Comparator circuit

[prprog]

I cannot make this circuit work. Please help. I took the circuit idea from a comparator WEB tutorial (that probably I did not undestood). The LED must turn on when I connect +5V to point A. Should it?

Thanks a lot.

 

[Mikebits]

Should work. Check your wiring, or perhaps your LED is backwards.

 

[prprog]

Thanks but I believed the LED is correctly oriented as it is. Anyway I change its orientation an it remain on all the time (even without any reference voltage).??? Any ideas??

Thanks.

 

[ericgibbs]

Thanks but I believed the LED is correctly oriented as it is. Anyway I change its orientation an it remain on all the time (even without any reference voltage).??? Any ideas??

Thanks.

Hi,
I assume you have pin 2 as the output.

Which pin are you usin as the reference voltage.?

You say the LED, glows which ever way round its connected.?

 

[ericgibbs]

hi,
This drawing shows the states of the LM339.

 

[blueroomelectronics]

Thanks but I believed the LED is correctly oriented as it is. Anyway I change its orientation an it remain on all the time (even without any reference voltage).??? Any ideas??

Thanks.

That's impossible, if the LED cathode was connected to +5 your LED will never light. You've wired up something wrong.

 

[prprog]

Comparator

Blueroomelectronics thanks for the reply.
Can you check the circuit diagram and tell me the correct connection of the LED? I try by connecting it the other way and it stay led all the time (even without the reference voltage or input voltage). I have seem many circuits on the web for comparator and the LED is always connected backwards???? This really confused me.

ericgibbs -> thanks eric . Pin 5 is the reference voltage (2.5 V) . Pin 4 is the input voltage, 5 volts in this case. Pin 2 is the output. So in this scenario the led must turn on when Pin 4 received the 5 volts.

Thanks

 

[ericgibbs]

ericgibbs -> thanks eric . Pin 5 is the reference voltage (2.5 V) . Pin 4 is the input voltage, 5 volts in this case. Pin 2 is the output. So in this scenario the led must turn on when Pin 4 received the 5 volts.
Thanks

hi,
When pin 4 goes higher than +2.5V the output will go low and the LED will light.

Try connecting pin 4 to about +3V, rather than directly to +5V, some comparator input/outputs will misoperate at voltages close to their supply voltage.

 

[Mikebits]

I just looked at the data sheet. I think you need a pull up on the output pin 2. In other words, put a 10K res from pin 2 to +5V.

 

[Mikebits]

Forget what I just said, I realize I was wrong. oops.

 

[prprog]

Comparator

Mikebits - thanks.
I will try that when I get home. The pull up will be from pin 2 to 5v and the LED will be tied to pin 2 and ground. Correct ? (This is very different to the circuits I have seem on the web and also on a very old radio shack F.Mims book I have.)
Thanks again.

 

[prprog]

Comparator

Mikebits .So I should not try this. Any suggestion? Should I try ericgibbs suggestion on the 3v instead of 5v.?

Thanks to all.

 

[ericgibbs]

Mikebits - thanks.
I will try that when I get home. The pull up will be from pin 2 to 5v and the LED will be tied to pin 2 and ground. Correct ? (This is very different to the circuits I have seem on the web and also on a very old radio shack F.Mims book I have.)
Thanks again.

hi,
If you connect the LED that way, it will work the other way around, that is the LED will go OFF when the Vin > +2.5V.

Try a lower voltage than +5V on pin4.

 

[Mikebits]

Is it possible that the circuit is breaking into an oscillation when the voltage is removed from pin 4. When the 5v jumper is pulled away, the pin 4 is left floating into a Wide band high gain circuit. Maybe a cap on pin 4 to ground would help??? What do ya think?

 

[Boncuk]

Why not kick off the LM339? Use an LM358 instead. Works reliably.

 

[audioguru]

The inputs are PNP transistors so they float high which will turn on the LED when pin 4 is disconnected.

 

[Mikebits]

I think it is safe to say then that a floating input is not a good idea; ergo, some impedance needs to be on the input with or without the jumper. I would say small series R with a 5k or so shunt (to ground after series R) on the input. Here is a pic of what I am thinking.
Data sheet recommends less than 10K impedance on inputs for stability. This should give a good solid 0v while wire is float. Comments?

**broken link removed**

 

[ericgibbs]

I think it is safe to say then that a floating input is not a good idea; ergo, some impedance needs to be on the input with or without the jumper. I would say small series R with a 5k or so shunt (to ground after series R) on the input. Here is a pic of what I am thinking.
Data sheet recommends less than 10K impedance on inputs for stability. This should give a good solid 0v while wire is float. Comments?

**broken link removed**

hi Mike,
My comment about the addition of the resistor to ground is how it would affect the driving source.
If the driving source was a 'high' impedance the 5K would load it.

The original circuit the OP posted should work, I agree that a floating input is not a good idea with most ic's.
Problem is we are seeing the circuit in isolation, how is he going to use it.?

 

[Mikebits]

The 5k would only present a 1 ma load. I agree however, knowing the drive source would be useful.

 

[AllVol]

Comparators are very simple to use, if you can remember which way round they go. The attachment from Paisley tells it all. I usually used the minus input for reference, applying my signal to the plus input. It seems to work better that way.