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Correct me if I am wrong but the following is what I think the assumptions we make for an ideal op-amp:

Negative feedback (output connected to the inverting input):

  • No current goes into either of the inputs
  • The voltage at each of the inputs is the same
  • We assume that \$A\$ in the formula \$V_{out}=A(V_+-V_-)\$ is equal to \$\infty\$. (Does this need to be infinity or can it just be very large?)

Positive feedback (output connected to the non-inverting input):

  • No current flows into either of the inputs
  • The output will always be at saturation (or heading there) (it is bistable) (This does however follow from the other assumptions, namely the one below)
  • We assume that \$A\$ in the formula \$V_{out}=A(V_+-V_-)\$ is very large. In this case however I don't think we have to assume that it is \$\infty\$.

In all other cases we assume that no current flows into the op amp.

If we have positive and negative feedback, all we can assume is that no current flows into the op-amp and the voltages depend on the circuit in question. In the specific case of the astable multivariate it happens that the output voltage changes between its two saturation values.

So here are my questions:

  1. Firstly, is the above correct and is there any further assumptions that I can add?
  2. Are there any exceptions to the above rules (staying of course with ideal op amps)? I was thinking about e.g. a capacitor in the negative feedback situation, like in an op amp integrator circuit, could the rule about the voltage across a capacitor not been allowed to change instantaneously mean that the assumption that the voltages at the inputs are the same does not hold?
    1. Is it possible to have a circuit with positive and negative feedback that does not go to one of its saturation points?

If you want me to expand on any point, just ask. Thanks.

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  • \$\begingroup\$ Depends on what you mean by "ideal". An ideal op-amp could mean infinite gain and no saturation. It's definitely possible to have a circuit with both positive and negative feedback that is stable, but it of course depends on the specifics. \$\endgroup\$ – John D Dec 10 '14 at 16:06
  • \$\begingroup\$ You need to add infinite open loop gain. Also, you don't need to have an assumption of saturation for the positive feedback config -- it comes out of the other assumptions. \$\endgroup\$ – Scott Seidman Dec 10 '14 at 16:17
  • \$\begingroup\$ @ScottSeidman I have edited it to inculde your points \$\endgroup\$ – user53915 Dec 10 '14 at 16:52
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    \$\begingroup\$ A negative capacitance amplifier does not saturate if the source capacitance is high, and that has negative and positive feedback -- en.wikipedia.org/wiki/Negative_impedance_converter \$\endgroup\$ – Scott Seidman Dec 10 '14 at 17:32
  • \$\begingroup\$ @ScottSeidman You need more assumptions than Iin = 0 and Vo = A(V+ - V-) for a positive feedback op-amp. Mathematically, all it does is flip the input terminals terminals: Vo = A (V- - V+), which has a mathematically finite solution when V- = V+, like the negative feedback case, which might be interpreted as the "correct" output for ideal positive feedback op-amps, but is completely unstable. \$\endgroup\$ – helloworld922 Dec 10 '14 at 18:05
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Positive feedback and negative feedback can be combined to create oscillation (astability). Aside from that, I think you're correct. In the case of the integrator, there's never a step change in the output, so there's never a discontinuity in the capacitor voltage. The only limit is the output voltage range of the op amp.

Assuming an infinite gain with negative feedback implies that the voltages at the two inputs will be equal. Assuming an infinite gain with positive feedback implies that the output will be saturated (unless V+ exactly equals V-).

Even in a real op amp, there's very little current into the input pins (microamps at most). Likewise, the open-loop gain is usually very large at low frequencies (on the order of 10^5 or 10^6 V/V). So the ideal op amp rules are often a pretty good approximation. But all real op amps are limited by their supply voltages, so sometimes that limit is included in an ideal model as well.

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  • \$\begingroup\$ Hi thanks for your answer but please could you explicitly answer the question of whether there are any exceptions? I think I now understand why we don't an exception with the integrator. Once again thanks. \$\endgroup\$ – user53915 Dec 12 '14 at 15:44
  • \$\begingroup\$ I don't think there are any exceptions. An ideal op amp can produce infinite voltage and current, so as long as there's any negative feedback path at all it should be able to keep the input voltages equal. \$\endgroup\$ – Adam Haun Dec 12 '14 at 16:38

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