1
\$\begingroup\$

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

NonInverting enter image description here

Inverting enter image description here

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!

schematic

simulate this circuit – Schematic created using CircuitLab


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

\$\endgroup\$
5
  • \$\begingroup\$ Can you be more specific on what your proposal does that the book doesn't. It's sort of hard to follow this description. \$\endgroup\$
    – Jaywalk
    Commented Mar 2, 2020 at 20:14
  • \$\begingroup\$ I think that the configuration in red (from the book) are wrong and I draw what I think are the correct configurations \$\endgroup\$
    – Ugo Mela
    Commented Mar 2, 2020 at 20:30
  • \$\begingroup\$ Yeah that's about where you lost me. Explain why the configuration in the book shouldn't work. \$\endgroup\$
    – Jaywalk
    Commented Mar 2, 2020 at 20:34
  • \$\begingroup\$ 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. \$\endgroup\$
    – Ugo Mela
    Commented Mar 2, 2020 at 20:46
  • \$\begingroup\$ 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. \$\endgroup\$
    – G36
    Commented Mar 2, 2020 at 20:52

2 Answers 2

1
\$\begingroup\$

There are a few errors in the book's schematics. Yours are better, but you switched the inverting and noninverting configurations. Also, unless the supply is negative, you want to connect the negative of the voltage source to ground and use the positive as the supply. Even with those two minor errors, yours are more accurate than the book's.

\$\endgroup\$
0
\$\begingroup\$

You are right.

However, R3 (in the non-inverting setup) and R2 (inverting) are very large compared to Rpullup. You would be forgiven for neglecting Rpullup, since the difference between 1.000MΩ and 1.003MΩ is 0.3%, far smaller than the tolerances of the resistors or voltage sources involved. In other words, the combined errors due to sources and resistors not having exactly their marked values is more significant than this 0.3% error.

Obviously, as R3 (or R2) get closer to Rpullup, your observation becomes more important, but in practice Rpullup should always be significantly smaller, to ensure that a "high" output from the comparator is as close to Vc as possible.

Edit: That book is chock full of errors, but ironically that can help readers like yourself who are paying attention, to ask the right questions. Yours was the right question, and you are certainly better off for having spotted this.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.