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I currently have a power supply that will be able to supply around -60 kV DC and 10 mA under the load that I will be hooking up to it. However, the internal resistance of the load will fluctuate and may not be pulling the full 10 mA at all times, thus, a means of monitoring the current flow will be needed. The tricky situation is that I will need to connect the negative power line from the power source directly to a multi meter that will measure the current. Please look to the graph below to reference the design:

schematic

simulate this circuit – Schematic created using CircuitLab

The negative side of the power source will provide the -60 kV DC, which must be manageable for the multi meter. Thus, I was curious what I could do to inexpensively allow a multi meter or ammeter to handle such high voltages and such a high power while still being able give an accurate measurement of the current.

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  • \$\begingroup\$ Is it arc proof? \$\endgroup\$ – Sunnyskyguy EE75 Sep 12 '18 at 2:18
  • \$\begingroup\$ What you show simply has R as a load, I'm sure that's not what you intend. What is the load? \$\endgroup\$ – Jack Creasey Sep 12 '18 at 2:19
  • \$\begingroup\$ Analog or digital? 50mV/50uA galvo with 10mA full scale uses 50mV/10mA= 5 ohm shunt. For 200.0mV DMM, you can use 200mV/10mA=20 Ohms shunt Min R load is 60kV/10mA=6MOhm unless you get Partial Discharge. (PD) \$\endgroup\$ – Sunnyskyguy EE75 Sep 12 '18 at 2:24
  • \$\begingroup\$ If these specs are true max, you can drop up to 10V for 100mW max shunt loss. \$\endgroup\$ – Sunnyskyguy EE75 Sep 12 '18 at 3:54
  • \$\begingroup\$ But what is the purpose of this tool if you already have a current meter. A ground shunt gives the same thing with voltage \$\endgroup\$ – Sunnyskyguy EE75 Sep 12 '18 at 4:00
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I am not sure where the problem is. If you use, say, 1-Ohm shunt resistor (of any size) as on the diagram below,

enter image description here

and connect a DMM in mV-measuring mode, you will have 1 mV/1 mA. The DMM is safely grounded, and no need for any kV-grade probes. The voltage applied to your fluctuating load will be just 10 mV less than the 60,000,000 mV, or just like 0% error. I don't see why you can't connect your current monitor this way.

You can add two regular diodes to protect DMM an shunt if you want in case or problems with parasitic capacitance on the LOAD during power-on/connect sequence.

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  • \$\begingroup\$ 1 Ohm gives too low a resolution for 3.5 digit DMMs try 20 Ohms for 199.9mV FSD \$\endgroup\$ – Sunnyskyguy EE75 Sep 12 '18 at 3:52
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    \$\begingroup\$ @TonyEErocketscientist, okay, make it 5 Ohm. Or 10 Ohms. Same difference. And the OP didn't mention any excessive requirements for accuracy of measurements... and there are better DMMs than 3.5 digits. \$\endgroup\$ – Ale..chenski Sep 12 '18 at 3:55
  • \$\begingroup\$ It could even be 1 kOhm for 100mW max shunt loss. The "tool" is simply a resistor. As you say, no problem. IF PD was an issue a bypass cap can be used with known cable inductance \$\endgroup\$ – Sunnyskyguy EE75 Sep 12 '18 at 3:56
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    \$\begingroup\$ @Transistor, it doesn't make much sense to have your instruments at deadly - 60 kV potential to ground, unless there is a very serious reason. Right now I really can't see that reason. \$\endgroup\$ – Ale..chenski Sep 12 '18 at 7:00
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    \$\begingroup\$ @Transistor, also looking at the full list of OP's questions, I am afraid we are dealing with X-Y problem. It looks like the best advice would be to stop all this playing with high voltages at 600-W power level before someone gets killed. \$\endgroup\$ – Ale..chenski Sep 12 '18 at 7:11
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enter image description here

The only sure fire way of measuring low current on a high-voltage line is with differential probes rated 80 KV (4 each 500M 20KV resistors, Caddock 0.025%) each across a precision (1%) 1K 100W-300W resistor which is in series with your load (the high wattage prevents arcing across the resistor). The source common mode voltage of 60KV is divided by 10,000 to get a safe 6 volt common mode voltage.

10mA would give you 10 volts across the current sense resistor and 1 mV at the TL052 buffer inputs and 10 mV at the AD524 inputs.

The AD524 gain is set to 1,000 and can be read with any DVM at the output pin as 1 volt/1 mA.

The AD524CD runs on +/- 15 volts. Because you are dividing the input by 10,000 the input impedance of the AD524 is an issue. I added a TL052 dual op-amp wired as a buffer with a gain of 10.00, to the AD524 inputs. The AD524 also has input and output offset trim and fine-gain trim.

Recommend using battery power to avoid common mode errors but safe for PC running LabVIEW if accumulating results.

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  • \$\begingroup\$ 10 mA current over 1k resistor gives you 10V across the shunt. Which arcing are you talking about? \$\endgroup\$ – Ale..chenski Sep 12 '18 at 3:05
  • \$\begingroup\$ There can be arcing across a small current sense resistor on a high voltage line if the current ramps up or down too fast, so best to have a larger resistor (longer) so less chance of arcing. It would not hurt the measuring devices, just the sense resistor. I have seen 300W 1K resistors explode all over the room because of a short while charging a capacitor bank. If the OP has a firm grip on the supply current then a 1K 1% 100W 75mm long sense resistor is ok. \$\endgroup\$ – Sparky256 Sep 12 '18 at 3:14
  • \$\begingroup\$ Why rated only at 40kV if the supply will give 60kV? BY the way the commercial available differential probe rated at 1.6kV has R1,R2 4Gohm \$\endgroup\$ – Marko Buršič Sep 12 '18 at 6:20
  • \$\begingroup\$ Greetings, how should I wire the TL052 as a buffer if I am looking to increase the precision? A drawing would be preferable if possible. \$\endgroup\$ – Super Nerds Team Sep 12 '18 at 15:20
  • \$\begingroup\$ Solder 4 500M resistors in series and cover with heat shrink tubing to make each 2G ohm resistor. Digikey has Caddock resistors. They are not cheap but they work. \$\endgroup\$ – Sparky256 Sep 12 '18 at 19:17
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Either use a shunt at ground potential (you can put a TVS or gas discharge tube across it to take care of surges) or use a battery powered meter at -60V that has Bluetooth or other RF communication of the voltage across the shunt at high potential. Again, the shunt resistor can have a TVS or gas discharge tube across it to protect the meter input in case of a short or surge.

If you want to put the meter at high potential you will likely want to surround it with an electric shield (as much as possible while still letting the RF out) to keep corona discharge from causing problems. Putting it deep in a box with one side open should work.

Here is an example of the kind of currents that can flow from a pointy electrode fairly well spaced from a grounded plane:

enter image description here

Graph from Diguang, Z., & Dexuan, X. (1990). Analysis of the current for a negative point-to-plane corona discharge in air. Journal of Electrostatics

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