# Noninverting and the inverting Comparator

I was studying the Noninverting and the inverting Comparator with Hysteresis from the book "Practical electronics for inventors".

I understand that when the comparator has an open collector then V_{out} is low (V_{out} = GND) and I agree with the relative schematic provided by the book.

The problem is related to the case when V_{out} is high. I think that the correct circuit are the following, not the one in the book. Am I wrong?

Thank you! simulate this circuit – Schematic created using CircuitLab

I've corrected the two mistake suggested by @Cristobol Polychronopolis

• Can you be more specific on what your proposal does that the book doesn't. It's sort of hard to follow this description. – Jaywalk Mar 2 at 20:14
• I think that the configuration in red (from the book) are wrong and I draw what I think are the correct configurations – Ugo Mela Mar 2 at 20:30
• Yeah that's about where you lost me. Explain why the configuration in the book shouldn't work. – Jaywalk Mar 2 at 20:34
• For instance, inverting case: V_out = HIGH, means that the transistor is off (V_out it's HIGH through the pull-up resistor), so it's not possible that resistor R2 is grounded because the transistor is off. If you follow the path from Vin to Vout you should see my schematic. – Ugo Mela Mar 2 at 20:46
• To me you are right. Clearly there is an error in the inverting equivalent circuit. and your corrected version looks good. As for the Non-Inverting version, notice that R3 is much much large than R_pull-up, hence we can ignore R_pull-up resistor influence on Vout voltage (do the math yourself to see). And it seems that the authors assumed that Vs = +Vc. So, no error here. But your equivalent circuit is also correct. So, good job. – G36 Mar 2 at 20:52