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As far as understand, a voltage multiple increases the voltage but decreases current.

Can someone please explain why current decreases?

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    \$\begingroup\$ Conservation of energy. \$\endgroup\$ Jan 15 at 20:47
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    \$\begingroup\$ P=V*I, if the voltage rises the current must decrese \$\endgroup\$ Jan 15 at 20:48
  • \$\begingroup\$ And on top of that, due to efficiency of any circuit cannot reach 100%, with the power you feed into a circuit, you get less out. \$\endgroup\$
    – Justme
    Jan 15 at 21:04
  • \$\begingroup\$ It doesn't so much decrease the current going out, as it increases the current coming in \$\endgroup\$
    – user253751
    Jan 15 at 21:54
  • \$\begingroup\$ Simply put the string capacitance decreases in series and thus impedance rises, for each Xc and diode Rs thus load Z must also rise \$\endgroup\$ Jan 15 at 22:19
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A voltage multiplier has a lot of capacitors. Capacitors have impedance. The capacitors are needed to add up the voltage, but the also limit the current that the multiplier can pass.

Conservation of energy plays a role. Since the voltage multiplier produces a continuous output, the power out can't be more than the power in. That means you have to get less current when the voltage goes up. That's a hard limit you can't get around.

You can get around the impedance problem to some extent. You can use larger value capacitors, or higher frequency drive, or both. The result is lower impedance and more current output.

The impedance of a Cockcroft-Walton multiplier as you showed in the question can be calculated from the following equation:

$$Z = \frac {4n^3 + 3n^2 - n }{2 \pi fC} $$

Where Z is in ohms, f is in hertz, C in farads, and n is the number of stages.

As you can see, the impedance goes up drastically as the number of stages increases.

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  • \$\begingroup\$ Thanks a lot for the comment, very informative. I played with the formula and seems that if you use a 1 µF condenser with a 200 MHz AC, a three stage multiplier will only cause one Ohm of impedance. Is this expected? It will preserve most of the current. \$\endgroup\$ Jan 16 at 3:56
  • \$\begingroup\$ Sounds reasonable. 200MHz is a rather high frequency. You'll have basically a radio transmitter if you use that frequency. You'll need diodes that can handle the frequency and the current. Good luck finding the right combination. \$\endgroup\$
    – JRE
    Jan 16 at 9:07
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There's the physics answer, conservation of energy.

Or there's the mechanistic answer. The output current flows in all the stages in series. The input current flows in all the stages in parallel. Therefore the input current will be larger than the output current.

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  • \$\begingroup\$ Could you please explain what you mean by "The output current flows in all the stages in series. The input current flows in all the stages in parallel"? I don't see it in the circuit. \$\endgroup\$ Jan 15 at 21:40
  • \$\begingroup\$ It might be good to simulate it for yourself and see. Then work it out for yourself with pencil and paper. Wave your hand up and down to help you visualize the varying voltage -- that always increases your reputation for sanity if you do it in public. \$\endgroup\$
    – TimWescott
    Jan 15 at 23:02

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