@Slyka Would you be interested to know why that's connected like that? We've seen similar stuff before. -F
@Felthry Yeah, I'd be very interested!
@Slyka so, the one thing that comes to mind first is that, until relatively recently, the junctions of common BJTs were better zener diodes than cheaply available zener diodes. They aren't designed to as tight a tolerance, but they have low zener impedance (i.e. very steep I-V curve). Though using them that way, you'd more likely leave either the collector or the emitter floating, not tie it to the base.
-F
@Slyka basically, on the assumption that this isn't a mistake, they're bound to be relying on some kind of breakdown behaviour of the device. This is the kind of thing that only the least or most experienced engineers would generally do, nowhere in between
there is also the possibility that it's a mistake, which is entirely reasonable. staying within the designed operating conditions of that transistor, it will do nothing significant
-F
@Slyka oh yes! another possibility is that they're using the transistor in reverse-active mode as a diode to speed turn-off of V23. this would be strange, but it would work.
-F
@Felthry Ohh, thanks so much for all the info and ideas!
I've had a similar suspicion that it might have something to do with being able to turn off V23 quicker, but I'm not sure why they wouldn't have just used a diode. They use them everywhere else, so it doesn't seem like an attempt to simplify the BOM.
According to the schematic X1/3 is the "Pre-scan impulse" so something timing/speed related would make a lot of sense here. But then they're only pulling up that output (X2/13) pretty weakly…
I'm gonna sit down tomorrow and actually scope out the waveforms and see if I can learn something from that and also if it isn't just a mistake in the drawing.
Thanks so much again for your help ^-^
