Timeline for How to protect OpAmp output from short circuit to +Vcc?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2019 at 8:21 | comment | added | Janka | At not supposed to be. You have to measure the intensity of the laser and check whether it's supposed to be that luminous in that very moment or not. And if not, cut the power supply. And you better do that in a different, simple circuit which fails open. | |
| May 17, 2019 at 7:22 | comment | added | Eye of Horus | @Janka I was afraid of that. Are there any starting points you could recommend to find a solution to this? | |
| May 16, 2019 at 9:04 | comment | added | Janka | In that case, you don't want to protect the output from short circuit but rather from being stuck. That's a whole other jar of pickles you just opened. | |
| May 16, 2019 at 9:00 | comment | added | Eye of Horus | Thank you for your answer. I want to protect against the output being at 5 V when it's not supposed to be. The output controls a laser, which would be permanently turned on in the case of a short. That should not happen and I'd rather have some parts break and have to replace them, as long as that would ensure the laser is not turned on. | |
| May 10, 2019 at 12:23 | history | answered | Bimpelrekkie | CC BY-SA 4.0 |