The Museum of HP Calculators

HP Forum Archive 08

[ Return to Index | Top of Index ]

Dumb Question - HP34 Charger
Message #1 Posted by scott on 10 Apr 2002, 9:47 p.m.

Ok, I have a nice HP34 that works great with fresh batt's charged outside the unit. The charger is apparently not working. Is it ok to just get a charger of the correct amperage, cut off the connector and splice it onto the new charger?

Your insights appreciated.

Re: Dumb Question - HP34 Charger
Message #2 Posted by Ellis Easley on 11 Apr 2002, 4:22 p.m.,
in response to message #1 by scott

There have been a number of discussions on that subject in this forum. The whole issue of battery chargers and/or AC adapters is the weakest point of these old calculators and other battery operated equipment. (Except the HP Classics, where HP provided a real constant current supply to charge the battery and a separate, constant voltage supply to operate the calculator at the same time.) I can only imagine that the battery manufacturers told the OEMs that charging NiCads wasn't very critical, judging by the instructions provided to the end-users.

The problem with substituting a different transformer is that the charging circuit in the calculator is just a diode to provide half-wave rectification and a resistor to limit the current. HP apparently depended on the resistance of the transformer winding to provide part of the total current limiting resistance. So the transformer secondary voltage and winding resistance is what you really need to match.

I like to have the proper HP charger for the sake of completeness, but I prefer to charge the battery externally and never operate the calculator from the adapter because of the risk of damage to the ICs due to over-voltage if the battery should happen to be making poor contact because of corrosion or whatever (except the Classics, where I trust the adapter).

If I was going to make a substitute charger, here is what I would do: First, I would only plan on charging the battery with the calculator off for a definite period of time. If you never turn the calculator on with the charger attached, you don't have to worry about over-voltage (this statement is probably not valid for the constant memory models). Next, I would try to use the lowest transformer voltage practical to provide enough charging current, just in case I accidentally turn on the calculator. Then, I would provide some external resistance to limit the charging current to a reasonable level.

Now, the last two items are working against each other: for charging batteries in a stand-alone setting (no concern about over-voltage) the best simple way to get a constant current is to use a voltage source considerably higher than the battery voltage, and a relatively high series resistance to drop the extra voltage. This way, as the battery voltage rises with increasing charge, the voltage across the limiting resistor changes by a small percentage, so the current changes by a small percentage. This scheme has two weaknesses: since the current limiting resistor is dropping more voltage, it is dissipating more heat, and if the connection to the battery should open or develop a high resistance, the voltage at the battery end of the current limiting resistor rises to a high voltage (which is why I wouldn't do this for an in-the-calculator charger).

The reason I would use some additional external resistance is that it might not be possible to find a transformer with the same output voltage as the HP transformer. If you use a transformer with less voltage, you will get less current and will need to leave it charging longer. It that doesn't bother you, and the transformer provides enough current to charge the battery in a reasonable amount of time, then you can be less concerned about accidental over-voltage. But if you want to charge the battery in the "normal" time, you will need a transformer with the same voltage or more. If all you can find is "more", then you will need some additional resistance to reduce the current.

I've just made a bunch of measurements of my AC adapters and the charging circuit in an HP25. I have two Woodstock chargers, one the older 115/230V type 82026A and one the newer 90-120V type 82041C, and two Spice chargers, 82087B and 82087D. All four are labeled 10VAC 1.8VA output (except the 82026A which only says "1.8W"). All four measure from 10.1 to 10.4VAC open circuit (my power line measured 119-120VAC during these measurements). The two Woodstock adapters measure 5.7 ohms, the Spice adapters measure 7 and 6.9 ohms. The HP25 charging circuit has an 8.2 ohm, 5%, 1 W resistor which measured 9.1 ohms (old carbon composition resistors are notorious for increasing in resistance over time). I would guess that the Spice models might have a lower value resistor since the adapters have more resistance. With a battery that is partially charged in the HP25, I connected the two Woodstock adapters and took measurements with the calculator off at all times. I verified that the power switch shorts out the 8.2 ohm resistor when the calculator is on, meaning the external resistance is the only thing limiting the current in that state. Both adapters measured 8.1VAC under load. The voltage across the "8.2" (9.1) ohm resistor was 1.5V (165 mA) with the 82026A and 1.4V (153 mA) with the 82041C. The battery measured 2.6V in both cases. I'm not sure how the accuracy of my meter is affected by the AC component of the voltage across the resistor but I trust it on the battery voltage and the transformer voltage since they should be just DC and just AC. According to the DigiKey catalog, the "quick charge" rate for a Panasonic standard 500 mA-h NiCad is 150 mA for 5 hours. HP recommends 6 hours for a full charge for the HP25 and 5-9 hours for the HP34C. If I were to substitute a different transformer, I would add some additional resistance to get about 150 mA. I would trust a current value calculated from the voltage measured across such a resistor (having verified the resistance) more than the current measured by a milliammeter in series, because in a ciruit like this the resistance of the milliammeter could have a sizable effect on the current.

An interesting side note: I also measured several 82059B and D adapters (the only difference between the two models seems to be the length of the cord) which are used with the HP71 and HP75 and many HPIL devices etc. They are labeled 8VAC 3W or VA max output. They measure 12.7-12.8 VAC open circuit and 6.5-6.6 ohms. I was surprised to see that while the nominal output voltage is lower than the Woodstock and Spice adapters, the open circuit voltage is higher. Since they have almost twice the output power, I guess the greater current would bring the output voltage down to the nominal value.

[ Return to Index | Top of Index ]

Go back to the main exhibit hall