Timeline for Why more voltage after full rectification
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 13, 2020 at 21:07 | comment | added | Kevin White | @GrahamNye - thanks. You're of course right, a slip of the keyboard. It should be 1.414. | |
| May 13, 2020 at 7:09 | comment | added | Transistor | @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. | |
| May 13, 2020 at 2:07 | comment | added | ks0ze | @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. | |
| May 13, 2020 at 2:06 | comment | added | Graham Nye | @Kevin White A typo here. 1.514 should read 1.414. | |
| May 13, 2020 at 1:09 | comment | added | Kevin White | @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. | |
| May 13, 2020 at 0:08 | history | answered | varun | CC BY-SA 4.0 |