Keeping it completely analog, is there a way to high-z an open collector output? An open collector always allows a path to or from a specific voltage potential and this is complicating a rather simple task for me. The only way I can think of to do this is to interject an analog switch IC into the mix, but is there a way to do this with passives? I could use a transistor but then I'd also have to add logic to control the transistor and at that point a switch IC would be just as easy. Thanks much!

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    \$\begingroup\$ Did you mean "totem pole" wherever you wrote "open collector"? \$\endgroup\$ Dec 31, 2015 at 8:19
  • \$\begingroup\$ Nope. I am using voltage comparators (LM319's and LM393's) with open collector outputs (requiring a pull-up resistor). \$\endgroup\$ Dec 31, 2015 at 8:20
  • \$\begingroup\$ What is your task? \$\endgroup\$
    – Marco
    Dec 31, 2015 at 8:21
  • \$\begingroup\$ I'm using the output of voltage comparators as an input to a positive-feedback latch made from another comparator. The circuit I'm following requires high-z, not just 0 volts and the open-collectors keep messing up the inputs. \$\endgroup\$ Dec 31, 2015 at 8:25

1 Answer 1


Open collector outputs are always hi-Z when they are high. The only way to source current from an OC output when they are high is via an external pullup.

  • \$\begingroup\$ :) Sure... but then you have unusable outputs. Wouldn't they always be high-z even if they were in their "low" state? \$\endgroup\$ Dec 31, 2015 at 8:27
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    \$\begingroup\$ No, an OC output is tied to ground through its low-side transistor when low. \$\endgroup\$ Dec 31, 2015 at 8:28
  • \$\begingroup\$ Wow. I hate when I learn something I should have known months ago. This could actually work for me since I don't actually need to ever drive the inputs high, just low(-5V in this case) or high-z. I'll give it a shot, but it sounds like it should work! \$\endgroup\$ Dec 31, 2015 at 8:31

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