PEN protection - phase shift

Hello All,

Just reading this nice article on PEN protection:

Looking at implementing PEN protection as per BS7671: 722.411.4.1 - section (iii), which I believe allows for deriving a virtual neutral, which I thought could be done by looking at the L -> N phase voltages.

Figure 7 in that article shows a phasor diagram where L1, L2, L3 are 239V, 287V, 177V respectively, and shows a neutral shift of 64V. However if those phases are vector added assuming 120 degree phase shift, it's more like 95.5V.  I had to tweak the phase shift somewhat to get close to the 64V (0, 232, 113 degrees for instance).

Just wondering if my thinking is wrong, or in the event of a PEN fault could phase shift occur?

Thanks in advance.


  • The virtual N (blue bob) is the shifted neutral as you have said. However, as I understand it, the blue bob voltage (disconnected N - not earth) is the geometrical vector addition of the 3-phases? So we have:

    That's not correct ... I think it's where you're going wrong.

    The virtual N (blue blob) is the vector sum of the voltages that develops across the impedances connecting each of the three phases to the PEN.

    The PEN is broken, so (worst-case) PEN current is zero, and therefore:

    iL1-PEN + iL2-PEN + iL3-PEN = 0

    (Lowercase i being a phasor current)

    If you know the actual impedances connected between each line and neutral, you can calculate iLx-PEN = (uLx-N - uPEN) / ZLx-PEN and calculate uPEN from the first equation (lowercase u being again a phasor voltage, so 230∠0 for L1, 230∠120 for L2 and 230∠240 for L3 - this is where the phasor vectors come from if the loads are resistive only)

    If the L-PEN impedances on each phase are identical, uPEN = 0 (i.e. same as transformer N)

  • Ah ok.. thanks Mike.

  • Ah that makes sense.