A simple yet effective circuit for a single LED on AC Mains....
Step 1: Parts & Assembly
1. An LED - 5mm or 10 mm of any colour
2. A diode, preferably 1N 4007
3. A resistor of 2 watts or higher rating, of value anywhere from 22 kilo ohms upto 100 kilo ohms.
4. A two pin male plug (hollow)
Lower resistor values will give more brightness and higher values will prolong LED life. So choose the tradeoff, depending on requirement.
Lower wattage of resistor like 1/2 watt or lower will not do and may burn since they are meant for 6v DC circuits, not for 220v AC Mains.
1. Connect black anode of diode to negative of the led.
2. connect the resistor to the positive of the LED.
3. Connect the free ends of the diode and resistor to the male pins.
Done. See the attached pic for clarity.
Another circuit with the diode connected "across" the LED is also attached. A bulb base adaptor is used instead of a male pin.
It should work on 110v AC as well as 220v AC Mains systems.
It will also work on any battery!!
Once again, read "Important" first line, in case you missed it....
Step 2: Edit: April 2014
This instructable is becoming popular and common sense tells me that despite the "important" disclaimer about being proficient in working with AC Mains, many new people may also venture into making one.
So few guidelines as below:
Since only two components are used, both of them have to be robust to withstand AC Mains load.
1. The diode that i have used is 1N4007.
This 1N4007 diode is rated for 1000 volts and hence is robust to withstand both 110 v AC as well as 220 v AC Mains that different countries follow. Its also readily available, i guess.
2. The resistor 22 kilo ohm and above, should be rated for at least two watts or higher.
This will ensure that it will not burn due to continuous operation. A resistor is a current limiter. It limits the AC Mains current and converts it to heat and thereafter dissipates it.
HEAT is a problem, since a capacitor is not used. However, since a higher wattage resistor is used, it will not fail. Lower wattage resistors will fail and can cause damage.
Please be responsible and thoroughly test your work.
3. I could not use a capacitor ( non-polarised ceramic capacitor for AC Mains ) since space was a constraint inside the male pin.
If possible, use one:
--> 0.22 uF or 224 K, rated for 400 volts, if you are on 220 VAC Mains
--> 0.47 uF, rate for 250 volts, if you are on 110 VAC Mains.
See my other Instructables, for projects that use more LEDs and therefore a capacitor. See also various other circuits in this website as well as internet.