So I'm an idiot and I need help interpreting a basic datasheet. This is for a Boost-mode LED Backlight Driver IC. https://media.digikey.com/pdf/Data%20Sheets/Diodes%20PDFs/AP3031.pdf
I need to drive an LED that draws 700mA with a typical forward voltage of 7V, and I want to drive it from a 3.6V Li-ion battery (worst case scenario, 3V when flat).
That's a voltage difference of 4V, so at 700mA, the heat dissipation would be 4V * 0.7A = 2.8W, not too bad. But that's only on the high side... Given that the LED draws 4.9W (7V * 0.7A), the current draw on the low side is actually 1.6A (4.9W / 3V battery voltage, and that's assuming 100% efficiency). 1.6A is above the max switching current, plus it would create a whole 6.4W of heat dissipation with the 4V voltage difference.
So from my understanding the chip wouldn't be able to handle that? Or am I doing something dumb with the math?