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I believe Bojan has the right idea, but question his math.
A fully charged '6v' battery is actually 6.8 v ( 1/2 of a nominal 12v battery's 13.6v) For a 400mA charge rate you need to drop 6.8v across the series resistance ( leaving 6.8v for the battery) which to me works out at 6.8/.4 = 17 Ohms, with a power rating of 17*.4*.4 = 2.72 Watts. Hence a 20 Ohm 5W resistor in series should do the trick.
One thing we've found with sealed lead-acid batteries over some years is that they do not like to be over-charged. When over-charging, the excess is converted to heat which cruels the battery. They seem to respond better to a little charge, but often. Hence, don't leave charging at 400 mA for too long.
In fact if it were me, I would use a few series resistor options; something like a 100 Ohm 1W for trickle charge rate, a 47 Ohm 2W for medium charge rate, and the 20 Ohm 5W for initial high charge rate. Start off on high for a couple of hours or so then switch to trickle for overnight or extended charging.
HTH,
BC
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