Here is simple circuit of 220VAC operated LED. So please can anybody explain the actual power consumption. And how to measure cos phi for capacitor and what to do with Resistor.

Well, a little arithmetic (from the school) ...

We are using the equates from the posts above.

The .27 uF (2.7e-7 Farad) capacitor has an impedance at 50 Hz:

Xc = 1/(2 * PI * f * C) = 1/(2 * PI * 50 * 2.7e-7) = 11,79 kOhm = 1.179e+4 Ohm

We calculate with the (rot) LED forward voltage = ~1.6v, the two Graetz bridge diodes = 2 * .7 V, so

the sum is 1.6 + (2 * .7) = 3 V. At 20 mA the circuit is assumed as a resistor

with value R = U/I = 3/.02 = 150 Ohm

The whole (sum) R in the circuit is 1k + 150 Ohm = 1.15e+3 Ohm = 1.15 kOhm.

Now we can calculate the complex impedance:

Z = SQRT ( R * R + Xc * Xc) = sqrt(1.15e+3 * 1.15e+3 + 1.179e+4 * 1.179e+4) = sqrt (1.403e+8) = 1.185e+4 Ohm

The effective current, I = U/R = 220/ 1.185e+4 = 1.857e-2 = 18.57 mA.

The effective power is dissipated on the resistive devices only (sum 1.15 k, see above) beacuse the C is really lossless.

So

**P =** I * I * R = 1.857e-2 * 1.857e-2 * 1.15e+3 = 0.397 W =

**397 mW**About the 1 k resistor in the circuit:

It is a

**must** beacuse it decreases the worst case peak current. If the circuit is switched on at the moment of the peak input voltage (in this case it is 310 v !!), and the capacitor is empty (working as a shortcut), the current is limited with the resistive devices only). Calculate with R = 1.15 k, the peak current:

Ip = U/R = 310 /1.15e+3 = 0.2695 A = 269.5 mA !!

About the cos PHI ... you can it calculate if needed :-)

An other thing: use a 470k.. 1MOhm resistor parallel with your circuit to discharge the capacitor after the switching off!

Hope this helps

zuisti