Summing op-amp circuit

Another form of an op-amp circuit is called the summing op-amp. We’re going to work through how this one works. What’s drawn here now is an inverting op-amp circuit with a single input. We’re going to call this V_a. We’ll call this A for now, and we have V_out. We worked out how this worked before, where V_out was a function of V_a times the ratio of these two resistors.

Now, I’m going to add a little twist to this and we’re going to analyze a different kind of circuit. So, I’m going to add another resistor right here like this and a second input. We’ll call this V_b. So, we’ll have R_a here; this will be R_b, and we’ll call this R_feedback. The question again is now, what is V_out in terms of the inputs V_a and V_b?

We’re going to use the idea of a 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. 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, you know, 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, these two voltages will be really close together.

They’ll be only microvolts apart at most. Because they’re so close, we can just say let's just say they’re the same for the purposes of analyzing this circuit. 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, 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 V_a and V_b. 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.

Right? So, we have zero volts on this node. We have zero current going this way. So 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 V_a and V_b and these two resistor values. So we can write I equals... now what is it? Well, we have I_a here and we have I_b flowing here. That means that I equals I_a plus I_b, right?

Let me write I_a and I_b in terms of Ohm’s law here. So we get I equals... what is I_a? I_a is this voltage divided by R_a, and this voltage is V_a minus 0 volts. So it’s V_a over R_a. Now let’s write down what’s I_b. I_b, we look at this voltage here; this voltage is V_b minus 0 or just plain V_b over R_b.

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 this current is I. And if I is going into R_f, we’ll label the voltage on R_f like that.

So now let’s write an expression for I in terms of R_f 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 0 minus V_out, so minus V_out divided by R_f. That’s Ohm’s 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 can write:

[
-\frac{V_{out}}{R_f} = \frac{V_a}{R_a} + \frac{V_b}{R_b}
]

Now we can get the final function. I’m just going to multiply through by R_f on both sides, and we get:

[
V_{out} = -\frac{R_f}{R_a} \times V_a - \frac{R_f}{R_b} \times V_b
]

So that’s our answer. That’s V_out as a function of V_a and V_b, the two inputs. You can see that the resistor ratio—there are 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 R_a equals R_b equals R_f. And just to pick a real number, we’ll just say they’re all equal to 10k ohms, and let’s see what this becomes now.

Now we have:

[
V_{out} = -\frac{R_f}{R_a} \times V_a - \frac{R_f}{R_b} \times V_b
]

We still have the negative sign, so this is some sort of inversion going on here. R_f over R_a is one, so it’s just V_a, and R_f over R_b is one, so this says V_b. So this gives us the nickname for this expression, which is called 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, -2 times V_a plus 3 times V_b. 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 (\frac{R_f}{R_a}) to equal 2, and I want (\frac{R_f}{R_b}) to equal 3.

So I can pick values; it’s the same R_f, and so I get to adjust R_a and R_b in here. I can pick component values: if I pick R_f equals 12k ohms, then I can pick R_a equals 6k and R_b equals 4k, and that would give me the ratios I want, and that would implement this function here.

So I’m going to 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:

[
R_f = 12k, \quad R_a = 6k \text{ ohms}, \quad R_b = 4k.
]

And we’ve designed a circuit that implements our summing function. So this is a pretty useful op-amp circuit.