HP67 protection



I'm in a great mood today as I've just got my hands on an HP67 (RED LEDS - yahoo! and 3 function keys!), but I've got a couple of queries:

i) What is the reason for 3 wires between the charger and the calculator? Is the calculator running off the batteries or the charger when the charger is connected?

ii) Is it safe to charge the batteries in the calculator, or would it be advisable to charge them outside the calculator?



The charger supplies both power to operate and power to charge. It is not necessary to charge classic batteries externally as it is totally safe to operate a classic with the charger only - with the notable exception that the 65 and 67 require a good battery in place for the card reader motor to operate.

See http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/articles.cgi?read=390 for charger internals.


Actually, it is NOT SAFE to operate the calculator using the charger alone and no batteries. While it is true that the calcualtor will work (except for the card reader as noted), your calculator may get damaged if you do (and this warning appears in the user manual). I have an HP-65 that got damaged that way: the capacitor in the charger burned out (spike from the outlet? I don't know...) and the calculator stopped working. I ended up having to replace the capacitor in the charger and a chip in the HP-65.

Edited: 14 Apr 2005, 9:32 a.m.


You're hanging your comments on a technicality. Your AC adapter failed, that's what damaged your 65, not the fact the battery was missing. This could happen with any device, not just HP calculators.

Once an adapter is plugged into a classic, the power supplied is from the adapter, NOT the battery as it is disconnected by shorting bar at the charger input. With the adapter plugged in, the battery is only connected to the card reader motor.

While there is a caution of running without batteries in place in the 65 and 67 manuals, it is generally regarded as a safe thing to do. There is no such caution in other classic manuals, in fact the 35 and 45 manuals specifically state that it is safe to operate without a battery in place.


It is _NOT_ safe to operate a 65 or 67 from the AC adapter without a good battery installed.

On these machines, the chips on the CPU board run off the regulated voltage output of the charger (as in all other classics) but the card reader sense amplifier chip runs off the battery pack directly (in order to supply enough current to run the motor) and if the battery pack is open-circuit or missing, this chip can be damaged, since the voltage across this chip then rises to about 16V

The 65 user manual contains an explicit warning about this. The chip was re-designed for the 67 (the new version is used in the 97 and 41 card readers too), and the new version is claimed to be able to withstand the charging voltage, but I don't like to risk it. It's an HP custom chip, and spares are not trivial to obtain. I charge the batteries either in a classic without a card reader, or in a 'reserve power pack'

This is all very clear from the schematics of the 65 and 67, BTW...



Thank you for the explaination, this is contrary to information previously posted by others, I stand corrected.

If I only had the schematics... the final word :)


Take a look at http://www.hpcc.org/cdroms/ , particularly the first item on that page...

OK, not as good as real HP schematics, and you will have to put up with my terrible handwriting, but both the 65 and 67 are on that CD-ROM...


Tony Duell wrote:

OK, not as good as real HP schematics, and you will have to put up with my terrible handwriting,

As someone who has just purchased the CD in question, I'd have to say that it is a tremendous value for anyone doing repair work, or just plain interested in learning about the hardware. While it's true that they don't look like official HP schematics, they are quite usable, and the handwriting is actually not too hard to interpret.
I've already gotten use out of the Tony's Topcat-series drawings (HP-91, HP-92) in my HP-97 reverse-engineering project.


What's wrong with charging the battery in the calculator?

(I can see why it can be bad to run from the charger without a battery.)

Do the batteries pull down the charger voltage to a level that the sense amp chip can handle? Do you think that the difference between the regulated voltage and even the charge voltage reduced by the batteries could damage the chip?

Incidentally, my HP-65 had no battery or charger, so I made a pack of three NiMH AAs. I'll have to make a charger before they run out! It'll probably just trickle charge at 50-100 mA, and I probably won't bother with a regulated supply.

Do you know whether this will place any nasty voltages on the calculator? Can I prevent damage by keeping the calculator switched off while charging?



The HP adapter has 2 outputs. A constant voltage output to run the logic, and a constant current output -- about 15V unloaded -- to charge the batteries.

When you plug the charger into the calculator, the battery is disconnected from the logic -- that's what the spring contact on the charger connector is for -- and each section is connected to the appropriate section of the adapater.

BUT in the 65 and 67, the sense amplifier chip is connected directly to the battery terminals. Not even the on/off switch is between them. If you connect an adapter to one of these machines, it relies on the battery pack acting as a shunt regulator to pull the voltage down. If the battery is missing or open-circuit, it won't do this, and the sense amplifer chip will get 15V across it. Even if the machine is switched off.

The original HP65 sense amplifier can be damaged by this. The later one, used in the 67, is supposed to stand it, but I don't want to risk it.

As to whether your home-made charger will damage the chip, it depends on the open-circuit voltage.

The schematic of the HP adapter is available on this site as ASCII-art, and is also on the HPCC CD-ROM.


Which chip did you replace, how did you get the spare part, and what where the symptoms of the fault?


Thanks for the response, guys. Looks like the best course will be to charge outside the calculator.

Thanks again

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