I have a theoretical question about op-amps. The diagram shows an inverting op-amp configuration with some resistors in the feedback path.

The non-inverting terminal of the op-amp is certainly zero since it is connected to ground. The inverting terminal is then forced to zero and becomes virtual ground (for ideal conditions).

Is it valid if I make R4 and R3 become parallel, and then use the formula of inverting gain?

So, it becomes Vo/Vin = [(R3//R4)+R5]/R. In other words, can we actually parallel two resistors that are NOT EXACTLY inbetween the same node, since I know that virtual ground and real ground are not the same nodes? My Professor sometimes make them parallel in some other case, but not this one. I am really confused why. The answer from my professor is: -R3/R4 [(R5+R3//R4) / (R3//R4)]. Note that // means parallel resistor.

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  • \$\begingroup\$ what do you mean with "Make R3 and R4 become parallel"? Could you draw the thus-modified schematic? \$\endgroup\$ Feb 6 at 19:00
  • \$\begingroup\$ @MarcusMüller question edited. \$\endgroup\$
    – ryan
    Feb 6 at 19:15

2 Answers 2


No, you can't. The reason is simple: by assuming \$R_3\$ and \$R_4\$ in parallel for the computation of the gain, you're actually assuming that the current injected into the inverting terminal from \$R_3\$ equals the current crossing the parallel of \$R_3\$ and \$R_4\$.

The above is not true because the current injected into the inverting terminal from \$R_3\$ is only a fraction \$R_4/(R_3+R_4)\$ (current divider's formula) of that of the parallel.

When you want to solve this kind of circuit, you have to proceed in the following way:

  1. Determine the voltage \$v_\mathrm{n}\$ at the inverting terminal as a superposition of the output voltage \$v_\mathrm{o}\$ and, possibly, the input voltage \$v_\mathrm{i}\$, assuming \$v_\mathrm{n}\$ open-circuited, and just that (that is, drawing no current, but don't assume any specific potential);
  2. Impose the relationship \$v_\mathrm{n}=v_\mathrm{p}\$, \$v_\mathrm{p}\$ being the voltage at the non-inverting terminal, to eliminate the variables \$v_\mathrm{n}\$ and \$v_\mathrm{p}\$, and determine \$v_\mathrm{o}\$ as a function of just \$v_\mathrm{i}\$. This step is justified by the fact that an ideal op amp, when there's (negative) feedback from the output to the input, steers the output voltage to keep \$v_\mathrm{n}=v_\mathrm{p}\$ (for an ideal op amp, it doesn't really matter whether the feedback is positive or negative, but indeed this distinction is fundamental when implementing circuits with real op amps).

Your approach to this can work but you're missing a step- you calculate the Thévenin equivalent of the network to the right of the R1/2/3 junction point.

Vt = Vo*R4/(R4 + R5) and Rt = R3 + (R4||R5).

Then the gain is Vt/Vin = -(R3 + (R4||R5))/R1 but that is the gain from input to the Thévenin voltage Vt. To relate that to Vo you need to multiply it by (R4+R5)/R4.

(I think you may have a transcription error in the class solution - it has to include R1)


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