# Understanding this peak detector circuit

Assuming ideal diodes, for $$\V_{in} \lt 0\$$ this is an inverting op amp with gain $$\\frac{R_3}{R_1}\$$, and $$\R_2\$$ keeps the non inverting input from floating. For $$\V_{in} \gt 0\$$, this is a buffer. The circuit probably makes the most "sense" if $$\R_3=R_1\$$, but if $$\R_3 \ll R_1\$$, than you are getting something akin to half wave rectification instead of full wave.

You are correct about the low pass filter, but since you are low-pass filtering a rectified signal, this is more like an rms-filter or envelope detector.

I wouldn't call it a "peak detector", as those usually store the peak value on a cap with no discharge path. Here, the cap discharges through a resistor, so the voltage is not stored.

The opamp circuit is a full wave rectifier with some level of noise rejection when the input peak amplitude starts to fall below about 0.6 volts. Positive voltages are amplified by D2 (D1 is blocking) and negative voltages are amplified and inverted via D1 (D2 is now blocking). R2 is needed to bias the non-inverting opamp input when the device is inverting i.e. D2 is blocked. R1 and R3 should be identical values.

The low pass filtering effect of C1 affects both positive and negative input voltages differently and is probably incidental to the whole circuit operating as a full wave rectifier.

Because it is a full wave rectifier, the envelope detector is fed with twice as many carrier cycles per second and hence it can deliver a better performance compared to when feeding the raw input signal directly to it.