I am creating a powerbase for an instrument that requires a potential difference that increases in discrete steps, and am attempting to power it with a Cockroft Walton (CW) voltage multiplier circuit.

With 15 stages and an 800V anode-cathode operating voltage the required input to the CW is 27V. However one of the design parameters is that the input to the PCB can only be 3.3V AC. I decided to bridge this gap using an LC tank resonator, as shown in the diagram:


simulate this circuit – Schematic created using CircuitLab

Without the LC tank resonator attached, the CW behaves as expected with acceptable efficiency dropoff, measured from the end of each resistor at the stages in the diagram relative to the ground at the end of the CW (positive polarity).

When L2 and C1 only from the diagram are added, driven by a frequency generator in series with them, the CW performs unexpectedly, whereupon at some frequencies the middle stages will be at a higher potential difference relative to ground than the earlier stages. The graph below shows how some of the different stages (stage X is labelled DYX, stage 15 is 'anode') behave around the resonance point of the LC circuit, which as expected from basic theory of LC resonates at 100kHz.

enter image description here

I understand that this behaviour is all related to the interactions of the capacitors and inductor but can't find a mathematical model to explain it. When I take a particular CW stage and treat all the capacitors between it and the inductor as the single capacitance of the LC tank 'seen' by that stage (e.g. at R3 combining C2 and C3 (series) with C1 (parallel) into a single value), the new resonance frequencies are well below what I see in the tests. Simulations with LTSpice do not show any of this unexpected behaviour, and only pick up on the resonance of the initial LC circuit.

Based on this I am sure that my assumptions about the different resonance points are incorrect or incomplete, and I am confused as to how the middle stages could at any time be at a greater potential difference than the 15th stage. Because of this I am unsure how to fix the issues in the next iteration of the design. Any suggestions would be greatly appreciated.

  • \$\begingroup\$ Exactly how did you measure the output voltages? What does p_d./v represent? \$\endgroup\$ Commented Jun 22, 2022 at 11:44
  • \$\begingroup\$ @BruceAbbott used a multimeter on dc setting between ground connection at end of CW, and test node above resistor, indicated on circuit diagram. The p.d./V is the potential difference between these two points for different frequencies measured in volts. \$\endgroup\$
    – Sellsy
    Commented Jun 22, 2022 at 15:02
  • \$\begingroup\$ All the voltages on your graph seem very low. What is the input resistance of your multimeter? (if you don't know then what make/model is it?). What is the coupling factor of your coils? \$\endgroup\$ Commented Jun 24, 2022 at 20:02
  • \$\begingroup\$ @BruceAbbott I'm using the fluke 289 multimeter, I had a scan of its datasheet and couldn't find a value. I had presumed (perhaps errouneously!) that for potential difference measurements this would be negligibly high. The coupling between inductors is interesting as I tried to estimate it by comparing to Spice sims with different values, however, as the circuit generally does not behave according to Spice (e.g. the p.d. measurements) I expect any estimates I make to have a large error. I've also found the mag fields are interfering with the rest of the board so I'm redesigning the coils. \$\endgroup\$
    – Sellsy
    Commented Jun 27, 2022 at 10:38
  • \$\begingroup\$ Manual says 10 Megohms <100pF. According to my LTspice simulation, 10M is a significant load on the higher voltage stages. I suspect the meter is damping/detuning the CW by different amounts depending on which stage is being measured. \$\endgroup\$ Commented Jun 27, 2022 at 10:55

1 Answer 1


Interesting approach. Here's a back-of-the-envelope calculation.

Think about the LC tank circuit you have designed. Empirically, we'll assume 15 volts on the resonant circuit since you have measured a peak voltage at about 15 volts. Using the equation for energy contained in a 680 pF capacitor, the peak energy in the tank circuit is 76.5 nanojoules.

So we'll be generous and pretend that we can deliver half of this energy 100,000 times per second at our frequency of 100 kHz. This yields 3.83 mW that can is available from the resonant circuit. At your peak voltage, you will have 250 microamperes to run your circuit.

This current must charge all of the capacitors in the string and charge them to a point where they can turn on the diodes. You simply do not have enough power to accomplish this.

An LC tank circuit does provide the option of providing higher voltage, but at a price - there is no way to create additional energy from resonance alone. There is really no advantage over a design that runs directly from a step-up transformer, either to get you to your 27 volts or all the way to your 800V . No matter what you do, you will have to provide the all of power at 3.3 volts capable of running your anode.

Good luck!

  • \$\begingroup\$ Thank you for the response, there is a lot here that I hadn't considered. Hopefully I can improve it at the next iteration. Will update when I've had a chance to build and test. \$\endgroup\$
    – Sellsy
    Commented Jun 23, 2022 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.