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I have an old Texas Instruments pocket calculator, for fun decided to get it working again. It is strictly a NiCd powered calculator, no alkaline recommended.

The charger is the 9130, says output 6 VAC, 175 mA. Tested with unloaded output measures 9 VAC, with a 20k ohm per volt VOM. Before I crack open the power supply case, is this "correct?" I don't like the sound of AC charging NiCds. Should be some kind of filtering in there that might need repair.

The SR-50 takes 3 NiCds, no circuitry in the very basic battery pack, will use new NiCds. Maybe I should get a new charger. The battery pack is required for operation of the calculator.

What is a good way to test the charger under load?

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    \$\begingroup\$ The charging circuit in the calculator converts the AC to DC for charging the NiCads. \$\endgroup\$ – Dave Tweed Feb 11 '13 at 1:44
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I have that calculator, and the charger. Circuit wise, there is a diode which ends up in series, and DC is presented to the battery and the calculator. 9V open and 6V +/- sounds right. I have rebuilt the packs putting in AA nicads. It helps to fit things in, if you can find slightly undersized cells. Hot melt glue can be your friend.

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The rating of the charger, 6VAC @ 175 mA indicates that it has a fairly high internal impedance. Also, NiCd cells are typically quite forgiving in their charging requirements: See this datasheet for instance, which states that for 1.2 Volt cells, "Maximum cell voltage should be considered to be 1.7 volts".

Even a single diode as a half-wave rectifier within the charging circuitry of the calculator would provide positive pulses every half-cycle, with peaks at 6 * 1.414 = 8.484 Volts approximately. A smoothing capacitor would bring this down to a charging voltage within the 5.1 Volt maximum cell voltage of 3 NiCd cells in series.

Thus, the 6 Volts AC charger should be fine, and the limited charging current (due to the presumed high internal impedance of the charger) would ensure that the batteries do not overheat due to too much current available while charging - no matter how much the batteries would tend to draw if available.

In such a situation using a charger with a higher current capacity might actually be a risk, in contrast to the typical advice that a power source rated at higher amperage works as a replacement for a lower amperage one.

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