RC time constant question

[eblc1388]

thank's for the help i really appreciate it, one last question before i go and buy the LM339, does the current coming out of the voltage divider need to be around a certain current (mA) to work properly on the input or is it just reliant on the volts???

The inputs of the voltage comparator takes practically very little current(nA ranges) and can in most cases be ignored completely.

You can also consider buying the LM393 which is a 8-pin device having exactly the same type of comparators as LM339(two instead of four) inside.

 

[Roff]

thank's for the help i really appreciate it, one last question before i go and buy the LM339, does the current coming out of the voltage divider need to be around a certain current (mA) to work properly on the input or is it just reliant on the volts???

The LM339 input current is very low (less than a quarter of a microamp max), so if you use the resistor values I mentioned, the error due to bias current will be less than 16 millivolts. This is 1/10 the error you could get if you use 5% tolerance resistors, which I wouldn't worry about. If you don't have 100k and 180k, you can use almost any other two resistors with approximately that ratio. I would keep the lowest value greater than 1kohms, just to avoid wasting current (and draining your battery?).
Your biggest source of possible error is the tolerance of your 470uF capacitor.
It is good that you are aware of the possible consequences of using resistors whose values are too high.

 

[shaneshane1]

i went and bought the LM339 and set it all up and it works great But i didnt take into account that when the power is off the capacitor takes forever to discharge, is there some way i can have the cap discharge fast when the power gets turned off?

 

[eblc1388]

Yes.

You can discharge the capacitor using a diode to connect the positive end of the capacitor to the +ve power supply rail. When you switch the power off, the LM339 supply will falls and when its voltage falls to below capacitor terminal voltage minus one diode drop, the diode will conduct and dump the charge of the capacitor to the LM339 circuit until the whole circuit reaches 0.6V.

This diode is reversed biased when the supply is healthy so plays no part in charging the capacitor and your RC time constant will remain the same.

 

[shaneshane1]

Yes.

You can discharge the capacitor using a diode to connect the positive end of the capacitor to the +ve power supply rail. When you switch the power off, the LM339 supply will falls and when its voltage falls to below capacitor terminal voltage minus one diode drop, the diode will conduct and dump the charge of the capacitor to the LM339 circuit until the whole circuit reaches 0.6V.

This diode is reversed biased when the supply is healthy so plays no part in charging the capacitor and your RC time constant will remain the same.

Thanks!!! I was expecting a lot more of a difficult answer than that, thats very helpfull, iv tested all different time delays using different formulars and changing resistors to see if they match my time calculations,and they all do.

so thanks to everyone for all the help on this topic, i now know alot more about RC time constant and how to put it to practice.