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i have following doubts about capacitive power supply or transformerless power supply...

what will be real power drawn, will it be 230 x (current flowing through C1) or 6 x (current flowing through LED)...and explain how it would be...

how it is different that if resistor is used of equal impedance instead of C1...i mean resistor will just take power as heat... but wont it be the same with capacitor which dont take power as heat but consuming the same amount as a resistor...

can i assume C2 and zener diode open circuited and why...enter image description here

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    \$\begingroup\$ this is a dangerous and potentially lethal circuit, because there is no isolation from the ac mains neutral. \$\endgroup\$
    – MarkU
    Sep 14, 2015 at 4:26

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If you put a capacitor across an AC supply it might take (say) 1A. If you put a resistor across the same supply and the resistor also took 1 A there would be real power dissipated in the resistor and your energy billing company would start charging you. With the capacitor, the current is exactly 90 degrees shifted so that the average energy entering the capacitor circuit is zero and your energy billing company would not charge you one penny.

That's how it is with resistors and capacitors. For resistors current is volts/resistance but for capacitors current is C.dV/dt and this means that if you differentiate a voltage you get a cosine wave for current (i.e. shifted 90 degrees).

If you dig around you will find out that a sine wave multiplied by a cosine wave has an average value of zero i.e. the average power taken is zero by a capacitor (or inductor for that matter).

Please understand this fundamental electrical principal before moving onto the other parts of the circuit.

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  • \$\begingroup\$ thank you sir for answering my doubt...can you answer my second question that can i consider zener diode and c2 as open circuit \$\endgroup\$
    – AkshayM
    Sep 15, 2015 at 12:49
  • \$\begingroup\$ You can't consider the zener as open circuit because this type of supply relies on the zener to clamp the voltages to the load. Without the zener C2 would continue charging until it exploded on over-voltage. \$\endgroup\$
    – Andy aka
    Sep 15, 2015 at 13:00

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