HomePowerSupply

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i have yet to actually build this up because the batteries last so long that's it's not really needed. Pretty much all you need to do is take any wall wart that can supply ~5V @ ~1A and feed it into this circuit.

if your wall wart voltage is a lot higher like 12V you will need a big heat sink too. The LM1086 dissapates the extra voltage so a 12V wall wart will force the LM1086 to pass off 6.5W as heat.

In general the GP2X draws about 0.75A so the power dissapated by the LM1086 is:

Power = Volatge Drop * Current

P_lm1086 = (Vin-3.3V)*0.75A

-- ChristopherPepe - 23 Dec 2006