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just can't wrap my head around this one for some reason. I am sorry if it's easy for you to see. Just confused on the duality of things in circuits, Is it a filter? Or is it a peak detector? How do you know which topology to apply?

I am only talking about the positive cycle of the circuit. Essentially the current runs from R1 -> R2 -> C1 and R3 -> GND This cycle charges the capacitor, however we could simplify the circuit knowing which path current is taking.

As D1 is R.B And D2 is F.B and The Op-Amp isn't doing anything.

schematic

simulate this circuit – Schematic created using CircuitLab

Which becomes this. My question is, isn't this a low pass filter? If it is why is that the output signal isn't attenuated then? Why is the input voltage at the output?

schematic

simulate this circuit

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Just my opinion: it's a pretty bad implementation of a full-wave rectifier followed by a poor implementation of a (sort of) peak detector.

It's a bad implementation of a full-wave rectifier because the gain of the input positive peaks is significantly less than the gain of the input negative peaks.

It's a poor implementation of a peak detector because of the forward voltage drop of diode D2 and the very short time constant of the output filter and load resistor (C1, R3).

You can do much better with some minor circuit changes.

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  • \$\begingroup\$ Agreed, but I need to understand the foundation in order to improve it \$\endgroup\$ – Pllsz Oct 19 '18 at 16:57
  • \$\begingroup\$ With the current path shown above it can be simplified to a circuit that looks like a low pass filter no? If so why doesnt it behave like one \$\endgroup\$ – Pllsz Oct 19 '18 at 17:00
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It's a half-wave precision rectifier, the precision of which is mitigated by the use of a simple diode to implement the peak detector. Full-wave response and precision peak detection will require more than one op amp, so I feel that low cost was a primary goal of the design.

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It's a AC-DC converter which is poorly designed. The output voltage will be the peak positive AC input voltage. Gain is 1 so R1 and R2 just act as feedback. The circuit compensates for the diode voltage drops by setting the output feedback voltage 0.7V higher than the input to counter the voltage drop of the rectifying output diode. FYI, that output capacitor is stupidly massive, and will take ages to charge up.

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  • \$\begingroup\$ Agreed its poorly design, just starting on a foundation to study it then improving it. Why doesn't it act like a low pass filter though? \$\endgroup\$ – Pllsz Oct 19 '18 at 16:56
  • \$\begingroup\$ Because the diodes act like a resistor when forward biased (turned on), so you effectively have a low pass filter. \$\endgroup\$ – sidA30 Oct 19 '18 at 17:47
  • \$\begingroup\$ What ways would you improve the circuit? My application is have it charge for every peak value and discharge for the next incoming peak value The application is used for Audio (Music), I want it to have follow every music signal's peaks. \$\endgroup\$ – Pllsz Oct 19 '18 at 17:53

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