|HP-97 battery pack substitute?|
Message #1 Posted by Mike McCauley on 3 Oct 2007, 12:13 a.m.
I have a HP-97 that I purchased some time ago off eBay. I use it almost daily, and it works great, save for one problem that pops up periodically. Every few years, it gobbles up a battery pack.
I use the machine on AC power only, never off the battery pack. As I think of it I unplug the wall wart and turn the machine on to run the battery down, but most times it is plugged in and turned off. As is well known, in time this trashes the batteries.
I chopped the battery pack open some time ago, and I rebuild it myself as required. This is not big deal, but the sub-C batteries aren't cheap, and NiCds are getting harder to find all the time.
After reading a number of posts on the site about how the charger/PS circuit works, I decided that I could substitute an electrolytic cap and a zener for the battery pack, the cap to provide the filtering to eliminate the "fluttering digit" problem, and the zener to insure that the machine would not be exposed to a high voltage spike resulting from the cap charging to the peak voltage of the wall wart.
I paralleled a 2200 uF 10V cap and a 5.1V 5W zener and attached it to the battery pack spring clips inside the battery compartment with clip leads. This works very well... for the most part.
When the machine is on, the wall wart/cap combo voltage gets pulled down to about 4.89V and the zener dissipates essentially zero power. When I turn the machine off, the cap voltage is clamped at 5.1V, and turning the machine on and off produces no obvious problems. All in all, the idea works great, except for one problem.
With the machine off, the zener heats up, quickly and to a relatively high temperature. Obviously, this indicates that the current limiting resistors on the output side of the unregulated supply in the machine are sized to allow too much current to flow though the zener.
I could use a higher voltage zener, but I'm concerned that a turn-on spike from the cap would harm the machine.
I could use a simple zener-transistor shunt regulator circuit with a small heat sink to spread the heat out. But the power being dissipated is still the same, and I am concerned about burning out the series resistors in the calculator
The idea behind all of this is to simply eliminate the "low power" LED being on when the next set of batteries gives out, and thus the buying of one set after another of batteries in the first place, batteries that, in my situation, are acting as no more than short-life filter caps.
Higher voltage zener? Shunt regulator? Bad idea, won't work? Something else?
Thanks in advance for your comments,