I just ordered a printed board from OSHPark. I'm done soldering all my components to it, and I tested it and found one small mistake. I have a "error" signal that is normally high, but goes low if anything goes wrong. I have an LED to indicate an error condition, and I tried to use a BS170 n-type mosfet to control the LED, because the 4043 latch I'm using to drop the error signal can't sink that much current.
The problem is that during an error condition, Q is LOW, hence the diode is off, whereas during normal operation, Q is HIGH, hence the diode is on. I want the opposite behavior, preferably while making minimal changes to my already soldered, printed circuit board. Writing over the silkscreen with a sharpie, for example, might be a last resort.
Unlike the latch that comes with circuitlab for the diagram above, the 4043 does not have a "not Q" output, or I'd wire that to the FET instead. I realize that this is a tall order, but what I'd really like is an "inverted" FET that I can just swap out the BS170. High -> closed, low -> open, three pins, TO-92 package if possible. My board footprint is such that it doesn't matter if the pin order is the same as the BS170, I can rotate and bend pins to accommodate any pinout. Just as long as it has three pins and is roughly TO-92 sized. Given the vast variety of specialized transistors out there, I figured it was worth an ask.
I looked into N-type depletion mode FETs, and learned that to switch them off you must bring the gate voltage below the source voltage, which won't work because I'm looking for a drop in to my existing circuit.
I consider getting a p-type FET, cutting traces, and adding flyovers to put the FET above the LED on the schematic, but I'd rather do this with minimal impact to the existing board.
Thank you for any help.