# Voltage Multiplier (Cockcroft Walton) Questions?

Discussion in 'General Electronics Chat' started by electronice123, Dec 4, 2009.

1. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
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If I use a voltage multiplier to charge a larger capacitor, and use a resistor to control the charging rate, will the discharge rate be the same as the charge rate?

Is it possible to keep the voltage at the large capacitor constant or prevent it from discharging? How?

What effect would internal arcing in the capacitor have on the circuit?

If I have a Inverter which outputs 1000V @ 5mA, at 35kHz to the multiplier circuit, how do I determine the capacitors and diodes I need to use to step up the voltage to say, 10kV?

2. ### SgtWookie Expert

Jul 17, 2007
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That's difficult to say. Will you be discharging it through the same resistor? Usually, CW multipliers are diodes and caps; resistors aren't in the circuit.

All capacitors have leakage current. Generally, the larger the capacitor, the greater the leakage rate.

A loud "bang", a bright flash, and possibly flying debris.

You realize that you're talking about potentially lethal power levels, right?

3. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
304
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The current flow does not change direction when the cap discharges.... So it should discharge at a rate equal to the time constant is what I'm thinking. The resistor is going to be used to prevent damage to the smaller caps and diodes when the larger cap discharges.

I am not going to build this circuit until I get to that point in school. I appreciate the warning about high voltage. I seem to get that alot as HV has been my greatest interest.

Basically what I'm trying to understand, is if it's possible to use a multiplier to charge a larger capacitor and keep it charged.

4. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
304
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OK, so if I have a multiplier circuit which uses several stages of .02uF caps. What would be the result if I used this circuit to charge a much smaller, 40pF cap?

5. ### studiot AAC Fanatic!

Nov 9, 2007
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One way to build high voltage circuits is to arrange a series 'stack' of components - diodes, capacitors, resistors so that the voltage across individual components is much lower.

This has the advantage of using cheaper more readily available components that trying to achieve the step up in one go.

Since energy is not added by the multiplier it follows that the current available will diminish in inverse proportion to the voltage increase - a bit like a transformer.

Oh and 40p is way too small to be useful.

6. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
304
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My last question, how do you determine the capacitance of the capacitors used in a multiplier?

I have read that the values are directly proportional to frequency.