#### Guy Lavoie

##### Active Member

Using a 12k resistor would work, but the resistor would be dissipating a bit of heat. Assuming that the 1.5 v voltage drop is negligable for now, 120 V x .01 A = 1.2 watts. Yet the label on the nightlight says 0.1 watts, so its not using a resistor.

The other way to drop the line voltage would be a series capacitor. This has the advantage of not dissipating heat, but by reflecting the unused power back to the power line. The impedance of a capacitor varies with frequency and the impedance decreases as frequency increases according to the formula:

impedence (ohms) = 1 / (2 * pi * f * C)

where f = frquency in Hz and C = capacitance in farads. with this forumula, a .22 uF cap would have an impedance of 12.057k at 60 Hz, to get about the same 10 mA current through the LED when it is forward biased:

1 / (2 * 3.1416 * 60 * .00000022) = 12057 ohms

Now if we calculate the impedance effect that this has on the 120 kHz frequency that X10 uses on the powerline:

1 / (2 * 3.1416 * 120000 * .00000022) = 6 ohms!

Amy wonder why the X10 signal is getting killed?