# Defined LM339 behavior when the (+) input is floating

I was troubleshooting a charge indicator circuit on a tire inflator/flashlight. The green LED is supposed to light up if the battery voltage is above a certain voltage level. The LED is connected to one of the outputs of an LM339 Quad Comparator. The corresponding (-) input to that output was connected to the circuit tracing back to the battery, but the corresponding (+) input just seemed to be floating. I couldn't see any traces off of that pin going anywhere. According to the 339's datasheet, output will go HIGH (with a pull-up resistor) if the (+) input voltage is higher than the (-) input voltage.

My question: With the LM339, if the (+) input is floating, does the IC have a set percentage of VCC to "make" the (+) input?

I'm not a pro at searching through datasheets, so maybe the answer is staring me in the face and I just missed it. I am as sure as I can be about that pin not being connected to anything on the PCB, though.

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"According to the 339's datasheet, output will go HIGH if the (-) input voltage is higher than the (+) input voltage". No. The output is open collector, and it can never go high by itself. With a pull-up resistor, it will go high when the (+) input is higher than the (-) input. See Vid in Ioh and Vol, on Page 6. –  Telaclavo May 21 '12 at 18:39
Thank you. I gladly welcome correction. –  Pete May 21 '12 at 18:46
Probably, that specific comparator is just used as an inverter / current driver. Whenever the (-) input voltage goes above a certain threshold (which, frankly, I still can't tell how much it is, when the (+) input floats), the output will pull low, turning the LED on (if cathode goes to output of comparator, and anode to positive or series resistance). –  Telaclavo May 21 '12 at 18:49
If the PCB has copper on the component side, is there a chance that a small copper trace comes out of that pin, and goes underneath the IC, so that you can't see it? –  Telaclavo May 21 '12 at 19:05
That's possible. If having an input on such a comparator not connected to anything leads to truly undefined behavior, I guess it must be connected to something, even if I can't tell what by looking at it. –  Pete May 21 '12 at 21:29
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So it seems that you have to apply a voltage lower than $V_{CC}$ minus two B-E junctions to have an effect on the left branch of the difference amplifier. Not connecting it will be the same as applying $V_{CC}$.
I don't think that the threshold is at $V_{CC}$. I would bet (though I'm not sure) for $V_{CC}-(2·0.7)$. The common-mode input voltage range includes GND (as the datasheet says), but not $V_{CC}$. –  Telaclavo May 22 '12 at 10:54