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I am trying to use a output signal to reset a counter, but the signal I have is "low when counting" and I need "low during reset".

I want to "invert" an open drain output to a new signal so that when the output is on (connected to ground), the "inverted" signal has voltage or is not connected and when the output is off (not connected) the signal is connected to ground. My source voltage is 12v in an automotive system.

I have been struggling with this for a while and I must be getting confused with the negative logic because each solution I have tried gives me the same signal. I would love to do this with resistors and one or more of these BJT transistors: 2N3904 2N3906 S8050 2N2907 S8550 2N2222 BC337 C1815 BC327 A1015 since that is what I have available already and would work best in the enclosure space I have.

Output: I have a hardware component (LCR-II) with output Aux 1 which is an open drain output common to the negative power input line. Sinking capability up to 1 Amp. Maximum circuit voltage is 12v.

Input: I have another piece of hardware (a digital counter model CUB5B000) with a user input that is internally pulled up to +12 V with 10 K resistance. The input is active when it is pulled low (<1 .0 V)

I usually just work in software, so I'm apparently rusty from my University electronics -- thank you!

Between the work I already did and the answer from @Elliot-Alderson, here is my current attempt at a solution:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Actually, I'm confused. You say the signals are "common to the negative power input line", but the document you linked to says the only power input is +9 to +28 V. \$\endgroup\$
    – The Photon
    Sep 5, 2019 at 17:32
  • \$\begingroup\$ @ThePhoton I think the OP sees the power supply like someone might, used to working on cars. The battery has a negative and a positive. So I think the OP's writing merely means that the open drain output is referenced to the negative side of the power supply. \$\endgroup\$
    – jonk
    Sep 5, 2019 at 18:17
  • \$\begingroup\$ Ross, are you willing to directly connect the negative side of your output hardware's power supply to your digital counter's chassis and/or ground reference? Or would you prefer some form of galvanic isolation to keep the two devices insulated from each other? \$\endgroup\$
    – jonk
    Sep 5, 2019 at 18:20
  • \$\begingroup\$ @ThePhoton Thank you. I quoted "common to the negative power input line" simply because I was trying to be complete, found on page 6 of the LCR-II installation manual. I am sure it is really meaning ground. Both of these devices will run from +9 to +28V DC, but they are being powered by +12V. The common/ground side of the counter and the LCR-II are connected to the vehicle chassis ground (and thus the 12V battery's negative terminal). \$\endgroup\$
    – Ross Bradbury
    Sep 5, 2019 at 18:52

1 Answer 1

Reset to default
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Connect a pullup resistor to your existing open drain output, so that it goes high when it is not actively pulled down.

Connect the existing open drain output to the gate of a MOSFET. A 2N7000 might be a good choice. Connect the source of the MOSFET to the negative power line. The drain of the added MOSFET will be the new open drain output to the counter.

When the original open drain output pulls low (is driving low) the new MOSFET will be disabled...open drain. When the original open drain output is disabled (pulled high through the pullup resistor) the MOSFET's drain will be driven low.

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  • \$\begingroup\$ Hmm, it may be that I was only missing this one pull-up resistor. I kept confusing that one with the one that is built into the counter. I will sketch this out and see if I can adapt that to one of the transistors I have available today. \$\endgroup\$
    – Ross Bradbury
    Sep 5, 2019 at 18:58
  • \$\begingroup\$ If you wanted it to drive high instead of low on the output, could you replace the MOSFET with a diode? \$\endgroup\$
    – robbie
    Apr 2, 2020 at 11:22
  • \$\begingroup\$ @robbie It's not clear what you are asking. The original question was about inverting the value of an open-collector output. If you don't want to invert the output value then you don't need a MOSFET or a diode, just the pullup resistor. \$\endgroup\$
    – Elliot Alderson
    Apr 2, 2020 at 13:53
  • \$\begingroup\$ @ElliotAlderson Sorry, I was asking about having a pullup resistor and not driving ground. I was kind of tired and I accidentally hijacked this question in a way. \$\endgroup\$
    – robbie
    Apr 2, 2020 at 16:29

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