Hello Meindert,

I agree with your analysis and go a step beyond: even if
there is no differential load across one of the batteries,
uneven charging is possible (or likely) because of
differences in the batteries themselves.

Regards,

Chuck

Meindert Sprang wrote:
"James Hahn" wrote in message
news
"Meindert Sprang" wrote in message
...

snip

There are two possible scenarios: one battery wears more quickly than

the

other, eventually reaching a state where the cell voltages reach higher
values due to increasing internal resistance. Thus the charges shuts off
too
early, leaving the better battery not topped up.

The charger does not monitor cell voltages. It monitors the voltage

across

all the cells in series.


That is exactly what I meant and exactly the reason why it can go wrong. So
I would like to invite you to shoot holes in my following argument:

The charge of a battery is the product of current x time.

Both batteries are in series and one load is connected to the set, operating
at 24V. Another load is connected across only one battery, operating at 12V.
So it is evident that one battery is discharged more than the other.

I start to charge the set in series, so the current through both batteries
is exactly the same. Since one battery is discharged more than the other and
the current throug both is the same, one battery must be charged longer that
the other. Exactly how am I going to achive that with the same current
through both batteries?
One battery will reach the full state before the other but is still being
charged with full current because the other battery hasn't reached the
voltage that corresponds with full charge.

Now, if you can come up with an valid argument why this shouldn't damage one
battery, I bow and take my hat off.

Meindert