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I would like to measure a current flowing on an RF power line in ICP (Inductively Coupled Plasma) system. The frequency of RF is 13.56 MHz.

So, I'm planning to construct my own CT (Current Transformer) for the current measurement. I originally thought that I have to choose some ferromagnetic material as a magnetic core in CT, But I'm curious what happens if I use a non-magnetic material such as a plastic or even an air. I read some paper which looks to use the air as the core. What are the advantage and disadvantage if the non-magnetic material is used in CT, instead of using the magnetic material?

The primary aim in constructing CT is that CT must not give non-ignorable influence on the power line under the measurement. As CT is nothing but a transformer, so many winding on the secondary side reduces an impedance seen by the power line (the primary side of the transformer) according to \$\tilde Z_L^\prime = {\left( {\frac{{{N_P}}}{{{N_S}}}} \right)^2}{\tilde Z_L} % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHb5MDXbpmVaibaieYlf9irVe % eu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-JfrVk % FHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaci % aacaqabeaadaqaaqaafaGcbaaeaaaaaaaaa8qaceWGAbWdayaaiaWa % a0baaSqaa8qacaWGmbaapaqaa8qacqGHYaIOaaGccqGH9aqpdaqada % qaamaalaaabaGaamOtamaaBaaaleaacaWGqbaabeaaaOqaaiaad6ea % daWgaaWcbaGaam4uaaqabaaaaaGccaGLOaGaayzkaaWdamaaCaaale % qabaWdbiaaikdaaaGcceWGAbWdayaaiaWaaSbaaSqaa8qacaWGmbaa % paqabaaaaa!4DCC! \$ where \${\tilde Z_L} % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHb5MDXbpmVaibaieYlf9irVe % eu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-JfrVk % FHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaci % aacaqabeaadaqaaqaafaGcbaaeaaaaaaaaa8qaceWGAbWdayaaiaWa % aSbaaSqaa8qacaWGmbaapaqabaaaaa!42A8! \$: secondary side load impedance, \$\tilde Z_L^\prime % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHb5MDXbpmVaibaieYlf9irVe % eu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-JfrVk % FHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaci % aacaqabeaadaqaaqaafaGcbaaeaaaaaaaaa8qaceWGAbWdayaaiaWa % a0baaSqaa8qacaWGmbaapaqaa8qacqGHYaIOaaaaaa!4439! \$: impedance seen by the primary side, \${{N_P}} % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHb5MDXbpmVaibaieYlf9irVe % eu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-JfrVk % FHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaci % aacaqabeaadaqaaqaafaGcqaaaaaaaaaWdbeaacaWGobWaaSbaaSqa % aiaadcfaaeqaaaaa!4263! \$: number of primary winding, which is 1 for CT, \${{N_S}} % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHb5MDXbpmVaibaieYlf9irVe % eu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-JfrVk % FHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaci % aacaqabeaadaqaaqaafaGcqaaaaaaaaaWdbeaacaWGobWaaSbaaSqa % aiaadofaaeqaaaaa!4266! \$: number of secondary winding. Is this idea correct?

Thanks for reading my question and I'm looking forward to replies.

Donggyu Jang
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  • Look at Rogowski coils. – pericynthion Mar 03 '17 at 06:30
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    An air core current transformer is pretty much a rogowski. But you will encounter errors due to the coupling coefficient not being 1 and more importantly due to RF capacitive currents and possible resonance in the coil. I would instead look at directional couplers, the output of which you can use to calculate the current on the line. – electrogas Mar 03 '17 at 06:45
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    For HF a better approach would be using a Rogowski coil. You can eliminate the DC bias with some PI controller, since you exactly know that mean value of your RF signal is zero. – Marko Buršič Mar 03 '17 at 10:48
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    Look a simple example without integrator circuit: http://www.aedie.org/9CHLIE-paper-send/252-argueso.pdf – Marko Buršič Mar 03 '17 at 11:02
  • Okay. People here suggest to use other scheme, instead of CT. I guess 13.56 MHz is rather high frequency so CT may not be suitable. If it is true, why? Could you give me some details of why CT is not good for high frequency like this? – Donggyu Jang Mar 04 '17 at 07:09

2 Answers2

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A good CT with a ferrite core is a challenge at this sort of frequency (and above) so I would definitely consider an air core especially if the only signal that can be picked up is the 13.56 MHz current. If you use a ferrite core it brings a different set of problems to the table like ferrite core losses and these might be excessive i.e. cause core saturation.

Another problem with using a ferrite core is the leakage inductance; a CT relies on good coupling in order to project the burden resistor's impedance (via the turns ratio squared which is a massive impedance reducer) so that the wire through the centre of the CT sees a very low impedance. With excessive leakage you will likely get core saturation and inaccuracies in measurement.

So, it's worth considering an air core. For sure, there will be much greater leakage inductance and this will mean the voltage out (that supposedly represents the primary current) will be inaccurate but there won't be any saturation non-linearities. Yes, it will need calibrating and this might be a big issue but not impossible.

You could go the whole mile and use a sniffer coil to receive the magnetic field and do the math to determine the actual current flow. In effect a sniffer coil is the natural extension of a CT: -

enter image description here

Andy aka
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  • Thanks for giving detailed comments. So...what you basically said is that a ferromagnetic material (I belive ferrite is one of kind of this) has problems of a magnetic flux saturation and energy loss due to eddy current, right? 13.56 MHz is rather high frequency so these problems may not be avoidable. So you suggested to use the air-core. Is my understanding clear? – Donggyu Jang Mar 04 '17 at 06:53
  • I didn't mention eddy current losses and these won't be top of the list of problems in this situation. The inability of the burden to project itself to the primary wire is the problem. This is due to excessive leakage inductance and is seen as a reactance in series with the burden. That reactance (due to operating at well-above 1 MHz) dominates and, might, in simple terms appear to be non-consequential but it isn't. You might think that OK it's in series with the burden but so what...... – Andy aka Mar 04 '17 at 09:43
  • .... current-out is (say) one-hundredth of current-in because that's how a CT works and that leakage inductance isn't going to alter that. And that is perfectly true providing the CT core doesn't saturate. But now, the projected secondary impedance onto the primary is much bigger and instead of just a few hundred micro volts being dropped by the primary (due to the flow of primary current), you might see several mV being dropped by the primary. Now, here's the tricky bit to understand.... – Andy aka Mar 04 '17 at 09:46
  • ... Any transformer has what is called magnetization inductance and this is seen as a component in parallel with the primary. For a voltage transformer this dominates but, for a CT it should not dominate at all - the reflected burden impedance should dominate i.e. all the primary current should mainly pass through the transformed impedance of the burden. If you have a 1 ohm burden and 100:1 CT ratio, the transformed impedance seen at the primary (due to the secondary burden) is one-ten-thousandth of an ohm (1E-4 ohms). This, in normal circumstances swamps the magnetization impedance...... – Andy aka Mar 04 '17 at 09:50
  • ..... However, that burden is now in series with a leakage inductance that might be several tens or hundreds of ohms. 100 ohms on the secondary will be seen as 0.01 ohms on the primary and now the small magnetization inductance takes significant current and the core saturates. A bigger core doesn't help either - inevitably there will be more leakage inductance as well as a greater magnetization inductance - no net gain! – Andy aka Mar 04 '17 at 09:54
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    See this answer by me that probably explains it better: http://electronics.stackexchange.com/questions/121363/why-do-current-transformers-have-a-minimum-frequency-rating/121376 – Andy aka Mar 04 '17 at 09:56
  • Are you understanding this answer now? You haven't made a comment for several days? – Andy aka Mar 08 '17 at 09:05
  • Well..actually not. I have a difficulty for an understanding of full circuit diagram of a real transformer, rather than the ideal one. I'm still on the track to follow you and it will take some time. I may give you some feedback once my study is done:) – Donggyu Jang Mar 08 '17 at 16:10
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You can do air cored, but your operating frequency is a little low to make it easy.

The usual approach is a 43 mix ferrite bead with say 33 turns on the secondary (Coupling factor -30dB), primary is a short piece of coax thru the center with the screen connected only at one end so that the screen forms an electrostatic shield. RF continuity for the line in formed by the surrounding box. At low frequency you can get away without the electrostatic screen, see the transformers in W6PQLs directional coupler for an example.

The trick is to make sure that the wire length in the secondary is short compared to the wavelength, not too hart at 13.56, but exciting if doing a very broadband one.

Dan Mills
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  • Hello. Could you please give me the details of "primary is a short piece of coax thru the center with the screen connected only at one end so that the screen forms an electrostatic shield. RF continuity for the line in formed by the surrounding box."? I don't understand what you said. I would like you to give more details of your statement. – Donggyu Jang Mar 04 '17 at 07:04