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enter image description here

I was just playing around with a few Op-Amp circuits, and suddenly got stuck on this simple one.
As the circuit shows, the Op-Amp output is connected to its non-inverting input terminal, so it should NOT act as a voltage buffer. However, as my image shows, its a different story (simulated with NI Multisim 14.0, however I got same results using EveryCircuit) We have virtual short. Output is 5V.
I expected -15V (Inverting input terminal would rise, differential voltage would go -ve, output would rush to -ve rail, and differential voltage still stays -ve).
I also cannot understand the output frequency of 50kHz. Why would it oscillate with such a small peak-to-peak voltage? I can't justify this observation. Is this a problem with the simulation software?

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    \$\begingroup\$ I simulated the same circuit in ISIS (a SPICE based simulation tool like the NI Multisim) and I got the same result except for the frequency (0Hz in my simulation). When I replace the opamp with an LM358 it gives -15V output as expected. Most likely, this should be about the simulation model of the opamps. By the way, try not to supply the opamp with higher than 22V total. I simulated with +/-9V but nothing changed, though. \$\endgroup\$ – Rohat Kılıç Aug 6 at 9:06
  • \$\begingroup\$ Yes when I simulated in EveryCircuit the freq was 0Hz too. \$\endgroup\$ – nn08 Aug 6 at 9:23
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    \$\begingroup\$ The output frequency of 50kHz is probably because the simulation is running at 100,000 time steps per second, and there is some small numerical rounding error that makes the results alternatiely rise and fall by a few nV. \$\endgroup\$ – alephzero Aug 6 at 18:59
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No - it is not a problem of simulation software. Surprisingly, the result is correct - but not realistic. What does this mean?

Theoretically, the result would be correct under two environmental conditions:

  • No power switch-on transients (power available since (t-infinity))

  • No external disturbances (noise, etc).

During simulation, the program automatically assumes these two conditions - unless you specify other conditions.

However, if you would switch-on both (or one) supply voltage at t=0, a transient analysis would reveal that the circuit is saturated due to positive feedback.

The situation can be compared with a mechanical model: A system of two balls - one lying upon the other one - could be stable as long as there is absolutely no external disturbance. But this is NEVER the case. The same applies to the real opamp usage: Power switch-on transients, supply voltage is not absolutely constant, noise,...)

I think, in 99.9% of all cases, the simulator does not fail but the user (misinterpretation of the results)

Comment: But there are two indications that the simulation result is "problematic" (not wrong, because the simulator was correct) and should be investigated in detail (under real conditions):

  • In your simulation, the output is positive - even with an input at the inv. node.

  • An AC analysis would show a rising phase function (which is also not realistic).

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  • \$\begingroup\$ Yes the positive output, that's what troubled me too. So does the simulator begin with an assumption of negative feedback (provided it nulled all other non-idealities you mentioned)? \$\endgroup\$ – nn08 Aug 6 at 9:31
  • \$\begingroup\$ No - the simulator assumes positive feedback because he does not "know" in advance that there might be a problem. By the way: When you calculate with pencil and paper (without knowing about instability) you will arrive at the same result (+5V) as the simulator. Remember the analogy with the two balls: If they are EXACTLY (but thats impossible in real life) placed upon each other, the calculation would show that it is a (marginally) stable system \$\endgroup\$ – LvW Aug 6 at 9:43
  • \$\begingroup\$ The deduction method that I wrote in my question, i.e., how to find the output voltage, is it then valid only for a step- kind of input? So then I could see a transient behavior from t=0? Actually this is how I deduce whether the opamp output goes to +ve or -ve rails, depending on the sign of its differential voltage. \$\endgroup\$ – nn08 Aug 6 at 9:54

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