2
\$\begingroup\$

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

\$\endgroup\$
  • 2
    \$\begingroup\$ Comparator op-amps can have an open-collector output. \$\endgroup\$ – Oldfart Jun 25 '18 at 6:32
  • 2
    \$\begingroup\$ @Oldfart I would give you the correct answer but can't do it if it's a comment. \$\endgroup\$ – EasyOhm Jun 25 '18 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 '18 at 6:37
8
\$\begingroup\$

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

\$\endgroup\$

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.