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Charging Batteries - Printable Version

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Charging Batteries - Wayne Stephens - 05-12-2004

An old, non-HP (gasp!) RPN calculator has recently come into my possession. It is powered by 3 "AA" sized rechargeable batteries wired in series. The output voltage for each battery is specified as 1.25 volts (They are brand new and I believe they are NimH?). There is about 3.8 volts present at the charging port when the calculator is turned off and the batteries are fully charged, which they were when I received the calculator.

The problem is, the original recharger unit was not with the calculator and the manual gives no information about it other than that it takes "about five hours" to reach full charge starting with dead batteries.

I have a bench-top variable voltage DC power supply which I intend to use to recharge the batteries. However, I do not have a lot of experience with rechargeable batteries (at least not beyond charging them with the original factory charger). Can someone who is more electrically/electronically astute than I recommend a charging voltage? The highest I have dared turn the power supply while connected to the charging port is 4.00 volts, but for all I know it will take a year to fully charge the batteries at that voltage. Is there a "rule of thumb" ratio of charging voltage to design output voltage? I just want to be able to fully charge the batteries without damaging them or the calculator.

Thanks in advance to any/all who can assist me with this.

Take care.


PS - The calculator is a National Semiconductor 4640 in truly outstanding condition (other than the missing battery charger).

Re: Charging Batteries - unspellable - 05-12-2004

The proper charging rate is in terms of current, not voltage. If your power supply has a current limit set it to 50 mA. Any AA sized Ni-Cd or Ni-MH cell will take at least 50 mA during charge. Do not set the voltage higher than 4 V. If your power supply does not have a current limit I would put a resistor in series with it to limit the current to 50 mA. In this case the voltage can be set somewhat higher to allow for the drop across the resistor.