Typically, a power supply's rating is a stable voltage and a maximum current draw. In other words, assuming the voltage is correct, the device connected to the supply can draw a current that is less than or equal to the maximum specified on the supply. So, you should be OK with a higher rated power supply.
Here's what I learned working on a battery powered charger for my iPhone and iPods:
The Voltage/current specified for USB is 5.1V @ 500mA. That's what you get if you charge the device by plugging it the USB port on your computer. (USB 3.0 supports 900mA, but that's a different story). Some of the iPhones/iPods can draw higher input currents from wall/car chargers to allow for faster charging. The older iPods also supported charging from the 11V power on FireWire which is why some of the older iPod chargers will not work with the non-Firewire chargers. For my scenario, I stuck with the 5.1V and 500mA and my charger works well on everything I've plugged into it.
Apple, being Apple does some funny things that may cause compatibility issues and they're not exactly forthcoming about how their devices and chargers work together. Basically, in the charger, they place a resistor circuit on the USB data lines and if the device doesn't sense the right resistance on those lines, it rejects the charger and you get a "Charging Not Supported on This Device" message they've also used different resistances to cause the device to allow different charging currents.
The good news is that people have figured out Apple's game to design more affordable chargers. The bad news is that depending on the vintage iPod/Phone, one charger that worked fine on one iPod, may not work on another. The compatibility isn't simply due to the current/voltage specs on the charger's label, it's in that little resistor circuit inside the charger so you can't always tell.
My advice is to buy it, try it and if it doesn't work, return it.
Oh, and the solid line above the 3 dashes indicates that the Voltage is positive. If the 3 dashes were on top it would be a negative voltage.
Message edited by author 2011-12-23 22:45:13. |
