hp67 power supply



#4

I recently took possession of a 67 that's been in the family since it was bought new. It might even be that the card reader still works! It was never used very much... However, we can't find the 110 transformer. Now I've looked up what is available here and still have a question or two. If I hook up a power supply are there issues to be aware of so I don't hurt the calculator? I understand I need a "good" battery pack installed. Is it right that that means it should be able to take a charge? I plan to use the calculator only when plugged in. How about damage to new batteries? Any other comments? Thanks very much!

Edited: 24 Feb 2004, 3:13 p.m.


#5

Not sure how much you know, so I’ll cover some of the basics. There are three posts in the charger connection well on the calculator, and three contacts in the charger plug. When you insert the charger plug into the calculator, it pushes two little spring contacts away from the outer two posts. These spring contacts normally short the outer two posts together. When they are pushed away, the battery supply to the logic circuits of the calculator is disconnected, and is replaced with a regulated voltage supply from the charger. The charger actually produces 2 voltages: one to supply the calculator logic circuits, and another that is intended to charge the battery pack. The source for charging the battery pack is produced between the center contact (marked with a -) of the charger plug and the contact marked with the +. Since you apparently do not have a charger, this would be between the center post and the left post in the charger plug receptor well, as viewed from the back of the calculator. This voltage varies as required to push the charging current through the battery. With no load, I have measured about 18 Volts between the center (-) contact and the contact marked with the +. The other voltage, produced between the center contact and the unmarked contact on the plug (between the center post and the right post on the calculator) is the regulated source intended to power the calculator logic circuits when the charger is plugged into the calculator. I have measured about 4.5 Volts between the center contact and the un-marked contact. The card reader requires a high amount of current compared to the logic circuits, which the regulated charger output would be unable to supply, so they hooked the card reader to the side of the circuit that stays connected to the battery when the charger is plugged in. With a “good” battery installed, the voltage on that side of the circuit will be clamped down around 4.5 Volts with the charger plugged in. When you read or write a card with the charger plugged in, the “slug” of current required comes from the battery, not the charger. Unless the battery is completely dead, it will be able to supply enough current in this situation. The HP-67 manual states “If a battery is fully discharged, it must be charged for at least 5 minutes before a card can be read.” The manual also states “Operating the HP-67 from the ac line with the battery pack removed may result in damage to your calculator.” This is because without a battery, the unregulated battery-charging voltage supply will rise to the unloaded maximum of around 18 Volts, which will be applied to the card reader. If I recall correctly, this is not an instant death sentence like operating a woodstock from a charger with no battery, but can damage components.

With the above said, since you do not have an original charger, there does not seem to be any danger that you could hurt your calculator. (Well, unless you go hooking up a voltage source any higher than 4.5 Volts to some input or another.) Some of your options might be as follows:

1. To insure that you have a good battery pack, re-build your pack, and rig up a way to charge it outside the calculator. Then operate only off of the battery.

2. Find a regulated 4.5 Volt DC power supply with a healthy output current capability (i.e. not rated in milliAmperes). Connect it to the battery contacts in the battery compartment, or to the posts in the charger plug well. You may connect between the center post (negative) and either outer post (positive). No battery pack will be required, since your power supply will basically “look” like the battery as far as the calculator is concerned.

3. Buy a charger on eBay or wherever, rebuild your pack, and operate as originally intended. The chargers seem to come up on eBay fairly regularly and are not that expensive.

Additional comments and/or corrections from more knowledgeable sources are invited and welcome.


Disclaimers – information provided is believed to be correct, but I am not responsible for any damage to your calculator or incidental loss. Use at your own risk.


#6

Thank for your time and help! I may take at least 2 of the 3 suggestions. Since I mostly use a 42s, especially for programing, I probably won't use the card reader. With one quick read of your notes it sounds like that is the tricky thing. I'll probably just get 4.5v to the battery compartment terminals, and continue to keep an eye on ebay. Thanks again.


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