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I think the circuit attached below is a peak detector because of the diode, resistor and capacitor at the op-amp's output. What I don't understand is the diodes on the input side. The input signal swings positive and negative. In the positive cycle, the signal goes through D2; what's the purpose of R2?

In the negative cycle the signal goes through D1 and R1. R3, C1, and R1 forms a low-pass filter network, right? Correct me if I'm wrong. I would appreciate any additional detail as to the operation of this circuit.

enter image description here

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  • \$\begingroup\$ Why don't you put in the effort to select an answer? \$\endgroup\$ – RoyC Aug 15 at 15:28
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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.

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    \$\begingroup\$ You are correct. R3 = R1 in the actual circuit. So the amplifier is unity gain. \$\endgroup\$ – Blue_Electronx Aug 12 at 16:20
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    \$\begingroup\$ Thanks for pointing out it's a envelope detector. It makes sense. \$\endgroup\$ – Blue_Electronx Aug 12 at 16:26
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    \$\begingroup\$ R1 and R3 have nothing to do with the filter, but the inverting amp component will simply not work without at least R1 in place (without which there is no effective feedback, and without R3, the gain of the inverting amp will be zero. \$\endgroup\$ – Scott Seidman Aug 13 at 12:44
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    \$\begingroup\$ I'm sorry -- R3, along w/ the feedback cap, determines cutoff freq of the op amp circuit. R1 does not. I thought you were referring to the output filter after the diode. \$\endgroup\$ – Scott Seidman Aug 13 at 13:40
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    \$\begingroup\$ In this case, you should think of the non-inverting portion as a voltage follower configuration with a filter in the feedback loop. You don't need a feedback resistor or R1 to get a unity gain. R1 is effectively invisible when the input is positive, because of the diode. \$\endgroup\$ – Scott Seidman Aug 13 at 15:50
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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.

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The diodes on the input are intended to compensate for the diode on the output. They face different directions so that each compensates for the positive drop on the output diode.

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  • \$\begingroup\$ In the circuit above, R3 and C1 right side are connected to the anode of the diode, but I've seen that in some peak detectors, this is connected to the cathode. What's the difference in connecting in a way or another? \$\endgroup\$ – Blue_Electronx Aug 12 at 15:15
  • \$\begingroup\$ That would work if the cathode was the rectified output, but in this case it's the peak-detected output. The op amp can't pull that node low to stay in its linear region. \$\endgroup\$ – Cristobol Polychronopolis Aug 14 at 14:14

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