Timeline for AC line indicator circuit - resistor gets fried
Current License: CC BY-SA 4.0
8 events
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| Jul 14, 2023 at 6:53 | comment | added | jpa | With D_add1 in series, there is no discharge path for C3 except through R8. This will result in much smaller current passing through the LED. | |
| Jul 13, 2023 at 22:28 | comment | added | marcelm | @SimonB I'd go as far to change "completely alter" to "render effectively useless". Also I challenge the notion that any change to the diodes is even necessary; I don't see how the diode would be too slow for the application (and that's ignoring the fact that 1n400x parts usually have pretty short forward recovery time). | |
| Jul 13, 2023 at 19:37 | comment | added | Simon B | @Kubahasn'tforgottenMonica Why do you need the extra diode? Having it there will completely alter the operation of the capacitive dropper. | |
| Jul 13, 2023 at 19:26 | comment | added | Kuba hasn't forgotten Monica | You can try two 1W 200ohm pulse rated resistors in parallel. | |
| Jul 13, 2023 at 19:26 | history | edited | Kuba hasn't forgotten Monica | CC BY-SA 4.0 |
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| Jul 13, 2023 at 19:24 | comment | added | Noel | Thank you. Hard to find a 3W pulse resistor. Mouser has 1W only. Will keep searching. Would you put the additional series diode in the same direction has the led then? | |
| Jul 13, 2023 at 16:29 | history | edited | Kuba hasn't forgotten Monica | CC BY-SA 4.0 |
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| Jul 13, 2023 at 16:22 | history | answered | Kuba hasn't forgotten Monica | CC BY-SA 4.0 |