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@fthies can you plz tell me what formula have you used...8.7/.020??
(Supply voltage - LED voltage drop) devided by desired current in amps (20 milliamperes = .020 amps)= resistance needed to limit the current through the LED. So for a 12volt system and a 2 volt drop LED the formula would be (12-2)/.020= 10v / .020= 50 ohms.
The power ratting for the resistor is calculated with this formula. Voltage (10v) multiplied by the current through it (.020). Which is (10v * .020a=.2w watts) so a 1/4 watt resistor would get pretty hot. I would error on the high side and go with a 1/2 watt resistor.
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Posted:May 31, 2010
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