I have a bunch of these:
This would make an excellent wall-plug case/supply for a Pro Mini project < 500mA.
Is there anything I need to look for on the circuit inside to make sure there's no auto-shutoff sort of stuff going on?
Not knowing what type of battery it is designed to charge I would still guess then the charge is completed the output will change. I think the best way to determine this is to charge a battery and monitor the voltage.
My bet would be its for NiCad which will usually drop to some low rate after is senses a certain terminal voltage.
You might just try connecting an LED (with a current-limiting resistor, of course) and run it for a few hours to see if something happens.
If you've got a mulimeter, I'd also recommend checking the voltage with and without a load to see if it's regulated. And, assuming you don't have an oscilloscope, put a load on it and measure the AC voltage to check the filtering. (Pure DC will measure zero on an AC scale.)
...Chargers don't have to be well regulated or well-filtered, but they should be current-regulated (and short-circuit protected).
If you don't have a multimeter, leave the LED connected and check to see if the LED dims when you add a load.
You know Ohm's Law, right? If you want a "known load", 8.4V with a 16.8 Ohm load would be 500mA. And, you'd need a 5 or 10 Watt resistor (8.4V x 1/2A = 4.2W).
Sweet. That makes sense...I'll test soon. This would be excellent usage of an otherwise scrap part.
I slid the other side out & soldered in a barrel pigtail to make use of my camera batteries for quick Arduino power.
The batteries are 7.2V Li-Ion
Googling the part number from the photo suggests it's a charger for a Canon 7.2 V Li-Ion battery pack.
Li-Ion are normally charged at a constant current until the battery reaches a set voltage and then constant voltage until current drops to a set level. If your circuit isn't drawing sufficient current (a few 10s of milliamps?), the charger may decide the charging cycle is complete and shut off. It's not very elegant, but you might be able to fool it by putting an additional load across the output.