Here is how I did it. Note the use of two resistors. Each dissipates ~0.52W, so 1W rated resistors are ok. Note that the voltage at the ends of the LED are near ground potential (in the US). This is what I had in mind when I posted this:
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Mike:
That circuit needs a series diode as well which will halve the power dissipation without losing any light output. With only one anti diode the resistors have to work twice as hard for the same light output. Adding another diode in series corrects this and so the power in the series resistor drops by half from about 240^2/R to about 240^2/(2*R)...
Oh, but the aggravation of having to get on the internet, or phone, and pay shipping... I get a perverse pleasure of doing a project out of what I have in the junk box...Hi Mike,
Yes, and they are about 7 cents each
There was another type (actually a subset of your "zero diodes version"), but the LED type required don't seem to be available anymore. I assume you're familiar with 2-lead bicolor component LEDs (usually red/green, but other combinations are available), which are actually two LED dice in anti parallel in a single package -- switch the polarity of applied current to change the displayed color, or apply AC to view a flickery mix of both colors. Some years ago there were available, LEDs made exactly the same way, but having two dice of the SAME color in anti parallel. One of those would also require only a series resistor to operate from mains AC power -- if you could find one.Then we have the zero diodes version. This requires two LEDs however, connected in anti parallel. The series resistor always conducts, but when it conducts it always produces light in one of the LEDs.
Sometimes capacitance is included to help filter the DC and so supply a smoother current to the LED which reduces motion flicker.
Also, sometimes a series capacitor is used to help reduce power consumption and get higher brightness. This is a bit more complicated and is usually not needed in an indicator if the LED is a high brightness type.