Powering devices from a USB port is something of a dark art. The USB specification says that devices can draw up to 500mA of current after negotiating with the host, but in reality, many USB devices just draw that much without asking for permission. Things get a bit more complicated with Apple devices, which use the two data lines (USB has two lines for power, two for data) to determine how much power to draw. But instead of going through proper negotiation (which would be complicated), it simply checks the voltages on the two data lines to determine how much power it can pull. Fortunately, others have documented (here and here) the combination of voltages and their respective current draws.
For my power pack, I’m using a circuit based on the ADP1613 chip to boost 3-4.2V power from the LiPoly battery to the 5V that goes out the USB jack. Since the chip supports outputs up to 2A, I’m configuring the USB jack to allow Apple devices to draw up to 1A (which, incidentally, means it’ll support iPads, in addition to iPhones — most competing products don’t support iPads so this is a key advantage). In theory, this is just a matter of picking the right set of resistors for the voltage dividers connected to the data pins. In practice, though, things are a bit more complicated because the voltage dividers draw power from the same 5V output that the device is charging off of, and when that output voltage drops under load, so do the voltages in the data pins.
So, today’s experiment was to use a breadboard (and some a hodgepodge of other hardware) to simulate the data-pin configuration on my circuit board, to make sure everything works. What I’ve discovered is that when the output voltage is 4.8V, my iPhone only drew 350mA, while at 5.2V, it drew 700mA. Since 500mA is the cut-off, I know that it was working as expected when it was drawing 700mA. Unfortunately, I don’t have a power source that’ll output exactly 5V, so I don’t know what’ll happen in that case. In any case, I think I’m going to modify my circuit (or rather, change one resistor in yet another voltage divider) so the output is 5.2-5.3V instead of 5.0V, just to be safe.
