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schematic

simulate this circuit – Schematic created using CircuitLab

From what I understand, the diagram above is of a differential amplifier and I understand them and how they work. But the question I have been given is to find the non-inverting gain and the inverting gain for the op-amp above. I'm struggling to understand how it can have two different gains so I would really appreciate it if someone could shed some light on this please. Going from what I have currently, I think the inverting gain is just the regular -Rf/Rg but I don't even have a guess for the non-inverting gain. This is my first time posting so sorry if did something incorrectly.

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First we must start with some basic assumptions, we assume that the inputs of the circuit are fed from ideal voltage sources and that the op-amp is an ideal op-amp. The latter means that the circuit is linear and we can apply the principle of superposition to calculate the two gains..

For the inverting gain we treat things as if the + input of the circuit (not the + input of the op-amp) is connected to ground. The bottom two resistors have no affect since the op-amps inputs are high-impedance and the top two resistors act just like a normal inverting amplifier.

-Rf/Rg = -20/5 = -4

For the non-inverting gain we treat things as if the - input of the circuit (not the - input of the op-amp) is connected to ground.

When we do this we find that the non-inverting gain is made up of two stages. First we have a voltage divider formed by the bottom two resistors in the diagram. Then we have the op-amp along with the top two resistors in the diagram acting as a non-inverting amplifier. So the overall non-inverting gain works out to

(30/(20+30)) * (1+(20/5)) = (3/5) * 5 = 3

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  • \$\begingroup\$ The above (correct) calculation is nothing else than an application of the well-known superposition theorem. \$\endgroup\$ – LvW Apr 26 '16 at 7:02
  • \$\begingroup\$ Yup, just added a bit of text to clarify that. \$\endgroup\$ – Peter Green Feb 17 '19 at 14:40

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