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Just in theory, please ignore the values...

Given the open loop gain of an opamp and its input voltage and a circuit similar to the figure below, how would I find the output voltage?

opamp circuit]([![CircuitLab Schematic b3e6ns]([![CircuitLab Schematic p5y8pf]

My thoughts were to just do nodal analysis and avoid the open loop gain all together since the circuit given is closed loop, I thought that it was irrelevant and just given to throw you off. Or would the correct steps be to assume: v_out = A(v_+ - v_-) v_out = A(-v_-)

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    \$\begingroup\$ Use ideal OpAmp assumptions. \$\endgroup\$ – Nick Alexeev Oct 20 '15 at 1:49
  • \$\begingroup\$ @NickAlexeev So then we would assume v+ = v-, and A = inf. And we would just find v_out using nodal analysis... right? \$\endgroup\$ – Shaun Dizzle Oct 20 '15 at 2:18
  • \$\begingroup\$ There is also the second ideal opamp assumption: The inputs of the ideal OpAmp draw no current. You are on the right track, though. \$\endgroup\$ – Nick Alexeev Oct 20 '15 at 2:31
  • \$\begingroup\$ @NickAlexeev Nice. Ok, yeah I am familiar with the assumptions for an ideal opamp, I was just curious about the assumption of what to do with the open loop gain. Like I said, I thought we should ignore that since our circuit is close loop (i.e. we have feedback). Thanks \$\endgroup\$ – Shaun Dizzle Oct 20 '15 at 2:51
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    \$\begingroup\$ The schematic doesn't show any op amps. Did you make changes to it in CircuitLab after posting the question? \$\endgroup\$ – scanny Oct 20 '15 at 3:41
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Answer reformulated:

As already noted, your circuit simplifies to: enter image description here

When you close the loop the open-loop gain is reduced to the ratio \$ \frac{\text R2}{\text R1}\$ so, since it's an inverting amplifier, the output voltage becomes: \$\ \ \text {Vout} = \ \ - \frac {\text R2}{\text R1} \times \text{Vin} \$

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