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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?

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  • \$\begingroup\$ 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. \$\endgroup\$
    – mhaselup
    Sep 20 '20 at 9:07
  • \$\begingroup\$ 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 \$\endgroup\$
    – Pecacheu
    Sep 23 '20 at 19:51
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Your maths is not quite correct. You are right that the LED will be drawing 4.9W, and most of this will be converted to heat (about 80%). You are also correct that the current draw from the battery to power the light will be 1.6A. However, you do not multiply the voltage difference to get the heat dissipated by the IC. This is a switching converter, so the heat dissipation to the chip is a function of the efficiency of the chip, which depends on the mosfet resistance, switching speed, and other things (plus the inductor ESR). Usually the datasheet will give you an idea of the efficiency at different currents and voltages. Looking at the graphs, its probably around 80%, so your heat in the chip will be about 1.2W, and your total current draw from the battery will be 2A.

You are also correct though that this chip will not be able to drive 700mA, since you need 2A inductor currents.

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  • \$\begingroup\$ Ahh, alright, that's making a bit more sense then cause my math made it seem like there was a lot of heat dissipation coming out of the chip. Though unfortunate that the chip won't be able to handle it because of that 2A, it seems like a nice one. Guess I'll either have to find another chip or use a second regulator between the chip and the battery. \$\endgroup\$
    – Pecacheu
    Sep 20 '20 at 1:26

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