Step 1: Supplies
An incandescent light bulb.
A .47 microfarad 200V capacitor.
1/4 watt 1kilo ohm resistor
a pair of leds
and miscellaneous things.
Step 2: Assembly
The circuit consists of two LED's wired in opposition, I ground down the LED's just short of the die and glued them together to make a single double LED. Twist the legs of the LED's together, on one side solder the capacitor, the other the resistor. Simple.
Step 3: Test
Step 4: The Bulb
Step 5: The How What and Why
Now, how does this work? We all know that to run an LED off a higher than rated voltage source we muct limit the current with a resistor. Indeed in this case we could limit the current with a resistor of value approx. 6.8K ohms, however that resistor would need to dissipate several watts!!! Not a good thing.
Since we are using an AC source we can take advantage of a property of a capacitor subjected to AC called Reactance. We can equate reactance to resistance. Calculating the reactance is a simple formula
R=1/(2*Pi*Freq*C) Solving this for C will give us the size capacitor we need to limit the current to the LED.
So why do we have a resistor at all? When the power is switched on there is an in rush of current and the 1K ohm resistor is there to limit that in rush current.
Finally, Why two LED's? Well an LED is a diode and since we are dealing with AC here we need to wire two led's in opposition so that the waveform can complete it's cycle. Essentially each LED is flickering at 60HZ but in opposite phase.