0
\$\begingroup\$

I am learning to use a comparator (LM393N) and running into trouble. No matter what I do, the output is always LOW. I am running off of a 9V battery, and I've added a voltage divider to send half of that to the V- (inverted) pin of the comparator. Yet, whether the non-inverted pin is connected directly to the positive rail (all the way HIGH) or to ground (all the way LOW), the Output pin always reads LOW (0V). What am I doing wrong? Here's a sketch of what I've set up where:

  • Pin 4 is inverted pin
  • Pin 5 is non-inverted pin
  • Pin 3 is positive power source
  • Pin 2 is output
  • Pin 12 is ground

Sorry for the weird pin numberings! I couldn't figure out how to change it in TinyCAD.

enter image description here

\$\endgroup\$
0

2 Answers 2

2
\$\begingroup\$

The device output is open collector. What this means it that it can either sink current (provide ground) or leave the output at a high impedance state (floating).
You need to add a pull-up resistor from output to the positive supply to force a high voltage level in that floating state.

The value of the pull-up depends on the needs of your circuit. The comparator can sink 6mA min so for 9v you can use a value of about 1.5k and higher.
If the circuit connected to the output doesn't need much current or you just want to experiment you can use a value like 4k7 to 10k (about 1-2mA)

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
0
2
\$\begingroup\$

The LM393 is open-collector output. You just need a pullup resistor to V+ to make it go high.

enter image description here

(from datasheet linked above)

\$\endgroup\$
0

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.