When the battery goes down from 12.6 to 11 this comparator will send a signal or a 0 voltage so I can take that output to turn on a charger.
I you have any idea how to build this I will appreciate a lot, my company is in risk so with this I can help.
Thank you beforehand for your time.
2006-11-30 01:01:00 · 2 answers · asked by pap3030 2 in Science & Mathematics ➔ Engineering
An idea would be to use a relay circuit to control the operation of the charger. The voltage out of the battery could be divided so as to latch a "normally off" relay to keep the charger off. The resistance of this circuit would be such that it would not excessively drain the battery. Once the relay is unlatched, the charger would come on.
Now the difficult part - the voltage applied to "charge" the battery would have to be in excess of the "high" voltage in your range. A suggestion would be to use a timer to then cycle off the relay.
There are commercially available products that already perform this function, if I am not mistaken.
2006-11-30 01:25:13 · answer #1 · answered by www.HaysEngineering.com 4 · 0⤊ 1⤋
Its really easy. Take an op amp, tie the negative input to a pot which goes from rail to rail. Take two identical resistors, tie one from power to + in, and the other from + in to ground. Add a feedback resistor to the positive rail and you're done. This might help you visualize.
http://home.cogeco.ca/~rpaisley4/Comparators.html
Who's your comany that they are having people with questions do the work anyway?
2006-11-30 10:13:47 · answer #2 · answered by Philip_Comer 3 · 0⤊ 0⤋