Peter,
Thanks for mentioning the power dissipation calculations when using a dropping
resister with a LED. Unfortunately, there is an error in your calculation.
You multiplied the current (0.02 amps) by the voltage drop across the LED (2
volts) to come up with 0.04 watts. This is power dissipated by the LED. For a
9 volt supply, the voltage drop across the resistor is 7 volts and the power
dissipated by the resistor is 7 x 0.02 = 0.14 watts. A 1/8 watt resistor will
get very hot, indeed, and will probably melt plastic. For safety's sake I
would probably use a 1/2 watt resistor. I have been looking for a good,
inexpensive LED driver which will not waste all the supply power in the
resistor.
Walt Jopke
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