All right.
Good morning.
Let's get going.
In today's lecture we continue
with the operational amplifier,
"op amp" for short.
And what we are going to do is
just build up a bunch of fun
building blocks using the op
amp.
As a quick review --
To quickly review what we've
seen about the op amp --
We represented the op amp as a
device that looked like this
where the amplifier had an
incredibly high gain.
So, if I had a small voltage
difference here --
I call this v plus and this v
minus with respect to ground.
And if I had a small voltage
difference then this gain here
would multiply the difference by
a large number and thereby
giving me an output that was on
the order of a million times
greater than this difference.
And because of that when I use
the op amp in a mode like this
without any negative feedback
the output would usually crank
up to the positive rail or the
negative rail.
We also saw that it had
infinite input resistance so
that the current flowing in here
or here was zero and also had
zero output resistance.
This is my ideal op amp where
irrespective of what load I
connect here the op amp would
supply pretty much any current.
Now, in practical op amps
that's not the case.
But suffice it to say that when
used as an ideal op amp the
output impedance,
the output resistance is going
to be zero.
The op amp is a huge workhorse
of the analog industry.
You will see based both on what
you've done on Tuesday and
Wednesday but also today that
it's very, very simple to build
circuits using the op amp.
When you use the amplifier,
you don't have to worry about
things like nonlinear analysis.
You don't have to worry about
am I really meeting the criteria
for saturation limits and so on?
To some extent you have to
think about that with the op
amp, too, because if the output
hits the positive rail or
negative rail it isn't going to
behave like you expect it to.
But fundamentally with this
primitive model,
this idea model it becomes
really simple to build circuits
with the op amp.
Therefore it has become a key
building block for circuits.
When circuit designers build
analog circuits very often their
primitive building blocks are
really an amplifier of this
sort, an op amp,
resistors, capacitors and some
of our other primitive building
elements.
If you look at the course notes
the readings are --
There are a bunch of examples
solved in Chapter 16.
And you will see that using the
op amp it is indeed possible to
build current sources that look
like more or less ideal current
sources.
It is also possible to build
voltage sources and so on.
It is an incredibly neat
building block using which you
can do all kinds of cool stuff.
In this course you will see a
whole bunch of example circuits
using the op amp.
In today's lecture you will see
things like a subtractor.
You will also see integrators
and a differentiator.
And then in your lab,
lab four, you will build a
really fun mixed signal circuit
involving both digital and
analog components.
And you will build what is
called a digital to an analog
converter using the op amp.
And of course I can build all
our good-old amplifiers and
circuits of that sort.
In a later lecture you will
also see how we can build
filters using an op amp.
This is going to be using the
knowledge you learn in terms of
connecting resistors,
capacitors and inductors
together and doing a frequency
domain analysis,
well we can throw the op amp in
there and build filters,
too.
This is just to give you a
preview of upcoming attractions.
For today I am going to focus
on these circuits.
I won't be covering any new
theory or any new set of
foundations but pretty much take
the simple properties that I
have explained to you about the
op amp.
And using those simple
properties very quickly build up
a bunch of circuits that you can
use to analyze signals in a
variety of ways.
Let's start with the following
circuit.
With op amps I start with this
little guy.
And what I am going to do is
use two voltage sources,
v1, and this is a resistor,
not an inductor.
And value R1,
value R2.
So, I have a voltage connected
by a divider,
voltage divider to the plus
input.
And I am going to provide some
negative feedback in the
following way.
This is going to be R2,
the same as this one here,
a resistor R1.
And then a voltage source v2
that I connect out here.
So notice that- Oh,
and I take the output vOUT out
here.
And that vOUT of course is with
respect to ground,
and R2, v1 and v2 are also
connected to ground.
What I am going to do is
analyze the circuit it two
different ways,
and as I analyze it describe
some other interesting
properties to you.
In the last lecture the
technique I used to analyze op
amps was one in which I replaced
the op amp with its ideal model
involving a dependent source and
so on with a large gain A and
showed that.
I wrote the expression and then
I let A increase to infinity to
the limits and got an expression
that was independent of A.
And then in recitation
yesterday you would have covered
another technique which makes it
much simpler to analyze op amps.
Let me very quickly review that
method.
We fondly call that technique,
there is no formal name for it,
but we fondly call that v plus
more or less equal to v minus
method.
This is also variously called
the virtual ground method and so
on, but we shall call it the v
plus more or less equal to v
minus method.
The insight here is that
whenever I use the op amp in a
way in which I am giving it
negative feedback,
so I am feeding some portion of
the output to its negative
input.
I am giving it negative
feedback.
That's one property.
Second property is that my
inputs, v1 and v2,
and my resistance values are
chosen such that the output is
not in saturation.
So, the op amp is not at the
plus VS rail or minus VS rail.
Rather it's somewhere in the
middle in its active region.
When that happens we claim that
the v minus and v plus for the
op amp are more or less equal.
And to give you some intuition
as to why that is so,
let's say the output is 6 volts
and my supply is plus/minus 12.
This is 6 volts and the
amplifier is a gain of a
million, ten to the six.
To sustain 6 volts at the
output all I need is a
difference of 6 microvolts here.
Six divided by ten to the six
is the difference between v plus
and v minus.
It's very, very,
very small.
It's so small as to make v plus
more or less equal to v minus.
All it takes is a very small
differential voltage here to
give you 6 volts at the output.
The key thing to observe is
under negative feedback,
when the op amp is not in
saturation the property that v
plus equals v minus holds.
And the way it works is that
it's not that it's a magical
property.
It is simply that when I apply
negative feedback the negative
feedback is such that it will
force this v minus node here to
be at more or less the same
voltage as v plus.
Remember the when in doubt
simply go back and think about
the anti lock brakes example we
did last time.
For example if v plus increases
the output will increase and so
will the voltage here and tend
to make these two equal.
What we can do,
being rather tricky here,
what we'll do is say look,
if we know for a fact that
under negative feedback the op
amp is going to engineer these
two node voltages to be more or
less equal then why don't I just
use that fact to begin with and
analyze my circuit assuming that
it's true.
This is just a bit of inverted
logic here that says look,
the circuit is going to make
that happen.
If the circuit is going to make
that happen to analyze the
circuit in its steady state,
why don't I just go ahead and
assume that to begin with?
This again goes back to us
wanting to be engineers here and
do whatever is simply and find
the simplest possible way of
getting some place.
I want to use that method,
the v plus equals v minus
method.
Let me just first write down
some values that I know about.
I know that v plus is simply a
voltage divider relation here.
That's v1 times R2 divided by
R1 plus R2.
And by the v plus equals v
minus method I know that this is
going to be equal to v minus.
And this is going to be true
because I am giving you negative
feedback here.
And we are going to engineer
the values of R1,
R2, v1 and v2 such that the op
amp is not in saturation.
So, we know that.
The next thing that we know,
let's say this is a current i.
This current i flows here.
Know that there is no current
going in here.
Op amp has an infinite input
resistance so there is nothing
going in there.
There is no current going in
there.
If there is no current going in
here, what must happen to i?
Remember, from the foundations
of the universe Maxwell's
equations and therefore KVL and
KCL hold.
KVL and KCL simply come
straight from nature.
You and I cannot mess with
that.
Bad things happen to you if you
do.
So, nature, Maxwell's
equations, KVL,
KCL.
It's simply nature.
So, KCL applies here.
Current comes in here.
Nothing goes there.
Don't argue.
The current has to go here,
period.
No if, ands or buts.
There is i coming in here,
nothing goes there,
so that current must flow here.
It has no choice.
It's from basic nature.
I can write down what my
current i is going to look like.
What is i going to look like?
Well, I know v2,
I know v minus.
v minus is the same as v plus.
And v plus is the i expression
given here.
So, I can write i as v2 minus v
minus divided by R1.
Let me keep track of those two
and then go ahead and compute
vOUT.
So, my goal in life is compute
vOUT as a function of the two
input voltages v1 and v2.
And just for kicks I have gone
ahead and computed some of the
intermediate node voltages and
currents.
How do I write vOUT?
What is vOUT?
vOUT is simply v minus from
KVL.
vOUT is simply v minus minus
the drop across this resistor.
So, the drop across that
resistor is simply iR2.
From good-old KVL from the
first lecture,
a voltage minus the drop across
the resistor is equal to vOUT.
Therefore it's simply v minus
minus iR2.
One thing to be very cautious
about, I will tell you right
now, is that the output here
relates to the inversion of the
voltage across this resistor R2.
Be very, very careful in that
if I have a voltage across this
resistor here that impacts vOUT
with a minus sign attached to
it.
Notice that iR2 is the voltage
across R2 and vOUT relates to
the negative of that.
Be very cautious.
That's one of the commonest
silly mistakes I have seen
people make in solving problems
like this.
Let's go ahead.
I know v minus and I don't know
i.
Let me substitute for i for
now, and that is v2 minus v
minus divided by R1 times R2.
Let me go ahead and collect all
the v minuses.
v minus, I get a one here,
minus minus becomes a plus,
and so I get R2 divided by R1
out there.
And then I minus v2 R2 divided
by R1.
That is vOUT.
Now let me go ahead and
substitute for v minus.
And that is simply v1 R2
divided by R1 plus R2.
That is v minus.
And this character here is
simplified to be R1,
R1 plus R2 minus v2 R2 divided
by R1.
What do we get?
I cancel these two suckers out
and what I end up with is v1 R2
divided by R1 minus v2 R2
divided by R1,
which is simply R2/R1(v1-v2).
What is interesting here is
that what I have ended up
building is a very primitive
subtractor.
So, my output relates to v1
minus v2 multiplied by the
constant factor given by R2
divided by R1.
Again, as I pointed out to you
at the beginning of this
lecture, no knew foundations
today, no new theories,
no new disciplines,
no new laws.
We are just going to take what
you have learned --
Three simple things,
infinite gain,
infinite input resistance,
zero output resistance,
plus this new thing v plus
equals v minus.
And just being armed with those
four principles we are just
going to charge ahead and
analyze a bunch of circuits.
It is purely intellectual and
pure applications today.
This is one way of doing it.
There is another way of solving
it.
We can solve the circuit.
Remember, whenever you see a
linear circuit and you see two
sources or three sources,
just think superposition,
right?
You see a linear circuit and
two or three sources,
think superposition.
We should be able to apply
superposition to this.
The op amp is simply another
building block.
It's a linear circuit.
So, let's see if we get the
same answer.
Let's try to solve the circuit
using superposition and see if
we get the same answer.
To do superposition what I am
going to do is build two
subcircuits.
One subcircuit in which v1 is
zero, and that subcircuit looks
like this.
If I set v1 to be zero then I
get R1 parallel R2 going to
ground.
So, if v1 is set to zero then
R1 goes to ground.
And I get R1 parallel R2 here.
And of course I have v2 as
before.
And this was R1,
this was R2,
and let me call that vOUT1.
Oh, I'm sorry.
Let me call it vOUT2
corresponding to that component
of the output that relates to v2
acting alone.
Remember superposition?
Build two subcircuits,
one that depends on v2 and
another one that depends on v1.
Let's do the second one,
too.
Second one is v2 going to zero.
Here is my little op amp.
And what I will do is simply
flip the op amp just to see if
you can identify some
interesting patterns.
Just flip the op amp around.
And this is v1 as before.
And recall that v1 was going to
the plus node through a resistor
R1.
And then I had a R2 to ground.
And then let me short v2 to
ground.
And when I short v2 to ground
what happens?
When I short v2 to ground what
happens is that the tail of R1
here goes to ground.
And so it is as if the output
is connected to the node v minus
through a resistor,
so it as if the output v R2 is
connected to the minus input
through a resistor.
We will draw it like this.
And the minus input goes
through a resistor R1,
to ground.
If you thought that patterns
were important in the earlier
part of the course doing voltage
divider patterns and current
divider patterns and amplifier
pattern, the source follower
pattern, op amps is all about
patterns.
You should remember two or
three simple patterns and be
able to write down the
expression for those just by
observation.
So, this is one common pattern
that you have seen before in the
very first lecture.
And I just wrote it down in
that manner.
Let me go ahead and solve this
circuit.
It turns out that this is also
a pattern.
I will analyze it today but in
the future v2 going to this node
through R1 and then R2 to the
output.
You have probably also seen
this in your recitation.
This one is called an inverting
connection and this one here is
called a non-inverting
connection.
Let's go ahead and do vOUT2.
vOUT2 is simply given by,
notice that since this is
ground, no current flowing here,
this voltage is zero.
If this voltage is zero,
this voltage is zero by the v
plus equals v minus method.
If this is zero,
the current that goes through
here is v2 divided by R1.
And that same current must flow
through the resistance R2 as
well.
If the current v2 divided by R1
flows through this resistor,
the drop across this resistor
is simply given by,
let me hide this for a second,
is simply given by v2.
So, v2 divided by R1 is the
current here.
This is zero.
So, the drop across this
resistor is v2 R1 multiplied by
R2.
That's a drop across this
resistor.
This voltage is simply zero
minus a drop across the
resistor.
So, it's zero minus the drop
across the resistor and that
gives me v2.
Again, remember this minus sign
comes in when I want to convert
this to get the output voltage
from that.
This is a very common pattern.
It's called an inverting
connection where the output is
some factor of the input voltage
and the factor is given by R2
divided by R1.
Let's go ahead and analyze this
guy now.
What is vOUT1 equal to?
I should have called this vOUT1
because it relates to v1.
vOUT1.
There is a v plus here.
From our first lecture I know
that vOUT1 relates to v plus in
the following way.
I know that it is v plus times
the sum of the resistances
divided by R1.
Based on the first lecture this
is true.
vOUT1 is simply an amplified
version of v plus where the
amplification factor is given by
R1 plus R2 divided by R1.
And I know v plus is simply a
voltage divider action here.
And I can take a simple voltage
divider action here because the
current going in is zero.
Looking in here this is as if
it's an infinite resistance,
so it is as if the element
simply does not exist.
The voltage here is simply v1
divided by R1 plus R2 multiplied
by R2, our voltage divider
pattern.
So, I get v1 times R2 divided
by R1 plus R2 times R1 plus R2
divided by R1.
These two cancel out which
gives me vOUT1 is simply v1 R2
divided by R1.
To get vOUT I add up the two.
vOUT is vOUT1 plus vOUT2,
which is my goal.
And that is simply v1 R2 by R1
minus v2 R2 by R1.
Thankfully what we have here is
the same as here.
Again, there is really nothing
new that I am going to cover
today.
Simply apply,
apply, apply,
four simple principles.
Here I have used superposition
and I am showing you a circuit.
So, it turns out with op amps
you should really remember that
pattern.
You will see it again and again
and again.
And each time you see it,
it will save you six minutes of
having to solve the circuit
without knowing the pattern.
So, remember this pattern.
You can pick up another three
or four minutes by remembering
this pattern here.
This pattern is simply v2 R2
divided by R1.
Imprint those two patterns into
your brains.
OK, so those are a couple of
simple circuits using the op
amp.
We built a subtractor.
The next step,
let's go ahead and try to build
an integrator.
Using this little building
block we can go ahead and try to
build a bunch of circuits.
We can build filters,
A to D converters and so on.
Let's build an integrator.
Abstractly I need to build this
box.
Which when fed a vI,
I want that box to integrate
and give me a vO which is vI
integrated over time.
That is what I want to build.
How do I go about building it?
What I would like to do next is
give you some flavor for design.
How do you go about designing
things with an op amp?
Knowing that you do not know
the pattern for this yet,
how do you go about designing
things?
Well, let's start with the
following intuition.
The intuition that I begin with
is that if I have a current i,
and remember that capacitors
and inductors related to,
you saw differentiation and
integration happening when we
dealt with capacitors and
inductors.
So, I think we have to invoke a
capacitor here or an inductor.
In this example I invoke a
capacitor.
Notice that if I stick a
capacitor in here this current
is i, capacitance C,
then my voltage vO is given by
what?
Voltage is simply the integral
of the current flowing through
it or vice versa i is C dv/dt.
If i is C dv/dt then v is
simply one by C integral.
If I can pass the current
through a capacitor then the
voltage across the capacitor
must be a current.
Notice then that vO is related
to i dt.
I have some multiplying
constants and so on,
but fundamentally what I have
found is if I can stick a
current through a capacitor then
the voltage across the capacitor
relates to the integral of the
current.
OK, that's interesting.
So, I have an integral in
there.
But I have a current.
Notice my goal was to integrate
a voltage.
What I figured out how to do
was if I can turn that voltage
into a current --
If I can turn that voltage into
a proportional current and then
pump that current through a
capacitor I will get the
integration that I want.
How do I convert my vI to i?
How do I do that?
Well, let's take a stab at it.
Here is my vI.
Let's take the resistor R.
And remember I need to stick
the capacitor here.
I have some current I here.
I don't know what the current
is yet.
And I stick a voltage here.
And what I am trying to do is
trying to see if I stick a
voltage and a resistance in
series then there is some
relationship between the current
and this voltage.
Recall that I am trying to make
this current be directly
proportional to the voltage vI.
But it turns out that i here is
not equal to vI divided by R.
If i was vI divided by R
somehow, I am done.
If i was vI divided by R,
by some magic,
then I have converted my
voltage to a current,
I feed that current through my
capacitor and vO is my integral
that I am looking for.
But unfortunately i is not
equal to vI divided by R.
You know that.
i relates to vI minus the
capacitor voltage divided by R.
So, i is not simply vI divided
by R for all time but i is
really vI minus the capacitor
voltage divided by R.
And, in fact,
when we did RC circuits you
wrote this equation to represent
the dynamics of the circuit,
RC dvO by dt plus vO equals vI.
We wrote down this circuit for
a first order RC,
wrote this equation for a first
order RC circuit.
Now, it does turn out,
to wrap up on this wild goose
chase that we went on,
it does turn out that if this
term here is much bigger than
that term.
If this term is much bigger
than that term then I can ignore
that term and write down RC dvO
by dt more or less equal to vI.
If that were true,
this would be true,
and then vO would be more or
less equal to one by RC integral
of vI dt.
Again, if this were true.
If this were true for all time
then vO would be integral of vI
dt.
Again, remember this is all a
wild goose chase.
Just write down WGC there just
so you don't get confused.
I am on this wild goose hunt
here trying to find a way to get
a current from a voltage which I
can then feed into a capacitor.
This was one thing I knew,
but this was not what I want.
But it does turn out to be what
I want when vO is very,
very small.
So, I see some glimmer of hope
but not quite.
It turns that in R and C,
if I make R and C very,
very big, if I have a huge time
constant, with a huge time
constant the voltage vO looks
like an integral of vI,
but only when I have a very
huge time constant.
So, I give up on that track.
Instead I try something else.
Another try.
I would like you to notice if
you take your op amp,
here is your op amp,
if you take this op amp and you
stick the positive terminal to
ground, under reasonable
feedback, under reasonable
negative feedback what do you
notice about the current?
If I had a current i flowing
here what did you notice?
Look at this picture.
I had a current i flowing in
here, v2 divided by R1.
And because this resistance was
infinite all the current went
through the upper terminal.
So, this is zero volts.
And by the v plus equals v
minus method this is also more
or less equal to zero.
And I have a current i flowing
in here, nothing goes here,
so then the i must flow up
there.
So, all I am doing here is
causing a reflection of the
current from this grounded node.
My current is being reflected
into, or deflected if you feel
like it, the upper edge here
after coming in through this
edge.
That is interesting.
We are just one step away from
the key insight.
I have an i coming in here,
an i going out there.
Notice that,
as I said before,
this is zero volts.
How do I get my voltage vI to
look like a current,
to become proportional to a
current?
It is simple now.
All I do is put a voltage vI
and put a resistor R out there.
If I do that,
and since this is zero,
the current i is given by vI
divided by R.
I have gotten to where I want
to be.
So, by using an op amp and
using the fact that the minus
node here, v minus is at the
same potential as v plus when
there is negative feedback then
I can stick a resistor here.
And because this is zero the
current here is simply vI
divided by R.
I have gotten to the first
place.
Now all I need to do is simply
pump this current through a
capacitor and I get the integral
of the, the voltage becomes an
integral of the current.
That is easy.
I stick my capacitor here and I
get my answer out there as vO.
Notice that when I do this,
let's say this is plus/minus
VC.
This is zero.
So, vO is minus VC.
Again, I will keep emphasizing
it maybe 17 times throughout
this course that if this is zero
then the output here is related
to the negative of this voltage,
common, common,
common mistake.
I will be very upset after
doing all this if I see this
mistake happen in any of the
future homeworks or finals or
whatever.
This should not happen.
So, vO is a minus sign here VC.
And I know that if I have a
current i through a capacitor
what is VC?
If I have current i through a
capacitor than this is simply t
i dt.
And i by design is --
So, I have my integrator.
It is a two-step process.
I stuck a resistor here,
so the current became equal to
vI divided by R.
Then I took that current and
pumped it through a capacitor
through this terminal here,
and the voltage across the
capacitor for a current i is
given by this expression.
This is Capacitors 101.
OK Capacitors 101 says that the
voltage across the capacitor is
simply one by C integral i dt.
Another way of looking at it is
the voltage across the capacitor
is C, I'm sorry,
the current through a capacitor
is C dv/dt.
This is simply the integral
form of that equation.
And I am done with my
integrator.
So, this is another very common
building block.
Remember this.
Most of the circuits we will be
seeing with op amps simply
involve something here and some
there.
And the output in this
inverting connection is the
output times,
if it is a resistance it is
simply R2 divided by R1,
if it's a capacitor I get the
integral form looking like this.
Yes.
Can someone tell me where the
negative sign went?
The blackboard ate it up.
Good catch.
After all that lecture about
watching the negative sign.
After this little bit of faux
pas here, now I will be doubly
mad if you guys make that
mistake.
All right.
Now that we have built the
integrator, I could give this
out as a homework problem.
And you should be able to
design a differentiator based on
what you've learned here.
You now have the tools to go
and do some design like this,
but we don't have any more
homeworks left so I guess I will
go ahead and solve this for you
right here and do the design for
you.
The building block that we need
looks like this,
d/dt here.
Let me take a vI and stick a vI
in there.
That's what I want to build.
And what I built here is that
different integrator box.
And what I would like to do now
is build a differentiator box.
How do I go about doing it?
I will go really slow here so
you will have some time to think
about it for yourselves and see
if you folks are crack op amp
circuit designers already,
if you have the right instincts
here.
Again, when you see
differentiation integration
think capacitors or inductors,
it doesn't matter.
In fact, as a homework
exercise, you may want to go
back and see how you can get a
similar effect using inductors.
Can you play with inductors and
get a similar effect?
So, inductors are devices that
are a dual of the capacitor.
Whatever we will do with
capacitors, there must be a
corresponding way with
inductors.
You can try it out in your
spare time.
Let's go back to this one here.
I will stick with the capacitor
way of looking at things.
I need a differentiation now.
Remember this.
If I have a vI and I stick this
across a capacitor,
I have a current C and some
voltage vc across the capacitor,
what does i relate to?
i is simply C dv/dt and vc in
this case is simply C dvI/dt.
If I can stick a voltage across
a capacitor, if my input voltage
is stuck across a capacitor then
the resulting current relates to
dvI/dt.
Here we have the opposite
problem.
By doing this simple trick,
I can obtain a current that has
the right form.
Now what I need to do is
somehow convert that current
into a voltage because the
abstraction that I need is a
voltage to voltage.
The next step,
what I need to do is somehow
convert a current to a voltage.
How do I go about doing that?
Again, remember for the op amp,
if I have a current i flowing
here then by the reflection
property i gets pushed up into
this edge, provided that the
whole circuit is working with
descent negative feedback.
Given this trick what I can do
is say look, suppose I did this.
Remember, my goal here is how
do I convert a current to a
voltage?
I have a current i coming in
here, and I can turn that into a
voltage because I know the
current must come out here,
I know this current must come
out there.
All I have to do is stick a
resistor in there.
If I stick a resistor in there
what is vO equal to?
vO is simply iR,
right?
That's right.
vO, I get i here,
so i pumps through here.
Remember, what comes in here
must get reflected up because
the current going in here is
zero.
All the i must come out here.
So, that i must pump through
this resistor.
The drop across this resistor
is iR.
That's the voltage drop across
that resistor.
And since this at a virtual
ground the output here is simply
zero minus this drop which is
minus iR.
So, I have gotten to where I
want to be.
I have my current i being
converted to a voltage.
I have taken my current,
and I have been able to convert
that into a voltage by sticking
a resistor in here.
As a final step,
I simply need to produce the
current.
And that is pretty easy to do.
Abstractly what I need to do,
again, this is design here so
we will talk about abstract
stuff.
If I had a voltage vI,
I need to produce a current
which relates to C dvI/dt.
And I know I can do that by
simply doing this.
By doing this I know my i is C
dvI, correct?
If I can get this effect,
I put this in quotes because
that's my pattern.
I am looking for a pattern,
where a voltage vI is directly
applied across a capacitor.
And when that happens the
current relates to C dv/dt.
Let's go back to our op amp
pattern here,
op amp circuit.
So far I have achieved --
I just repeated this out there.
And so somehow I need to take
this pattern here and learn from
that pattern and apply the
pattern here.
So, what I can do is,
this is a ground node,
correct?
Now, the poor little capacitor,
what does it care,
whether it's a ground node or a
virtual ground node?
As long as it's a zero volt
node down here what does it
care?
What I am going to do is stick
this point, not here but into a
virtual ground node.
I am going to grab that point,
take it here and stick it here.
The poor little capacitor
doesn't know the difference.
I have really suckered the
little beast.
This is vI.
Remember this.
My i through the capacitor is
proportional to C dv/dt.
Instead what I have done is
taken this guy and stuck it here
to get something like this.
Just remember these four or
five little tricks.
And you apply them in op amp
circuits again and again and
again and again.
So, this is vI,
this is my virtual ground.
As far as this poor little
capacitor is concerned,
it is chugging along merrily
thinking that it is connected to
ground.
Little does it know it is only
a virtual ground,
all right?
But the current i here is
simply C dvI/dt.
And that current,
the C dvI/dt,
that current flows through here
and gives me vO as iR.
So, vO is simply minus R.
Let me substitute for i there,
C dvI/dt.
OK, so notice then that my vO
is now proportional to dvI/dt.
So, vO is some RC time constant
times dvI/dt.
Therefore, I have my
differentiator circuit.
Remember this as a closing
thought.
Remember this v plus more or
less equal to v minus trick.
And to the extent possible
simply use that trick to analyze
op amp circuits under feedback
and not in saturation.
Just remember these two.
Very quickly for the demo,
I have a square wave input here
to the op amp,
that's my vI to the integrator.
And this is the output vO.
The integral of a square wave
is a triangular wave,
as you can see.
And we will do the same thing
for a differentiator.
And for the differentiator,
I input the square wave to this
differentiator circuit.
And I get this,
wherever there is a sharp rise,
I get this huge negative spike
and a positive spike because of
the minus sign.
So, this is the differentiator
circuit.
Then I feed this into the op
amp.
OK.
Thank you.
