I have a following circuit with an operational amplifier. It is a configuration I haven't seen before and I am not quite sure how to calculate the gain in terms of its inputs A and B.

enter image description here

The input B is just the configuration of Inverting Amplifier $$Gain_B = -\frac{470}{240}$$ Then I am puzzled by the 4k3 resistor. If the resistor wouldn't be there, the configuration would be similar to the differential amplifier and the gain would be $$ G = \bigg(\frac{240+470}{240}\bigg)\bigg(\frac{3k6}{3k6+510}\bigg)A-\bigg(\frac{470}{240}\bigg)B$$ However, that is neglecting the 4k3 resistor, which somehow changes the gain - it reminds me of the non-inverting amplifier configuration, which would have gain of $$G_\text{part} = 1+\frac{470}{4k3}$$ and the two resistor in front of it would just act as a potential divider \$\frac{3k6}{3k6+510}A\$

The correct solution should be $$G \approx 2.687A-1.9583B$$

Just to clarify, I am assuming and ideal op-amp here (i.e. infinite gain, etc.). This should not be a hard problem, however, I have failed to find even after a hour of googling (it might be caused by the fact that I do not know how to call this circuit configuration).

  • \$\begingroup\$ Have you considered using the currents to calculate it? \$\endgroup\$ Jan 1, 2016 at 23:30
  • \$\begingroup\$ Is your first statement (GainB = -470/240) true? It would be true if the + input was tied to GND - but it's not. So, for example, if VB is the same as the + input voltage the output voltage would also be VB. \$\endgroup\$
    – Transistor
    Jan 1, 2016 at 23:30
  • \$\begingroup\$ it's just a supeposition of two common circuit, ground b and evaluate the gain you get from A, then ground A and evaluate for B \$\endgroup\$ Jan 1, 2016 at 23:36
  • \$\begingroup\$ @ignacio-vazquez-abrams How could I use the currents? \$\endgroup\$
    – Pter
    Jan 2, 2016 at 10:41

2 Answers 2


It's just a supeposition of two common circuits, ground B and evaluate the gain you get from A, then ground A and evaluate for B.

you've got the B term right "-470/240" = -1.9583

for the A term ground the B input so the 4.3K is parallel with the 240

$$ \bigg(\frac{240||4k3+470}{240||4k3}\bigg)\bigg(\frac{3k6}{3k6+510}\bigg)A $$


Let's denote the voltage at the positive and negative inputs of the opamp with \$P\$ and \$M\$ respectively.

Then we get the following equations (should be solved for \$I\rightarrow \infty \$ for the open loop gain (I think its called)):

(1) \$Z = I(P-M)\$ (opamp)

(2) \$P = \frac{3k6 A}{3k6+510}\approx0.87591A\$ (voltage divider)

(3) \$\frac{B-M}{240} = \frac{M}{4k3}+\frac{M-Z}{240}\$ (currents at negative input)

From (3):

(4) \$M = \frac{Z}{470K}+\frac{B}{240K}\$ where \$K=\frac{1}{4k3}+\frac{1}{470}+\frac{1}{240}\approx0.0065269\$

From (1) and (4):

\$Z = I(P-\frac{Z}{470K}-\frac{B}{240K})=IP-\frac{IZ}{470K}-\frac{IB}{240K}\$

divide both sides by \$I\$ and observe that \$\lim\frac{Z}{I}=0\$ and we get:

\$0 = P - \frac{Z}{470K}-\frac{B}{240K}\$

From this follows:

\$Z=470KP-B\frac{470}{240}\approx2.687A - 1.9583B\$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.