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Re: Behind the eight (volt) ball
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Posted by ILB in MO on January 24, 2000 at 15:47:53 from (216.89.172.186):
In Reply to: Behind the eight (volt) ball posted by Fred Martin on January 20, 2000 at 16:10:02:
I read with interest the messages a couple of weeks ago and stayed quiet. But I take my tail in both hands and will jump in on this discussion. You can't use a resistor in series with a 12V charger to drop the voltage to 8V unless you carefully monitor the voltage AT the battery. If you just hook up the charger and resistor then walk away, eventually the battery will have a full 12V charger running against it. As the battery charges, it's voltage increases reducing the current through the resistor which in turn reduces the voltage drop which in turn increases the voltage applied to the battery. In English, as the battery charges, the resistor will drop less and less voltage until it doesn't drop any. By that time, the battery will be overcharged and probably damaged. Now I know that many of you have done it with and without a resistor and gotton away with it. So much for theory (:> There is a better solution than the heavy duty resistor or a lite bulb. A silicon diode drops about .7V regardless of the current through it. They are available in rather high current ratings rather cheaply. If you string 3 of them in series, you will drop the voltage by 2.1V. I would go for 7 15 Amp diodes to reduce my 12V, 10 Amp battery charger (14.5V), to about 9.6V. That should give your 8V batteries a full charge without any danger of overcharging.
ILB
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