Timeline for Resistor in my AC LED circuit gets fried
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 15, 2021 at 21:47 | history | edited | winny | CC BY-SA 4.0 |
added 3 characters in body; edited title
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| Nov 15, 2021 at 20:32 | answer | added | Transistor | timeline score: 2 | |
| Nov 15, 2021 at 20:17 | answer | added | hacktastical | timeline score: 5 | |
| Nov 15, 2021 at 20:13 | vote | accept | nikoss | ||
| Nov 15, 2021 at 20:12 | answer | added | Enrico Migliore | timeline score: 2 | |
| Nov 15, 2021 at 19:40 | comment | added | Alex Hajnal | Try doing the same calculation for the 309 V across the 11kΩ resistor. (The peak voltage for 220VAC is actually quite a bit higher than 220VDC: VDCpeak = VACrms * √2 ≅ VACrms * 1.414 ⇒ 220VAC * √2 ≅ 311 VDC ) | |
| Nov 15, 2021 at 19:39 | comment | added | user263983 | 2.5W resistor should dissipate. 218*0.02=4.36. And current flows only half wave. But another halfways will burn your LED, because a max negative voltage 5V, you have 310V. | |
| Nov 15, 2021 at 19:34 | comment | added | Bimpelrekkie | This is not the way to operate an LED at 220 V AC. Indeed the resistors will get too hot, that could be solved by using resistors with a higher power rating. Another problem is that an LED doesn't like it when you're applying a reverse voltage across so it will breakdown and might suffer damage. Watch this video and let BigClive show you what circuit to use: youtube.com/watch?v=Q23uh7AjjXw | |
| Nov 15, 2021 at 19:33 | answer | added | nanofarad | timeline score: 12 | |
| S Nov 15, 2021 at 19:27 | review | First questions | |||
| Nov 15, 2021 at 19:38 | |||||
| S Nov 15, 2021 at 19:27 | history | asked | nikoss | CC BY-SA 4.0 |