0
\$\begingroup\$

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!

\$\endgroup\$
  • 1
    \$\begingroup\$ Did you mean "totem pole" wherever you wrote "open collector"? \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 31 '15 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\$ – Rory O'Hare Dec 31 '15 at 8:20
  • \$\begingroup\$ What is your task? \$\endgroup\$ – d3L Dec 31 '15 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\$ – Rory O'Hare Dec 31 '15 at 8:25
5
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ :) Sure... but then you have unusable outputs. Wouldn't they always be high-z even if they were in their "low" state? \$\endgroup\$ – Rory O'Hare Dec 31 '15 at 8:27
  • 2
    \$\begingroup\$ No, an OC output is tied to ground through its low-side transistor when low. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 31 '15 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\$ – Rory O'Hare Dec 31 '15 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.