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I am designing a circuit with both signal ground and earth. As far as I know, a TVS should be connected to the earth.

What if there is a huge voltage different between the ground and the earth?

Let say I am protecting a 5 V signal (with respect to the ground) with a uni-directional TVS diode which is connected to the earth.

When ground = 0 V, earth = -1000 V, signal = 5 V, in this case, the voltage difference between the signal and the earth is 1005 V. Will the TVS diode be always conducting and pulling the signal pin very low?

When ground = 0 V, earth = 1000 V, signal = 5 V and an 800 V ESD event happened, the voltage across the TVS is still -200 V. Does that mean the TVS diode is never protecting the signal pin?

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    \$\begingroup\$ "As far as I know, a TVS should be connected to the earth." Why? What is your reference? \$\endgroup\$
    – RussellH
    Commented Mar 1 at 9:29
  • \$\begingroup\$ What is the signal connected to? What is being protected? What sources of surge are you concerned about, just ESD or--? What relevance is earth with respect to ESD, a very fast pulse that affects local circuitry? \$\endgroup\$ Commented Mar 1 at 9:46
  • \$\begingroup\$ In what context? Ground = earth in electronics. In an AC mains context with safety ground, we might use different terms to separate earth (plain old ground) from safety ground (connected to earth fault breaker). I take it that's the case here? \$\endgroup\$
    – Lundin
    Commented Mar 1 at 10:18
  • \$\begingroup\$ By the "ground" you mean PE( protective earth) from you primary feeder? Using proper terminology may help a lot. In electronics common wire some times called ground. \$\endgroup\$
    – user263983
    Commented Mar 1 at 11:10

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You protect the signal to the ground you need. If you want to limit some ESD event between chip data pin and chip ground, it has nothing to do with earth potential, as the chip may be battery powered and have no connection or reference to earth potential.

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