I have a little 3.6V electric screwdriver that came in handy when it was new. After a while though, it started having trouble holding a charge, and is currently pretty useless. So I've opened it up and removed the battery pack, and I've been poking around with my multimeter. So here's what I know:
The battery pack is made of 3 batteries in series. From the labels (BYD D-AA800P 1.2V K38) I'm pretty sure they are NiCd 1.2V AAs, with 800 mAh each. If I understand my battery physics right, in series that's 3.6V, 800 mAh total. Well, its a 3.6V drill, so that makes sense so far...
Two of the batteries are measuring 1V, and the third is measuring 0V. I suppose a dead cell would explain why its not charging.
The wall adapter has an output of 4.2VAC, 100 mA. That seems odd to me, but the multimeter confirmed at the barrel connector, 4.6VAC, 0VDC.
So I hooked it up and measured directly at the tabs that connect the battery pack in case there's a rectifier hidden in there somewhere. Still got 4.6VAC, but also got 2.3VDC.
Well, I'm confused as to how my battery pack is actually getting charged. What does it mean that I'm getting half of the AC as DC, but still the AC? Does that mean I have a diode but no rectifier? Is 2.3VDC enough to charge these batteries? Is the 4.6VAC damaging them? Looks like there may be a diode hidden under the shrinkwrap between the barrel connector and the + battery tab?
I plan to get new batteries to replace the old ones, but I'd like to be able to understand this stuff, so I can be sure not to wreck the new batteries.