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The original charger for my HP19C had its calculator plug repaired (or better: rebuild) by the original owner. This new plug is build in order to secure a given polarity (a cut-out groove engages with the post between the two contact pins in the calculator), however, it appears that the charger output is AC (8VAC) - is there a reason for the chosen polarity. It appears that my spare charger (type 82066, 220V) with the original plug can be connected either way. Further, the original charger has an output of 3VA, the more recent one 2,5VA - is there a reason for this? Perhaps the latter, made in 1987, was intended primarily for the HP75.

Any comments would be appreciated.

Regards, Bo

There is no reason to concern about polarity with AC adapters that deliver AC on their low voltage side.

The "polarity" may matter, in theory, if the adapter had an autotransformer (I don't know any real-world case); and as a safety issue (ground connection, etc.) Again, this should not be the case.

In the Woodstock series (HP21, 22, 25, 27, 29C, the low-voltage AC is rectified by a diode inside the calculator, and the pulsating DC is applied, via a current limiting resistor, to the batteries. In such manner, the batteries receive a charging current, and also act as a filter and voltage regulator. That is why you should not operate such models with an AC adapter if rechargeable batteries are not present.

The 2.5 VA vs. 3.0 VA issue may have to do with the particular transformer used to build the AC adapter. A 300 mA current at 8 VAC (rms) gives 2.4 VA. I don't think your calculator takes that much current. So the 2.5 VA transformer seems to be OK.


Dear Andrés, thanks for the detailed explanation. Regards, Bo Kristoffersen.