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.

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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\$ – Ignacio Vazquez-Abrams Jan 1 '16 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 '16 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\$ – Jasen Jan 1 '16 at 23:36
  • \$\begingroup\$ @ignacio-vazquez-abrams How could I use the currents? \$\endgroup\$ – Pter Jan 2 '16 at 10:41

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\$


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