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# Physics/voltage multiplier versus a van de graaff

Question
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

ANSWER: A voltage multiplier uses a ladder of capacitors and diodes to store and transfer charge in an AC circuit to a terminal.  A Van de Graaf generator accomplishes exactly the same task using a belt to physically transport the charge.  A Van de Graaf does not complete a circuit, either, charge is put on at one end of the belt and taken off at the other end.  They're not completed circuits, so much as one-way charge transport devices.  The Van de Graaf uses the principle that charge will exit the belt and move to the surface of the high-voltage terminal to accomplish this one-way transport, the voltage multiplier takes advantage of the one-way nature of diodes.

Perhaps you can be more specific?

---------- FOLLOW-UP ----------

QUESTION: thank you,
to be more specific:
if one connects a wire to one terminal of a low voltage DC generator and extends that wire out away. there would not be any charge build-up in the wire unless it is brought near a grounded terminal, as in a capacitor.
now, regarding the differences between a van da graaff and a voltage multiplier. in the case of a >25kV voltage multiplier this would remain the same case (no charge build up unless a grounded terminal is near). but with a van da graaff, the charge does build up regardless of the grounded terminals proximity.
is this correct so far?
THANK YOU,
Gene

ANSWER: No, there's always a grounded terminal.  The surrounding environment provides that in either case.  There's no difference.

---------- FOLLOW-UP ----------

QUESTION: thank you,
so, i can not see why (in a voltage multiplier) the current would go through the whole circuit all the way to the end electrode, to produce the top voltage build up of charge on the end terminal (in an open circuit). the current has no reason to flow there through all the capacitors to the end; not unless there is another terminal near the very end tip. the omnipresent ground that you refer to would be there, but it would pull the current evenly out at every single point of the circuit, and least at the end. so yes the ground effect would build up some charge on the end tip but a tiny fraction of the maximum.  Whereas with the va de geraaff, the current is forced into the end terminal and so it is at maximum regardless of ground proximity. is this correct?
THANK YOU,
Gene

http://www.rmcybernetics.com/images/main/pyhsics/voltmultip22.jpg
No, see above, there's an AC source.  The charge is transported in "buckets" from capacitor to capacitor up to the top.  Wikipedia has a thorough description.  http://en.wikipedia.org/wiki/Voltage_multiplier  The current cannot flow backwards through the diodes.  You need to think about it in stages from when it goes from high to low, and see where the charge goes.  Actually, having read it, the wikipedia page spells out the steps under the "operation" section quite well.

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