> In Dan's scenario, and assuming the charger is set to slow, what
> causes the voltage on the battery
> charger to rise above 14.6 V and force additional amperage into the
> battery? Is it only the
> variability of the AC voltage source?
I can't really answer about Dan's charger, as I don't know enough about it or
how he made those measurements.
But I can say that my Schumacher "10/30/50 amp" battery charger is almost
identical in construction to Dan's description (except it only has two switch
positions and transformer taps, the "30 amp boost" and "50 amp start" are the
same position). I just checked the output voltage in the "10 amp" position with
3 different DMMs, they all gave different values (as Frank mentioned before),
but all readings were less than 12 volts ! Yet it charges batteries just fine
...
The joker is that high ripple (rapidly varying voltage) we mentioned before.
The output of the battery charger (with no battery connected) varies from almost
zero to a relatively high value 120 times a second. Most voltmeters read some
approximation of the average voltage, but the peak voltage is higher (by a
factor of 1.414 if anyone cares). For my charger (again on the 10 amp setting
and with my current line voltage, etc.), the peak voltage is 16.9 volts (all 3
DMMs agree on this point, +/-.01 volts which is pretty good for comparing a
Fluke to a couple of HF cheapies).
Obviously (I hope) current will flow into the battery any time the instantaneous
output voltage of the charger is higher than the battery voltage. So even with
the battery at 15 volts, my charger will continue putting current into it. The
average current will be fairly low compared to the 10 amp rating, but still
enough to shorten the life of the battery.
If the numbers Dan reported were RMS averages, then even on the "slow" setting,
his charger's peak output voltage would be 13.4 * 1.414 = 18.9 volts !
In case you are wondering, the diodes in the battery charger prevent current
flow in the opposite direction during the times when the charger's open circuit
output would be less than the battery voltage.
Randall
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