LibraryReferenceEarthing

Grid current

The maximum grid current $I_{G}$ is the part of an earth fault's current that flows from the earth mat into the soil, raised for the DC offset of the fault current (IEEE Std 80-2013 §15.1). It drives the rise of earth potential (GPR) and the mesh and step voltages.

1The split of the fault current

During an earth fault at the substation, the fault current returns to its sources. Part of it flows from the earth mat into the soil, and the rest returns on the shield wires of the lines and the sheaths of cables. The split factor $S_{f}$ is the share that flows through the mat. The design takes 0.6 by default, set on the Protection page, unless a current-division study gives a value.

2The DC offset

An asymmetric fault current carries a DC component that decays with the time constant of the network. The decrement factor $D_{f}$ raises the symmetrical current to the effective value of the asymmetric current over the fault's duration.

D_{f} = 1 + \frac{τ}{t _{f}} (1 - e^{- 2 t_{f} / τ}), τ = \frac{X}{ω R}

(1)

IEEE Std 80-2013 §15.10, Eq. 79

I_{G} = D_{f} S_{f} 3 I_{0}

(2)

IEEE Std 80-2013 §15.1, Eq. 63 and 64

where
$D_{f}$	decrement factor
$τ$	DC time constant of the fault current, which IEEE 80-2013 writes $T_{a}$	s
$t_{f}$	duration of the fault	s
$X / R$	ratio of reactance to resistance of the fault path
$ω$	angular frequency, $2 π f$	rad/s
$S_{f}$	split factor
$3 I_{0}$	symmetrical earth fault current	kA
$I_{G}$	maximum grid current	kA

$D_{f}$ is largest for a short fault and a high X/R ratio, and falls towards 1 as the fault lasts longer (Figure 1). The design takes the fault duration as the shock time and an X/R ratio of 20 by default. The earth fault current is the design fault level of the HV level, the larger of the two levels at a switching substation.

Fault duration0.50 s · X/R 40: 1.120 · X/R 20: 1.062 · X/R 10: 1.031

Figure 1.The decrement factor falls towards 1 as the fault lasts longer, and is higher for a higher X/R ratio, at 50 Hz. The design service calculated the values by Eq. (1).

Example 1Grid current of the default 220 kV project

Given a design fault level of 50 kA at 220 kV, an X/R ratio of 20 at 50 Hz, a fault duration of 0.5 s and a split factor of 0.6.

Time constant and decrement factor by Eq. (1):

τ = \frac{20}{2 π \times 50} = 0.0637 s, D_{f} = 1.0618

Grid current by Eq. (2):

I_{G} = 1.0618 \times 0.6 \times 50 = 31.85 kA

Result The maximum grid current is 31 853 A, 0.6 of the fault current raised by 1.0618 for its DC offset.

Sources

[1]IEEE Std 80-2013, IEEE Guide for Safety in AC Substation Grounding. §15.10, Eq. 79; §15.1, Eq. 63 and 64.

CitationSubstationLayout Reference, Grid current, revision 1, 8 Nov 2026, https://substation.solarlayout.app/library/reference/grid-current

Grid current

1The split of the fault currentCopy link to The split of the fault current

2The DC offsetCopy link to The DC offset

1The split of the fault current

2The DC offset