Why does common mode current flows on outer surface of the coaxial cable?

Discussion in 'General Electronics Chat' started by Dong-gyu Jang, Feb 10, 2016.

  1. Dong-gyu Jang

    Thread Starter Member

    Jun 26, 2015
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    Hello.

    I've read that common mode current (maybe noise) flows on outer layer of the outer shield of the coaxial cable and normal current as differential mode current flows on central conductor and inner layer of the shield in coax. I don't understand why the common mode selects outer layer of its path and why normal current is 'closely' confined?

    I'd uploaded similar question before but...I didn't get clear answer.
     
  2. nsaspook

    AAC Fanatic!

    Aug 27, 2009
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    It's easier to visualize 'why' by thinking about how the electromagnetic field internal and external to the coax moves as a waveguide.
    [​IMG]
    TEM00 (Transverse Electric Magnetic) mode https://en.wikipedia.org/wiki/Transverse_mode
    [​IMG]
    Then you have to understand Skin Effect on how the shield has an inner and outer surface conductor: http://www.sigcon.com/Pubs/edn/modelingskineffect.htm

    Now we can examine the complete field structure.
    [​IMG]
    http://blog.lamsimenterprises.com/2011/02/22/coaxial-transmission-line-geometry/
    When this field geometry is disturbed by common mode signals the fields no longer cancel on the outer shield surface, this gives rise to a electric field and current flow on the outer shield surface conductor circuit as a noise current.

    This external field shield current can be useful if used in the correct way.
    http://www.antenna-theory.com/definitions/infinite.php
     
    Last edited: Feb 10, 2016
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  3. Dong-gyu Jang

    Thread Starter Member

    Jun 26, 2015
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    4
    Hello.

    Thanks to give me very kind explanation.

    I'd known skin effect and your referred materials gives some fresh aspect of this.

    So in normal operation, return current should be on the inner layer of the shield in coax. If they're on the outer layer of the shield, because of the skin effect which limits field-penetrating depth, the return current is no more associated with the center conductor in terms of the field. It is as if there is no return current in view of signal current on center conductor and is not physically right situation. I guess this is what you want to say.

    However, still, I don't understand why noise current flows only on outer layer of the shield. I imagines that this current flows on center conductor and the magnetic field associated with it will be confined within inner layer of the shield according to your reference. No problem at all.

    Did I miss something else?

    Beside, I'm now confused with my original terminology, "common mode current", The typical definition of this may be the currents flowing both hot and neutral lines. Maybe my original meaning of "common mode current flows only on outer layer of the shield" is wrong and this noise flows on both conductors in current?
     
  4. nsaspook

    AAC Fanatic!

    Aug 27, 2009
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    How and where the 'noise' current flows is a complex subject (EMC) that depends on the source/load connections and a host of other factors.

    The primary methods to eliminate unwanted signals are redirection, dissipation, reflection and cancellation (generation of an equal but opposite masking field) of energy. Each of these might have different effects on a "common mode current" path.

    http://www.analog.com/media/en/technical-documentation/application-notes/41727248AN_347.pdf
    http://www.analog.com/media/en/training-seminars/tutorials/MT-095.pdf
     
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  5. Dong-gyu Jang

    Thread Starter Member

    Jun 26, 2015
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    4
    Thanks. I'll read your reference.

    So...basically noise can go anywhere rather than going along certain side of the circuit like outer layer of the shield, right? I'm happy to at least break my wrong knowledge.
     
  6. nsaspook

    AAC Fanatic!

    Aug 27, 2009
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    A perfectly terminated coax with perfect shielding will exclude the noise signal from the inner conductors so you're not wrong but things are never perfect.
     
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