## LED 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 **LED 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.

## Connecting the LED diode 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).

## Connecting the LED diode 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

Where:

- 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)

- For the 100/120 VAC case a 0.47 uF non-polar capacitor, 200 or more volts.

- For the 220/240 VAC case a 0.22 uF non-polar capacitor, 300 or more volts.

- 2 common **Light-Emitting Diodes**

- 1 common **rectifier diode** (1N4001)

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