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I am reading this website and don't understand why this circuit doesn't blow up. If Vin = Vcc then seems to me top opamp A1 would drive low (Vcc > 2/3Vcc and inverting comparator) and the bottom one A2 would drive high (Vcc > 1/3Vcc and non-inverting comparator). So now you have two opamps driving against each other and whichever one overheats first looses.

https://www.electronics-tutorials.ws/opamp/op-amp-comparator.html

enter image description here

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    \$\begingroup\$ Comparator op-amps can have an open-collector output. \$\endgroup\$
    – Oldfart
    Jun 25, 2018 at 6:32
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    \$\begingroup\$ @Oldfart I would give you the correct answer but can't do it if it's a comment. \$\endgroup\$
    – EasyOhm
    Jun 25, 2018 at 6:35
  • \$\begingroup\$ I am now fully reading the full text in detail to see if they mention it somewhere. \$\endgroup\$
    – Oldfart
    Jun 25, 2018 at 6:37

1 Answer 1

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There is no issue if the output is an open collector type.

I found the text in the article you where referring to:

As a result the output stage of the voltage comparator is generally configured as a single open collector (or Drain) transistor switch with open or closed states rather than actual output voltages as shown:

enter image description here

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