.
It will be convenient for me,

266
00:19:13,000 --> 00:19:19,000
as I show you how to build an
amplifier, to introduce a new

267
00:19:19,000 --> 00:19:24,000
device, a new element,
called the dependent source.

268
00:19:24,000 --> 00:19:30,000
OK, let me introduce a new
device for your arsenal of

269
00:19:30,000 --> 00:19:38,000
devices, along with resistors,
You learned about a MOSFET,

270
00:19:38,000 --> 00:19:45,000
a switch, voltage source,
current source,

271
00:19:45,000 --> 00:19:54,000
and now a dependent source.
So, a dependent source looks

272
00:19:54,000 --> 00:19:59,000
like this, OK,
has an output port,

273
00:19:59,000 --> 00:20:07,000
and has a control port.
So, a dependent source in its

274
00:20:07,000 --> 00:20:12,000
simplest form has two ports:
an input port and an output

275
00:20:12,000 --> 00:20:14,000
port.
Remember, a port is a

276
00:20:14,000 --> 00:20:19,000
convenient pairing of terminals,
and I apply signals to such

277
00:20:19,000 --> 00:20:23,000
terminal pairs.
But this is a abstract diagram

278
00:20:23,000 --> 00:20:27,000
for a dependent source,
and to get a little bit more

279
00:20:27,000 --> 00:20:34,000
specific, let me show you an
example of a dependent source.

280
00:20:34,000 --> 00:20:36,000
So, let's say,
here's my input,

281
00:20:36,000 --> 00:20:40,000
and I label the terminal
variables for the input.

282
00:20:40,000 --> 00:20:45,000
VC is the voltage applied to
the input, and IC is the current

283
00:20:45,000 --> 00:20:50,000
into this terminal here.
And, here is the symbol for the

284
00:20:50,000 --> 00:20:54,000
dependent source.
Much like a current source or a

285
00:20:54,000 --> 00:20:59,000
voltage source has a circle
around it, the corresponding

286
00:20:59,000 --> 00:21:04,000
symbol for a dependent source is
like so.

287
00:21:04,000 --> 00:21:06,000
So this example,
for instance,

288
00:21:06,000 --> 00:21:09,000
is a dependent,
current source.

289
00:21:09,000 --> 00:21:14,000
I can apply the corresponding
output variables,

290
00:21:14,000 --> 00:21:20,000
I0, OK, and I can say that the
current, I, is some function.

291
00:21:20,000 --> 00:21:24,000
In this example,
I've designed the example that

292
00:21:24,000 --> 00:21:30,000
the current through the current
source, I, is some function of

293
00:21:30,000 --> 00:21:36,000
the input voltage or the control
voltage, VC.

294
00:21:36,000 --> 00:21:40,000
OK, so notice that the current
through a current source,

295
00:21:40,000 --> 00:21:43,000
the current through this
current source,

296
00:21:43,000 --> 00:21:45,000
I, is some function of another
variable.

297
00:21:45,000 --> 00:21:49,000
OK, in this example,
it's the voltage across its

298
00:21:49,000 --> 00:21:51,000
control port.
Not surprisingly,

299
00:21:51,000 --> 00:21:57,000
this device is called a voltage
controlled current source --

300
00:22:10,000 --> 00:22:14,000
-- or a VCCS.
So, in like manner I can also

301
00:22:14,000 --> 00:22:19,000
devise other forms of sources.
You can think of this is a

302
00:22:19,000 --> 00:22:24,000
device where a voltage controls
an output current.

303
00:22:24,000 --> 00:22:27,000
You can think of all other
combinations,

304
00:22:27,000 --> 00:22:33,000
current controlling current,
voltage controlling voltage,

305
00:22:33,000 --> 00:22:38,000
current controlling voltage,
and so on.

306
00:22:38,000 --> 00:22:43,000
So, another example,
I give you another dependent

307
00:22:43,000 --> 00:22:50,000
source, and in this situation,
my output current is controlled

308
00:22:50,000 --> 00:22:52,000
by an input current,
VC.

309
00:22:52,000 --> 00:22:57,000
IC rather.
And I claim that I for this one

310
00:22:57,000 --> 00:23:02,000
is some function of a current,
IC.

311
00:23:02,000 --> 00:23:06,000
OK, it's another dependent
source where the output current

312
00:23:06,000 --> 00:23:09,000
for its output port is related
to the current,

313
00:23:09,000 --> 00:23:11,000
IC.
And, this is a current

314
00:23:11,000 --> 00:23:15,000
controlled current source.
OK, it's a current controlled

315
00:23:15,000 --> 00:23:18,000
current source.
And, if I had lots of time on

316
00:23:18,000 --> 00:23:22,000
my hands, and I was wanting to
kill time, I'd sit around

317
00:23:22,000 --> 00:23:25,000
drawing for you,
other types of dependent

318
00:23:25,000 --> 00:23:28,000
sources.
I would draw for you a current

319
00:23:28,000 --> 00:23:33,000
controlled voltage sourced,
and I could also draw for you a

320
00:23:33,000 --> 00:23:37,000
voltage controlled voltage
source.

321
00:23:37,000 --> 00:23:42,000
OK, so that's an abstract
diagram for such a source.

322
00:23:42,000 --> 00:23:48,000
And so, let's do a few examples
involving elements like this.

323
00:23:48,000 --> 00:23:53,000
To begin, just so you can build
up your intuition,

324
00:23:53,000 --> 00:23:57,000
let me start by doing a very
simple circuit,

325
00:23:57,000 --> 00:24:01,000
involving an independent
current source,

326
00:24:01,000 --> 00:24:09,000
OK, just so we can relate back
to what we've been doing so far.

327
00:24:09,000 --> 00:24:14,000
So, let's say I have some
resistor, and I have a standard

328
00:24:14,000 --> 00:24:17,000
current source with current I
nought.

329
00:24:17,000 --> 00:24:20,000
This is an independent current
source.

330
00:24:20,000 --> 00:24:24,000
Remember the circle?
And, some resistor,

331
00:24:24,000 --> 00:24:29,000
R, and let's say I care about
the voltage across the resistor.

332
00:24:29,000 --> 00:24:35,000
OK, so I have a current I
nought flowing through it.

333
00:24:35,000 --> 00:24:39,000
So, I can very quickly write
down VR as, simply,

334
00:24:39,000 --> 00:24:43,000
I0 R.
OK, it's the drop across the

335
00:24:43,000 --> 00:24:48,000
resistor when a current I nought
flows through it.

336
00:24:48,000 --> 00:24:53,000
OK, so this is what you've been
used to doing.

337
00:24:53,000 --> 00:24:57,000
Correspondingly,
I can do an example with a

338
00:24:57,000 --> 00:25:03,000
dependent current source.
And, as an example,

339
00:25:03,000 --> 00:25:07,000
I'll use a voltage controlled
current source.

340
00:25:07,000 --> 00:25:13,000
OK, a voltage controlled
current source is a dependent

341
00:25:13,000 --> 00:25:20,000
current source whose output
current depends on the voltage

342
00:25:20,000 --> 00:25:25,000
applied at the control port of
the current source.

343
00:25:25,000 --> 00:25:30,000
So let me build a little
circuit.

344
00:25:30,000 --> 00:25:34,000
OK, so here's my current.
And let's say it's VC IC for

345
00:25:34,000 --> 00:25:37,000
the control port,
and similarly,

346
00:25:37,000 --> 00:25:41,000
let's say my current I here is
some function of the control

347
00:25:41,000 --> 00:25:44,000
port voltage.
And let's say,

348
00:25:44,000 --> 00:25:47,000
to be specific,
there is some K over VC,

349
00:25:47,000 --> 00:25:50,000
some function.
OK, there are a variety of

350
00:25:50,000 --> 00:25:55,000
dependent sources that can be
built, and here's a hypothetical

351
00:25:55,000 --> 00:26:01,000
device where the output current
is mathematically related to the

352
00:26:01,000 --> 00:26:07,000
input in the following manner.
So, let me build a circuit of

353
00:26:07,000 --> 00:26:11,000
the following form.
So, let's add the resistor,

354
00:26:11,000 --> 00:26:14,000
R, and here's my circuit,
OK?

355
00:26:14,000 --> 00:26:18,000
And, as before,
let me look to figuring out

356
00:26:18,000 --> 00:26:21,000
what VR is.
So, notice that I have to

357
00:26:21,000 --> 00:26:27,000
supply some voltage at the input
so that the output can depend on

358
00:26:27,000 --> 00:26:34,000
the input because right now I
don't know what the input here.

359
00:26:34,000 --> 00:26:38,000
So what I'll do is let me apply
VR over here.

360
00:26:38,000 --> 00:26:41,000
OK, so let me make this
connection.

361
00:26:41,000 --> 00:26:46,000
OK, let me make the connection
from here to here.

362
00:26:46,000 --> 00:26:52,000
What I've done is I've applied
VR at the control port of the

363
00:26:52,000 --> 00:26:58,000
dependent current source.
OK, and I often draw a circuit

364
00:26:58,000 --> 00:27:03,000
like this.
This looks pretty messy.

365
00:27:03,000 --> 00:27:10,000
I will often draw the circuit
like so: R, VR.

366
00:27:26,000 --> 00:27:30,000
OK, short form circuit drawing
would look like this.

367
00:27:30,000 --> 00:27:34,000
This is a complete drawing that
I show you the explicit

368
00:27:34,000 --> 00:27:38,000
connections of the control port,
but oftentimes,

369
00:27:38,000 --> 00:27:43,000
when the control port does not
have any other impact in the

370
00:27:43,000 --> 00:27:48,000
circuit, you can eliminate,
don't explicitly show the

371
00:27:48,000 --> 00:27:51,000
control port.
Rather, you can simply show the

372
00:27:51,000 --> 00:27:57,000
dependence of the output current
on whatever circuit variable you

373
00:27:57,000 --> 00:28:00,000
have in mind.
So, you can draw the diamond

374
00:28:00,000 --> 00:28:04,000
like this, and see its current
is some function of VR.

375
00:28:04,000 --> 00:28:10,000
VR in this is case is K divided
by VR, OK?

376
00:28:10,000 --> 00:28:14,000
OK, so let's go ahead and
analyze this little circuit

377
00:28:14,000 --> 00:28:18,000
here, and look at what this
might give us.

378
00:28:18,000 --> 00:28:21,000
Our goal, as before,
is to find out the value,

379
00:28:21,000 --> 00:28:23,000
VR.
So, in this case,

380
00:28:23,000 --> 00:28:28,000
let's apply the Node method to
this node, and sum the currents

381
00:28:28,000 --> 00:28:34,000
into that node to be zero.
OK, so sum the currents going

382
00:28:34,000 --> 00:28:39,000
into that node to be zero.
The current going down is

383
00:28:39,000 --> 00:28:44,000
simply VR divided by R.
OK, and that is equal to the

384
00:28:44,000 --> 00:28:47,000
current that is going out of the
node.

385
00:28:47,000 --> 00:28:50,000
And so that is equal to F of
VR.

386
00:28:50,000 --> 00:28:54,000
And I know that F of VR is
given by K divided by VR.

387
00:28:54,000 --> 00:29:00,000
OK, a simple application of the
Node method.

388
00:29:00,000 --> 00:29:04,000
So then, I collect VR's on the
left hand side,

389
00:29:04,000 --> 00:29:10,000
and I get VR squared is K times
R, OK, and VR is simply the

390
00:29:10,000 --> 00:29:13,000
square root of KR.
There you go:

391
00:29:13,000 --> 00:29:17,000
I'm done.
OK, I've gone ahead an applied

392
00:29:17,000 --> 00:29:22,000
the Node method to this,
and when have to figure out the

393
00:29:22,000 --> 00:29:27,000
current here,
I simply reflect the fact that

394
00:29:27,000 --> 00:29:32,000
it depends on VR like so,
and I just go ahead and solve

395
00:29:32,000 --> 00:29:37,000
the circuit.
Remember, the workhorse of the

396
00:29:37,000 --> 00:29:40,000
circuit industry,
the Node method,

397
00:29:40,000 --> 00:29:42,000
when in doubt,
apply it.

398
00:29:42,000 --> 00:29:46,000
It simply works.
And notice, this is a nonlinear

399
00:29:46,000 --> 00:29:48,000
circuit.
OK, the dependence is

400
00:29:48,000 --> 00:29:53,000
nonlinear, and I get the
response like so.

401
00:29:53,000 --> 00:30:00,000
So, to plug in some numbers,
supposing K was 10 to the minus

402
00:30:00,000 --> 00:30:05,000
3 amperes per volt,
and R was one kilo ohm,

403
00:30:05,000 --> 00:30:12,000
then I can plug the numbers in
and the kilo here cancels with

404
00:30:12,000 --> 00:30:18,000
the 10 to the minus 3,
and I get VR equals 1 V.

405
00:30:18,000 --> 00:30:24,000
OK, this simply says,
if I build a circuit like this,

406
00:30:24,000 --> 00:30:30,000
then this voltage here will be
1 V.

407
00:30:30,000 --> 00:30:34,000
So, again, as long as you
remember that the dependent

408
00:30:34,000 --> 00:30:38,000
source is simply another little
circuit element,

409
00:30:38,000 --> 00:30:43,000
OK, and you usually draw just
the output port for dependent

410
00:30:43,000 --> 00:30:48,000
sources, and reflect the way
that the control affects the

411
00:30:48,000 --> 00:30:52,000
current, that'll suffice,
and you get,

412
00:30:52,000 --> 00:30:57,000
through the application of the
Node method, the variable you're

413
00:30:57,000 --> 00:31:02,000
interested in.
Let's do another example,

414
00:31:02,000 --> 00:31:08,000
OK, of another fun current
source, a voltage controlled

415
00:31:08,000 --> 00:31:12,000
current source,
and look at it this way.

416
00:31:12,000 --> 00:31:17,000
So, let's say I have a
resistor, and I have a current

417
00:31:17,000 --> 00:31:22,000
source, a resistor,
RL, and this goes to some,

418
00:31:22,000 --> 00:31:26,000
I apply a VS here.
Remember this short form

419
00:31:26,000 --> 00:31:33,000
notation; that's simply applying
a supply VS between that node

420
00:31:33,000 --> 00:31:39,000
and the ground.
OK, and let us say the current

421
00:31:39,000 --> 00:31:45,000
IV through the device is some
function of the current at its

422
00:31:45,000 --> 00:31:49,000
control port.
OK, so I'm not going to show

423
00:31:49,000 --> 00:31:53,000
you that.
But remember that the device

424
00:31:53,000 --> 00:31:58,000
already looks like this,
that there is a control port

425
00:31:58,000 --> 00:32:02,000
here.
I'm not showing that to you.

426
00:32:02,000 --> 00:32:05,000
And let us say that I apply
some voltage,

427
00:32:05,000 --> 00:32:09,000
VI, to the input port.
The reason we often don't show

428
00:32:09,000 --> 00:32:13,000
the input port is for many
practical dependent sources,

429
00:32:13,000 --> 00:32:17,000
the input has no other effect
on the circuit.

430
00:32:17,000 --> 00:32:19,000
So, for example,
in this case,

431
00:32:19,000 --> 00:32:22,000
the input has infinite
resistance looking in.

432
00:32:22,000 --> 00:32:25,000
So therefore,
if I apply a VI here,

433
00:32:25,000 --> 00:32:28,000
it doesn't draw any current
from VI.

434
00:32:28,000 --> 00:32:32,000
I simply apply the voltage,
VI.

435
00:32:32,000 --> 00:32:38,000
It doesn't affect the circuit
in any other way except in terms

436
00:32:38,000 --> 00:32:42,000
of how it controls the current
ID.

437
00:32:42,000 --> 00:32:49,000
So let's say the current ID is
some function of VI because VI

438
00:32:49,000 --> 00:32:55,000
is applied at the control port.
OK, and as I pointed out

439
00:32:55,000 --> 00:32:59,000
before, I oftentimes,
just for clarity,

440
00:32:59,000 --> 00:33:06,000
just to show this dependent
source explicitly.

441
00:33:06,000 --> 00:33:10,000
OK, so let's work the example.
So as I said,

442
00:33:10,000 --> 00:33:16,000
I'm going to choose ID to be F
of VI, and let's pick some

443
00:33:16,000 --> 00:33:22,000
specific parameters here.
Let's say it's K by two VI

444
00:33:22,000 --> 00:33:28,000
minus one, both squared.
OK, and let's say this is true

445
00:33:28,000 --> 00:33:33,000
for VI less than equal to one
volt.

446
00:33:33,000 --> 00:33:38,000
And let us also say that ID
equals zero for VI less than one

447
00:33:38,000 --> 00:33:40,000
volt.
OK, it's a dependent source,

448
00:33:40,000 --> 00:33:45,000
and it can have various forms
of dependences on the input.

449
00:33:45,000 --> 00:33:49,000
And, I just picked an example
of some hypothetical,

450
00:33:49,000 --> 00:33:53,000
or as yet, hypothetical
dependent source,

451
00:33:53,000 --> 00:33:58,000
the current through which is
related to the input using a

452
00:33:58,000 --> 00:34:02,000
square law relation,
VI minus one all squared as

453
00:34:02,000 --> 00:34:08,000
long as VI is greater than one.
And if VI is less than one,

454
00:34:08,000 --> 00:34:11,000
then the current is simply
zero, it shuts off.

455
00:34:11,000 --> 00:34:16,000
So, I can go ahead and apply.
So, let's say I want to find

456
00:34:16,000 --> 00:34:19,000
out V0 versus VI.
So, I care about finding out

457
00:34:19,000 --> 00:34:22,000
V0.
V0 is the voltage of this node

458
00:34:22,000 --> 00:34:26,000
with respect to ground.
OK, so it's a slightly more

459
00:34:26,000 --> 00:34:32,000
complicated circuit than you saw
up here, than you saw up there.

460
00:34:32,000 --> 00:34:35,000
So, let's go ahead and do this
example.

461
00:34:35,000 --> 00:34:39,000
Start by applying the workhorse
of the circuits business,

462
00:34:39,000 --> 00:34:43,000
the Node method,
and let's start with doing this

463
00:34:43,000 --> 00:34:46,000
for VI.
Let's first do it for VI

464
00:34:46,000 --> 00:34:50,000
greater than one,
notice the behavior of this is

465
00:34:50,000 --> 00:34:53,000
different for different ranges
of VI.

466
00:34:53,000 --> 00:34:58,000
So let's first do it for VI
greater than or equal to one and

467
00:34:58,000 --> 00:35:02,000
apply the Node method.
Node method says sum the

468
00:35:02,000 --> 00:35:06,000
currents going into this node;
we know the voltage at this

469
00:35:06,000 --> 00:35:07,000
node.
It's VI.

470
00:35:07,000 --> 00:35:09,000
We know the voltage at this
node.

471
00:35:09,000 --> 00:35:11,000
It's VS.
OK, the only unknown is V

472
00:35:11,000 --> 00:35:14,000
nought.
And so, let's go ahead and

473
00:35:14,000 --> 00:35:16,000
write the node equations for
that node.

474
00:35:16,000 --> 00:35:20,000
So, the current going up,
let me simply equate the

475
00:35:20,000 --> 00:35:24,000
current going up to the current
that has been supplied by this

476
00:35:24,000 --> 00:35:27,000
particular node here.
And, that should equate that

477
00:35:27,000 --> 00:35:31,000
the two of them should sum to
zero, the current going up plus

478
00:35:31,000 --> 00:35:36,000
the current going down should
sum to zero.

479
00:35:36,000 --> 00:35:38,000
So, I get V0 minus VS divided
by R.

480
00:35:38,000 --> 00:35:43,000
That's the current going up.
Plus, the current going down

481
00:35:43,000 --> 00:35:46,000
must sum to zero,
plus ID must sum to zero.

482
00:35:46,000 --> 00:35:51,000
And ID is going to be K divided
by two VI minus one all squared.

483
00:35:51,000 --> 00:35:56,000
That must equal zero.
Straightforward application of

484
00:35:56,000 --> 00:36:00,000
Node method, current going up
plus the current going down at

485
00:36:00,000 --> 00:36:05,000
this node should equal zero
because the total current

486
00:36:05,000 --> 00:36:09,000
leaving the node must be zero,
OK?

487
00:36:09,000 --> 00:36:13,000
So I can go ahead and simplify
this, multiply it throughout by,

488
00:36:13,000 --> 00:36:17,000
I call this RL here.
So, multiply it throughout by

489
00:36:17,000 --> 00:36:22,000
RL, and move all of this to the
other side, so I get VS divided

490
00:36:22,000 --> 00:36:24,000
by RL, multiply it throughout by
RL.

491
00:36:24,000 --> 00:36:28,000
I get VS at this side.
I take this term to the other

492
00:36:28,000 --> 00:36:31,000
side.
This becomes a minus.

493
00:36:31,000 --> 00:36:34,000
RL multiplies here,
so I get KRL.

494
00:36:34,000 --> 00:36:39,000
That's the expression I get.
V nought is VS minus KRL

495
00:36:39,000 --> 00:36:43,000
divided by two times VI minus
one all squared.

496
00:36:43,000 --> 00:36:48,000
Let me put a box around this
because I will be referring to

497
00:36:48,000 --> 00:36:54,000
this more times in 6.002 for a
variety of reasons than probably

498
00:36:54,000 --> 00:36:59,000
any other equation on Earth.
OK, this is the first time you

499
00:36:59,000 --> 00:37:03,000
saw it.
You saw it here.

500
00:37:03,000 --> 00:37:06,000
OK, mark it down.
You'll smile every other time

501
00:37:06,000 --> 00:37:10,000
you look at it in quizzes,
and you will find out why this

502
00:37:10,000 --> 00:37:14,000
comes up very often in 6.002.
So, I'll just give you a few

503
00:37:14,000 --> 00:37:18,000
seconds to savor this big moment
in your 6.002 life.

504
00:37:18,000 --> 00:37:21,000
All right, OK,
so it's pretty simple actually.

505
00:37:21,000 --> 00:37:23,000
I mean, there's really not
much.

506
00:37:23,000 --> 00:37:28,000
A lot of this stuff is just a
plain old, simple application of

507
00:37:28,000 --> 00:37:31,000
the Node method,
and things just fall out.

508
00:37:31,000 --> 00:37:35,000
It's just so simple.
So, the V nought,

509
00:37:35,000 --> 00:37:41,000
I apply the Node method,
I get V nought for this

510
00:37:41,000 --> 00:37:47,000
nonlinear circuit.
I can also it for VI less than

511
00:37:47,000 --> 00:37:49,000
one.
For VI less than one,

512
00:37:49,000 --> 00:37:54,000
when VI is less than one,
what happens?

513
00:37:54,000 --> 00:37:58,000
ID is zero.
OK, since ID is zero,

514
00:37:58,000 --> 00:38:01,000
think of this as an open
circuit.

515
00:38:01,000 --> 00:38:06,000
OK, so there's no voltage drop
across RL.

516
00:38:06,000 --> 00:38:13,000
And, this voltage V nought is
equal to VS.

517
00:38:13,000 --> 00:38:15,000
So, I like to see things in
pictures.

518
00:38:15,000 --> 00:38:18,000
I'm not an equations kind of
person.

519
00:38:18,000 --> 00:38:21,000
I'm much more of a graphical
person.

520
00:38:21,000 --> 00:38:25,000
So, let me draw a little graph
to show how V nought,

521
00:38:25,000 --> 00:38:29,000
to see the form of V nought,
and then let's study that

522
00:38:29,000 --> 00:38:34,000
little system a little bit more
carefully.

523
00:38:34,000 --> 00:38:39,000
So, this is page seven,
and we plot V nought versus VI

524
00:38:39,000 --> 00:38:42,000
for you.
And let's take a look at how

525
00:38:42,000 --> 00:38:46,000
this really simple circuit
looks.

526
00:38:46,000 --> 00:38:50,000
This has got nothing.
It's got an RL resistor

527
00:38:50,000 --> 00:38:55,000
connected to a supply,
and a dependent current source,

528
00:38:55,000 --> 00:39:01,000
and I apply some voltage VI at
the input.

529
00:39:01,000 --> 00:39:04,000
It's a very,
very simple circuit.

530
00:39:04,000 --> 00:39:08,000
So, let's see.
So as long as VI is less than

531
00:39:08,000 --> 00:39:14,000
one, the output stays at VS.
OK, that makes intuitive sense,

532
00:39:14,000 --> 00:39:18,000
right?
As long as the current here is

533
00:39:18,000 --> 00:39:21,000
zero, this is like an open
circuit here.

534
00:39:21,000 --> 00:39:26,000
If this is an open circuit,
then effectively,

535
00:39:26,000 --> 00:39:31,000
V nought is simply the voltage
VS.

536
00:39:31,000 --> 00:39:35,000
V nought simply appears here.
If you want to grunge through

537
00:39:35,000 --> 00:39:39,000
KVL and KCL, go ahead.
VS minus RL times the current

538
00:39:39,000 --> 00:39:43,000
is V nought, and the current is
zero so it's,

539
00:39:43,000 --> 00:39:45,000
yes.
So, this is simply VS.

540
00:39:45,000 --> 00:39:50,000
When VI goes above one volt,
fun stuff begins to happen.

541
00:39:50,000 --> 00:39:54,000
OK, when V nought goes above
one volt, then this equation

542
00:39:54,000 --> 00:39:58,000
applies because VI is greater
than one.

543
00:39:58,000 --> 00:40:02,000
This equation applies.
And, when VI is a one,

544
00:40:02,000 --> 00:40:06,000
one minus one is zero.
This term cancels out,

545
00:40:06,000 --> 00:40:08,000
so this is VS.
OK, phew!

546
00:40:08,000 --> 00:40:12,000
So, I start off here.
As VI increases,

547
00:40:12,000 --> 00:40:15,000
what happens now?
As VI increases,

548
00:40:15,000 --> 00:40:19,000
this term here becomes
increasingly negative,

549
00:40:19,000 --> 00:40:23,000
OK, subtracting from VS.
OK, so I get some behavior like

550
00:40:23,000 --> 00:40:26,000
this.
V nought begins to drop.

551
00:40:26,000 --> 00:40:31,000
And it makes intuitive sense,
right?

552
00:40:31,000 --> 00:40:35,000
As ID begins to increase,
the voltage here will begin to

553
00:40:35,000 --> 00:40:40,000
drop because I'm drawing more
and more current through RL.

554
00:40:40,000 --> 00:40:43,000
I'm dropping more and more
across RL.

555
00:40:43,000 --> 00:40:48,000
So more and more drops across
RL, so V nought begins to drop

556
00:40:48,000 --> 00:40:51,000
too.
So, it looks something like

557
00:40:51,000 --> 00:40:53,000
this.
I'll show you a little demo,

558
00:40:53,000 --> 00:40:58,000
but my claim is that you have
just seen an amplifier.

559
00:40:58,000 --> 00:41:03,000
Whoa.
You just saw an amplifier.

560
00:41:03,000 --> 00:41:07,000
So, I snuck an amplifier by
you, OK?

561
00:41:07,000 --> 00:41:11,000
So, I just snuck an amplifier
past you.

562
00:41:11,000 --> 00:41:18,000
I'll show you why in a second.
So, let's take a look at this

563
00:41:18,000 --> 00:41:23,000
waveform here.
Let's not worry about what

564
00:41:23,000 --> 00:41:29,000
happens way down here.
We'll talk about that a little

565
00:41:29,000 --> 00:41:36,000
later.
But, look at this curve here.

566
00:41:36,000 --> 00:41:44,000
I claim there is amplification
in the following sense.

567
00:41:44,000 --> 00:41:51,000
Focus on some change in the
input voltage,

568
00:41:51,000 --> 00:41:57,000
delta VI, OK,
and for that change in input

569
00:41:57,000 --> 00:42:06,000
voltage, I get some change in
the output voltage.

570
00:42:06,000 --> 00:42:09,000
OK, for some change in the
input voltage,

571
00:42:09,000 --> 00:42:13,000
delta VI, I get some change in
the output voltage.

572
00:42:13,000 --> 00:42:17,000
And guess what?
In this, at least the way I

573
00:42:17,000 --> 00:42:21,000
have drawn it,
delta V nought divided by delta

574
00:42:21,000 --> 00:42:26,000
VI, if I can find regions of the
curve where this is greater than

575
00:42:26,000 --> 00:42:32,000
one, then I have amplification.
OK, so what's that saying?

576
00:42:32,000 --> 00:42:37,000
What that's saying is that if I
apply some voltage here,

577
00:42:37,000 --> 00:42:42,000
OK, and I change that voltage
by a small amount from,

578
00:42:42,000 --> 00:42:46,000
let's say, 2 V to 2.1.
OK, I am going to find the

579
00:42:46,000 --> 00:42:50,000
output voltage.
Let's say I go from 2 V to 2.1

580
00:42:50,000 --> 00:42:53,000
here.
OK, abstractly out there,

581
00:42:53,000 --> 00:42:57,000
I might have an output that
goes from three to,

582
00:42:57,000 --> 00:43:03,000
let's say, two V perhaps.
OK, so for a 0.1 change here,

583
00:43:03,000 --> 00:43:07,000
I'm going to get a bigger drop
here, so from 3 V to 2 V,

584
00:43:07,000 --> 00:43:10,000
giving me an amplification in
this little circuit.

585
00:43:10,000 --> 00:43:14,000
OK, so we'll see this again and
again, and you'll really

586
00:43:14,000 --> 00:43:17,000
understand it.
So, I have a small change in

587
00:43:17,000 --> 00:43:21,000
the input, and I have a
corresponding larger change in

588
00:43:21,000 --> 00:43:23,000
the output.
So, I've shown you an

589
00:43:23,000 --> 00:43:26,000
amplifier.
I haven't shown you a linear

590
00:43:26,000 --> 00:43:29,000
amplifier.
There's an extra charge for

591
00:43:29,000 --> 00:43:32,000
that.
OK, that'll happen later.

592
00:43:32,000 --> 00:43:35,000
OK, all I've shown you so far
is an amplifier,

593
00:43:35,000 --> 00:43:37,000
and this happens to be a crummy
amplifier.

594
00:43:37,000 --> 00:43:40,000
It's a nonlinear amplifier
because, notice,

595
00:43:40,000 --> 00:43:43,000
this is not linear.
It's a nice little curve,

596
00:43:43,000 --> 00:43:45,000
and so it's not linear.
But, I promised you an

597
00:43:45,000 --> 00:43:49,000
amplifier, and I'm cheap,
and that's all you get for now.

598
00:43:49,000 --> 00:43:52,000
OK, we'll see linear stuff
later, but for now,

599
00:43:52,000 --> 00:43:55,000
I have a little amplifier.
So, let's do some real numbers,

600
00:43:55,000 --> 00:44:00,000
and plot some numbers down,
and also look at a demo.

601
00:44:00,000 --> 00:44:04,000
So, let's do an example.
Let's say VS is 10 V,

602
00:44:04,000 --> 00:44:10,000
that the K is two milliamps per
V squared, and let's say RL is

603
00:44:10,000 --> 00:44:12,000
five kilo-ohms,
OK?

604
00:44:12,000 --> 00:44:18,000
So, let me substitute these
values into that equation,

605
00:44:18,000 --> 00:44:21,000
and I get V nought is,
VS is ten.

606
00:44:21,000 --> 00:44:25,000
So, it's ten minus,
KRL divided by two.

607
00:44:25,000 --> 00:44:31,000
So, K is two milliamps.
Two milliamps times five

608
00:44:31,000 --> 00:44:36,000
kilo-ohms is ten divided by two
gives me five,

609
00:44:36,000 --> 00:44:40,000
and VI minus one squared.
That's what I have.

610
00:44:40,000 --> 00:44:46,000
I just plug in a bunch of
numbers, and that's what I get.

611
00:44:46,000 --> 00:44:52,000
So, what I'll do is let me just
do a little table for you,

612
00:44:52,000 --> 00:44:57,000
and plot using real numbers,
simply plot those values for

613
00:44:57,000 --> 00:45:00,000
you.