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# Physics/voltage multiplier "vs" a van de graaff

Question
hello,
voltage multiplier versus a van de graaff
please tell me, in the case of a voltage multiplier, i can see no reason why the current should go to the terminal/electrode, unless there is a way for the current to complete the circuit; is this correct? that is, if a voltage multiplier (of >25kV) has one terminal in air and the other terminal is not near there will be no build up of charge on the end electrode; since why would the current go there. whereas with a van de graaff the current (using equivalent voltage) has to go to the terminal and builds up the charge regardless of where the other terminal is and does not need the complete circuit. is this correct?
thank you,
gene

Hello gene,

In the usual version of a Van de Graff generator, one sphere attains a negative charge and the other a positive charge. Electrons are transported by the moving belt. Check out the following web site:
http://en.wikipedia.org/wiki/Voltage_multiplier

The section titled Operation does a good job of explaining how an example of a voltage multiplier works. In the example the output terminal attains a positive charge. The voltage multiplier's equivalent of the Van de Graff generator's other sphere is maintained at neutral voltage. If you reversed the direction of the diodes in the voltage multiplier, the output would attain a negative charge. As wikipedia's explanation indicates, it would take a few cycles of the AC source to attain maximum voltage. Charges are flowing through the diodes during that time. Once equilibrium is reached, there is no current. (Except to replace the charge lost by the same type of leakage as the Van de Graff generator.) However, if you connected a load between the output and the neutral, current would flow. In order to maintain the output voltage, the process described in wikipedia's "Operation" section would work to replace the charges carried away by the current.

I hope this helps,
Steve

Physics

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#### Steve Johnson

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