Description
Source - http://www.bowdenshobbycircuits.info/
The circuit below illustrates powering a LED (or two) from the 120
volt AC line using a capacitor to drop the voltage and a small resistor
to limit the inrush current. Since the capacitor must pass current in
both directions, 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
subsituted 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 value, 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 average. 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.
Circuit DiagramSource - http://www.bowdenshobbycircuits.info/
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