Assume that in our case, for instance, the rated current of the
GFPD (fuse) is 1A, there are 5 parallel strings in the PV array
(n=5), and Isc of each module is 5A. Simple calculation by the
equations above will give us Ig=25A in the worst case. According
to the melting time vs. current diagram in Fig. 4 (next page), the
1A fuse takes less than 0.01s to clear the ground fault. In this
instance, the fault path is detected by the fuse and the fault is
interrupted successfully.
According to the NEC Article 690.5, if a grounded conductor
is opened to interrupt the ground-fault current path, the gridconnected
PV inverter fed by the faulted PV array shall
automatically cease to supply power to the grid. Meanwhile, an
indication of the fault should be provided. After the shutdown of
the PV inverter, the whole PV array goes into the open-circuit
condition, waiting for maintenance personnel to fix the problem.