Originally posted by joe shmo
Just trying to put together a analytic model for how long it may take to drain battery by means of a simple resistor...hitting some "walls" so to speak, and was hoping for some input.
So im assuming the work a battery can do is given by (1)
W= Work the battery can do
Q = charge the battery holds
V = voltage across the terminals
W = Q*V (1)
...[text shortened]... jack about circuit analysis and MAYBE the whole thing needs to be scrapped...I don't know?
The interesting thing about modern batteries like Li ion and so forth is you charge them up at a certain rate, say 100 ma @ 4.8 volts. Or about 0.48 watts for however long it takes to get to the rated amp hour or milliamp hour rating. What is interesting is these newer batteries convert relatively little of the energy into heat so it is essentially 100% charge and discharge cycle, so if you have a 2000 ma hour battery and you charge it at 100 ma, then after 20 hours you will stuff the battery with 2000 ma hours of energy and that is just about all you used so the energy in Vs out is very close to 1 to 1. The dW/dt is just the change in watts per unit time. When the battery first starts up it has a certain voltage to start with and that will change somewhat from start to finish, so for an efficient battery charger you need a constant current supply not a constant voltage supply.
I just did some experiments in that regard and was using a constant voltage power supply that could supply a LOT more energy that that needed to charge up those AA sized Metal hydride batteries I was using. As the battery charged up with the constant voltage, the current drain constantly went down in response to the changing effective internal resistance so in order to maintain a charge, the voltage has to be constantly adjusted upwards, picture you with a DVM measuring the voltage of that power supply and an ammeter to measure the current going into the battery.
You will find you have to constantly up the voltage slightly to maintain the same current which is a pain in the butt since you have to baby sit the charging process.
To make it more user friendly, instead you use a constant current power supply, where the voltage gets automatically adjusted to maintain the same current. That way you don't have to keep staring at a DVM all day.
Our ancient computer used to run a semiconductor sputtering machine uses 3 AA sized batteries, NiCad's originally but I tried metal hydride ones with about 3 times the milli amp hour rating. The voltage comes out to about 3.6 volts for a fully charged battery, somewhat more when the battery if fully charged. To charge that battery, you need about 30% higher charging voltage to overcome internal resistance so a 3.6 volt battery needs something like 4.8 volts. Of course you need some nice power supplies, constant current and constant voltage to do it justice as an experiment. You don't need to worry too much about the math since for the most part, you put in 1 watt you get out 1 watt, within 95% anyway.