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I'm doing a project with a Cockcroft-Walton multiplier that will output up to 40 kV DC at up to 10 mA. The the circuit takes an input of 10 kV AC at 60 mA and 60 Hz (from a neon sign transformer) and passes it through a two-stage full wave Cockcroft-Walton multiplier with 50 nF capacitors.

I'd like see the current waveform my load is drawing, and due to the nature of what it's powering (a fusor), it's not possible to measure the current to ground. It might be possible to measure the AC current at the neon sign transformer (or before the neon sign transformer) but I fear the current waveform could be "mangled" if I measure before the Cockcroft-Walton. the voltage droop of the multiplier is also nonlinear, so I can't just measure the output voltage and calculate the output current.

Normally, I would just add a 100Ω resistor to the output of the Cockcroft-Walton, and measure the voltage drop across the resistor (since a voltage drop of 1V would correspond to a current of 10 mA), which I would then feed into an Arduino.

But there's my problem - to make this work, I'd have to let the Arduino's internal ground float to the ~40 kV of the output, which makes it really hard to power and read data from it. While I could power the Arduino with a battery and read data from it over Bluetooth, this is a pretty gross solution. I also thought about using an op-amp, but I don't think any op-amp could survive a 40 kV difference between its inputs and its outputs.

So my question is this: How can I measure the current waveform being supplied by my Cockcroft-Walton multiplier in a way that can be read by an Arduino at ground?

Edit: Added more details about the Cockcroft-Walton specifications

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    \$\begingroup\$ A hall current sensor? Those are inherently isolated. Not sure if you can find a 40kV one though.You could try and make a wireless current sensor that glows or blinks an LED proportionally and read its brightness or frequency or pulse width or duty cycle. Or some fancier method of IR communication with current measurement. \$\endgroup\$
    – DKNguyen
    Oct 16, 2019 at 1:23
  • \$\begingroup\$ I assume just clipping a battery multimeter to the shunt resistor wont work for you? You need waveform data? \$\endgroup\$
    – DKNguyen
    Oct 16, 2019 at 1:30
  • \$\begingroup\$ Yeah, I need waveform data (I'll update the question). As for the LED idea, what you're describing sounds a lot like an opto-isolator, but I couldn't find any that support more than 10 kV. A hall current sensor would also work in principle, but again I've been unable to find any that would work for 40 kV. \$\endgroup\$
    – Gavin Uberti
    Oct 16, 2019 at 1:39
  • \$\begingroup\$ Use discrete leds and photodiode spaced apart in air, like a TV remote, far enough so 40kV cant jump \$\endgroup\$
    – DKNguyen
    Oct 16, 2019 at 1:40
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    \$\begingroup\$ As @DKNguyen, a standard method is to use an optoisolator with fiber optics -- be sure the light-pipe does not have metallic reinforcement, and that the photodetector is well grounded. See instructables.com/id/Opto-Isolator-Homemade \$\endgroup\$
    – DrMoishe Pippik
    Oct 16, 2019 at 1:42

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I came up with this way of current sensing on the secondary 20:1 step-up transformer ground side with a shielded non-inductive 10mOHm shunt.

enter image description here

Although 1mohm arc current is a major crosstalk issue with <50ps rise time, this output arc discharge to air gap gives 10% of the voltage secondary 10mOhm shunt. HV diodes and plastic caps with wise RF/HVDC clean construction are essential. 0.3m BIL 200kV x2 in series, ceramic bushings minimum over a teflon sheet baseplate is what I used. Even with a grounded safety cage, the paint gave off static discharge 5m away and you could feel the hair on your neck rising. Much more risky than 50kV AC due to dust.

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  • \$\begingroup\$ Would you mind explaining more how this shunt can be used to measure the current? I don't quite understand the circuit above. \$\endgroup\$
    – Gavin Uberti
    Oct 16, 2019 at 4:10
  • \$\begingroup\$ @GavinUberti The shunt is at he low side. If the supply is referenced (not floating) the voltage here will be V = I x R instead of 40kV. \$\endgroup\$
    – Jeroen3
    Oct 16, 2019 at 6:43
  • \$\begingroup\$ the ESR of an arc in a dielectric is inversely proportional to the current density to the discharge path. I chose a low Voltage ESR gap that might resemble the arc of boosting a dead. Car battery while starting with 12V and 129A spark density to the electrode. Except 50kV can creep 5 cm even alcohol clean insulation. \$\endgroup\$
    – Tony Stewart EE75
    Oct 16, 2019 at 11:47

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