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I want to make a voltage multiplier to gain a high voltage, I understand that you must input AC for it to work and you get DC out. I also understand that the multiplier is made up of capacitors and diodes.

What I don't understand is how you calculate what size capacitors you use. I have often just seen a value of 10nF; but what does that do, and can you use different value and what effect does it have? Are there any equations for voltage multipliers, even for calculating total output voltage? Say I wanted to make a voltage doubler to start off with; would I use 10nF?

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  • \$\begingroup\$ "I also understand that the multiplier is made up of ..." - Where do you understand these things from? \$\endgroup\$ – Anindo Ghosh Dec 12 '12 at 20:07
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    \$\begingroup\$ From research i have done, but through my research i haven't found the answer to the question i am asking. \$\endgroup\$ – Jack Trowbridge Dec 12 '12 at 20:19
  • \$\begingroup\$ Perhaps my comment was too round-about. My point was that if you are looking for good answers, your question needs to have better information and source links than just "I understand that ..." \$\endgroup\$ – Anindo Ghosh Dec 12 '12 at 21:12
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Higher the capacitance, more energy is stored, so be carefoul. As the roule of The rule of thumb is to take a range from 10nF to 100nF. And don't forget to add the load of at least 40MOhm.

Output voltage (once when it is settled) is 2*Vmax*N where N is the number of diode-capacitor sections.

You can find more explanation on:

http://www.cirvirlab.com/index.php/tutorials/90-diode-voltage-multiplier-circuit.html

and insted of calculating, you can simulate voltage multiplier on

http://www.cirvirlab.com/simulation/diode_voltage_multiplier_circuit_online.php

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    \$\begingroup\$ Great answer, but one question, why do you need a load of at least 40MOhms? \$\endgroup\$ – Jack Trowbridge Dec 13 '12 at 19:18

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