How do comparators work?

[jayrare]

Sorry if this is the wrong forum, I'm really bad at judging that.

I've just started to get into electronics and am already confused. How do comparators work (in layman terms)? Doesn't it just compare whether pin 2 or 3 (I have LM393) is getting more electricity, and then turn on pin 1 is pin 2 is more? Thats how I understood it, but when I tried connecting two different ohm resistors (which should create differences, right?) nothing interesting happened to the LED I had in the circuit.
If it was already on (which happens if the negative end is the side connected, which I also don't understand why thats the only way it works) it stays on, and if it was off (positive side) it stays off.
The ONLY way I can get the comparator seem to be able to have any affect is if I connect a wire from pin 2 to negative power, which turns the LED off, which I also don't understand.

Can someone please help me understand what I'm doing wrong? I've tried every wire combo I can think of.

[hgmjr]

Here is an example of a Comparator experiment located in the AAC Tutorial section that you can try.

You might want to sketch your circuit hookup and post it here. Then the members can comment on hook-up problems you may be experiencing.

hgmjr

[lightingman]

So far you have spoken of linear comparators..... There are also digital comparators such as the CD4063 and the 7485.....These will accept two 4 bit binary numbers, and compare them giving a result of hiher, lower or equal to. Thay can be cascaded to any number of bit's.....Daniel.

[Dave]

This is probably better suited in the General Electronics Chat.

Dave

[techroomt]

the lm393 package actually holds 2 comparators inside (a & b). first of all, this type of comparator will use bi-polar supplies (e.g. +10vdc and -10vdc). then the two inputs, + (pin 3) and - (pin 2), are constantly compared. whichever input has the more positive voltage applied (the farthest right on a number line), the output will swing to within a volt of that polarity Vcc. say 6vdc is applied to pin 3 and 5vdc is applied to pin 2, the output (pin 1) will swing to +Vcc minus 1 volt or +9vdc. if 7v were applied to pin 2, the output would swing to -9vdc. since the comparator has only 2 output states, + or - Vcc (minus 1volt), it is often called an analog to digital converter.

[GonzoEngineer]

And don't forget the most important part.....most comparartors have an open collector output......don't forget the pullup

It would really help if you post a schematic.

[jayrare]

Okay, at the risk of sounding like a complete idiot, I really didn't understand what anyone said. I -tried- drawing a schematic, but I'm not really sure how its done, so I had to guess. Also, since I don't get comparators, I decided not even to try and use the symbol.

I've tried many different combinations of connecting the different pieces to either negative or power, and many different values for resistors 2 and 3. After several dead LEDs and two smoking comparators, I've decided against trying anymore combos until I know the correct way.

[hgmjr]

You have the basic conections correct.

Try taking a resistor of the same value as R2 and connect it between pin 2 and ground. This will give you a reference voltage of half the power source or 4.5 volts. Then change the voltage you apply to pin 3 first to 9V then to ground and see that the LED lights up when pin 3 is at ground and it does not light when pin 3 is tied to the power supply.

If you have a potentiometer, you can use it to vary the voltage at pin 3 and see what happens as you vary the voltage up and down.

You can next swap the inputs so that pin 3 is tied to the 4.5V reference voltage and the potentiometer is tied to pin 2. You will see that the behavior of the LED will be reversed. That is, when you adjust the voltage at pin 2 below 4.5 volts the LED will be off and when you adjust the voltage at pin 2 above 4.5V the LED will turn on.

hgmjr

[jayrare]

I must be doing something wrong. I switched R2 to ground, and the voltage at it was 0 so how do I get the "reference voltage"?

[hgmjr]

Originally Posted by jayrare

I must be doing something wrong. I switched R2 to ground, and the voltage at it was 0 so how do I get the "reference voltage"?

I probably did not explain it that good.

What I meant for you to try was to keep the resistor R2 in its original connection state while taking a third resistor whose value was the same as R2 and connect one end of the resistor to the input that is presently connected to R2 and the other end of the resistor to ground. That will form a voltage divider that will provide the reference voltage of about 4.5 volts. The 4.5V is the reference voltage to which I was referring.

hgmjr