Introduction: LED on AC Mains
Important: Attempt only if proficient in working with AC Mains
A simple circuit diagram for a single LED on AC Mains 110v or 230 volts or even DC battery!!!
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.
Step 3: My YouTube Videos:
4 People Made This Project!
We have a be nice policy.
Please be positive and constructive.
Hi, You wrote:
>The resistor 22 kilo ohm and above, should be rated for at least two watts or higher.
I'm in 120v country. According to another YouTube video (not yours) a 1N4003 diode and a 47K ohm resistor will work (with 120V) for small "power-on" indicator light.
I don't need terribly bright light...and I want to avoid heat. I'm adding indicator lights to light switches, similar to the attached picture...only brighter.
Can you please respond back with your thoughts on resistor rating? Thank you.
1/4 wattage you will definitely risk burning, since you are connecting to AC Mains. Also, 2 wattage or higher will not affect brightness. They are just for higher heat tolerance, during continuous use, like TV, Projector, etc...So for safety reasons use higher wattage, even with 47Kilo-ohms resistor with lower light intensity... hope it helps...
another clarification: The factory-made indicator light, as pictured uses a zener diode which I cannot read the small print on, and a 820K Ohm resistor, NOT a 82K Ohm resistor as I stated previously. My bad!
Clarifying the above question: The other YouTube video to which I refer above suggests using a 47K ohm resistor __at only 1/4 watt__! You're suggesting use of at least a 22K ohm resistor of at least 2 watts.
SO my questions are:
Is a 1/4 watt ok, or will I risk burning it up.
And if I go with at least 2 watts, will I reduce light output by 16x?
FWIW, the factory version uses a zener diode and a 82K ohms resistor, but it's not nearly bright enough.