- [Voiceover] Another form of an op-amp circuit is
called the summing op-amp.
We're gonna work through how this one works.
Let's draw in here now is an inverting
op-amp circuit with a single input.
We're gonna call this
this will be V.
We'll call the VA for now.
And we have V out.
And we worked out what how this worked before
where V out was a function of
VA times the ratio of these two resistors.
And now I'm gonna add a little twist to this
and we're gonna analyze a different kind of circuit.
So I'm gonna add another resistor right here.
Like this.
And a second input.
We'll call this VB.
And so I have RB, or sorry, RA here.
This will be RB.
And we'll call this R feedback.
And the question again is now,
what is V out in terms of
the inputs, VA and VB?
We're gonna use the idea of a virtual ground.
Virtual ground.
The idea of a virtual ground applies to most
op-amp circuits, and it's a really useful way
to simplify the analysis.
And when we use a virtual ground
one of the things I like to do is just
draw a little symbol here
like that, for my virtual ground.
The virtual ground, as a review,
if the voltage coming out of this op-amp
is in a reasonable range, sort of a plus or minus
10 volts, or something like that.
And because the gain of this op-amp is so enormous
on the order of 100,000, or a million
that means that, when this is working properly
that these two voltages will be really close together.
They'll be only microvolts apart, at most.
And, because they're so close
we can just say, let's just say they're the same
for the purposes of analyzing this circuit.
And, since this input is at zero volts
this is at zero volts.
That means that this input is very close to zero volts.
So this node here
this node here, we say
is at a virtual ground.
So let's move ahead, let's analyze this circuit
and what we want to find is we want to find
V out as a function of
VA and VB.
Let's see if we can do that.
So looking at my circuit the other thing
I know about op-amps that's really important
always is, that the current
going into an op-amp input
is zero, or practically zero.
And for the purposes of what we're doing here
we can treat it as zero.
All right.
So, we have zero volts on this node.
We have zero current going this way.
So let's go after
let's go after this current right here.
Let's figure out what that is.
So we'll call that I.
And we can express I in terms of VA
and VB in these two resistor values.
So we can write I
equals, now what is it?
Well, we have IA here.
And we have IB flowing here.
That means that I equals
IA plus IB.
All right.
And let me write IA and IB in terms of
Ohms Law here.
So we get I equals, what is IA?
IA is this voltage
divided by RA.
And this voltage is VA
minus zero volts.
So it's VA
over RA.
And now let's write down what's IB?
IB
we look at this voltage here.
This voltage is VB minus
zero.
Or just plain
VB over RB.
All right, so that's one step.
We've written this current in terms of
what's going on on the input side over here.
Now one thing we know, because this current is zero,
we know, what?
We know
this current is I.
And, if I's going into RF
we'll label the voltage on RF like that.
So now let's write an expression for I
in terms of RF and this voltage here.
What is this voltage here?
Well, let's look at the two ends.
This is zero volts on this node here.
And on this side it's
V out.
So, the current there, the current
on that side
equals
it's zero minus V out.
So minus V out
divided by RF.
That's Ohms Law for this resistor right here.
Now we know that this current equals this current.
So let's just set these two expressions together.
So I'll do that over here.
I can write minus V out
over RF
equals this one
VA over RA
plus VB over RB.
Now we can get the final function
I'm just gonna multiply through by RF on both sides.
And we get
V out equals
take the minus sign with us
minus
RF
over RA
times VA
plus
RF
over
RB
times
VB.
So that's our answer.
That's V out
as a function of
VA and VB, the two inputs.
And you can see that the
resistor ratio, there's two resistor ratios here,
that are participating in the answer.
So it helps to do a little special case of this.
Let's let all the resistors be the same.
So we'll say that
RA equals RB
equals RF.
And just to pick a real number we'll just say
they're all equal to 10 K ohms.
And let's see what this becomes now.
Now we have V out
equals, we still have the negative sign.
So this is some sort of inversion going on here.
RF over RA is one.
So it's just VA.
And RF over RB is one.
So this says
VB.
So this is what gives us the nickname
for this expression
which is called a
a summing
op-amp.
Let's say I want to use my summing op-amp
in an application where what I want is
I want V out to equal say
minus two times VA
plus three times VB.
Okay, this is what I want.
How do I do that?
So, these two coefficients here are
functions of the resistor values.
That was our original expression up here.
So what I want is
RF over RA
to equal two.
And I want RF
over RB
to equal three.
So I can pick values
it's the same RF
and so I get to adjust RA and RB in here.
So I can pick component values.
So if I pick
RF equals 12 K ohms.
Then I can pick
RA equals six K.
And RB
equals four K.
And that would give me the ratios I want
and that would implement
this function here.
So I'm gonna go fill those out
on the top schematic.
We'll just write those in.
And we'll go fill these in here.
So what we said was RF was 12 K.
RA was six K ohms.
RB would be four K ohms.
And we've designed a circuit
that implements our summing function.
So, so this is a pretty useful op-amp circuit.