# Noninverting and the inverting Comparator

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

NonInverting

Inverting

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. Commented Mar 2, 2020 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 Commented Mar 2, 2020 at 20:30
• Yeah that's about where you lost me. Explain why the configuration in the book shouldn't work. Commented Mar 2, 2020 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. Commented Mar 2, 2020 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
Commented Mar 2, 2020 at 20:52