Apple - Is it OK to charge my iPhone with my iPad charger?

Your assumption is correct. Your iPhone has an intelligent charge controller on-board and this controller will only draw as much as it needs.

Sure, the iPad power supply can supply twice as much power as the iPhone needs, but it should not be dangerous by any means.

I don't know if there is an "official" word on this, but I would be genuinely surprised that it voids your warranty. Apple engineers are very smart and know that users will interchange powersupplies if they recognize that the connectors are physically compatible (as is the case here.)

Long story, short: you should be fine.


The UK Apple store says the iPad charger is fine with iphones / nano

http://store.apple.com/uk/product/MC359B/A?fnode=MTc0MjU4NjA&mco=MTc3Mzc5MDU


TL;DR: Yes, you can use an iPad charger to charge an iPhone.


Basic electrical laws apply. The USB specification is a nominal 5Volts. That applies to ALL USB devices and chargers. It has to be so for a UNIVERSAL (U) SERIAL (S) BUS (B) to be UNIVERSAL!! (Duh sounds obvious and it IS).

So your charger may deliver 2Amps at 5Volts => 10Watts (W=V*A) Your charger may be capable of delivering 3A. At 5V that implies it is capable of handling a (3A * 5V = 15W) load.

Ohms Law states that Current = Volts/Ohms. Therefore for a given resistance (Ohms) and a given Voltage there WILL be a given current.

So let's say your phone has a terminal resistance of 10 Ohms. the USB specification says that the USB supply voltage is 5V therefore the current your phone will draw is 5V divide by 5 Ohms = 1A

Your 10W charger CAN (ideally) supply 2A at 5V which is 2 times what your phone needs, BUT the phone will ONLY draw a maximum of 1A BECAUSE that's all it needs and can draw due to its internal resistance and the charging circuitry which will limit it even more as the battery charges up.

So for the guy whose charger wires burned: the ONLY way that this could have happened is firstly if the charger wires were severely underrated and the current being drawn was more than they were capable of. This could only happen if the resistance at the end where you'd plug into your device got shorted out. The charger will not be at fault. UNLESS the charger was faulty - then it would have damaged any USB device you plugged it into.

As regards the phone that supposedly burned black: again the same thing. The charger can supply current only according to its capability and the resistance of the load (phone/ ipad, etc). The load device has a given idle resistance and charging circuitry to control and limit the charge current. The charger cannot supply more current than Ohm's law allows and also than what it is capable of delivering.

So the conclusion is that either your charger was faulty and its output voltage was greater than 5V, which would cause (push) a greater current to flow than the load device was capable of; OR your load device (phone, etc.) charging circuitry was faulty, which caused it to draw (pull) more current than it was supposed to.

Bottom line is that when all devices are functionally sound and non-faulty, it is impossible for you to burn it out by using a charger that has a higher rated wattage.

Using a charger with a lower rating can cause the charger to overheat and maybe fail as the load device will attempt to draw the charge current that it needs but the charger can only supply the maximum it is designed to supply.

So, for example, if you have a charger that can supply only 500mA @ 5V (i.e., 2.5W of power) then if your iphone, ipad, etc., tries to draw 1A or 2A, etc., the charger can supply max of ONLY 0.5A. So, Ohm's Law applies again.

Let's once again assume that your load device is 5 Ohms. Remember Ohm's Law: V = I*R. Therefore, if I=0.5A and R=5 Ohms, then the charger voltage will fall to V = 5 * 0.5 = 2.5V because the charger can deliver max of only 500mA. So, if you draw too much current, the voltage will drop.

Take the extreme: if you shorted out the charger terminals, the short circuit voltage is 0V and the short circuit current is 0.5A because the short circuit resistance is 0 Ohms.