Figure 1:8 inch LCD module backlight circuits.
First calculate how much processing 2W load switching current . Assuming 80% efficiency , the input current is equal to Vout × Iout / Vin × efficiency = 10 × 0.18 / 5 × 0.8 = 450mA.CAT4139 inductive boost LED driver with 750mA ( min ) drive capability , it is very suitable for this application .
Inductor current rating should be able to handle peak switch current without LED driver into saturation . Once saturation occurs , there will be current surges, because the function has become an inductive resistor , the circuit no longer work as expected . Suitable inductor current rating should be greater than or equal 80mA.
LED during operation the maximum output voltage should be below the rated maximum output voltage. Since three LED series , in cold temperatures , the total could be as high forward voltage 11.4V (3 × 3.8V). 24V LED open circuit detection threshold is much higher than its limit. If the LED is off , the output voltage will rise and remain at 30V, time components in low power mode , the current from the power absorbed by only a few milliamps. 30V nominal output voltage capacitor is suitable.
Now consider a powered by a 12V 6W LED lights. You can use six high-brightness white LED connected in series to achieve with a 300mA constant current to drive a typical forward voltage of 3.3V.
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