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Help with rebuilding a PowerBook "Type 1" battery?

Since the battery for my PowerBook 230 Duo wouldn't hold a charge, I decided to take it apart. Now I'm staring at 10 NiMH cells that I need to replace. Unfortunately the cells have no markings on them. They are simply metal cylinders wrapped in green plastic coating.

On this website I found some minimal information about the "Type 1" battery I am trying to rebuild:

http://support.apple.com/kb/TA32393?viewlocale=en_US

and this site gives general rebuilding tips:

http://www.fenestrated.net/~macman/DuoBattery/index.html

but what I need is the cells with which to rebuild the battery. Does anyone here know of a source for these NiMH cells? Or does anyone perhaps know the specs for each individual cell... voltage, power capacity, etc.?

 
Each of the 10 individual cells is a cylinder slightly shorter and narrower than a standard AA battery:

standard "Duracell" AA: 2" long, 9/16" diameter

cell in PB battery: 1-15/16" long, 1/2" diameter

The PB battery cells notably lack the little protrusion of the positive end of the battery which AA cells have... the PB battery cells are thus essentially flat on both ends.

The fit of the cells in the case is so tight that I doubt I could make AA cells fit.

The battery also has within it four "devices" in addition to the cells. Each device has two wires, and each device looks different from the others. Does anyone know what these devices are, and what their purpose is?

As you look at the battery, it has 5 contact pads. Here is what I can visually determine to be the connectivity of the devices and cells within the battery:

pad 1 to +cell1

-cell1 to +cell2

-cell2 to device1

device1 to +cell3

-cell3 to +cell4

-cell4 to +cell5

-cell5 to +cell6

-cell6 to +cell7

-cell7 to +cell8

-cell8 to device2

device2 to +cell9

-cell9 to +cell10

-cell10 to pad5

pad2 to device3

device3 to pad3

pad3 red wire to device4

device4 black wire to pad4

No two of the four devices looks the same. Device1 looks sort of like a diode. Device2 is a rectangular metal case (approx 1 mm by 2 mm by 12 mm, with leads coming out of one small end). Device3 looks a bit like a very tiny blue capacitor. Device4 looks a bit like a small black transistor (but only has two leads going to it).

 
This site had useful info about replacing a "PRAM" battery:

http://macfaq.org/fullfaq.html

But otherwise I wasn't able to find much about the "main" battery in my PowerBook 230 Duo. I think at this point I want mostly to know:

- Why does the battery have five contact pads, if all it needs to provide to the PowerBook is +12VDC and ground?

- Can I just use any NiMH cells (or other power technology!) that will fit in the battery case to rebuild it, or do I need to exactly match the charge/discharge specs of the original cells?

- How much power does the PowerBook 230 actually draw from the battery when it is in operation?

 
Why does the battery have five contact pads, if all it needs to provide to the PowerBook is +12VDC and ground?
Because then it looks more complicated from the outside, so people won't try to take it apart and replace the cells...a technique pioneered by the same guy that came up with "planned obsolescence" ;)

 
Actually, the case was itself plenty of deterrent to being taken apart. I got it open only by essentially destroying it.

Seriously, I have seen the devices described as for "safety". Two of the devices are clearly in series with the cells. Could the devices possibly be in the battery in order to prevent overcharging, or even a fire?

 
There is a link to an article by James Garry (which is in the 2nd article linked in your original post). That website no longer exists, and the webarchive version was missing most of the pics/graphics. But a little googling turned up James Garry's new website, and he still has that article online, which has a lot of good information: http://www.bipbip.info/jg/mod1.html It's for Type II batteries, but answers a few questions that the other articles didn't.

 
OK, so it appears that rebuilding the battery is my only way to go, unless I am willing to shell out over $100 for a new battery, which I'm not. Thanks to everyone in this thread for the help on this issue!

 
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