dank
Seasoned Member
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pair of dual 18 Imperials
Posts: 423
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You will need to put a diode and a resistor in series with the LED. The diode needs to be there to prevent the LED from seeing too much reverse voltage (PIV) as LED's usually have a low PIV rating. The diode should have a PIV of 400 or more. 1n4007 is more than adequate. Note: if you put the diode in backwards the LED won't light and may blow from excessive PIV. Connect the diode anode (non banded end) to the 120 vac source, the resistor to the diode cathode (banded end), the LED + terminal to the resistor, and the LED - terminal back to the 120 vac source. You can find the + and - LED terminals with your ohm meter...on lowest ohms scale if you touch the leads directly to the LED, the LED should light when the + voltage from the ohm meter (red lead usually) is on the + LED terminal.
The resistor value is dependent on the LED efficiency and how bright you want the LED. LED efficiency varies considerably. I have a couple of these switches and my green one needs a 10k resistor while the blue one is quite happy with 100k indicating the blue is 10 times more efficient than the green.
I would suggest you get a few resistors and a 1n4007 diode and start trying different resistor values. If you get resistors that are 2 times each other (10k, 22k, 47k, 100k) you can build most any value you want. These 4 can produce most any value between 5.6k (all in parallel) and 179k (all in series). Start with a higher value of ohms and work down to go from dim to bright.
If you PM me your address I can send you these 5 components. They are just a few pennies but if you have to buy them quantity 1 you will spend 100 to 1000 times that.
Dan
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