Wednesday, February 16, 2011

BUCK CONVERTER for HIGH-POWER LED 48mA to 90mA

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This circuit is a "Buck Converter" meaning the supply is greater than the voltage of the LED. It will drive 1 high-power white LED from a 12v supply and is capable of delivering 48mA when R = 5R6 or 90mA when R = 2R2.

The LED is much brighter when using this circuit, compared with a series resistor delivering the same current. But changing R from 5R6 to 2R2 does not double the brightness. It only increases it a small amount.

The inductor consists of 60 turns of 0.25mm wire, on a 15mm length of ferrite rod, 10mm diameter.   Frequency of operation: approx 1MHz. The circuit is not designed to drive one 20mA LED.This circuit draws the maximum for a BC 338

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