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Shielding - Need to ground an electrostatic shield?

  1. Apr 3, 2014 #1
    Shielding -- Need to ground an electrostatic shield?

    I am reading this app note by analog devices:

    At the bottom of the first page it says: "An electrostatic shield, to be effective..." "But grounding the shield is useless if the signal is not grounded"

    Why is grounding the shield useless if the signal is not grounded?

  2. jcsd
  3. Apr 3, 2014 #2


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    did you not go on to continue reading the top of the next page ?

    basically ... if the shield and the signal ground ( the reference point ) are not connected together
    you can develop a potential difference between the shield and the signal gnd
    rendering the shield ineffective

    The text for fig8 should be in huge bold letters.

  4. Apr 3, 2014 #3
    I read it, twice. But I'm not sure that Fig 8 applies. Fig 8 has a series source that is still referenced to the circuit ground.

    Would a Faraday cage, floating in mid air still provide a shield for electronic equipment inside? If I leave a shield floating, not connected to anything, does the shield still do its job?
  5. Apr 3, 2014 #4


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    Figure 3:
    What is the reference point for the return path of the signal? For the shield to be effective it must shunt (redirect the noise currents from the signal loop path to a low resistance alternative loop path instead) the noise away from the signal conductor to a point that's also at the effective ground/return path of the signal (D) so no voltage can develop on the shield in reference to signal ground. Without a hard wired common 'ground' path for the return of noise energy the shield would just become another path in the noise current loop path to the signal loop.
  6. Apr 3, 2014 #5
    As a thought experiment, if I had an electronic circuit (powered by batteries) sitting in sealed metal case and that metal case was not electrically connected to the circuit's ground, would the shield not provide shielding from EM radiation/noise to the electronic circuit?
  7. Apr 3, 2014 #6


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  8. Apr 3, 2014 #7
    I've read that, but I then have a dilemma with the textbook examples of Faraday cages saying that internal (to the case) electric fields are zero if the case is placed in an external electric field.

    Quote from reply #2:
    Now for electromagnetic waves, a perfectly electrically conducting sphere will block all incoming and outgoing waves. This is regardless of the grounding of the shield. So the grounding of the shield helps it with static charge buildups. Although it wouldn't surprise me if grounding a Faraday cage does improve performance due to real world limitations.
    Last edited: Apr 3, 2014
  9. Apr 3, 2014 #8


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    The textbook examples with perfect conductors, perfect isolation and static charges are correct but that's not the typical environment for an electrical circuit with shielding requirements for power and signal cables snaking around in all sorts of directions close to noise sources and connected to an enclosure full of holes for external wiring.
  10. Apr 5, 2014 #9

    jim hardy

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    The premise of shielding is that you want capacitively coupled currents to flow in the shield wire, not in the signal wire(s).
    To that end, the shield should be at same potential as the signal wires, or as near to it as is practical.
    That way there is near zero potential difference between the signal wire and its shield;
    which means there's near zero voltage available to drive current through the shield-to-signal capacitance,
    so capacitively coupled current does not flow into the signal wires (as desired).

    Now - if your (sealed metal case/Faraday cage/shield) is at local earth potential,
    and your electronic circuit inside it is 1 kv above local earth potential,
    does a potential difference exist between the inside of your Faraday cage and your electronic circuit?
    Surely it does. So capacitive current can flow between them, in amount [1kv/Xc].
    Tying your Faraday cage not to earth but to the measuring circuit's signal common removes that potential difference.

    In a practical world, that would place the Faraday cage at dangerous potential,
    so one would either surround it with another cage that's at local at earth potential,
    or re-arrange the gizmo to remove that 1kv elevation above local earth.

    Any help?

    Try a google on "driven shield", guarding.

    old jim
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