Capacitive power suplply

capacitivepowerpower supply

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

Best Answer

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.