Hello,
I have a battery of NiCd's which is 6V nominal.
I wish it to supply 0.714 Amps for 3 hours.
-this is 3*60*60*0.714 Coulombs = 7714 Coulombs.
If i am to charge this battery up for this service, and i have a 19 hour charge time
then would i be right in saying that i need a constant charger current of 112mA?
Hello there,
That works if your battery or battery pack is around 1.5 ampere hour. But if it is 2.1 ampere hour then it doesnt work as MikeMI said. So if you havent already figured in the charge acceptance of a NiCd then you have to do that first.
It's very simple but the way to start is with the ampere hour rating of the battery or battery pack. You take the ampere hour rating, multiply by 1.4, then divide by the current. So a 1.5 ampere hour battery being charged with a 100ma power supply would be calculated as:
x=1.5*1.4=2.1 ampere hours (including the charge acceptance for the battery)
t=x/i=2.1/0.1=21 hours
So you'd have to charge for 21 hours with 100ma.
Notice that in the above we start with the Ampere Hour rating of the battery and use that as the basis for calculating the charge time with any current source.
Also, you mention that you want to use a "constant charger current" of 112mA. That fine, but you dont really need a constant current. The current can vary over time so the circuit can be simpler than an actual constant current circuit would be. The only difference is that instead of the simpler division by the constant current, we have to sum all the shorter time charging with a current that changes gradually from some min to some max. Why we go through all this is so that we dont have to actually build a constant current charger which is much more complicated than a simple DC wall wart.
The new calculation is similar to the way we did it before, with a slight modification...
First, x is still equal to the ampere hour rating of the battery times the charge acceptance of a NiCd:
x=1.5*1.4=2.1 ampere hours.
That doesnt change. But because the current now starts out higher and ends up lower we have to sum up all of the incremental ampere hours that change over time. Luckily, doing this results in a very simple formula if the ampere hour increments change linearly over time. Unfortunately, they dont, but lucky for us again they dont change by that much in most cases so we can still use a simpler formula to compute the time to charge:
t=2.1/Iavg
where
Iavg is the average current during the total charging period. Iavg is easy to calculate too:
Iavg=(Imax+Imin)/2
So what we end up with is:
t=2.1*(2/(Imax+Imin))
A good question at this point is how do we determine Imax and Imin. Luckily again this isnt hard to do either. We have to charge one time and measure the current when we first start to charge, then the current once the batteries are surely charged (after a long enough time to ensure a full charge). So the starting current may be 150ma and the ending current 50ma, which makes the average current 100ma over time. Using these two in the formula we have:
t=2.1*(2/(0.150+0.050)=2.1*2/(0.200)=2.1*10=21 hours.
So you dont need a constant current charger if you dont mind doing just a little more work for the calculation.