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Good morning.
Today we move in the direction

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that takes a big turn from the
direction we have been going in

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so far.
All the devices we have had up

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until now, resistors and voltage
sources, and even your digital

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devices like the AND gate or the
inverter and so on had a very

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specific property.
We didn't dwell on that

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property, but that property was
that these were not what are

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called memory devices.
In other words,

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the outputs at any given time
are a function of the inputs

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alone.
In other words,

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if you took your inverter or
your NAND gate for that matter

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and you build a circuit
comprising 50 NAND gates

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connected in structures that we
have talked about,

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you apply an input and boom you
get an output.

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And your output is a function
of the inputs alone,

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right?
The same thing with your

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resistors and voltage sources.
At any given point in time your

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output VO of T was some function
of the input VI of T.

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What we are going to do today
is discuss a new element which

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will introduce a whole new class
of fun stuff for all of us to

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deal with.
And that is called storage.

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In other words,
the output of a circuit is now

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going to depend not just on the
inputs but it is going to depend

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on the background or it is going
to depend on where the circuit

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has been in the past.
So past is going to matter.

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It is a very fundamental
difference.

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And what I would like to do is
start by giving you folks a

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little bit of a surprise.
I am going to do a little demo

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taking two of your inverter
circuits.

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I am going to start by taking a
couple of inverters.

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Remember, I am using this
structure here as an inverter.

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And I am going to couple this
to another inverter and take an

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output C, some VS,
some load resistance RL,

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my B terminal and my A
terminal.

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So I'm going to apply some
input between ground and my A

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terminal.
And for fun I want to apply a

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square wave at the input.
A square wave between zero and

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5 volts.
And this is how my time goes.

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Let's assume that VS is 5
volts.

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So what I am going to do is
plot for you the behavior of

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this inverter.
I am going to plot for you A,

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which would look like this.
I am going to plot for you B,

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which would be the inverted
wave form.

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And then plot C,
which would be a wave form that

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looks like this again.
Let me do a plot here.

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So this is A.

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-- and so on.
Time goes this way.

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And let's say this is between
zero and 5 volts.

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And B should be an inverted
wave form that should look like

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this.

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If all that we believe of the
world so far is true then this

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is how things should behave,
so C should look like this.

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This is what the world should
look like and if everything that

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you learned about is true and
correct and all of the good

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stuff.
Let me show you a little demo

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and see if I can try to pull the
rug out from under all that you

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have learned so far and show you
some surprising stuff.

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Here are the three wave forms
that I showed you up here.

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This is my A.
This is my A wave form.

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This is the B wave form.
Notice that B,

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as you expect,
is an inverted form of A.

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And this is C.
We all expect this,

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correct?
But what I am going to do is

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let me expand the time scale on
this so that I can look at these

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transitions a little bit more
carefully.

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I am just going to expand the
time scale.

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There you go.
All I have done is expanded the

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time scale and spread that out a
little bit.

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And what you see there is quite
different from what you expect.

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A is a square wave as expected,
but B is stunningly different.

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It is a zero as expected
because this is a one.

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But here I get some really
strange behavior,

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behavior that is like nothing
on earth.

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Like nothing you have seen
before.

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And then, of course,
it becomes a one eventually,

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but there's some really,
really shady stuff going on

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here.
And so far you are not prepared

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to deal with this.
We have not given you the

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facility to deal with his issue.
What is the problem with this?

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We could say who cares?
What is the problem with this?

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Let's look at the result.
I am looking at this,

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I am focusing on this piece
here.

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And notice that instead of
being a sharp rise it looks like

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this.
It is going up a little bit

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more slowly.
What kind of problem would that

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create?
The problem that it creates is

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the following.
Let me play around with this

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graph a little bit more.
What I am going to do is just

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take this output here,
the C output and line it up

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against the A output.
And so I am going to line up

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the C wave form on top of the A
wave form.

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So you can see for yourself if
something really,

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really strange and nasty is
happening, I am just going to

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move up the C wave form and line
it up.

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What is happening out there?
If you look carefully,

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what you observe is that the C
wave form transitions just ever

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so slightly later than the A
wave form.

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Look here.
And I claim that it is because

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of this.
Because of this,

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the C wave form falls just a
little bit later,

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and that little thing we see
out there is a delay.

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So nothing you have learned so
far prepares you for this.

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Suddenly, instead of the output
exactly following the input,

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my output is following the
input but a little bit later.

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And it is this fact of life
that things happen a little bit

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later, is really the reason why
each of you and all of us needs

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to buy new computers every
couple of years.

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This simple basic fact.
If this fact of life didn't

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exist, you would buy one
computer and be done with it for

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life.
Intel would make gobs of money

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one year, and so would Dell and
Gateway and so on,

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and then no more.
That's it.

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This is it.
But because of this a little

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itty-bitty difference here the
entire semiconductor technology

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is charging along trying to do
something about that.

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You buy newer and newer
computers each year.

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It turns out this little
itty-bitty thing here,

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that is called the delay,
the inverter delay.

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And it happens because of a
specific element that has been

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introduced here that we have not
shown you so far.

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And a large part of the
semiconductor industry and

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follow-on courses and design and
so on focuses on how could I

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make my delay smaller,
how can I get to be faster and

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faster and faster?
This relates to how fast we can

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clock your Pentium IV.
Remember it came all the way to

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1.3 gigahertz?
What's the fasted Pentium money

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can buy today?
What is the fastest P4?

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Oh, 3.2 have come out?
I don't know.

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Ken claims 3.2.
But, yeah, there you go,

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3.2 gigahertz.
It all has to do with this

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little itty-bitty thing.
You saw it for the first time

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here.
When some of you become CTOs at

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Intel and so on,
just remember that it all began

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on October 16th with this little
rinky-dink thing here.

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What you are going to learn now
is some really cool stuff that

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has huge implications for life.
So why does that happen?

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Why did this transition happen
just a little bit later?

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The reason is that remember
when this wave form reaches VT,

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the threshold voltage of this
MOSFET, this guy is going to

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switch, right?
So because of the slower rise

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of the voltage,
the VT is going to be reached a

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small amount of time later.
So I am going to hit VT

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slightly later.
And because of that this guy is

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going to transition just a bit
later because this intermediate

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wave form B is slower.
It hits VT just a little bit

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later than if it would have made
an instantaneous transition.

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And therefore my output falls
just a little bit later and this

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gives rise to my delay in the
inverter.

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We can call that d if you would
like, some delay.

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In your course notes,
this material is covered in

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Chapters 9 and 10.
That was to kind of motivate

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why we are going to be doing all
that you we will be doing.

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Don't anybody come within a
foot of this even by mistake.

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I mean it.
It is pretty deadly stuff.

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Today we will talk about the
capacitor.

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And in the next couple of
lectures I am going to tie it

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all together and show you how
this relates to that.

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I will show you exactly how the
delay happens.

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You can compute it based on
some simple principles that you

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will learn about in the next
couple of lectures.

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What I am going to do is first
of all show you,

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I claim that that delay happens
because of the presence of a

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capacitor somewhere in there.
What I will do now is take you

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into a closer look,
take a closer look at the

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MOSFET and show you were the
capacitor is.

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This is the MOSFET that you
have seen so far,

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drain, gate and source.
This is called an n-channel

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MOSFET.
And what I am going to do is

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dissect this and show you what
is actually happening,

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what this looks like on
silicon.

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So here is my slab of silicon.
It is very thin.

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And let's say this is,
I won't go into details here.

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You will learn a lot more about
this in future device classes

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like 301 and so on,
but suffice it to say I will

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just introduce it here to give
you a sense of where the

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capacitor is.
This is p-type silicon.

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And the way you build a MOSFET
is you create a couple of tubs

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in which you dope to be n-type.
The basic silicon is dope

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p-type.
And this guy here is n-type.

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And what you do is a thin oxide
layer is placed on top of that

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and then on top of that a thin
metal layer.

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This is a metal layer.
This is a thin piece of oxide,

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silicon dioxide.
And this is my P substrate.

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Now this is a little metal
layer that is really a wire on

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top of the silicone.
This metal layer could be some

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sort of a wire that meanders
around on the surface of

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silicone.
And this is a wire that

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connects to the gate.
This is the gate of my MOSFET.

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And this guy here is the drain.
And this guy here is the

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source.
And this is my gate.

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So there is a little piece of
metal here.

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This is this piece of metal
here.

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And there is a piece of oxide
and then my silicone substrate.

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Notice that this is my oxide.
When I apply a positive voltage

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to the gate here with respect to
the substrate,

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what happens is that I draw up
negative charges.

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I draw up electrons here into
this channel region and I have

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corresponding plus type out here
so that I get a view here that

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looks like a couple of plates.
And I end up with an oxide in

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the middle.
There is no connection.

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Two plates separated by a small
distance with plus q and minus q

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on the plates.
And, because of that,

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what ends up happening here is
that this piece behaves like a

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capacitor.
So a capacitor has two plates

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with a thin insulating material
in the middle with some

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permittivity epsilon.
And so I get a little piece of

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a capacitor here.
That is the capacitor that is

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forming.
I did not set out to build that

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capacitor, but there is a
capacitor nonetheless.

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So when I apply a positive
voltage at the gate,

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negative electrons are pulled
up here which forms a channel,

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and then a current can then
flow.

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And that is how the MOSFET
turns on.

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So n-type electrons back to
n-type, and I get electron flow

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here and that gives me my
channel.

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This is just kind of devices in
four minutes or less.

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You will do an entire course on
this, if you like,

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if you take 301.
What we do is to be able to

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capture the behavior that we
just saw, the funny delayed

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behavior, we have to augment our
model.

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We have to introduce a new
element.

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So what we do is here is a
MOSFET, gate,

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drain and source.
And notice here we model this

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by putting a little capacitor,
CGS between our gate and the

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source.
So this becomes a simple model

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for our MOSFET device which is
the good old gate drain source

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device from the past with a
little capacitor CGS having some

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value for CGS in maybe ten to
the minus 14 or thereabouts

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farads.
So that is a little capacitor

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that has come about in this
device that we fabricated here.

231
00:17:03,000 --> 00:17:07,000
It is that capacitor that is at
between node B and ground

232
00:17:07,000 --> 00:17:12,000
because it is between the gate
and the source of the second

233
00:17:12,000 --> 00:17:15,000
inverter.
And it is that capacitor that

234
00:17:15,000 --> 00:17:20,000
is playing the games that we saw
out there.

235
00:17:27,000 --> 00:17:31,000
So let's look at some of the
behavior of an ideal linear

236
00:17:31,000 --> 00:17:34,000
capacitor.
A capacitor,

237
00:17:34,000 --> 00:17:38,000
as I said, has a couple of
plates.

238
00:17:38,000 --> 00:17:44,000
There are a couple of plates.
Between the plates is some

239
00:17:44,000 --> 00:17:51,000
dieletric, permittivity epsilon.
Let's say the area of the

240
00:17:51,000 --> 00:17:57,000
plates is A, and let's say the
plates are separated by a

241
00:17:57,000 --> 00:18:02,000
distance D.
I get some charge here,

242
00:18:02,000 --> 00:18:06,000
let's say q.
So q and minus q on the

243
00:18:06,000 --> 00:18:10,000
capacitor.
And the capacitance C is given

244
00:18:10,000 --> 00:18:15,000
by epsilon A divided by D.
Epsilon, as I said,

245
00:18:15,000 --> 00:18:19,000
is the productivity of the
dielectric.

246
00:18:19,000 --> 00:18:25,000
So if it is free space then it
would be epsilon zero which is

247
00:18:25,000 --> 00:18:32,000
the permittivity of free space.
That is the capacitance in

248
00:18:32,000 --> 00:18:35,000
farads.
And the symbol looks like this.

249
00:18:35,000 --> 00:18:38,000
Capacitor C.
Voltage v.

250
00:18:38,000 --> 00:18:40,000
Current i.
So this, much like the

251
00:18:40,000 --> 00:18:46,000
resistor, voltage source and so
on, this now becomes a primitive

252
00:18:46,000 --> 00:18:52,000
element in your tool chest of
elements like the voltage source

253
00:18:52,000 --> 00:18:56,000
and so onn.
Capacitance with the voltage v

254
00:18:56,000 --> 00:19:01,000
across it and a current i.
And I have assigned the

255
00:19:01,000 --> 00:19:05,000
associated variables here
according to the associated

256
00:19:05,000 --> 00:19:08,000
variable discipline.
A question to ask ourselves is

257
00:19:08,000 --> 00:19:13,000
remember we said we are all now
in a playground from all of

258
00:19:13,000 --> 00:19:17,000
nature, in this playground where
the lumped matter discipline

259
00:19:17,000 --> 00:19:20,000
holds?
And also remember that we said

260
00:19:20,000 --> 00:19:23,000
that for the lumped matter
discipline to hold we have to

261
00:19:23,000 --> 00:19:29,000
make a couple of assumptions.
One of those assumptions was

262
00:19:29,000 --> 00:19:33,000
that dq/dt, for all their
elements should be zero for all

263
00:19:33,000 --> 00:19:36,000
time.
So right now what about the

264
00:19:36,000 --> 00:19:39,000
capacitor?
It has got some charge q.

265
00:19:39,000 --> 00:19:42,000
So charge must have built up
somehow.

266
00:19:42,000 --> 00:19:47,000
Does that mean that I lied all
along, that we are no longer in

267
00:19:47,000 --> 00:19:51,000
this playground,
that we have been ejected from

268
00:19:51,000 --> 00:19:56,000
the playground because of the
capacitor, or are we still in

269
00:19:56,000 --> 00:20:01,000
the circuits playground in which
the lumped matter discipline

270
00:20:01,000 --> 00:20:06,000
holds and all good things happen
and so on?

271
00:20:06,000 --> 00:20:09,000
It seems like a contradiction,
doesn't it?

272
00:20:09,000 --> 00:20:12,000
I took you from Maxwell's
playgrounds to the EECS

273
00:20:12,000 --> 00:20:17,000
playground where I said the
lumped matter discipline holds.

274
00:20:17,000 --> 00:20:21,000
And one of the foundations of
the LMD was that dq/dt should be

275
00:20:21,000 --> 00:20:26,000
zero for all time inside the
elements that we are going to

276
00:20:26,000 --> 00:20:28,000
deal with.
And right now boom,

277
00:20:28,000 --> 00:20:32,000
it's not four weeks into the
course and Agarwal introduces an

278
00:20:32,000 --> 00:20:38,000
element and it has q in it.
It turns out that the capacitor

279
00:20:38,000 --> 00:20:41,000
also adheres to the lumped
matter discipline.

280
00:20:41,000 --> 00:20:45,000
Remember the discipline says
that dq/dt is zero for all time

281
00:20:45,000 --> 00:20:48,000
within elements.
So I am going to be clever.

282
00:20:48,000 --> 00:20:52,000
What I am going to do is I want
to choose element boundaries in

283
00:20:52,000 --> 00:20:55,000
a very cleaver way.
Notice that if I have q here on

284
00:20:55,000 --> 00:21:00,000
this plate then I get minus q on
the other plate.

285
00:21:00,000 --> 00:21:04,000
So if I take the whole element,
the element as a whole,

286
00:21:04,000 --> 00:21:09,000
if I am careful in terms of how
I package my boundaries,

287
00:21:09,000 --> 00:21:14,000
if I put both my plates inside
my element boundary then I still

288
00:21:14,000 --> 00:21:17,000
do get the net charge being
zero.

289
00:21:17,000 --> 00:21:22,000
So dq/dt is indeed zero for all
time provided I make sure that

290
00:21:22,000 --> 00:21:27,000
my element has both the plates.
Therefore, if you come across

291
00:21:27,000 --> 00:21:32,000
somebody else that gives you an
element that says I have an

292
00:21:32,000 --> 00:21:36,000
idea.
Let's create a new branch of

293
00:21:36,000 --> 00:21:40,000
electrical engineering in which
we model the capacitor not as

294
00:21:40,000 --> 00:21:44,000
one element for two plates,
but let's build a capacitor by

295
00:21:44,000 --> 00:21:48,000
combining two new elements,
two garbage elements called G1

296
00:21:48,000 --> 00:21:51,000
and G2.
G1 is like the top plate.

297
00:21:51,000 --> 00:21:55,000
G2 is the bottom plate.
I put them together and I get a

298
00:21:55,000 --> 00:21:58,000
capacitor.
But notice if I just pick one

299
00:21:58,000 --> 00:22:03,000
plate then the element G1 will
not adhere to the LMD.

300
00:22:03,000 --> 00:22:08,000
It adheres to the LMD because I
choose my element boundaries in

301
00:22:08,000 --> 00:22:11,000
a way that both plates come
within it.

302
00:22:11,000 --> 00:22:14,000
So it is very fundamental and
key.

303
00:22:14,000 --> 00:22:18,000
And you can read a lot more
about it in the course notes.

304
00:22:18,000 --> 00:22:23,000
I purposely dwelt on that
simple point because I think it

305
00:22:23,000 --> 00:22:29,000
is foundational and important.
And you really need to

306
00:22:29,000 --> 00:22:33,000
understand that the capacitor
does satisfy LMD.

307
00:22:33,000 --> 00:22:37,000
We are still in the good old
playground.

308
00:22:37,000 --> 00:22:41,000
A few simple facts here.
These are in the notes.

309
00:22:41,000 --> 00:22:45,000
And you have also seen this
before, I am sure.

310
00:22:45,000 --> 00:22:51,000
I can relate the charge to the
capacitance and the voltage as q

311
00:22:51,000 --> 00:22:55,000
is equal to Cv.
And q is in coulombs,

312
00:22:55,000 --> 00:23:00,000
this is in farads and this is
in volts.

313
00:23:00,000 --> 00:23:06,000
So there is some charge q
stored on the capacitor and it

314
00:23:06,000 --> 00:23:10,000
is in coulombs and q is equal to
Cv.

315
00:23:10,000 --> 00:23:17,000
So I can differentiate this
with respect to time to get the

316
00:23:17,000 --> 00:23:21,000
current, and that becomes
i=dq/dt.

317
00:23:21,000 --> 00:23:28,000
So the current at any given
time is dq/dt.

318
00:23:28,000 --> 00:23:32,000
And so I substitute for q in
terms of Cv here.

319
00:23:32,000 --> 00:23:36,000
That is what I get.
So the current i=d(Cv)/dt.

320
00:23:36,000 --> 00:23:41,000
A 6.002 assumption,
capacitance in general can be

321
00:23:41,000 --> 00:23:44,000
time-varying.
I can get time-varying

322
00:23:44,000 --> 00:23:48,000
capacitors.
In fact, there are some sensors

323
00:23:48,000 --> 00:23:51,000
which are capacitive.
And, as I talk,

324
00:23:51,000 --> 00:23:57,000
my sound waves can change the
pressure on the top plate of the

325
00:23:57,000 --> 00:24:02,000
capacitor.
And move the top plate of the

326
00:24:02,000 --> 00:24:08,000
capacitor, thereby changing the
capacitance by moving the plate.

327
00:24:08,000 --> 00:24:13,000
Remember d here,
as the plate moves closer I get

328
00:24:13,000 --> 00:24:17,000
a higher capacitance.
So we won't be dealing,

329
00:24:17,000 --> 00:24:23,000
unless explicitly said so,
with time-varying capacitances.

330
00:24:23,000 --> 00:24:29,000
So what we can do is 6.002
allows us to write Cdv/dt.

331
00:24:29,000 --> 00:24:33,000
So my current source capacitor
is Cdv/dt.

332
00:24:33,000 --> 00:24:39,000
I can also write down the
energy, capacitors store energy.

333
00:24:39,000 --> 00:24:42,000
E=1/2Cv^2.
I am sure you have seen all

334
00:24:42,000 --> 00:24:46,000
this before in physics and so
on.

335
00:24:46,000 --> 00:24:52,000
That is the amount of energy
stored in the capacitor if it is

336
00:24:52,000 --> 00:24:56,000
holding a charge q.
Let me do a little

337
00:24:56,000 --> 00:25:02,000
demonstration for you.
They don't make glasses like

338
00:25:02,000 --> 00:25:07,000
they used to.
Our friend Lorenzo has charged

339
00:25:07,000 --> 00:25:11,000
up this capacitor.
It is a huge capacitor.

340
00:25:11,000 --> 00:25:15,000
It is a 250 volt capacitor so
it is nasty.

341
00:25:15,000 --> 00:25:19,000
He has charged it up and has
kept it there.

342
00:25:19,000 --> 00:25:25,000
And to show you that it does
contain stored charges it has

343
00:25:25,000 --> 00:25:30,000
been sitting there holding
charge.

344
00:25:30,000 --> 00:25:36,000
Maybe the first row should go
backwards, just step back for a

345
00:25:36,000 --> 00:25:40,000
second.
I think you guys would be safe

346
00:25:40,000 --> 00:25:44,000
but I just don't want to take
any chances.

347
00:25:44,000 --> 00:25:48,000
This is holding a bunch of
charge.

348
00:25:48,000 --> 00:25:54,000
It is kind of sitting there.
If I short the terminals it

349
00:25:54,000 --> 00:25:58,000
should try to say oh,
I've got a path,

350
00:25:58,000 --> 00:26:02,000
let me get my charge out.
All right.

351
00:26:02,000 --> 00:26:05,000
Let's do it.
This is always a scary moment

352
00:26:05,000 --> 00:26:08,000
for me.
And I say a little prayer

353
00:26:08,000 --> 00:26:10,000
before I do this.

354
00:26:19,000 --> 00:26:19,000
Good?
OK.
Gee, you guys would love to see
me getting fried,

355
00:26:21,000 --> 00:26:22,000
huh?
All right.

356
00:26:22,000 --> 00:26:23,000
Let's see.

357
00:26:45,000 --> 00:26:47,000
So it did contain charge.

358
00:27:00,000 --> 00:27:04,000
So there is a reason why
Lorenzo puts one hand inside his

359
00:27:04,000 --> 00:27:08,000
pocket when he shorts it,
because there is a natural

360
00:27:08,000 --> 00:27:12,000
tendency to hold the wire with
both hands, and la,

361
00:27:12,000 --> 00:27:16,000
la, la, la, la and put it
across the capacitor.

362
00:27:16,000 --> 00:27:21,000
By doing this you are
guaranteed that you will just be

363
00:27:21,000 --> 00:27:25,000
touching it with one hand.
Hopefully you folks will

364
00:27:25,000 --> 00:27:29,000
remember for life that a
capacitor can sit around and

365
00:27:29,000 --> 00:27:33,000
hold its charge for a while.
All right.

366
00:27:33,000 --> 00:27:35,000
That is enough of fun and
games.

367
00:27:35,000 --> 00:27:39,000
Let's get on with our business
of building circuits.

368
00:27:39,000 --> 00:27:41,000
What I am going to do is,
as I promised you,

369
00:27:41,000 --> 00:27:46,000
I am going to close the loop on
that example by halfway through

370
00:27:46,000 --> 00:27:49,000
the next lecture.
I'm going take you on a bit of

371
00:27:49,000 --> 00:27:53,000
a journey involving capacitors
and resistors and involving some

372
00:27:53,000 --> 00:27:57,000
analysis, and then we will close
it all up for you at about the

373
00:27:57,000 --> 00:28:02,000
middle of next lecture.
What I would like to do next is

374
00:28:02,000 --> 00:28:06,000
here is a new element.
And let's do some fun stuff

375
00:28:06,000 --> 00:28:10,000
with elements.
Well, you know about voltage

376
00:28:10,000 --> 00:28:14,000
sources, you know about
resistors, let's put them

377
00:28:14,000 --> 00:28:17,000
together and see how they
behave.

378
00:28:17,000 --> 00:28:21,000
Let's have a capacitor here,
C, vc(t) and some current i.

379
00:28:21,000 --> 00:28:25,000
What I am going to do,
in general, whenever I have

380
00:28:25,000 --> 00:28:30,000
something new or something
strange, let's say like a

381
00:28:30,000 --> 00:28:36,000
capacitor or some other device.
It is interesting to model the

382
00:28:36,000 --> 00:28:41,000
rest of the circuit behind it if
it contains only resistors and

383
00:28:41,000 --> 00:28:46,000
voltages and linear elements as
a Thevenin equivalent.

384
00:28:46,000 --> 00:28:50,000
So let me do that.
This is R and this is vi.

385
00:28:50,000 --> 00:28:54,000
This stuff in the back is my
standard pattern,

386
00:28:54,000 --> 00:28:59,000
voltage source in series with a
resistor, and I connect that

387
00:28:59,000 --> 00:29:03,000
across my capacitor.
But remember,

388
00:29:03,000 --> 00:29:07,000
although you saw those funny
wave forms and so on,

389
00:29:07,000 --> 00:29:10,000
the capacitor is a linear
device.

390
00:29:10,000 --> 00:29:15,000
Because you can see from here
that the current relates to

391
00:29:15,000 --> 00:29:18,000
dv/dt.
That is a linear operation.

392
00:29:18,000 --> 00:29:23,000
You don't see V squareds and
Vis and things like that in

393
00:29:23,000 --> 00:29:25,000
there.
It's is a linear device.

394
00:29:25,000 --> 00:29:32,000
Let's go back to our trusty old
method, the node method.

395
00:29:32,000 --> 00:29:36,000
If you just blindly apply the
node method and simply grunge

396
00:29:36,000 --> 00:29:40,000
through a bunch of math,
you should be able to get to

397
00:29:40,000 --> 00:29:44,000
the answer, that is for some
voltage v or some form of

398
00:29:44,000 --> 00:29:49,000
voltage vi, I should be able to
figure out what vc looks like.

399
00:29:49,000 --> 00:29:52,000
So let's do that.
This is the node that is of

400
00:29:52,000 --> 00:29:56,000
interest here with the unknown
node voltage vc.

401
00:29:56,000 --> 00:29:58,000
So let me apply the node
method.

402
00:29:58,000 --> 00:30:03,000
(vc-vi)/R is the current going
this way.

403
00:30:03,000 --> 00:30:09,000
That plus the current through
the capacitor should equal zero.

404
00:30:09,000 --> 00:30:14,000
And what is the current through
the capacitor?

405
00:30:14,000 --> 00:30:20,000
The node method tells me that,
get the current in terms of the

406
00:30:20,000 --> 00:30:25,000
element values.
We know that the current is

407
00:30:25,000 --> 00:30:31,000
given by CdvC/dt.=O.
Just shuffling things around a

408
00:30:31,000 --> 00:30:35,000
little bit, I can write RC
dvc/dt+vc=vi.

409
00:30:35,000 --> 00:30:41,000
We are writing the node
equation and then getting the

410
00:30:41,000 --> 00:30:47,000
equation that characterizes this
little circuit.

411
00:30:47,000 --> 00:30:51,000
Notice here that this has units
of volts.

412
00:30:51,000 --> 00:30:58,000
And since I have time here,
this also must have units of

413
00:30:58,000 --> 00:31:00,000
time.

414
00:31:06,000 --> 00:31:12,000
Let's go about solving this
little circuit and understanding

415
00:31:12,000 --> 00:31:16,000
how it behaves.
The specific example that we

416
00:31:16,000 --> 00:31:22,000
will look at looks like this.
Let's say the capacitor voltage

417
00:31:22,000 --> 00:31:25,000
at time T=0 is V0.
This is given.

418
00:31:25,000 --> 00:31:30,000
So at time T=0,
I am telling you that the

419
00:31:30,000 --> 00:31:36,000
capacitor contains a charge.
And because of that there is a

420
00:31:36,000 --> 00:31:40,000
voltage V0 across it.
That capacitor had a voltage of

421
00:31:40,000 --> 00:31:45,000
250 volts across it and most of
the devices we deal with in

422
00:31:45,000 --> 00:31:48,000
laptops and so on today,
like the Pentium IV,

423
00:31:48,000 --> 00:31:53,000
voltages are on the order of
1.5 volts, very small voltages.

424
00:31:53,000 --> 00:31:57,000
So that is the value in the
capacitor, the voltage.

425
00:31:57,000 --> 00:32:02,000
That is called a state.
This is called the state,

426
00:32:02,000 --> 00:32:05,000
capacitor state.
It is the state of the

427
00:32:05,000 --> 00:32:08,000
capacitor.
And I also give you that

428
00:32:08,000 --> 00:32:10,000
vi(t)=VI.
So my voltage is VI.

429
00:32:10,000 --> 00:32:14,000
And somehow,
I am not telling you how,

430
00:32:14,000 --> 00:32:19,000
but some how it arranged to
have the capacitor voltage be V0

431
00:32:19,000 --> 00:32:22,000
at time T=0.
Now I want to look to the

432
00:32:22,000 --> 00:32:28,000
solution to this for t greater
than or equal to zero.

433
00:32:28,000 --> 00:32:35,000
And in that time my voltage vi
is at some capital VI,

434
00:32:35,000 --> 00:32:41,000
some DC voltage VI.
So I am going to solve the

435
00:32:41,000 --> 00:32:48,000
differential equation RC
dvc/dt+vc=vi given these two

436
00:32:48,000 --> 00:32:53,000
values.
Input is DC voltage VI and VC0

437
00:32:53,000 --> 00:33:00,000
is V0, the initial charge in the
capacitor.

438
00:33:00,000 --> 00:33:03,000
So from now until almost to the
end of the lecture,

439
00:33:03,000 --> 00:33:08,000
it is just going to be math by
solving this very simple first

440
00:33:08,000 --> 00:33:12,000
order differential equation.
And the key here will be that

441
00:33:12,000 --> 00:33:16,000
throughout 6.002 we will be
following one method to solve

442
00:33:16,000 --> 00:33:19,000
these.
There are many methods to

443
00:33:19,000 --> 00:33:23,000
solving differential equations,
and we will follow one method.

444
00:33:23,000 --> 00:33:27,000
That method is called the
method of homogenous and

445
00:33:27,000 --> 00:33:31,000
particular solutions.
In 1802, I believe,

446
00:33:31,000 --> 00:33:36,000
you would have learned maybe
this, and certainly other

447
00:33:36,000 --> 00:33:40,000
methods.
You can use any method to solve

448
00:33:40,000 --> 00:33:43,000
it.
We will just stick to one

449
00:33:43,000 --> 00:33:46,000
method.
And this is also used in the

450
00:33:46,000 --> 00:33:50,000
course notes.
In this method what we do is

451
00:33:50,000 --> 00:33:55,000
take the solution VC by finding
two other components.

452
00:33:55,000 --> 00:34:00,000
One is called the homogenous
solution.

453
00:34:00,000 --> 00:34:03,000
And summing that up with the
particular solution.

454
00:34:03,000 --> 00:34:08,000
And that is the total solution.
So total solution is the sum of

455
00:34:08,000 --> 00:34:11,000
the homogenous and the
particular solutions.

456
00:34:11,000 --> 00:34:15,000
And the method has three steps.
As I said before,

457
00:34:15,000 --> 00:34:19,000
we will be using this method
again and again with every

458
00:34:19,000 --> 00:34:23,000
differential equation that we
encounter in this course.

459
00:34:23,000 --> 00:34:26,000
And you won't encounter a while
lot.

460
00:34:26,000 --> 00:34:31,000
The first step we find the
particular solution.

461
00:34:31,000 --> 00:34:39,000
The second step,
find the homogenous solution.

462
00:34:39,000 --> 00:34:47,000
The total solution is the sum
of the two.

463
00:34:47,000 --> 00:34:51,000
And then find ---

464
00:34:56,000 --> 00:34:58,000
There will be some unknown
constants depending on the

465
00:34:58,000 --> 00:35:02,000
equation that you have.
And in the end we simply find

466
00:35:02,000 --> 00:35:06,000
the unknown constants by
applying the initial conditions

467
00:35:06,000 --> 00:35:09,000
that we have.
Boom, boom, boom.

468
00:35:09,000 --> 00:35:10,000
Particular.
Homogenous.

469
00:35:10,000 --> 00:35:13,000
Find constants.
Three things.

470
00:35:13,000 --> 00:35:17,000
So let's go about solving this
equation and apply those three

471
00:35:17,000 --> 00:35:19,000
conditions.
Again, remember,

472
00:35:19,000 --> 00:35:24,000
what I am doing now for the
next 10 minutes or 15 minutes is

473
00:35:24,000 --> 00:35:29,000
using math that you know about
to simply solve this first order

474
00:35:29,000 --> 00:35:35,000
of differential equations.
There is nothing really new

475
00:35:35,000 --> 00:35:39,000
that I am going to talk about
here.

476
00:35:39,000 --> 00:35:43,000
One is to find the particular
solution vCP,

477
00:35:43,000 --> 00:35:49,000
which will then be added into
the vCH to get me the solution.

478
00:35:49,000 --> 00:35:54,000
So the way you find the vCP is
you find any solution that

479
00:35:54,000 --> 00:36:00,000
satisfies this equation.
This is the equation.

480
00:36:00,000 --> 00:36:03,000
You find any solution that
satisfies it.

481
00:36:03,000 --> 00:36:07,000
And find the simplest possible
solution that money can buy.

482
00:36:07,000 --> 00:36:10,000
Find it.
That's the particular solution.

483
00:36:10,000 --> 00:36:13,000
Any solution is fine.
In this case,

484
00:36:13,000 --> 00:36:16,000
a really simple one would be
vCP equals VI.

485
00:36:16,000 --> 00:36:21,000
Let's see if a constant works.
One thing you will realize in

486
00:36:21,000 --> 00:36:25,000
differential equations is that
they are actually much simpler

487
00:36:25,000 --> 00:36:30,000
than they seem.
And the reason is that almost

488
00:36:30,000 --> 00:36:33,000
every time you have to assume
you know the answer,

489
00:36:33,000 --> 00:36:37,000
and then you are checking to
see what you assumed was

490
00:36:37,000 --> 00:36:40,000
correct.
Assume the answer is this like

491
00:36:40,000 --> 00:36:44,000
you are really smart,
and then check it out and say

492
00:36:44,000 --> 00:36:46,000
oh, yeah, that must have been
the answer.

493
00:36:46,000 --> 00:36:51,000
So here we assume that I think
VI is going to work so let's try

494
00:36:51,000 --> 00:36:53,000
it out.
Substituting in here.

495
00:36:53,000 --> 00:36:55,000
RC dvc/dt is 0.
vi is a constant.

496
00:36:55,000 --> 00:36:59,000
So I get vi equals vi,
so therefore this is a

497
00:36:59,000 --> 00:37:02,000
particular solution.
Done.

498
00:37:02,000 --> 00:37:05,000
I substitute vi here.
So dvi/dt=0.

499
00:37:05,000 --> 00:37:08,000
This vanishes and vi=VI.
Bingo.

500
00:37:08,000 --> 00:37:12,000
Therefore, VI is a solution to
this equation.

501
00:37:12,000 --> 00:37:15,000
So I am done with my vCP.

502
00:37:22,000 --> 00:37:23,000
And in general what you have to
do is use trial and error.

503
00:37:23,000 --> 00:37:24,000
By trial and error try out a
bunch of solutions until you get

504
00:37:24,000 --> 00:37:24,000
lucky.
In general, again,
in all of 6.002 for many of the
excitations a simple constant

505
00:37:25,000 --> 00:37:26,000
usually suffices.
Our second step is to find the

506
00:37:26,000 --> 00:37:27,000
homogenous solution.
And we can also do that very

507
00:37:27,000 --> 00:37:27,000
quickly.
And to do that we have to find
a general solution to the
homogenous equation.

508
00:37:29,000 --> 00:37:52,000
The homogenous equation is the
same differential equation but

509
00:37:52,000 --> 00:38:04,000
with the drive set to zero.

510
00:38:11,000 --> 00:38:15,000
We want to follow a set pattern
to solve the differential

511
00:38:15,000 --> 00:38:19,000
equations here,
and the set pattern is find

512
00:38:19,000 --> 00:38:23,000
vCP, vCH, find constants.
And to find vCH we are also

513
00:38:23,000 --> 00:38:29,000
going to follow a set pattern to
find the homogenous solution.

514
00:38:29,000 --> 00:38:34,000
So we set the drive to zero,
so vi is set to be zero.

515
00:38:34,000 --> 00:38:38,000
And I need to find a general
solution to this.

516
00:38:38,000 --> 00:38:43,000
As I promised earlier,
diff equations are really,

517
00:38:43,000 --> 00:38:49,000
really simple because the way
we are going to solve them is we

518
00:38:49,000 --> 00:38:55,000
are going to assume we know the
answer and then go check it.

519
00:38:55,000 --> 00:39:00,000
So let's try Ae^st.
Let's try and see if this can

520
00:39:00,000 --> 00:39:05,000
solve this particular equation
for some values of A and S.

521
00:39:05,000 --> 00:39:10,000
I am telling you that the
solution is going to be of this

522
00:39:10,000 --> 00:39:12,000
form.
Assume it.

523
00:39:12,000 --> 00:39:16,000
And then simply go ahead and
find me A and S,

524
00:39:16,000 --> 00:39:21,000
and do that by substituting it
back into the equation and find

525
00:39:21,000 --> 00:39:27,000
out the corresponding As and Ss.
So let's go ahead and do that.

526
00:39:27,000 --> 00:39:36,000
I get RC.
I substitute this back up so I

527
00:39:36,000 --> 00:39:50,000
get dAe^(st)/dt+Ae^st=0.
And let me plug that in and see

528
00:39:50,000 --> 00:40:00,000
what comes.
I get RCAse^st+Ae^st=0.

529
00:40:00,000 --> 00:40:05,000
I want to discard the trivial
solution of A being 0.

530
00:40:05,000 --> 00:40:11,000
That is a trivial solution so I
will discard that.

531
00:40:11,000 --> 00:40:16,000
And what I will do is cancel
out the As from here,

532
00:40:16,000 --> 00:40:21,000
assuming A is not zero,
and cancel e^st here.

533
00:40:21,000 --> 00:40:28,000
And what is left is RCs+1=0.
What this is saying is that if

534
00:40:28,000 --> 00:40:34,000
I can find an S such that this
is true then Aest is a general

535
00:40:34,000 --> 00:40:40,000
solution to my homogenous
equation.

536
00:40:40,000 --> 00:40:44,000
This is easy enough.
And so S=-1/RC.

537
00:40:44,000 --> 00:40:51,000
If I choose my S to be -1/RC
then the simple math that I have

538
00:40:51,000 --> 00:40:57,000
gone through shows me that this
must be the solution to the

539
00:40:57,000 --> 00:41:02,000
homogenous equation.
Or in other words

540
00:41:02,000 --> 00:41:07,000
vCH=Ae^(-t/RC).
All this is saying is that

541
00:41:07,000 --> 00:41:12,000
Ae^(-t/RC) is a solution to my
homogenous equation.

542
00:41:12,000 --> 00:41:16,000
A is an unknown constant.
A is some constant.

543
00:41:16,000 --> 00:41:21,000
I don't know what that is yet.
Notice RC has popped up again.

544
00:41:21,000 --> 00:41:26,000
And the cool thing about RC is
that, this is time,

545
00:41:26,000 --> 00:41:33,000
this also has units of time.
We commonly represent RC as

546
00:41:33,000 --> 00:41:38,000
some time constant tau,
as units of time.

547
00:41:38,000 --> 00:41:45,000
Associated with that circuit is
the time constant tau,

548
00:41:45,000 --> 00:41:50,000
which is simply RC.
I commonly write this as

549
00:41:50,000 --> 00:41:53,000
Ae^(-t/tau).

550
00:42:02,000 --> 00:42:08,000
I am very the end here.
I have the particular solution

551
00:42:08,000 --> 00:42:11,000
here.
I have got the homogenous

552
00:42:11,000 --> 00:42:16,000
solution there.
I need to tell you about

553
00:42:16,000 --> 00:42:21,000
something else.
The way I found the homogenous

554
00:42:21,000 --> 00:42:27,000
solution was in four steps.
I assumed a solution of the

555
00:42:27,000 --> 00:42:32,000
form Ae^st.
I created this equation here in

556
00:42:32,000 --> 00:42:34,000
S.
This is called the

557
00:42:34,000 --> 00:42:38,000
characteristic equation for that
circuit.

558
00:42:38,000 --> 00:42:44,000
We will see this time and time
again for RC and other forms of

559
00:42:44,000 --> 00:42:47,000
circuits.
Assume a solution of this form.

560
00:42:47,000 --> 00:42:51,000
Construct the characteristic
equation.

561
00:42:51,000 --> 00:42:55,000
Find the roots of the
characteristic equation.

562
00:42:55,000 --> 00:43:00,000
In this case it is an equation
in S.

563
00:43:00,000 --> 00:43:05,000
So this is the root.
And then form the solution

564
00:43:05,000 --> 00:43:08,000
based on that root.
Four steps.

565
00:43:08,000 --> 00:43:15,000
Ae^st, characteristic equation,
root and then write down the

566
00:43:15,000 --> 00:43:21,000
general homogenous solution.
Four steps there.

567
00:43:21,000 --> 00:43:28,000
And finally I want to write
down the total solution.

568
00:43:28,000 --> 00:43:33,000
And the total solution is
simply vCP+vCH.

569
00:43:33,000 --> 00:43:38,000
And vCP was VI and vCH was
Ae^(-t/tau).

570
00:43:38,000 --> 00:43:43,000
tau was simply RC.
That is my solution.

571
00:43:43,000 --> 00:43:50,000
Now, remember the last step.
The last step was form the

572
00:43:50,000 --> 00:43:59,000
total solution and find out the
remaining constants.

573
00:43:59,000 --> 00:44:05,000
Find out the remaining
constants by using my initial

574
00:44:05,000 --> 00:44:10,000
conditions.
At t=0, I know that vC=V0.

575
00:44:10,000 --> 00:44:15,000
I know that.
And so therefore I can

576
00:44:15,000 --> 00:44:20,000
substitute t=0 to find the
constant.

577
00:44:20,000 --> 00:44:26,000
So I know that VO=VI+A.
t=0, this thing becomes 1,

578
00:44:26,000 --> 00:44:35,000
and so I get this equation from
which I get A=V0-Vi.

579
00:44:35,000 --> 00:44:38,000
In other words,
my solution vC is simply

580
00:44:38,000 --> 00:44:43,000
VI+(VO-VI) e^(-t/tau).
So the last 15 minutes have

581
00:44:43,000 --> 00:44:47,000
just been math.
No electrical engineering here,

582
00:44:47,000 --> 00:44:53,000
but electrical engineering
stopped at the point where you

583
00:44:53,000 --> 00:44:58,000
wrote this differential equation
down, went through a bunch of

584
00:44:58,000 --> 00:45:03,000
math and came up with a
solution.

585
00:45:03,000 --> 00:45:07,000
Purely mathematically.
So here I simply used math to

586
00:45:07,000 --> 00:45:11,000
get you the solution.
And, as I have been promising

587
00:45:11,000 --> 00:45:16,000
you throughout this course,
in the next lecture I will give

588
00:45:16,000 --> 00:45:19,000
you an intuitive EE method of
doing it.

589
00:45:19,000 --> 00:45:23,000
Real electrical engineers,
real EECS folks don't do it

590
00:45:23,000 --> 00:45:26,000
this way.
Real EECS folks do it

591
00:45:26,000 --> 00:45:30,000
intuitively.
And I will show you how to do

592
00:45:30,000 --> 00:45:35,000
it in four easy seconds in the
next lecture.

593
00:45:35,000 --> 00:45:41,000
But you need to understand the
foundations of how this comes

594
00:45:41,000 --> 00:45:44,000
about, and so this is the
answer.

595
00:45:44,000 --> 00:45:49,000
You can also get the current iC
is simply Cdvc/dt.

596
00:45:49,000 --> 00:45:53,000
I won't do that for you,
but you can simply

597
00:45:53,000 --> 00:45:57,000
differentiate it and get the
current.

598
00:45:57,000 --> 00:46:02,000
So I can plot for you vC,
time t, vC.

599
00:46:02,000 --> 00:46:08,000
The intuitive way of looking at
this is I have VI which is the

600
00:46:08,000 --> 00:46:15,000
final value of the voltage.
When t is infinity this part

601
00:46:15,000 --> 00:46:19,000
goes to zero so the vC is simply
VI.

602
00:46:19,000 --> 00:46:26,000
And then there is a component
V0-VI which decays according to

603
00:46:26,000 --> 00:46:32,000
this starting out at an initial
value of V0.

604
00:46:32,000 --> 00:46:38,000
Notice when t is zero vC is V0,
you can see that in the

605
00:46:38,000 --> 00:46:45,000
equation, and so it starts out
at V0 and ends up at VI.

606
00:46:45,000 --> 00:46:49,000
I start here,
I end up here.

607
00:46:49,000 --> 00:46:55,000
And this portion V0-VI decays
out over time like this.

608
00:46:55,000 --> 00:47:04,000
And this decay is governed by
the RC time constant or tau.

609
00:47:04,000 --> 00:47:10,000
I am going to show you very
quickly a couple of examples of

610
00:47:10,000 --> 00:47:17,000
wave forms, one that goes like
this and one that looks like

611
00:47:17,000 --> 00:47:21,000
this.
This is when I start with some

612
00:47:21,000 --> 00:47:28,000
value V0 and I don't apply any
input, it should decay down to

613
00:47:28,000 --> 00:47:32,000
zero, t, t, vC,
vC.

614
00:47:32,000 --> 00:47:37,000
If I apply zero for VI then
this should simply decay down to

615
00:47:37,000 --> 00:47:41,000
nothing over time.
And if I apply some VI but

616
00:47:41,000 --> 00:47:45,000
there is no state in the
capacitor then that same

617
00:47:45,000 --> 00:47:48,000
equation is going to look like
this.

618
00:47:48,000 --> 00:47:53,000
You can go and confirm for
yourselves that when I apply

619
00:47:53,000 --> 00:47:58,000
some input but the capacitor has
zero state, I start at zero,

620
00:47:58,000 --> 00:48:04,000
I finish up at VI and my wave
form looks like this.

621
00:48:04,000 --> 00:48:07,000
There you go.
That's the first one.

622
00:48:07,000 --> 00:48:13,000
The second one where I have 5
volts on the capacitor and no

623
00:48:13,000 --> 00:48:16,000
input.
Assume that at time equals zero

624
00:48:16,000 --> 00:48:21,000
I take away an input,
short the input voltage to

625
00:48:21,000 --> 00:48:25,000
ground for example,
apply zero volts.

626
00:48:25,000 --> 00:48:31,000
You will see the decay from 5
volts to 0 volts.

627
00:48:31,000 --> 00:48:37,000
And in the first case I start
with zero volts in my capacitor,

628
00:48:37,000 --> 00:48:42,000
I apply input of 5 volts,
and notice that at t=0 the

629
00:48:42,000 --> 00:48:45,000
capacitor rises up to that
level.

630
00:48:45,000 --> 00:48:51,000
So notice that these circuits
with capacitor and resistors are

631
00:48:51,000 --> 00:48:56,000
typified by wave forms that are
exponential rises and

632
00:48:56,000 --> 00:49:01,000
exponential decays.
We will see more of that next

633
00:49:01,000 --> 00:49:04,000
time.