thanks eric,
please enlighten me again on circuit LM339, R4 68k and R8 18k is a voltage divider, and the current flowing through this (in my own understanding) is about .162mA
what is the basis of electing a total 86k resistance?
The selection of resistors is to give the required voltage ratio from the voltage applied to the top of the divider and the junction of the two resistors. Usually high values are chosen so as not to draw too much current from the battery being monitored.
Obliviously the 68K and 18K add up to 86K, which when the top of the divider is at 14V a current of 0.162mA flow flow the two resistors, so the voltage across the lower 18k will be 2.93V.
An easier way is: - 14v * 18k/86k
i try to calculate using ohm's law: (please correct me if i'm wrong!)
* 68k will have about 11.0697V
* 18k will have about 02.9302V OK
* R4 & R8 is 20.9% of 14VDC which is about 2.8V on their junction, then there is 15k R3 connected to this junction.
if the source of R3 is 2.8V with a current of .162mA, what is the current on junction R3 and -inps?
Dont think in terms of current inputs to the LM339, think voltages.
This divided voltage of 2.93v is on one pin of the LM339 and the Vref voltage on the other LM339 input.
So when the divided voltage is greater than the Vref the charge relay is OFF and as the 14V battery voltage falls the divided voltage will fall to less than the Vref and the charge relay will be on.
will the Voltage on junction R3 and -inps drop?