Voltage multiplier output power

[Darius Baronas]

Hi! I have 880mV 8Hz AC signal and doing two experiments : diode bridge and voltage multiplier rectification and output power measuring depending on load. So i connected diode bridge and precision potentiometer as a load. Measured current and voltage and got maximum power with 500Ohm load (240mV * 0.97mA = 232.8mW) . The did same experiment using voltage quadrupler scheme https://upload.wikimedia.org/wikipe...adrupler.svg/240px-Voltage_quadrupler.svg.png . Soldered same diodes as in diode bridge. With that voltage doubler i got max power with 2k load , and power was (1,16V*0,58mA=672.8 mW) . Why the power is much bigger compared diode bridge ? Please help, thanks!

 

[Vbase]

If you add the power on the resistor to the power on the diodes you may get more similar total power. You cannot ignore the power on the diodes.

 

[Darius Baronas]

Yes, but diode brigde have 4 diodes and one capacitor, voltage multiplier have 4 diodes and 5 caps. So why the power difference is so big ? How to know what is maximum power from my AC source ? And what is maximum power after rectification ?

 

[Vbase]

What I was saying is add the power on the diodes, not the number of diodes.

 

[Darius Baronas]

But voltage multiplier have 4 diodes too..so here there are power on diodes in bouth diode bridge and multiplier. Do not understand..

 

[Vbase]

If your source has 0 output resistance then the power you take from it depends on the load resistance only. If the voltage doubler has lower total resistance than the bridge rectifier then the power it takes will be higher. It is as simple as ohms law.
The total power taken from the source is the power on the load resistor + the power on the diodes. You measured the power on the resistor because it is easy, on the diodes it is more difficult to measure the power. To get more accurate measurement you have to calculate the power on the diodes which is voltage on the diode X current.