Guy Lavoie
Active Member
I just got my Elk ESM-1 signal tester today and I wanted to solve one nagging problem I had with a specific circuit. After testing signal strength and disconnecting various devices, it turns out that the culprit was a nightlight that uses a LED and a large lens. It then dawned on me that this was not so surprising after all. To drop the 120 volt line voltage down to a voltage that won't burn out a LED, you must add series impedance. A quick calculation with ohm's law shows that to get about 10 mA through the LED (assuming RMS voltage and half wave rectified), you would need to put about 12k of resistance in series with the LED (120 v / .01A = 12000).
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?
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?
Thanks for this posting. I was having a hard time with an LED nightlight. I hooked it up to a lamp X-10 unit and was using it for a little light in the room in the morning. (I get up very early and my wife continues to sleep.) I was never able to get it to work in the plug that I really wanted it in. This past weekend I got a normal (christmas tree light) night light. Now it works great and where I wanted it. 
