Operational Amplifiers with feedback resistors

Question

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.

.

Answer 1

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).

Answer 2

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)