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I have a circuit: Opamp circuit

Assuming the op-amps are ideal, how do I derive an expression of Vout in terms of A and B?

I set B to 0 and worked out that the output of the opamp connected to A is 9A, and by setting A to 0, I worked out the output of the opamp connected to B is -2B.

I then merged the two expressions together, forming Vout = 9A-2B. I think this is incorrect. Please help me understand where I went wrong.

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  • \$\begingroup\$ Put voltages on each input and calculate the output - for the inverting input attached to B, keep in mind that the feedback will servo to maintain that point at the same voltage provided by the non-inverting amplifier. \$\endgroup\$ – Peter Smith Dec 2 '20 at 13:38
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Your expression is correct. Let's put some values in to prove that.

If the expression is 9A - 2B then we can state with a +1V input at both A and B, the output should be +9 - 2V = +7V.

Annotated amplifier

You will see I have identified the inputs of the second amplifier with points X and Y.

Apply 1V at A. The voltage at point Y must therefore be +3V. The output will servo to maintain point X at this +3V. Apply +1V at B. Point X is therefore 2V more positive than B. To achieve this, the output must go 4V more positive than this point * so the output = 4+3 = 7V

  • As the input resistor from point B will have 2V across it and there is no current into the inverting input, then the same current must flow in the feedback resistor. The ratio of voltages is the same as the ratio of resistances so the voltage across the feedback resistor must be 2 * 2V = 4V. As the voltage at the inverting input is 3V, then 3V + 4V = 7V.
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  • \$\begingroup\$ Yes, this makes a lot of sense, thank you very much! I was a bit confused when I built this circuit in a simulator and it read Vout as 12V for any value of A or B larger than 2V \$\endgroup\$ – SidS Dec 2 '20 at 15:56

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