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[Moved] Help Regarding Peak To Peak Detector Circuit

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Gursahib Singh

Newbie level 5
Sep 11, 2015
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Hey guys need some help from you...

I have made a small circuit (Peak to Peak Detector Circuit). but i am confused what should be the voltage ratings of the capacitors and wattage ratings of the resistors. Maximum AC voltage will be 550V AC.

When i supply 550 V AC, the DC output is about 800V Untitledert.png

The schematic shows a peak detector but onl one polarity. Not peak-to-peak.
In the schematic the internal source impedance is missing. If you have 550 VAC source, then the capacitor (100 uF) will charge to a peak voltage of SQRT(2) x 550 V, or 780 V. I doubt you can find such capacitor on the market. As the capacitor would discharge by more than 1MOhm in the voltage divider on the left, I would use a 0.1 to 1.0 uF capacitor with 1 kV rating, easier to find.
With 800 V max peak, use a 3kV reverse diode, and use a transformer from AC circuit, not the ground shown, for safety.
Resistors in the voltage divider should be rated to U squared/ R, therefore 0. 64 W, the other resistors only 1/8 Watt.

As the peak detector will accept AC voltage at 50 or 60 Hz, the easier solution (also for safety), use a small AC transformer, 220 to 6 or 12 V, then you can use a low-voltage diode like 1N4001 and the capacitor rated for 25 V. Adjust the resistors in voltage divider to get a suitable voltage for your ADC. The mentioned transformer can process also 50/60 Hz harmonics, so the peak detector can accept also "abrupt" peaks coming on mains line.
Use the ground only on transformer secondary as shown in the schematic.

The client has strictly restricted me from using any transformer.

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