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In a plasma application (Helium or Argon Plasma), we would like to measure the current. The voltage is around 10kV AC and modulated with a function generator.

Currently we only have one of those rings that you put around the cable, but I read that those only work for high-ish amperage >1A. While using this device, we get an output of around 12mV peak to peak, put I think that this is very inaccurate.

Is there any cheap way (we are students) of measuring the current more precisely? Preferably without killing equipment and/or operators :)

For a low voltage circuit I would use a high accuracy resistor and measure the voltage drop across, but I think that this is too dangerous in such high-voltage applications.

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  • \$\begingroup\$ Coil up the cable 10x, and clamp your current meter through the loop. This will produce a 120mV p-p signal. \$\endgroup\$ – Rocketmagnet Feb 22 '13 at 11:49
  • \$\begingroup\$ A good idea but sadly this ring thing is only big enough for the cable to go through once (approx 1.5cm diameter). It is an super old device I have to add (people hwo read PhD comics will know: doing cutting edge research with equipment from the mid 80s rocks). \$\endgroup\$ – Martin H Feb 22 '13 at 11:59
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    \$\begingroup\$ Do you know roughly how much current you're expecting? Is it AC? \$\endgroup\$ – Rocketmagnet Feb 22 '13 at 12:11
  • \$\begingroup\$ sorry forgot to mention it. It is AC, I've edited it. We expect something in the ~10mA range peak to peak \$\endgroup\$ – Martin H Feb 22 '13 at 12:25
  • \$\begingroup\$ Curious what the frequency of operation is? 13.5MHz? 27MHz? I had some experience with plasma many years back and had to instrument a plasma head in a very similar fashion to what @Rocketmagnet describes in his answer. I'll see if I can find some old notes. \$\endgroup\$ – Jason Feb 22 '13 at 13:02
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You could use a transformer to multiply up the current.

Current transformer

The image shows a current transformer working in the normal way for measuring very high currents. The measured current will be 1/N times that in the red cable. Basically you want to do the opposite.

Get a large iron ring, and loop your cable through it 10x. Then loop another small wire through the ring and measure the current in that.

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  • \$\begingroup\$ Ah ok, so this is basically reversing the ring that I am using now. Currently it is just like in the picture where the red cable is going through the ring and the winding gives me 0.1V per Amp. \$\endgroup\$ – Martin H Feb 22 '13 at 13:06
  • \$\begingroup\$ Will this be affected by insulation? \$\endgroup\$ – Douglas Feb 21 '18 at 19:20
  • \$\begingroup\$ @Douglas - No. The flow of current is not normally affected by insulation. \$\endgroup\$ – Rocketmagnet Feb 22 '18 at 12:54
  • \$\begingroup\$ I meant more the transfer of current from the primary to the secondary. \$\endgroup\$ – Douglas Feb 22 '18 at 16:47
  • \$\begingroup\$ @Douglas - Again no. The transfer of current from one coil to the other is by magnetic field. As long as the insulation doesn't affect the magnetic field, you'll be fine. \$\endgroup\$ – Rocketmagnet Mar 4 '18 at 23:35
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12 mVpp is a decent signal, and well above noise floor of reasonable electronics. Using a current transformer as you are doing is a good idea since it inherently isolates the current signal from the high voltage. As long as that signal has good signal to noise ratio, you are fine. 12 mVpp is high enough that it should not be hard to amplify it as neeed without adding significant noise. Since this is AC, you don't have to worry about DC offsets.

This is really not a hard problem. You want to detect the amplitude of a AC signal. Consider that AM radios routinely amplitude demodulate much weaker signals. It would help to know the frequency range of both the AC signal (the carrier in AM terms) and what you need the answer to be (the demodulated signal). Without knowing those, it is difficult to make any concrete suggestions.

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protected by W5VO Feb 24 '13 at 1:23

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