Timeline for Led on breadboard burned out why?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 25, 2018 at 20:27 | comment | added | Transistor | @jsotola: The second LED has been permanently prevented from lighting already! | |
| Dec 25, 2018 at 19:52 | comment | added | jsotola | if the two circuits are the same, then removing the resistor should prevent the LED from lighting in each of the circuits .... try it | |
| Dec 25, 2018 at 19:31 | vote | accept | Bob G. | ||
| Dec 25, 2018 at 9:14 | answer | added | Transistor | timeline score: 9 | |
| Dec 25, 2018 at 8:40 | comment | added | Bob G. | Ignore the math in the background btw. | |
| Dec 25, 2018 at 8:20 | comment | added | Bob G. | I finally got enough charge. So here are the pictures. This seems like the same circuit but the bad one burns led out for some weird reason. There is the resister coming first, so its not bypassing. | |
| Dec 25, 2018 at 8:19 | history | edited | Bob G. | CC BY-SA 4.0 |
Added pictures
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| Dec 25, 2018 at 8:17 | history | undeleted | Bob G. | ||
| Dec 25, 2018 at 8:10 | history | deleted | Bob G. | via Vote | |
| Dec 25, 2018 at 7:53 | comment | added | deathismyfriend | If you connected the led directly to +9v power then your bi-passing the resistor. The resistor needs to be in between the +9v and the + side of the led. Or the ground and the ground side of the led. | |
| Dec 25, 2018 at 7:51 | comment | added | Bob G. | Yes you're right and then I connected a wire from the led to the positive line. I will see if I can upload a picture or two. | |
| Dec 25, 2018 at 7:43 | comment | added | deathismyfriend | From what it sounds is you placed the resistor on the positive line only and not in between the positive and the led. It would help a lot more if you had pictures or a diagram of how you are wiring it. As it is a little unclear | |
| Dec 25, 2018 at 7:39 | history | asked | Bob G. | CC BY-SA 4.0 |