Hi all! It is good practice to remove batteries from calculators when they're not used for longer periods of time in order to avoid corrosion of the contacts. But: Do all batteries exhibit this corrosion problem? Or is it only rechargable NiCds? What about the small button-sized cells in the Voyagers/Pioneers?
Because of this battery issue I'd like to operate calculators only with the power supply - but some models (ie. 21, 31E, 65) have problems running from the power supply alone. Can this be fixed? Ie. by using special power supplies that give off DC instead of AC (for 21, 31E) or more current (for the 65 magnetic card reader).
Thanks for help, Tony



I experienced corrosion in many HP41's BATT/IO assy and their batteries were neither NiCad, nor other rechargeable.

You are right about button-sized cells, they are much more stable. Some Silver-Oxide are better. But regular alcalines deserve more care and attention.

I use my calcualtors, and from time to time, I check batteries compartment (at least once a month; better if it is twice a month)

Hope this info helps you.

Best regards.


I haven't seen button cells leak much in things that I have owned (i.e. things that I was the original owner) but as others have reported, they do leak. In my own experience, my 15C that I bought at a sidewalk sale worked but went through batteries in about a week, it actually drew more current off than on. This turned out to be due to battery leakage. After diassembling the machine and cleaning the corrosion, it has worked for several years on one set of batteries. But here's the thing: the Voyagers and Pioneers aren't made to be taken apart, so if their batteries leak, you have a lot more work to do to clean it up.

It would be interesting to know whether alkaline or silver oxide button cells have more of a tendency to leak.

Also, I've had alkaline N cells and AAA cells leak in a 41C and a 71B.

On the power adapter guestion, I haven't done what I am about to describe, but I have thought about building a module to fit in the battery compartment that contains a zener diode or a properly biased IC shunt regulator so I can safely operate a Woodstock or Spice model from the standard AC adapter for extended periods without worrying about overcharging the batteries - no batteries present, just using semiconductors to replace the voltage regulation function of the NiCads. I haven't measured the actual current requirement of the calculators - if they need more than the limited charging current, my idea won't work. A large capacitor along with the shunt regulator might provide enought peak current to make it work reliably, although a large capacitive load might make an IC regulator unstable.

The thing about the 65 and 67 (as I understand it) is that the card reader circuitry is connected directly to the battery because of its greater current requirement. The Classic charger has two outputs, a voltage regulated one for the calculator circuitry so you can use the calculator while the battery charges, and a current regulated one to charge the battery. Normally, the metal spring in the charger receptacle connects the battery to the calculator circuitry, then when you plug in the charger, that connection is opened. The non-card reader Classics ar OK to operate from the charger without the battery in place since the battery is out of the circuit. But the current regulated output has a high output voltage when the battery isn't there (same problem as Woodstock and Spice) which can damage the IC in the card reader. However, although I don't know how much current the card reader motor needs (with no load my 67 motor only drew 13 mA!), since the problem is essentially the same as the Woodstock and Spice, a shunt regulator in place of the battery might work there too.

On the subject of building a new DC power supply to safely operate a Woodstock or Spice with no batteries, you might have a problem because of the current limiting resistor in the calculator. It is in series with the rest of the calculator so it will drop a voltage proportional to the instantaneous current. To the extent that the current drain changes from moment to moment (typically a lot in the case of digital circuits - for instance, changing from a few to a lot of LEDs on), this voltage drop of the resistor would show up as noise on the DC supply of the calculator chips, some of which would normally be filtered out by the batteries (and some of which will be filtered out by the capacitors on the circuit board). If there is too much noise, the result might be erratic operation.

The biggest problem with building a new DC adapter is the plug that fits in the calculator. Unless you have some absolutely un-repairable AC adapters to cannibalize!


Thank you Ellis and Luiz!

>>The biggest problem with building a new DC adapter is the plug that fits in the calculator. Unless you have some absolutely un-repairable AC adapters to cannibalize!

Well, not really: I would take a working HP AC adaptor, cut the cable and insert some off-the-shelf connector. That way I can either use the HP supply or my "special" one.


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