I was simulating a circuit similar to the one shown in the schematic below (educational, but not for school), and noticed the peak in the Bode plot at around 100 kHz. The circuit shown has been slightly reduced from the original, which is why it seems somewhat pointless, but still displays the same unexpected behaviour (details of original don't seem relevant, but are available upon request).

At low frequencies, I expected the capacitors to act as opens, so the circuit would reduce to a unity gain non-inverting amplifier. At high frequencies, I expected the capacitors to act as shorts, reducing the circuit to a voltage follower. Both of these extremes match the simulation, but I'm failing to come up with an explanation for what appears to be resonance at 100 kHz.

Switching to an ideal single-pole op amp eliminates this artifact, but it is present other op amp models, so it is not unique to the LT1001. Other models yield different frequencies and magnitudes, but the spike is still present. For reference, the most realistic model I have prior exposure to is a finite-gain single-pole op amp.

Can anyone shed some light on this simulation result? What nonideality is responsible?



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    \$\begingroup\$ Notice how the oscillation occurs at 90deg phase shift. Can you identify a source of delay in the feedback path? \$\endgroup\$ – user133493 Apr 7 '17 at 3:42
  • \$\begingroup\$ I think I see. At that frequency, the phase shift of the op amp and the phase shift of the RC feedback network give a total of 360 degree shift, so the negative feedback turns to positive feedback. Is that about right? \$\endgroup\$ – andars Apr 7 '17 at 4:50

You cannot expect an opamp to perform when you have input and feedback capacitors at 0.5uF. Work out the equivalent impedance at 100 kHz (3 ohms) and you should eventually conclude that the current needed from the opamp output is far greater than what it can supply for even very moderate input voltages.

  • \$\begingroup\$ Thank you for your answer. I understand that the impedance goes way down so the op amp would have to supply too much current, but wouldn't that effect be even worse at 200 kHz? I don't quite see why overcurrent would cause such a distinct spike. \$\endgroup\$ – andars Apr 7 '17 at 14:45
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    \$\begingroup\$ As the frequency rises past the resonant point the dominant signal is the input that is coupled to the output directly via two very low impedance. At this point and above you no longer have an inverting opamp but basically a short from output to input. Remember this is a sim you have done and sims won't behave like real life when pushed this way. How will they behave exactly is pure speculation. \$\endgroup\$ – Andy aka Apr 7 '17 at 17:28

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