Hi,
I would like to use the LM339 as a basic comparator and I want to check that I respect the common mode voltage range.
My circuit is very basic, the LM339 is powered from a single +15V supply and I set the (+) input with a voltage of 7.5V (so my trip point will be 7.5V).
On the (-) input, I have a square wave signal 0V / 15V. So the output of the comparator should also be a square wave signal.
My concern is: "can I use the input signal directly with an amplitude of 15V or should I use a resistor network to lower it before (-) input of comparator ? or in another words, do I respect the common mode input voltage range ?"
In the datasheet it is said that input common voltage range is 0 -> 13V with a 15V supply and if my understanding of common mode voltage is right, I have a Vcm = 11.25 V ((15+7.5)/2).
Thank you for your help !
Regards
Oliver,
On the LM339 comparator (VCC=15V), only one input needs to be between 0V and 13V. The other input can be 0V to 30V.The TL331 and LM393 also have this capability.
Regards,Ron Michallick
Regards,Ronald MichallickLinear Applications
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Hi Ron,
Thank you for your answer. Ok for this particular device. For my understanding, if I was using another device that does not have this special capability, is my understanding of common mode voltage range correct (i.e. Vcm = 11.25 with 7.5V voltage on (+) input and 15V on (-) input) ? It is ambiguous in some App Note, they say that each input shall respect common voltage range, meaning that none of the input shall go above 13V if my supply is 15V.
So I would like to make the common mode voltage rande definition for a comparator very clear in my mind.
Best regards.
The common mode for each input must be considered separately. The comparator does not use the average of the two inputs.