Posts: 850
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Joined: Mar 2009
The answer is not straightforward. If you were to use the calculator without batteries, then 4.5V should work fine.
However, to charge the batteries, you need some voltage overhead (hence the 6V charger). If you are not using an original charger this is not straightforward either.
The original chargers were unregulated and of a relatively low current output (100mA to 200mA). But it is in the "unregulated" that the trick lies. When the battery is is low it draws a higher current from the charger, and because it is unregulated the output voltage drops thereby automatically serving as a limiter. Now when the battery gets full, the voltage increases and now because the charger is unregulated, the current it delivers drops - thereby once again acting as an automatic limiter.
In contrast, modern regulated power supplies will attempt to keep the voltage the same and deliver their maximum current if necessary (usually several times higher than the original power supply). This can have the effect of giving a high current to the battery when it is low and not reducing the current to a safe limit (usually <0.1C) when the battery is charged. An excessive current when the battery is full or faulty can lead to destructive overheat.
I recommend either using an original charger (in GOOD condition) or building in a current limit circuit (keeping it to ~0.1C) into the calculator for the battery charging. I usually do the latter and include a PTC fuse.