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I am wondering which kind of circuit is that and what is does when +-Vin is applied on the input. The output of the circuit is cathode of D26/R30/C52 node. Thank you.

enter image description here

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  • \$\begingroup\$ It's a precision rectifier with LPF. If D26 was not present then I'd say the cutoff frequency is around 63kHz. But I think that the cutoff frequency is chip-dependent when D26 is present. The main problem with this rectifier is the crossover (or zero-cross) distortion (i.e. the circuit does not output zero when the input is zero). \$\endgroup\$ – Rohat Kılıç Oct 8 '20 at 13:49
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It's a precision half wave rectifier.

It can be considerably improved by connecting another diode with cathode to the op-amp output and anode to the inverting input. This stops the output slewing towards the negative rail when the input is positive.

Note that the output impedance is not symmetrical. It's low for positive going output but 10k otherwise. This can cause unexpected problems if unbuffered.

A possible better solution is a full wave precision rectifier using 2 op-amps.

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This is a full-wave precision rectifier. When a negative voltage is applied to the input it acts as an inverting amplifier, so Vout = -Vin. When a positive voltage is applied, the op-amp rails near the negative supply rail and the positive voltage finds it's way to the output, Vout = +Vin (if loading is minimal on the output).

The op-amp output has to slew all the way from near the negative rail to +0.6V at the negative-going zero crossings so it's not going to be great at high frequency. The capacitor also doesn't help with high frequency response but it may help with stability. Op-amps which are railed don't necessarily come out of saturation, and start slewing, as quickly as one might hope- and any measure of that is often omitted from the datasheet information, particularly if it is bad.

Also, some op-amps behave differently in this configuration because of protection diodes between the inverting and non-inverting inputs, so current flows through R31.

Edit: See below for simulation of this working.

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

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  • \$\begingroup\$ Hey, Spehro. It's half wave. Sure you meant to say that, \$\endgroup\$ – Graham Stevenson Oct 8 '20 at 14:50
  • \$\begingroup\$ @GrahamStevenson Full wave. \$\endgroup\$ – Spehro Pefhany Oct 8 '20 at 14:52
  • \$\begingroup\$ Ahh .. when unloaded. I see what you mean. Really needs a buffer though. \$\endgroup\$ – Graham Stevenson Oct 10 '20 at 11:42
  • \$\begingroup\$ @GrahamStevenson Yes, it does, and before the averaging filter. But it can just be a voltage follower, one of the remaining 3 amplifiers in that quad package. \$\endgroup\$ – Spehro Pefhany Oct 10 '20 at 11:54

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