Timeline for How do I light an LED on a ground signal when I don't have enough voltage left after the PNP?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 19, 2024 at 21:00 | comment | added | MOSFET | @TimWilliams Thanks, Tim. Wasn't familiar with how the history was managed and didn't want the answer to be inconsistent. | |
| Jul 19, 2024 at 20:59 | history | edited | MOSFET | CC BY-SA 4.0 | Erased incorrect circuit |
| Jul 19, 2024 at 4:37 | comment | added | JkingNH | Of the 3 solutions in the edited solution, the one on left is the simple, robust solution. The center and right solutions can work under certain conditions, however, the right solution as mentioned above, requires a particular min Vth in order to turn of the LED, the center solution - the negative TC of Vbe (2x for the Darlington configuration), makes it a sensitive balancing act to make the circuit work. The solution on the left - functionality is incredibly insensitive to device parameters - it will just work. | |
| Jul 18, 2024 at 22:32 | comment | added | KJ7LNW | Are the N-FETs in the first revised circuit all enhancement mode? | |
| Jul 18, 2024 at 19:29 | comment | added | Tim Williams | You should remove the original (erroneous) content -- doing so materially improves the answer, and we already see edit history below so don't need to include any history and "edit" sections in the post. Cheers! | |
| Jul 18, 2024 at 14:49 | comment | added | MOSFET | @BK303 Yes, I agree. My original reasoning was based on a 13.8V spec that was changed in the original question. 6.5V would have been the ground reference for the circuit, 13.8V would have been the LED power supply. Therefore, you can get Vgs = -6.5V with respect to system ground. But that's all irrelevant now. | |
| Jul 18, 2024 at 7:51 | comment | added | BK303 | @KJ7LNW, In my opinion the first circuit of this revised answer is the best, which has the N-FET inverter, followed by N-FET low-side switch. That one works well even with the revised PTT high voltage of only 6.5V. The third circuit, which is simple with just high side P-FET switch controlled directly by PTT signal was the MOSFET equivalent of my original PNP answer, but with the revised question stating that PTT might only pull high to 6.5V instead of 13.8V, then when PTT is not active (high), the M2 P-FET Vgs is 6.5V-8V =-1.5V and could be very close to turning on accidentally. | |
| Jul 18, 2024 at 6:54 | comment | added | BK303 | @MOSFET, do you agree with assessment that suggested depletion mode N-FET as low-side switch would be ON when PTT line is at ground (Vgs=0), and still be ON when PTT line is at +12V (Vgs=+12). | |
| Jul 18, 2024 at 6:48 | history | edited | MOSFET | CC BY-SA 4.0 | Added circuits that actually work this time! |
| Jul 18, 2024 at 5:59 | comment | added | Tim Williams | I'm afraid this won't work: the D-NMOS never turns off. Pin5 would have to go negative for that to happen. It mostly works with PMOS (as a source follower), but the LED may still turn on while line is high. | |
| Jul 18, 2024 at 5:45 | comment | added | BK303 | Is the original request turn LED on when PTT line is pulled high (not active) or when PTT is grounded? This is good circuit for LED turns ON when PTT not active, and my answer is good when LED turns on when PTT is active (at ground). And by the way, the N-FET needs to be Enhancement Mode (not Depletion mode), otherwise PTT has to be -3V to turn off LED, correct? | |
| Jul 18, 2024 at 5:08 | vote | accept | KJ7LNW | ||
| Jul 18, 2024 at 4:59 | history | answered | MOSFET | CC BY-SA 4.0 |