PNP transistor leaking B-C in really simple circuit - Page 1
Forum section: electronics / beginners

author: puzzledtinkerer
author_rank: Newbie
timestamp: January 29, 2024, 11:29:57 pm
content: I think this question is probably below beginner level, but much internet searching and head scratching has left me non the wiser, so here I am.I have a photoreflective sensor to be used as a flywheel sensor for a tachometer drive. It's NPN, but the tachometer wants a positive pulse. PNP equivalents with sensitivity adjustment are hard/expensive to come by, so I thought a simple circuit with a PNP transistor would flip the output to give me what I needed. At 12V, the circuit below turns the tachometer signal terminal on and lights the LED when the sensor signal terminal is grounded. Connecting it up to the sensor and tachometer, it works perfectly up to 4000 RPM (~66 Hz) which is what I need it to do. Unfortunately, as I raise the supply voltage through 13 to 14V which it will be with the alternator running, the LED comes on permanently, albeit dimly to start with (it gets brighter when the sensor signal terminal is grounded, but dims and remains on when that reverts to open circuit).I have tried changing various resistor values, but to no avail and it's reached the point that I'm guessing rather than calculating. I thought that everything should be within limits as designed.If I disconnect the emitter of the transistor, the LED stays on, implying that there must be leakage from base to collector because there's no other route to power it.R1 = 5.6 kΩR2 = 200 ΩR3 = 5.6 kΩR4 = 510 ΩTransistor is a BC327According to the transistor'sdatasheet, 14V is well within its collector-emitter limits, and I thought that the pull-up resistor R1 should be keeping the base-emitter difference under control when the sensor signal pin is open. Whether the combination of R2 and R3 does that when it's grounded could be another question, but that's not where I have my problem. I don't think I've blown the transistor because if I drop the voltage back down to 12 it starts working again.I'm not sure what I've got wrong and would really welcome some guidance. I could just chuck a potential divider in to drop the input voltage down to 12V, but I'm not sure how it would then perform when the supply side was only 12V, and also I wouldn't learn anything that way!Huge thanks in advance if anyone can point me in the right direction.

author: retiredfeline
author_rank: Frequent Contributor
timestamp: January 30, 2024, 12:28:04 am
content: Tried a different specimen of the transistor? Maybe you have a dud?

author: Manul
author_rank: Super Contributor
timestamp: January 30, 2024, 12:48:50 am
content: The best fitting explanation would probably be that you have emitter and collector swapped.

author: Jwillis
author_rank: Super Contributor
timestamp: January 30, 2024, 01:24:22 am
content: If you increase the voltage you have to also change the biasing of your amplifier don't you? On a fixed base bias circuit the Q point can vary greatly with changes to temperature. So raising the voltage would also change the temperature of the transistor throwing the Q point closer to saturation.Help me out here because I'm not sure.

author: puzzledtinkerer
author_rank: Newbie
timestamp: January 30, 2024, 01:14:43 pm
content: I confess to carelessness. Somehow, when I rebuilt the circuit from its initial test implementation, I got the transistor back to front. Thank you, Manul!I guess it was running in reverse and appeared to work at 12V because the resistances on the low side were sufficient to drop the currents with the phototransistor open low enough to constitute an off state for the transistor and / or LED, or something like that... ?Anyway, having turned the transistor back round, all is well. I just need to source an enclosure for it and mounting bracket for the sensor, and I'll have a working tacho.

author: Manul
author_rank: Super Contributor
timestamp: January 30, 2024, 04:42:39 pm
content: It might be advisable to replace the transistor with a new one. In some cases, transistors get damaged or degraded because of base emitter brakedown. And your original transistor has certainly experienced a fair bit of that.BJT transistors actually work with CE swapped, just not very well. And another problem is that BE junction has quite low reverse brakedown voltage, contrary to BC junction. So if you have CE swapped, at some voltage, typically in the range of 7-10 volts, BE junction starts to conduct similar to zenner diode. So when you operate a transistor like that, what you get is essentially a low voltage, crappy transistor.

author: berke
author_rank: Frequent Contributor
timestamp: January 30, 2024, 05:48:31 pm
content: Quote from: Manul on January 30, 2024, 04:42:39 pmCan confirm this can easily kill a transistor, I just had a small deadbug-mounted differential amplifier show unexpectedly low gain, turns out I had boneheadedly mounted the BC547 backwards (E and C swapped). Upon rewiring it, it still wasn't working. Everything was fine after I replaced that transistor. The meter thinks the broken one has an hFE of ~10 (was about ~300 before).

author: puzzledtinkerer
author_rank: Newbie
timestamp: January 30, 2024, 06:47:57 pm
content: Thank you. I've got a box of the things so will replace it to be on the safe side.