以下是解釋
BUCK CONVERTER for HIGH-POWER LED
48mA to 90mA
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.
能否詳細說明此電路的作用和此電路的分析!!
小弟感激不盡!!! |
楼主
标题置顶
标题高亮
