I've been analyzing the schematic for an old commercial linear power supply, the Farnell LT30/2 (schematic). The schematic contains a couple of examples of an op amp arrangement that seems very popular in power supply regulation circuits.

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The op amp has a diode at the output and can only sink current. I understand that the op amp is essentially a comparator and the diode is there to reduce the drive to the pass transistors. In this case, the op amp (1/2 of IC2) is comparing the output of the Howland current source (0 - 1mA) across R12 (33k) to the output voltage at +O/P.

What I don't understand is the purpose of R16 and C4.

I'll make a stab at it and say it's to reduce the effect of spikes on the output. My rationale is that the time constant of R12 and C4 is quite high. If the output suddenly spiked, the leading edge of the spike would be slowed to allow the op amp to catch up. That's what I'm thinking, but I have no idea if this is true or not.


1 Answer 1


That's a filtering element in the feedback path: High frequencies pass easily through the capacitor, low frequencies don't.

Thus, the effect of this is that high frequencies get attenuated by this. That can have a lot of purposes – usually, noise robustness.

  • \$\begingroup\$ Right, that's what I figured. Good to get the confirmation. \$\endgroup\$
    – Buck8pe
    Commented May 31, 2017 at 15:30
  • \$\begingroup\$ sorry for the delayed comment, but thinking about this configuration some more and seeing the answer to "Negative feedback capacitor in Op-Amp comparator circuit" (Related), I still have some questions. Isn't the gain of the high freq transient 1+(680k/10k)? And how does the diode affect the op amps ability to equalize its inputs - does this gain equation even apply any more? \$\endgroup\$
    – Buck8pe
    Commented Jun 1, 2017 at 12:45

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