0:00:00.800,0:00:04.110
Okay, now we're going to work on our
first op-amp circuit.
0:00:04.110,0:00:08.610
here's what the circuit is going to look
like watch where I put the plus sign is
0:00:08.610,0:00:22.890
on the top on this one and we're going
to have a voltage source over here this
0:00:22.890,0:00:31.529
will be plus or minus Vin that's our
input signal and over on the output will
0:00:31.529,0:00:40.800
have Vout and it's hooked up this way
there's a resistor another resistor to
0:00:40.800,0:00:48.300
ground and this goes back to the
inverting input and we're going to look
0:00:48.300,0:00:52.170
at this circuit and see what it does
now we know that connected up here the
0:00:52.170,0:00:58.760
power supplies hooked up to these points
here and the ground symbol is zero volts
0:00:58.760,0:01:02.940
and we want to analyze this
circuit and what do we know about this
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we know that Vout equals some gain. We'll write the gain right there, a big big
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number times v-, sorry v+ minus v-.
And let's label that v+ is this
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point right here and v- is this
point right here. And we also know that
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the currents, let's call them i+ and
i- equals zero and that's the
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currents going in here this is i-
here and that's i+. And we know those
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are both zero. So now what I want to do
is I want to describe our amplifier with
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some sort of a circuit model with these
properties, alright? And the circuit model
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for an amplifier looks like this: We have
v- here v+ here and over on
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this side we haven't here's a new symbol
that you haven't seen before it's
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usually drawn as a diamond shape and
this is a voltage source
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but is this it's a special kind of
voltage source it's called a voltage
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dependent voltage source
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and it's the same as a regular ideal
voltage source except for one thing, it
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says that the V in this case Vout
equals gain times v+ minus v-
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so the voltage here depends on the
voltage somewhere else and that's what
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makes it a voltage dependent that's what
that means
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so we've just taken our gain expression
here and drawn a circuit diagram that
0:02:59.549,0:03:06.150
represents our voltage expression for
our circuit now specifically over here
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we draw on an open circuit on v+ + v -
so we know that those currents are zero
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so this model this circuit sketch
represents our two properties of our of
0:03:18.449,0:03:25.349
our op-amp. So I'm going to take a second here and I'm going to draw the rest
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of our circuit surrounding this model
but I need a little bit more space.
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So let's put in the rest of our circuit
here we had our voltage source
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connected to v+ and that's Vin, and
over here we had Vout let's check V out
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was connected to two resistors and the
bottom is connected to ground and this
0:04:01.979,0:04:10.229
was connected there. So what our goal is
right now we want to find Vout as a
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function of Vin. That's what we're
shooting for so let's see if we can do
0:04:19.139,0:04:24.630
that. Let's give our resistor some names,
let's call this R1 and R2, our favorite
0:04:24.630,0:04:30.690
names always. And now everything is
labeled. Oh, and we can label this point
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here and this point we can call v-, v-
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So that's our two unknowns. Our unknowns are Vo (Vout), and V-. So let's
0:04:44.830,0:04:49.240
see if we can find them. So what I'm
going to do is just start writing some
0:04:49.240,0:04:54.520
expressions for things that I know are
true. For example, I know that I know that
0:04:54.520,0:05:06.460
Vout equals A times v+ - v-. Alright, that's what
0:05:06.460,0:05:11.380
this op-amp is telling us is true. Now
what else do I know? Let's look at this
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resistor chain here. This resistor chain
actually looks a lot like a voltage
0:05:16.390,0:05:20.320
divider. And it's actually a very good
voltage divider. Remember we said this
0:05:20.320,0:05:26.850
current here, what is this current here?
It's zero. I can use the voltage divider
0:05:26.850,0:05:31.600
expression that I know. In that case I
know that v-, this is the voltage
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divider equation, equals Vout times what?
times the bottom resistor. Remember this?
0:05:42.960,0:05:52.810
R2 over R1 plus R2. So the voltage
divider expression says that when you
0:05:52.810,0:05:56.860
have a stack of resistors like this with
a voltage on the top and ground on the
0:05:56.860,0:06:06.190
bottom, this is the expression for the
voltage at the mid point. Okay, so what
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I'm going to do next is I'm going to
take this expression and stuff it right
0:06:11.140,0:06:19.840
in there. Let's do that see if we get
enough room okay let's go over here now
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I can say that Vout equals a times v+ minus Vout times R2 over R1 plus
0:06:36.430,0:06:38.760
R2.
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Right, so far so good. Let's keep going,
let's keep working on this Vo
0:06:45.300,0:07:00.500
equals a times v+ minus A Vo
R2 over R1 plus R2.
0:07:01.430,0:07:06.660
All right, so now I'm going to gather all
the V naught terms over on the left hand
0:07:06.660,0:07:21.300
side. Let's try that so that gives me V
naught plus A V naught times R2 over R1
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plus R2. And that equals A times v+,
and actually I can change that now,
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v+ is what? v+ is Vin. Okay, let's
keep going. I can factor out the V naught.
0:07:43.730,0:08:01.470
V naught is 1 plus A R2 over R1 plus R2, and that equals A Vin. Alright, so
0:08:01.470,0:08:06.990
we're getting close and our original
goal, we want to find Vout in terms of
0:08:06.990,0:08:12.120
Vin. So I'm going to take this whole
expression here and divide it over to
0:08:12.120,0:08:16.050
the other side, so then I have just V
naught on this side and Vin on the
0:08:16.050,0:08:27.800
other side. Let's make some more room. I
can do that. I can say V naught equals A
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Vin divided by this big ol'
expression 1 plus A R2 over R1 plus R2.
0:08:43.250,0:08:50.790
Alright, so that's our answer. That's
the answer, that's Vout equals some
0:08:50.790,0:08:54.800
function of Vin.
0:08:54.949,0:08:58.980
Now I want to make a really important
observation here. We're going to, this is
0:08:58.980,0:09:03.899
going to be a real cool simplification.
Okay so this is the point where op-amp
0:09:03.899,0:09:08.730
theory gets really cool. Watch what
happens here. We know that A is a giant
0:09:08.730,0:09:12.899
number. A is something like 10 to the
fifth or 10 to the sixth, and it's
0:09:12.899,0:09:17.160
whatever we have here, if our resistors
are sort of normal sized resistors, we
0:09:17.160,0:09:21.389
know that a giant number times a normal
number is still going to be a very big
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number compared to 1. So this 1 is
almost insignificant in this expression
0:09:27.029,0:09:31.889
down here. So what I'm going to do, bear with me, I'm going to cross it out, I'm going to say
0:09:31.889,0:09:36.509
no, I don't need that anymore.
So if this if this number here if A is a
0:09:36.509,0:09:40.970
million, 10 to the sixth, and this
expression here is something like 1/2
0:09:40.970,0:09:47.100
then this total thing is 1/2 of 10
to the 6 or a half a million, and that's
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huge compared to 1, so I can pretty
safely ignore the 1, it's very very small.
0:09:52.250,0:09:57.839
Now when I do that, well look what
happens next. Now I have A top and bottom
0:09:57.839,0:10:03.480
in the expression and I can cancel that,
too. So the A goes away. Now this is
0:10:03.480,0:10:07.649
pretty astonishing. We have this
amplifier circuit, and all of a sudden I
0:10:07.649,0:10:12.029
have an expression here where A doesn't
appear the gain does not appear. And what
0:10:12.029,0:10:20.910
does this turn into? This is called V
naught equals Vin times what? times R1
0:10:20.910,0:10:31.740
plus R2 divided by R2. So our amplifier,
our feedback circuit, came down to Vout
0:10:31.740,0:10:37.490
is Vin multiplied by the ratio of the
resistors that we added to the circuit.
0:10:37.490,0:10:41.790
This is one of the really cool
properties of using op amps in circuits,
0:10:41.790,0:10:46.379
really high gain amplifiers. What we've
done is we have chosen the gain of our
0:10:46.379,0:10:51.029
circuit based on the components that we
pick to add to the amplifier. It's not
0:10:51.029,0:10:55.379
determined by the gain of the amplifier
as long as the amplifier gain is really
0:10:55.379,0:11:01.380
really big. And for op amps that's a
good assumption it is really big. So this
0:11:01.380,0:11:06.779
expression came out with a positive sign,
right, all the R's are positive values, so
0:11:06.779,0:11:12.950
this
is referred to as a non-inverting op-amp
0:11:15.019,0:11:25.829
circuit, amplifier. So just to do a quick
example, if R1 and R2 are the same then
0:11:25.829,0:11:30.779
we end up with an expression that looks
like this Vout equals R1 plus R2, R
0:11:30.779,0:11:42.779
plus R over R is equal to 2, so the gain
is 2 times Vin. So just to do a quick
0:11:42.779,0:11:51.420
sketch, just to remind ourselves what
this looks like, this was Vin and we had
0:11:51.420,0:12:05.399
what out here? We had a resistor and a
resistor to ground, and this is Vout.
0:12:05.399,0:12:09.329
So this is the configuration of a
non-inverting amplifier built with an
0:12:09.329,0:12:15.240
op-amp. The two resistors in this voltage
divider string connected to the negative
0:12:15.240,0:12:19.110
input, and it's going to be one of the
familiar patterns that you'll see over
0:12:19.110,0:12:24.649
and over again as you read schematics
and you design your own circuits.