I don't understand the output of the window comparator. How does the window comparator switch from HIGH to LOW? There is no transistor or anything else. Only two OPAMP which outputs are put together.

Can somebody explain it to me?

Here is a image of a window comparator:

Window Comparator

Here is the description of the window comparator and the source of the image:


  • 3
    \$\begingroup\$ This question has been asked before. If you read the article: they are open collector op-amps. \$\endgroup\$
    – Oldfart
    Mar 8, 2019 at 9:39
  • \$\begingroup\$ I dont see there open collector op-amps in this article. Do I need a special op-amp? I simuated it in orcad and used a LM324. And it works. But i dont know why... \$\endgroup\$
    – otto
    Mar 8, 2019 at 9:54
  • \$\begingroup\$ @TobiasK you say there are no transistors or anything... What do you think is inside the op-amp? You used an LM324 in a simulation, the datasheet has an internal schematic. You can see the transistors in there. You don't need a special op-amp to make a window comparator, but for this particular configuration, you want an open-collector type. All of these terms can be Googled and then with the explanation of the window comparator in your linked page, it should all make perfect sense \$\endgroup\$
    – MCG
    Mar 8, 2019 at 10:05

2 Answers 2


They are not op-amps - they are comparators and comparators of this type have an open-collector output hence, the two outputs are wire-ORed i.e. they form a wired logic function when sharing the same pull-up resistor (\$R_L\$)

  • 1
    \$\begingroup\$ but why there are no diodes in the circuit? And how is logic? Because the Output is only HIGH when Comperator No1 is LOW and Comperator No2 is HIGH. It is like an XOR Gatter. \$\endgroup\$
    – otto
    Mar 8, 2019 at 13:02
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    \$\begingroup\$ No, the output is only high when the respective open-collector transistors in both comparators are off. If either is on, then the output is pulled low irrespective of the other being off. \$\endgroup\$
    – Andy aka
    Mar 8, 2019 at 13:11

Case1: vinboth reference voltages opamp1 produces LOW(-Vsat) and opamp2 produces high(+Vsat) so the output cancel each other and net output =0 Note: Opamp1=inverting comparator and Opamp2= Noninverting comparator

  • 1
    \$\begingroup\$ Welcome to the site. Your answer needs to contain some explanations rather than just firing off the solution. It should be written clearly enough to teach future readers. Please can you edit it and greatly improve it. Thanks. \$\endgroup\$
    – TonyM
    Apr 7, 2020 at 10:23

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