Hi All,

I have built a simple working comparator circuit as shown below

View attachment comparator.bmp

IC1 is an LM358AN Op amp, the two 10K pots allow me to change the voltage at the two input pins with the two 100n caps to provide a bit of filtering.

the issue comes when I replace the LM358AN Op Amp with an LM393N Comparator IC, with the rest of the circuit the same I get a constant 0 Volts on the output, are there any changes that need to be made to the circuit when using a comparator IC when compared to an Op Amp?

I have double checked the pinning is the same between the two IC's and I have tried multiple LM393N IC's in case I had one faulty one, therefore I can only assume that there is something simple wrong with my circuit.

Any help or hints would be greatly appreciated.

 

The comparitor is an "open collector" meaning that it is not a push-pull output like the op amp. You need to connect output to a resistor then to the positive rail. The output is a simple transistor type and you must complete the circuit. You can then measure voltage from the output pin of the comparitor.

Current draw will be essentially zero amps in HI state (transistor leakage current for a HI output (since it is connected to the Positive rail through a resistor)), current draw shoud be in the range of 1 to 5 ma (select yor resistor based on your circuit voltage) for low state (transistor is ON and allowing current to flow, measuring at collector essentiall inverts the signal at the transistor). Check the chip's internal schematic on the DATASHEET to count how many times the signal is inverted. Absolute Max is 20 mA according to datasheet

 

Cheers GopherT,

I had a feeling it would be something simple, this is the first time I've built a circuit like this and I was under the impression that both ways would require the same circuit, obviously this isn't the case. I have soldered a resister to Vcc and voila it works.

 

Excellent! That was fast.

 

just the way I like it.