The bode plot of the loop is as shown here

Bode Plots

The transient step response looks like this

Transient Response

The block diagram looks something like this - Basically a single stage opamp(OPAMP in fig) with high gain(cascode structure) biasing a fet(M1) such that its drain is equal to Vbias. Small signal step is applied at I1 Ibias and transient current through the fet M1 drain is observed. Block Diagram

Am I missing something here?

  • \$\begingroup\$ What kind of system are we dealing with? Is it completely linear over the operating range? \$\endgroup\$ – John D Sep 12 '19 at 20:30
  • \$\begingroup\$ It’s a second order system with an opamp in it. Used some miller compensation to split poles apart. When a step current is applied at a point in the loop, I’m observing an overshoot inspite of PM being close to 90 degrees \$\endgroup\$ – SBO Sep 12 '19 at 21:06
  • \$\begingroup\$ Show your circuit? It certainly seems like it should settle nicely. \$\endgroup\$ – TimWescott Sep 12 '19 at 21:58
  • \$\begingroup\$ Also would be good to show how you're measuring the open loop response. How you're "breaking" the loop, injecting the disturbance and where you're measuring the response. \$\endgroup\$ – John D Sep 12 '19 at 22:09
  • 1
    \$\begingroup\$ It doesn't look to me like your loop gain measurement technique is accounting for the FET's gain. \$\endgroup\$ – Andy aka Sep 13 '19 at 7:06

I suspect you're looking at an artifact of the propagation delay through the op amp. The output overshoots because the effects of the negative feedback aren't felt until a finite time period after the response to the step.

If there's a propagation delay of N ns, the step response won't produce an effect on the output until T=N, and thus the feedback won't produce an effect until T=2N.

  • \$\begingroup\$ Surely, loop analysis characterizes precisely all of that..? \$\endgroup\$ – Adil Malik Sep 13 '19 at 14:38
  • \$\begingroup\$ @AdilMalik - No, loop analysis essentially looks at the response to a CW sine wave after all perturbations have died out. You are looking at a step response, where the effects of the higher-frequency non-linearities are still in force. \$\endgroup\$ – WhatRoughBeast Sep 13 '19 at 17:32
  • \$\begingroup\$ Yes ofcourse. I agree that frequency response is done on a linearised model. However, this is not what this answer is about and what I am questioning. Christobol refers to delay thru the opamp which should be characterised in the linear model etc? \$\endgroup\$ – Adil Malik Sep 13 '19 at 18:28

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