Running LED by 220v

This circuit is simple way to running some parallel LEDs by direct 220v.
The output current depend on the capacitor and the resistor and can calculate from following equations:


The series capacitor in AC current work as a resistor called (Capacitive reactance)
in first you have to find it from following formula:


Xc = 1 ÷ (2π * f * C)


Where:

Xc is the Capacitive reactance in ohm
f is the frequency of mains voltage in Hz(usually 50 or 60Hz)
C is the capacitor value in Farad
V is mains voltage


Then add this Capacitive reactance to the resistor to find total resistor (Z):


Z= √( (Xc^2)+(R^2) )


Now you can use Ohm's law for finding output current:


I = V ÷ Z




EXAMPLE:

Find the output current for following circuit?


Xc = 1 ÷ (2π * f * C)

Xc = 1 ÷ (2π * 50 * 0.0000001)

Xc = 31847.1 Ohm


Z= √((Xc^2) + (R^2))

Z= √((31847.1^2) + (150^2))

Z= 31847.4 Ohm


I = V ÷ Z

I = 220 ÷ 31847.4

I = 0.00697A = 7mA

Since 5mm LEDs need to 5 to 10mA for running so this capacitor and resistor value is enough to running one LED, but if you want to running more parallel LEDs need to bigger capacitor for passing more current.

categories : AC 110v-220v Projects,circuits,LED projects