Light-Emitting diode connected to 120/240 VAC
Caution: This circuit is connected directly to the power line (120 / 240 volts AC), so you must take special care to test it.
This circuit shows one or two LEDs (Light-Emitting Diode) directly connected to the outlet (120 VAC or 240 VAC). The reduction of the AC input voltage to one that is suitable to use in a LED diode is achieved using a capacitor and a resistor.
When two Light-Emitting Diodes are connected, the first LED diode will conduct in the negative semicycle of the wave and the second one will conduct in the positive semicycle of the wave. If you want to connect only one LED diode you must replace the other one with a common rectifier diode. If this is not done, the LED diode, will be burned.
The 1 Kilohm resistor is used to avoid possible current peacks.
LED diode connected to 110/120 Volts, 60 Hertz
With a 0.47 uF non-polar capacitor the reactance will be 5643 ohms, and the current through the LED (or LEDs) will be 21.3 mA (milliamps).
LED diode connected to 220/240 Volts, 50 Hertz
With a 0.22 uF non-polar capacitor the reactance will be 14.468 ohms, and the current through the LED (or LEDs) will be 16 mA (milliamps)
The formulas that were used are:
- Xc = 1/(2 pfC). Capacitive reactance formula.
- I = V/Xc. Ohm’s Law for the capacitive reactance
- p = 3.1416
- f = frecuency (50 or 60 hertz)
- C = capacitor value (farads)
- V = voltage
- I = current
- Xc = capacitive reactance
Note: the effect of the 1 kilohm resistor is negligible as most of the voltage drop is in the capacitor.
List of components
- 1 1K resistor (0.5 watts)
- 1 0.47 uF non-polar capacitor, 200 volts or more for the 100/120 VAC case.
- 1 0.22 uF non-polar capacitor, 300 volts or more for the 220/240 VAC case.
- 2 LEDs
- 1 1N4001 rectifier diode