# Am I doing this ideal op-amp problem correctly?

1. Apr 20, 2014

### pinterest

1. The problem statement, all variables and given/known data

2. Relevant equations
$$Vp=Vn$$
$$Ip=In=0$$

3. The attempt at a solution
By voltage division:
$$Vp = 8/(8+8) V3$$
$$Vp = 1/2V3$$
$$Vn = 1/2V3$$

By node-voltage method at the inverting input terminal:
$$(Vn-V1)/4 + (Vn-V2)/4 + (Vn- V0)/8 = 0$$
Multiply equation by 8
$$2(Vn-V1) + 2(Vn-V2) + Vn-V0 = 0$$
$$V0 = Vn+2Vn-2V1+2Vn-2V2$$
$$V0 = 5Vn-2V1-2V2$$
Substitute in 1/2V3 for Vn

$$V0 = 5/2V3-2V1-2V2$$

Is this the correct answer? I'm pretty sure about the V1 and V2 part, because it matches the summing amplifier circuit equation of -(Rf/R1*V1+Rf/R2*V2). However, for V3, I don't see how its resistor composition would yield an answer of 5/2. Could someone clarify?

Thanks

Last edited: Apr 20, 2014
2. Apr 20, 2014

### rude man

One easy way to work problems with multiple inputs is to take one input at a time, get the corresponding output, then add all three outputs. So for your worrisome V3 input the output is

Vo3 = V+(1 + Rf/Ri) = 0.5V3(1 + Rf/Ri) = 0.5V3(1 + 8/2) = 2.5V3

Also,
Vo1 = -8/4 v1
Vo2 = -8/4 v2

Then by superposition,
Vo = Vo1 + Vo2 + Vo3.

Note that when you do that the other voltage sources are set to zero volts (grounded), not left 'floating'.