Electronic – Cockcroft-Walton generator producing unexpectedly low voltages

high voltage

I'm trying to build a high voltage DC power supply from AC 220V mains.

At the moment I'm just trying with one stage of the depicted CW generator. The components I'm using are:

  1. 1N4007 diodes
  2. Panasonic ECWHC3F103J capacitors (3kV, 10,000pF)
    Or:
  3. 400V 0.1uF polycarbonate capacitors

When I use capacitors (3), the output DC voltage across one stage is 158V. Two stages is still 158V. What could I be doing wrong here?

When I use capacitors (2), the output DC voltage is 258V. What causes the difference in output voltages when I change the capacitors?

Shouldn't the theoretical output of one stage be 220xsqrt(2), and of 2 stages 2x220xsqrt(2)?

Also, how do I increase the rate at which the high voltage DC output can charge a capacitor bank? The maximum short term forward bias current of the 1N4007 diodes is 30A which seems to be sufficient.

Please ask if you need any photos of the circuit to diagnose the problem. Thanks in advance.

Three stage CW multiplier, from Wikipedia

Best Answer

The thing most people don't realize is that the current through a voltage multiplier is greatly limited by the impedance of the capacitors.

Your 30 nF capacitors have an impedance of over 100 kiloohms at 50 Hz. The effective impedance will be higher because there's more than one capacitor in each stage.

With that impedance, pretty much any load will pull the output down.

Even the leakage current of the diodes and the capacitors will be a problem.

Even the 100 nF capacitors have a high impedance at 50Hz - it'll be over 30 k per capacitor.

The total impedance of a Cockcroft-Walton multiplier goes up as the cube of the number of stages.

The impedance of a half wave Cockcroft-Walton multiplier is given by:

$$\frac {1}{2 \pi fC} (4n^3 + 3n^2 - n)$$

Where:

  • \$n\$ is the number of stages (1 for a Greinacher doubler.)
  • \$C\$ is the capacitance in farads.
  • \$f\$ is the frequency in hertz of your input signal.

I don't have an equation for a full wave multiplier. I think it'll be lower because some of the capacitors are effectively in parallel.

At any rate, the total impedance goes up quite rapidly with the number of stages.

For a single (half wave) stage, the impedance is six times the impedance of a single capacitor.

To get higher current and get closer to the theoretical output voltage despite the load, you have really just two choices:

  1. Use a higher frequency.
  2. Use larger capacitors.

Both reduce the impedance the capacitors.

Since you are working with line frequency, you'll have to use larger capacitors.

You also want to use the lowest possible number of stages.

  1. Use a transformer to get a higher AC voltage going in to the multiplier to reduce the number of stages.
  2. Use larger capacitors to reduce the impedance of the individual stages.
  3. Figure out the impedance of the load, and see how much current you need to reach the required voltage in the required charge time.

I spent some time playing with voltage multipliers a while back, and learned a good bit about them.

I suggest you do your experiments first with low voltage until you understand how things work before you move on to high voltage.

Once you have an understanding of how current and voltage play together, you can build your high voltage circuit safely - and have a good chance of it working properly on the first try.

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