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I used 105j 400 V capacitor. Connected it to 230 V AC. Then I measured voltage after capacitor and it was reduced to 116 V AC. Then I checked voltage after 4 diode rectification, here measured 204 V DC.

Why the measured voltage is higher? Is it safe or or dangerous to touch this 204 DC volt?

This is first time I'm posting in this forum. I don't know how to upload pics Actually the capacitor I'm referring is AC voltage drop capacitor. So o for I have not connect anything after bridge rectifier

Thank you.

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  • \$\begingroup\$ As pointed out in the answer below, the problem is that your capacitor is on the wrong side of the rectifier. Therefore, on every negative cycle, it discharges and if you are measuring with a multimeter you will get an averaging effect that decreases the displayed voltage. \$\endgroup\$
    – ks0ze
    Commented May 13, 2020 at 0:32
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    \$\begingroup\$ A picture is worth a thousand words. Can you show us your circuit? There is a great tool on this site to draw them. \$\endgroup\$
    – Warren Hill
    Commented May 13, 2020 at 12:04
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    \$\begingroup\$ It is dangerous to touch 204V DC. \$\endgroup\$
    – user16324
    Commented May 13, 2020 at 12:36
  • \$\begingroup\$ Did you remember to try your meter in both AC and DC? It is likely this power has both an AC and a DC component. \$\endgroup\$
    – Harper - Reinstate Monica
    Commented May 14, 2020 at 6:01

1 Answer 1

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Could you please share the circuit diagram. It might help me understand better.

After rectification, the voltage increases by 1.414 times because it gives out peak to peak voltage. If you are trying to rectify 12V AC, you would get 17 V.

Also, you would want the capacitor after the bridge rectifier. You need to get rid of the negative pulse before you pass it through the capacitor.

It's not safe to touch high DC voltage. Also not safe to directly plug into 230 VAC main voltage. Hence people suggest that one use an Isolation transformer.

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  • \$\begingroup\$ @ks0ze - AC voltages are commonly measured as RMS. If you rectify that you will get approximately 1.514 times that DC voltage on the capacitor after the rectifier. (sqrt(2)). At medium to high voltages, you can usually ignore the forward voltage of the diodes. \$\endgroup\$
    – Kevin White
    Commented May 13, 2020 at 1:09
  • \$\begingroup\$ @Kevin White A typo here. 1.514 should read 1.414. \$\endgroup\$
    – Graham Nye
    Commented May 13, 2020 at 2:06
  • \$\begingroup\$ @KevinWhite, yep... too used to dealing w/ vpp. I still don't think this clearly answers the question because the OP is seeing a lower DC voltage than the input AC due to the cap being on the wrong side of the rectifier and therefore not decreasing the ripples for dc measurement. I know it's stated that the cap is in the wrong place but not explained why it is impacting results. \$\endgroup\$
    – ks0ze
    Commented May 13, 2020 at 2:07
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    \$\begingroup\$ @varun: "After rectification, the voltage increases by 1.414 times because it gives out peak to peak voltage." No, the peak voltage stay the same as the AC's peak at \$ \sqrt 2 V_{RMS} \$ and the DC voltage will remain at peak if a smoothing capacitor is added (and no load). "If you are trying to rectify 12V AC, you would get 17 V." Same as above. You can edit your answer to improve it. \$\endgroup\$
    – Transistor
    Commented May 13, 2020 at 7:09
  • \$\begingroup\$ @GrahamNye - thanks. You're of course right, a slip of the keyboard. It should be 1.414. \$\endgroup\$
    – Kevin White
    Commented May 13, 2020 at 21:07

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