As a summing circuit or an op-amp summer.

A two input summer where one of the inputs is V2 and

let me label the second input, this no voltage as Vx.

Now this technique of identifying subcircuits within more complicated

circuits can greatly simplify the analysis of the more complicated circuit,

because we can use the known results for

the subcircuits to speed up our overall analysis.

So for example, the inverting amplifier.

We know that the output voltage is related to the input voltage for

this inverting amp by Vx, the output voltage is equal to the input

voltage times minus R2, the feedback resistor over R1.

So no analysis was required, we just used our known result to relate V1 to Vx.

Now let's look at the summing circuit alone and

analyze its output voltage versus input voltages.

So let me redraw the summing circuit, like this.

Here is our resister R3 with our input voltage V2.

Here is the resister R4 with input voltage V1.

They're connected together and

connected to the inverting terminal of the op-amp and

I can draw the feedback resistor R5 output voltage and

this should be Vx, the Vx input is applied to R4.

So what I want to is use superposition of V2 and Vx to solve for

the output voltage of Vout for the summing circuit.

Now remember, when we use superposition,

we turn one of the input sources on with all of the other sources off and solve for

the output voltage, then we repeat that for every other input voltage source.

Then once we've determined the contribution to the output voltage for

each source individually,

we add all the contributions together to determine the total output voltage.

So, I'm going to begin by turning the V2 source on.

V2 on and Vx source off.

Now Vx is a voltage source.

When we turn a voltage source off, its voltage becomes zero volts or ground.

So, I can, for this condition, rewrite the circuit, like this.

Here's our resistor R3.

Here's our resistor R4 with Vx now grounded.

Here is V2.

Then I connect the rest of the circuit, like this.

R5, Vout and I want to solve for

a Vout in terms of V2.

Now the first thing to notice here in the circuit is that R4 has no effect

on the circuit and the reason for that is the voltage on this side of R4 is equal

to the voltage on this side of R4, so no current flows through R4.

This is an ideal op-amp, so the voltage at the non-inverting terminal is equal to

voltage at the inverting terminal.

This voltage is ground, this voltage is also ground.

So we have ground on this side, ground on this side.

So the voltage difference across R4 is equal to 0.

So the current through R4 is equal to 0.

Another way to see that is you could actually write

the Ohm's Law equation, V equals IR.

In this case, V, the voltage across R4 is equal to 0.

0 minus 0.

So that IR must be equal to 0.

R is a non-zero quantity, so the current I must be equal to 0.

So we can replace the resistor R4 by an open circuit.

So let me redraw the circuit one more time.

Here is V2.

Here is a resistor R3.

R 4 is an open circuit.

Or in other words is just left out.

Ground the non-inverting terminal and here is the feedback resistor R5, Vout.