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Op amp rectifier : transfer function and output impedance


Hello guys,

I'm working on an EE student project which uses a rectifier to keep the peak value of a sine wave.
You can find a PSpice pic of the schematics below (Ve is the input (1kHz), Vs is the output).

I really have problems to analyse this pic. It is quite easy when I replace the diode by a short (and forget about the capacitor for now), then I just have Vs = (2+R3/10k)Ve (the potentiometer R3 gives a variable gain) by considering the op amp is ideal. Until now, it is okay.
When Ve is positive, the diode is on and the upper relation between Ve and Vs is correct, knowing that the 0.6V voltage drop accross the diode is compensated by the op amp.
When Ve is negative, the diode is off and the op-amp saturates at -Vsat behind the diode.
Until now, is it correct ?

Everything becomes more complicated when we put on the capacitor. It works perfectly to keep the peak value but I don't understand everything yet.
When the diode is on, the time constant should be very small, isn't it ? The output impedance of the op amp should be near zero, so the time constant tc = RC is very small, right ?
And when the diode is off, all resistors in the feedback loop (R1, R2 and R3) come in the time constant and tc becomes high, isn't it ?
Well, then why isn't the capacitor just following the signal when it is positive ? And stay at zero when it is down ?
I'm sure I'm missing something simple, but I don't know what...

And, when the time comes to find the total transfer function Vs/Ve (for both cases when the diode is on and off), I'm completely lost.
Just the same when it comes to find output impedance...

Hope I expressed myself clearly (and my English wasn't too bad).
Any help would be greatly appreciated.

Thanks a lot,

The diode will only conduct if the anode is ~0.6V more positive than the cathode (not than ground, since the output is not grounded), so the op-amp will never sink current, only source. The voltage across the cap should follow the op-amp on rising slopes, but since the discharge of the cap is very slow, the voltage won't decrease as the signal from the op-amp decreases. At this point, the diode becomes reverse-biased until the cap is discharged through R1+R2+R3.


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