jpp412 Posted April 4, 2005 Report Share Posted April 4, 2005 I'm trying to understand the function of the 339. I have read all of the datasheets (which don't help very much as far as basic understanding) and think that the basic idea is to output high when the + input is > the - input and a low otherwise. Is this the correct basic idea?Beyond that - when I look at the example datasheets which include application examples - I find that most of the examples elude me in terms of what they will do.Any help or pointers to resources that I could study would be much appreciated.Thanks in advance.Joe Quote Link to comment Share on other sites More sharing options...
audioguru Posted April 4, 2005 Report Share Posted April 4, 2005 Hi Joe,You almost understand the operation of an LM339.Since the output is only the collector of an NPN transistor, the output becomes open-circuit (and is pulled high by an external resistor) when its + input is more positive than its - input. When its - input is more positive than its + input then the output transistor conducts and pulls any voltage provided by an external pull-up resistor to ground.The datasheet applications show that this comparator is just a switch to ground that is controlled by the very accurate voltage match of its inputs. ;D Quote Link to comment Share on other sites More sharing options...
jpp412 Posted April 5, 2005 Author Report Share Posted April 5, 2005 Thanks Audioguru. Your explanation helped alot. I'm going back through their ex apps with this in mind to see if I can make sense of it now.Thanks again!Joe Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted April 7, 2005 Report Share Posted April 7, 2005 A comparator litterly amplifies the logic into oblivian. No, not really oblivian for the fact that the unknown zone exists. So basically you have a modest type amplifier I think. I still think these things have higher gain. I know that the output is always good provided the input. It's funny but they actually have a gain of 1. But in reality, and this is the point I am trying to make, the voltage curve rises to it's uppermost voltage very quickly because the gain is high. So no it is not accurate to potray the voltage as not doing enough to make the transition. It does quite a bit, but the voltage will always charge with the capacitor. So although the voltage flies immediately, the RC time constant will hold onto the voltage keeping it slewed. Quote Link to comment Share on other sites More sharing options...
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