# Measuring small currents at DC high voltages without floating meter

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?

• 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. – DKNguyen Oct 16 '19 at 1:23
• I assume just clipping a battery multimeter to the shunt resistor wont work for you? You need waveform data? – DKNguyen Oct 16 '19 at 1:30
• 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. – Gavin Uberti Oct 16 '19 at 1:39
• Use discrete leds and photodiode spaced apart in air, like a TV remote, far enough so 40kV cant jump – DKNguyen Oct 16 '19 at 1:40
• 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 – DrMoishe Pippik Oct 16 '19 at 1:42