chrispbass
Well-known member
wthrelfall said:hi again good people.
So I used up the LED output on the board to power an LED in my meter, meaning I'm taking power for my Power indicator LED from the unregulated supply after the rectifier. My question is, how do I calculate the current limiting led that's required to put before the LED? I don't know the rating of the led in my switch I'm afraid, but when I put a 9 volt battery across it it glows, not particularly brightly.
thanks in advance for your advice!
Basically, you need the operating voltage of the led and its current draw and the voltage of the supply you intend to use.
ohms law says V = I x R, rearranging you get R = V/I ...
in this case you subtract the forward voltage of the led, so V(supply) - V(led) / I(led current) = your required resistor in ohms
e.g if your supply was 12V, your led operating voltage was 2.2V and your led draws 20mA (0.02A) you get
12-2.2/0.02 = 490 ohms so use closest value to 490 you have, best to go bigger rather than smaller.
You can check the wattage of resistor you need by squaring the current and multiplying by your resistor.
In this case = 0.02 x 0.02 x 490 = 0.196 watts ,so a 1/4 w 490 ohm resistor will suffice.
Try and find the data sheet for your switch, to get specs for led. Failing that, measure the current draw with a multimeter.
hope that helps!!