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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.

Schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

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.

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    \$\begingroup\$ Why not swap SET and RESET? \$\endgroup\$ – EM Fields Jun 7 '15 at 8:45
  • \$\begingroup\$ The error handling circuit needs Q to be low on an error condition. It's not realistic to change the error handling circuit. That is, the LED isn't the only thing I'm using that signal for. \$\endgroup\$ – Ed Krohne Jun 7 '15 at 10:15
  • \$\begingroup\$ The SR latch is symmetric, so if you swap S & R (set and reset) it will be the same as using a "not Q" output. It's two traces to cut and jump, but no parts to add. I will probably be the simplest solution. \$\endgroup\$ – Austin Jun 7 '15 at 15:08
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Assuming the latch runs from a 5V supply, you can use a PNP transistor connected as an emitter follower .. the LED + resistor will see 0.6V less voltage but that should not be a big deal.

schematic

simulate this circuit – Schematic created using CircuitLab

Oh, and you don't need to do anything else- the pinout of the BS170 is D-G-S and the pinout of the 2N3906 (or 2N4403 etc.) is E-B-C, so it's a drop-in solution.

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    \$\begingroup\$ This is a nice solution. \$\endgroup\$ – Michael Karas Jun 7 '15 at 12:04
  • \$\begingroup\$ Clever. A PNP transistor works where a P-type mosfet wouldn't, because "ON" means current is flowing, but "ON" doesn't change the voltage much at the source of a hypothetical P-type mosfet. I need to remember that. \$\endgroup\$ – Ed Krohne Jun 7 '15 at 12:36

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