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AFE3010: N-G fault detection

Part Number: AFE3010

Hi Team,

I have a question regarding N-G fault detection, if there is no additional current injected into L, N, then it is difficult to detect N-G fault because the leak current will flow back to N cable from ground. 

I am wondering how AFE3010 works to detect N-G fault, when NG_OUT injects a pulse current, transformer's secondary winding will create common-mode reflected current on both N & L cables, then how should current transformer senses differential current under double fault condition(both L and G have leakage path to GND)? 

Regards,

Charles

  • Hey Charles,

    Sorry for the delay. We are looking at this and will respond shortly.

    Best,

    Peter

  • Hey Charles,

    In you first image, you show the directions of I_LEAK as opposites of each other, this is not correct. Both leakage from the load terminals will flow in the same direction into Earth ground during positive line phase and out of Earth ground and back into CT during negative phase.

    For the second image, this scenario is not a concern. In you drawing you show two fault paths. One from load Hot to Earth ground(I_inject(L)) and the other is load Neutral to Earth ground (I_inject(N)). I_inject(L) is the "hot-ground fault" and I_inject(N) is the "neutral-ground fault". 

    In the scenario where there is only a neutral-ground fault present, the AFE3010 will typically detect all faults resistances of <= 5-Ω (this is the resistance between load neutral and Earth ground.

    In the case of a double fault, the AFE3010 will have already sensed any hot-ground fault that > 5mA (~24kΩ fault resistance), thus system does not need the neutral-ground detection. 

    Even if there was an on-going hot-ground leakage of < ~4mA (30kΩ fault resistance) and thus AFE3010 thus was not detecting this, the AFE3010 would still easily detect any valid neutral-ground fault <=5-Ω. In this worst-case scenario, the amount of lost (un-sensed) differential injection current (due to this 30kΩ H-G pathway) is 100*5/(30k+5) = 0.016%, which is negligible.

    Hope this helps.

    Sincerely,

    Peter