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russ_watters

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NascentOxygen

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Suppose your voltage doubler is powered from a 12v source, and is drawing 2A from that source. Even if it were 100% efficient, then at 24v output it could deliver only 1A. Otherwise, it would be outputting more power than it drew from the source, and efficiency would exceed 100%.

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jim hardy

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yes, the question should be addressed backward from OP.

Since at best, power in = power out,

and power = Volts X Amps,

to double Volts out,, you must either halve (OOPS ! edit - not volts, but ) Amps out or double Amps in.

Since at best, power in = power out,

and power = Volts X Amps,

to double Volts out,, you must either halve (OOPS ! edit - not volts, but ) Amps out or double Amps in.

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if you are discussing the doubling rectifier such as bridge (Delon) doubler

http://en.wikipedia.org/wiki/Voltage_doubler

then the maximum current depends on the value of capacitors, and indeed the current through load will be half of the current that flows from AC source into the doubler as half of the cycle the charge from the input waveform (current*time) is put into one capacitor and other half of the time into another*, while both capacitors will be discharging through the load simultaneously; if you consider the net charge, the net charge that will flow through the load is half of the net charge that flows into the capacitors from the input (the charge that flows through the load is equal to charge added into one capacitor, or the charge added into another), and so is the averaged current (which is simply total charge per time).

*not so neatly in practice as it will only consume spikes of current at the peaks of the original waveform.

edit: with regards to current through a specific resistor with and without doubler, that is not relevant. If you have a high value resistor, without the doubler the power consumption (from the AC source) will be lower than with the doubler. In particular the current through a high value resistor will double, while the current consumed from the source, and the power, will quadruple, if you add a doubler. The current after the doubler will still be half of the current consumed from source, but the current consumed from source will quadruple. (for low values the doubler may fail to actually double the voltage, depending to the capacitance and frequency)

http://en.wikipedia.org/wiki/Voltage_doubler

then the maximum current depends on the value of capacitors, and indeed the current through load will be half of the current that flows from AC source into the doubler as half of the cycle the charge from the input waveform (current*time) is put into one capacitor and other half of the time into another*, while both capacitors will be discharging through the load simultaneously; if you consider the net charge, the net charge that will flow through the load is half of the net charge that flows into the capacitors from the input (the charge that flows through the load is equal to charge added into one capacitor, or the charge added into another), and so is the averaged current (which is simply total charge per time).

*not so neatly in practice as it will only consume spikes of current at the peaks of the original waveform.

edit: with regards to current through a specific resistor with and without doubler, that is not relevant. If you have a high value resistor, without the doubler the power consumption (from the AC source) will be lower than with the doubler. In particular the current through a high value resistor will double, while the current consumed from the source, and the power, will quadruple, if you add a doubler. The current after the doubler will still be half of the current consumed from source, but the current consumed from source will quadruple. (for low values the doubler may fail to actually double the voltage, depending to the capacitance and frequency)

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