I opened an old DC supply. I tried to replace its filter capacitor, which was 47uF/16V, and I plugged 1000uF/25V capacitor and checked its output. Its output was 23v which is strange because I measured the Pulsating DC via a multimeter, it supplies 25v only for just a second and then it drops to ~16v but the capacitor (1000uF/25v) somehow retains that potential and that's strange!.

I didn't get that phenomenon because the cap voltage should also drop to 15v but it doesn't. why ?

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    \$\begingroup\$ By "Pulsating DC", do you mean AC? Also, have you considered the memory effect that a capacitor may have? \$\endgroup\$ – user2813274 May 25 '17 at 6:44
  • \$\begingroup\$ By pulsating DC I means "DC without being filtered" after the rectification. I don't know about the memory effect. \$\endgroup\$ – Nouman Tajik May 25 '17 at 7:07
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    \$\begingroup\$ 16V (rms) x 1.4 = 22.4V (peak voltage) - larger cap will loose less voltage between peaks. (google 'ripple voltage') \$\endgroup\$ – JIm Dearden May 25 '17 at 10:37
  • \$\begingroup\$ That's why the DMM was reading ~23v. The cap was holding the peak \$\endgroup\$ – Nouman Tajik May 25 '17 at 11:18

Better replace your 1000/25 with 47uF as was in the original design. Be sure to use the same type - ESR, temperature, etc. Unless it is a very low power supply (less than 1-2W) I really doubt this capacitor is a "filter" one on the secondary output. It may have a different function and changing its value so radically you change how device handles its startup and protection. If you cannot find 47/16 you can put 47/25V or 35V, but be sure the capacitance is 47uF and temperature and ESR is like on the original.

The strange thing you are observing actually is not that strange. Probably the power supply tries to turn on for a second, goes to 23V, its feedback dont return the right signal and its shuts it self down. After a couple of seconds when its startup capacitor has been recharged it tries to start again.

If the supply's output was 12V it is very likely that you have a feedback fault issue. This feedback fault may have rasulted in output overvoltage and output capacitor failure. The fault may be a loose solder joint, some component in the feedback loop has failed - TL431, optocoupler, some resistor, and finally it may be the main IC.

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  • \$\begingroup\$ Actually, the supply doesn't have any feedback mechanism. It only contains Transformer, Rectifier, and Power relay (which purpose I don't know yet). The transformer is center tapped and I had just replaced the rectifying diodes. On the front of the supply, it says it can supply 10-30 Amps so it is a high wattage supply. \$\endgroup\$ – Nouman Tajik May 25 '17 at 7:11
  • \$\begingroup\$ @NoumanTajik Strange. Better draw the schemtic then. 47uF in a this kind of power supply should have some other function. It is too small for filtering 50/60Hz and 30Amps. \$\endgroup\$ – Todor Simeonov May 25 '17 at 7:17
  • \$\begingroup\$ You maybe right here. When I got the supply, it's connection were broken So I wired it from the scratch. But it contains only 3 components A Transformer, Rectifying diodes, and power relay (JQX-76F), on the power relay there was a capacitor (47uF). I don't know what was the purpose of "power relay" in this scenario so I just pulled it out and connect the rectifying diodes to the transformer. I thought there must be a cap for filtration so I used 1000uF/25V but it seems like 47uF (in actual design) was just used for eliminating debouncing for the relay and I guessed it wrong. \$\endgroup\$ – Nouman Tajik May 25 '17 at 7:30
  • \$\begingroup\$ @NoumanTajik Yes, it could be for the relay only, but there should be an additional diode to charge this cap and to prevent discharging it from the high current path. \$\endgroup\$ – Todor Simeonov May 25 '17 at 7:43
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    \$\begingroup\$ There could be plenty of reasons. If it switched the high current path it would be some kind of undervoltage lockout, it may be used to switch some other signal depending on the state of power supply an so on. I understand that you don't have the schematic and the PSU is already disassembled, but without the full schematic I doubt that anybody could answer you :) \$\endgroup\$ – Todor Simeonov May 25 '17 at 7:51

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