The circuit in the figure illustrates to power a LED from the 120 volt AC line using a capacitor to drop the voltage and a small resistor to limit the inrush current. This circuit can operate one or two LED. There are three concept to operate the LEDs. Operation of this circuit is when the capacitor must pass current in both direction, a small diode is connected in parallel with the LED to provide a path for the negative half cycle and also to limit the reverse voltage across the LED. A second LED with the polarity reversed may be substituted for the diode, or a tri-color LED could be used which would appear orange with alternating current.
The circuit is fairly efficient and draws only about a half watt from the line. The resistor value (1K / half watt) was chosen to limit the worst case inrush current to about 150 mA which will drop to less than 30 mA in a millisecond as the capacitor charges. This appears to be a safe values, I have switched the circuit on and off many times without damage to the LED. The 0.47 uF capacitor has a reactance of 5600 ohms at 60 cycles so the LED current is about 20 mA half wave, or 10 mA averages. A larger capacitor will increase the current and a smaller one will reduce it. The capacitor must be a non-polarized type with a voltage rating of 200 volts or more.
The lower circuit is an example of obtaining a low regulated voltage from the AC line. The zener diode serves as a regulator and also provides a path for the negative half cycle current when it conducts in the forward direction. In this example the output voltage is about 5 volts and will provide over 30 milliamps with about 300 millivolts of ripple. Use caution when operating any circuits connected directly to the AC line.
The circuit is fairly efficient and draws only about a half watt from the line. The resistor value (1K / half watt) was chosen to limit the worst case inrush current to about 150 mA which will drop to less than 30 mA in a millisecond as the capacitor charges. This appears to be a safe values, I have switched the circuit on and off many times without damage to the LED. The 0.47 uF capacitor has a reactance of 5600 ohms at 60 cycles so the LED current is about 20 mA half wave, or 10 mA averages. A larger capacitor will increase the current and a smaller one will reduce it. The capacitor must be a non-polarized type with a voltage rating of 200 volts or more.
The lower circuit is an example of obtaining a low regulated voltage from the AC line. The zener diode serves as a regulator and also provides a path for the negative half cycle current when it conducts in the forward direction. In this example the output voltage is about 5 volts and will provide over 30 milliamps with about 300 millivolts of ripple. Use caution when operating any circuits connected directly to the AC line.
no capacitor needed
ReplyDelete1 diode across led
3 100k 1/4 watt resistors in parallel on one leg of led...
simple works...