@Junior posted:...
Okay....I believe I finally found what I was looking for....sort of.
Like you say, "sort of" 
. What's described is a simple technique (a 5-cent diode) to reduce the voltage applied to an incandescent bulb. The lamp is dimmer, the bulb lasts much longer, and operates much cooler so nothing melts. See if these pictures help.
On the left is the original situation. An incandescent bulb is not polarity sensitive and it "lights up" on both the positive and negative halves of the AC voltage. A diode blocks half of the AC signal so only one polarity goes through; in the example above with the diode oriented as shown ("band" on the right) only the positive voltages get to the bulb. So the bulb is powered essentially half of the time as before so it is dimmer and cooler. BTW, in the pictures I only show the "hot" side of the circuit; the power supply common is assumed to be connected to the common of the bulb.
Your situation is slightly different.
You have an LED (instead of an incandescent bulb). An LED is a Light-Emitting-Diode so it is a diode in itself. Hence, unlike a bulb which lights up on both halves of the AC voltage, the LED only lights up on the positive halves of the applied voltage. So the negative voltages do not contribute to the LED brightness. And to add insult to injury the negative voltages do not play well with the LED and eventually destroy it. So in your case, the external diode is providing the function of blocking the negative voltage from reaching the LED. And since the LED only lights up on the positive pulses anyway, the lack of negative pulses does not affect the brightness.
The two cases are similar in many ways in that the external diode is extending the life of the light...but via slightly different electrical "mechanisms."
Let's take your situation at "face value." That is, everyone agrees that there are LED lights out there that have the diode built-in ... but that is not the matter at hand. But if you have extracted the "hot" wires to your "unprotected" LEDs then you could indeed bus them and apply a DC power supply so that the LEDs only see positive voltages.
I don't know how difficult it is to perform surgery on the switches to extract the "hot" wire to your LEDs. That is not what I'm commenting on. What I was saying is a single 5-cent diode (e.g., 1N4001, 1N4002, 1N4003, etc.) would allow you to use whatever AC voltage you had been using.
Hope this helps...
