I am trying to build an approximate opamp with a differential amplifier and add a feedback loop to get the approximate wanted gain.

I used a differential amplifier with a current source and a current mirror, and got open loop gain of 3k.

When I add the feedback loop resistors I expect to get:

$$ A_{cl} = \frac{A_OL}{1 + A_OL \beta}\cdot K =\frac{3k}{1 + 3k\cdot\frac{1}{1+20} }\cdot\frac{20}{1+20} \approx 19.8$$

Instead I get $$ A_{CL} \approx 70$$

I don't understand what I'm doing wrong, and if what I want to do is even possible.

EDIT: In the schematic below I placed by mistake a feedback resistor of 50k instead of 20k like my original circuit.

I also forgot to mention, for the NPNs and PNPs: $$\beta = 100, V_A = 100[V]$$


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


Your feedback is positive. As the voltage at the base of Q7 goes more positive Q7 turns on more, and it's collector voltage goes down. The voltage at the collector of Q6 will do the opposite, going more positive. This voltage is what you are feeding back, so as the input goes more positive the feedback goes more positive as well, which should either drive it into saturation or oscillation.

By taking the output from Q6 you are making the base of Q7 the non-inverting input and the base of Q6 the inverting input. You can see this if you remember that a common emitter amplifier inverts the signal, and a differential amplifier is just two common emitter amplifiers tied together.

In an op-amp there are usually 3 stages, the differential input, the voltage amplifier, and the output stage. The feedback is taken from the output stage which will be 180 degrees out of phase with the inverting input.

  • \$\begingroup\$ Hi, thanks, I added a voltage buffer (I used a common-collector with the same current mirror connected to it's emitter) after the differential amplifier and then connected its output as feedback (through the resistors, and got exactly what I need. \$\endgroup\$ Jun 3, 2022 at 8:24

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