Using a 97 without batteries


I'd like to use my HP97 strictly as a desktop without having to fret about the condition of the battery pack by running it off the recharger. It doesn't seem to want to power up without the batteries in, however. Is there some way to hack this problem (dummy load or something)?


One solution would be to fill the battery compartment with a bunch of large capacitors (10V and tens of thousands of uF) and a shunt regulator.

This would essentially mimic nicads (except with much lower capacity :-)

A suitable shunt regulator is a zener diode (get one rated at 6V at couple of watts) or alternatively use a zener and a transistor with the transistor shunting most of the current.

(thinking about it, you could *probably* get away without the shunt regulator)

Just be careful to use the correct charger...

A problem you're likely to have is that high current demands must be met from the capacitors. Thus the card reader may be a little difficult to get working without *LOTS* of capacitors.

It may actually be cheaper (and easier) to use nicads. If you're going to leave them permenantly on charge you'd be best advised to reduce the charge current to (about) C/50 or so.

The nicads in the HP97 currently charge quite a bit faster (C/10?) which will cause problems if they are left on charge indefinitely.

approx C/200 is the rate that you need to charge nicads to counter their self-discharge characteristics, but as you'd actually be using the nicads for short periods you'd need a higher rate to charge them again.

An even cheaper solution is to use AA nicads in conjunction with reducing the charge current (use a pair of diodes and a resistor if you don't want to butcher the HP97). The lower capacity of the AA nicads will not be an issue since you're not using them to power the HP97 for any length of time.

the ascii circuit diagram below is intended to show a diode and a resistor used to limit charge current, and a reverse diode to remove the resisitor from the discharge path. I don't thing the additional 0.6V drop will cause major problems.

---|>|--^v^v^v^--- + battery - -----------


The resistor value should be chosen to limit charge current to C/50. The diodes could be 1N4001 or any similar type.

C/50 refers to Nicad capacity (in mAH) divided by 50 hrs

so C/50 for 800 mAH batteries would be a 16mA charge current.



Thanx for the fast response! If I come up with a tested, bulletproof scheme for this, I'll post it. The 97 makes an ideal replacement for the typical Casio/Sharp paper-tape desktop when tax time rolls around because I don't have to drop back to algebraic mode to fill out my 1040!

All for now, Doug


I have ruined an otherwise good HP25 by substituting the capacitor inplace of the battery pack. The capacitor raises the DC voltage supplied to the calculator


Thats why you need to shunt it with a zener....

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