# LED Driver IC off a single-cell Li-ion battery (Too large of a voltage difference?)

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?

• I think you're getting confused with linear regulators where power dissipation is a function of the voltage drop. This is a switching convertor, the waste heat is a function of the efficiency of the device, a good device might be 95+% efficient i.e. very little heat dissipation. You might be snookered by the max cycle-to-cycle current limit of 1.5A on the spec. You can likely find another beefier device to address this. Sep 20, 2020 at 9:07
• Yeah, I took from the answer that I am confusing linear voltage regulators with switching when I did my math so I see where I went wrong XD Sep 23, 2020 at 19:51