Maximizing Efficiency: Comparing AC and DC Sources for Charging Capacitors

[physicsming]

A sine wave source charge a capacitor through a diode.

Is it more efficiency than a DC source charging a capacitor ?

base on this

https://www.physicsforums.com/attachment.php?attachmentid=17653&d=1235114409

 

[DragonPetter]

Seems the conclusions being made aren't quite right. You cannot transfer energy through a resistor to the capacitor without loss in the resistor no matter how much of the sine wave period you pass through it.

A DC source charging a capacitor will have less loss than a Sine source charging through a diode. You have a voltage drop across the diodes in the forward bias, and so energy is lost. Not exactly sure why you're comparing this to the figure you attached because that example is charging through a resistor, which holds true for sine and DC sources either way.

 

[physicsming]

thanks for your answer

i need to analyse the efficiency of a AC/DC charge pump,

so i need to know the power loss of a sine wave charging a capacitor firstly.

but the general thesis and data is only analysis the DC/DC capacitor charging loss.

In this paper "A new visit to an old problem in switched-capacitor converters"

propose that the while Vin from voltage source keep close to the voltage of capacitor

it will have high efficiency. the formula in the paper for analysis DC capacitor charging

through resistor is efficiency=1/2(1+Vci/Vin). Vci is the capacitor initial voltage.

Can i use this formula to analysis the efficiency of AC/DC capacitor charge by replace

Vin to Vrms ? thanks

 

[Dmytry]

usually there's inductor in series (before the diode bridge) for power factor correction purposes. With inductor you can avoid the loss.

When you are charging capacitor off sine wave through diode and resistor the loss will depend on the time when the circuit is turned on, as well as voltage and resistor and capacitor values. If the capacitor is small and resistance is very small and you turn circuit on at zero voltage, you have no losses (besides the diode's voltage drop). If you turn same circuit on at max phase, you can burn out the diode.