Am I doing this ideal op-amp problem correctly?

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SUMMARY

The discussion centers on solving an ideal op-amp problem involving voltage division and the node-voltage method. The user initially calculated the output voltage (V0) as V0 = 5/2V3 - 2V1 - 2V2, but sought clarification on the contribution of V3. A response clarified that the output for V3 can be computed using the formula Vo3 = 0.5V3(1 + Rf/Ri), resulting in Vo3 = 2.5V3. The superposition principle was also emphasized, indicating that other voltage sources should be grounded during calculations.

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Homework Statement


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Homework Equations


Vp=Vn
Ip=In=0

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:
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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'.
 

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