Someone asked me how voltage multipliers work, and I realized I didn't really have a clear understanding of them myself. I also wanted to learn how to draw diagrams in Inkscape, and I also wanted to further refine my brain's water analogy for electricity, so I've combined them all into this page.
So wires, of course, are modeled as pipes. In the water analogy, no pipe ever just opens up to the outside world. If you disconnect the power source from the rest of the circuit, the pipe ends will suddenly have caps on them, since a wire is always surrounded by air; an insulator. (Notice that capacitors are often abbreviated "cap", and the thing you stick on the end of a pipe is also called a "cap". This is probably the only place in the universe that the two definitions will be used side by side, so I will try not to use the abbreviation. Try not to get confused.)
The output of a transformer is an AC source, and I am going to model it as a piston that moves up and down.
A capacitor does not store charge. It holds an imbalance of charge. I've represented this as a spherical tank with a big rubber sheet across the middle.
As the piston oscillates up and down, the diaphragm in the capacitor moves back and forth. This is simply because the piston is pulling water from one half of the capacitor, and pushing the same amount of water into the other half.
The rubber sheet stretches as this happens, storing energy.
Although the water on one side of the capacitor can't go through the rubber and come out the other side, the same amount of current that goes into one side must come out the other, so we say that a current is going through the capacitor. This current can only flow while the rubber sheet is stretching, though. You cannot have a constant direct current flow through the capacitor unless the rubber sheet breaks, in which case the capacitor is dead.
If you then disconnected the capacitor (each end of which would then magically have caps on it, containing the pressure), and then connect the two ends together, the rubber sheet will snap back to the middle position, and the extra water in one hemisphere will rush into the other as fast as it can. This is what happens when you short-circuit a real capacitor. The excess charge on one plate rushes across the short as fast as it can to fill up the empty spaces on the other plate.
When I first learned about capacitors, I knew they were used for timing circuits, and somehow got this idea in my head (*cough cough Forrest M. Mims III cough*) that they charged up until dielectric breakdown, and that is how they were used to time things. This is not correct in the slightest! This would be equivalent to timing something by charging up the water capacitor until the rubber sheet rips in half. This is breaking the capacitor! You could never use it again after that! But silly little me didn't get it.
Diodes, obviously, are represented as little one-way flaps. You can push current through them one way, but not the other.
When the piston moves upwards, the capacitor's rubber sheet is stretched out.
But after the piston reaches its peak and starts to move back downwards, the one-way valve closes and the capacitor stays pressurized.
Now the piston can move up and down as often as it wants, but the valve won't open again and the capacitor will just sit there with the same pressure across it.
(The little gaps around the piston represent the source resistance, which is where the pressurized water flows when the valve is closed and the piston is moving.)
Now we add a second diode and capacitor to get a voltage doubler.
See how the second capacitor now has twice the stretch in the rubber sheet? The pressure from the first capacitor was stacked on top of the pressure from the AC source, and both "charged up" the second capacitor in series.
It turns out that the second diode now has a voltage that is switching the same as the AC source. We can then add another voltage doubler across that diode and do the same thing again.
Now it's easy to get confused. We aren't charging each successive capacitor to a larger voltage. Instead, capacitor 1 has 1x, capacitor 2 has 2x, capacitor 3 has 1x, capacitor 4 has 2x, and so on. each separate stage has the same voltage across its components. The reason that we get a quadrupling effect is because these stages are stacked on top of each other. The pressure difference across capacitor 2 is the same as the pressure difference across capacitor 4. But the pressure difference across both capacitors together is the sum of the two individual pressure differences. This is now quadruple the original peak voltage.
To get higher voltages, you keep stacking more stages on top. There are other details about such devices that I will not go into. You should now understand the analogy well enough to extrapolate yourself...
Mistakes or comments; email me.