i wan't to measure the maximum and minimum output swing of a Comparator. I'm using the circuit which you can find attached.

However, since the Output of the Comparator (TI LM2903) is open drain, I need the Pull up. I also need the Load Resistor to measure the output swing with a load.

To trigger the Comparator, I'm using a +-1V voltage with a given 0V reference voltage.

With the -1V at the non-inverting input I get -9.8V at the output, but with +1V I only get 4.9V because of the voltage divider (Pullup to Load 1:1).

How can i achieve a full positive output swing to nearly 10V?

Best regardsenter image description here

  • \$\begingroup\$ Isn't it clear that when you're dealing with an open-collector or open-drain output, the concept of "positive swing" is not a characteristic of the comparator, but rather a characteristic of your external circuit? \$\endgroup\$
    – Dave Tweed
    Sep 4, 2014 at 18:40

2 Answers 2


The simplest fix is to use a opamp or comparator that has bi-directional output drive. You're not asking for much unusual from the opamp or comparator, so there should be plenty of options out there.

If you really really want to use a comparator with open collector output, then you invert the logic by swapping the inputs, and have the comparator turn on a high side PNP, which then drives the load.

  • \$\begingroup\$ I thought about that, but the circuit is a test measurement (servo loop) for all kinds of opamps/comps so it needs to be compatible with non-bi-directional output drive as well. \$\endgroup\$
    – Bommelchen
    Sep 4, 2014 at 12:59
  • \$\begingroup\$ The OP hasn't defined what he means by "a full positive output swing to nearly 10V?", so until that's cleared up it's all just guesswork. \$\endgroup\$
    – EM Fields
    Sep 4, 2014 at 13:00
  • \$\begingroup\$ @EMFi: "Full positive output swing to nearly 10 V" is clear. He wants the load to be driven to nearly the full +10 V supply level when the comparator output is high. How else would you interpret that? \$\endgroup\$ Sep 4, 2014 at 13:08
  • \$\begingroup\$ Olin: "to nearly 10V" is clearly qualitative and vague, and in order to be clear enough for quantitative work to be done, the actual load voltage and its limits of error should be specified. \$\endgroup\$
    – EM Fields
    Sep 4, 2014 at 13:39
  • \$\begingroup\$ @EMFi: You are nitpicking. \$\endgroup\$ Sep 4, 2014 at 14:03

Take the load resistor out of the circuit. ;)


With the transistor (the output switch of an LM2903) completely turned off,

enter image description here

  • 1
    \$\begingroup\$ Presumably the load is what needs to be driven for overall operation, and is the one thing that therefore can't be eliminated. \$\endgroup\$ Sep 4, 2014 at 12:02
  • \$\begingroup\$ Well, although my answer was intended to be tongue-in-cheek and jocular, I see it clearly missed its mark. Now that the OP has so kindly provided the level of detail needed to formulate a proper response, see my humorless edited answer arrived at using the nit-picking details you seem to disdain. \$\endgroup\$
    – EM Fields
    Sep 4, 2014 at 21:38
  • \$\begingroup\$ If you want to joke arouned like that, do it in a comment. Someone not very familiar with the subject can't know if you're serious or not. Answers here need to be actual answers. \$\endgroup\$ Sep 4, 2014 at 22:21
  • \$\begingroup\$ If answers here need to be actual answers, then you ought to take your own advice and refrain from posting opinionated, nebulous fluff like: "so there should be plenty of options out there." \$\endgroup\$
    – EM Fields
    Sep 4, 2014 at 22:38
  • \$\begingroup\$ With reference to the equations on the right in the image above, should the the formula on the second line start "R2>=" not "E2>="? \$\endgroup\$
    – David1
    Nov 29, 2018 at 8:01

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