On a somewhat different note: Looking through some experimental literature on oil/water separation using using electric fields. There is AC, sine and square, 50Hz to 50kHz, DC and pulsed DC, so I think the starting experiment is with AC, 60Hz. Experiments showed little affect of frequency on separation efficiency. I'm seeing glass vessels wrapped with a copper foil and an electrode in the center of the glass vessel in the solution. Applying the HV to this sets up the electric field, but I expect very little current. The AC out of the neon sign transformer isn't to bad to measure as the transformer output impedance is 295K, although 1000V open circuit measures 971V when loaded with a 10MΩ meter. ( I did build a 90MΩ prob e) I'm curious to know what current the Vessel with the electric field is drawing? So, I'm thinking of doing the usual, putting a resistor in series with the load and measuring the voltage drop across the resistor. In this case the current is so low, I'm thinking 1MΩ resistor in series will only drop 1V at 1ua, and I suspect current to be lower than that. Does that seem like the way to do the current measurement? Living vicariously through my son's experiments at his work, Mikek
Fun fact: I've seen the neon sign transformer say 30ma, impedance protected. The transformer we're using is 9000V at 30ma. I measured the output impedance at 295kΩ.
9000V / 295kΩ = 30.5ma, that could not have come out better! :-)