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Step voltage

The step voltage is the difference in surface potential that a person bridges with the feet 1 m apart, without touching an earthed object, while fault current flows from the earth mat into the soil (IEEE Std 80-2013 §3).

Read first: Tolerable body current · Surface layer and Cs

1The surface potential during a fault

During an earth fault, part of the fault current flows from the earth mat into the soil and returns through the earth to the source. The soil around the mat rises in potential. The potential is highest over the conductors and falls with distance from them, so two points of the surface 1 m apart are at different potentials. A person standing across two such points carries a current from one foot to the other (Figure 1).

Figure 1.The section shows the edge of an earth mat during an earth fault, with a person standing on the surface layer over the perimeter conductor. The drawing is schematic and not to scale.

The potential falls fastest outside the mat's perimeter, where no conductor lies under the surface (Figure 2). Inside the mat it rises over each conductor and dips between them.

Figure 2.The surface potential, as a share of the rise of earth potential (GPR), rises from 10 m outside the west edge of the default 220 kV mat to a near-constant value 20 m inside it. The soil is the example two-layer soil of 90 Ω·m over 30 Ω·m at a depth of 3.0 m. The numerical analysis of the design service calculated the values on the mat it chose for this soil, with a mesh spacing of 2.98 m.

2The body in the circuit

The current enters at one foot, passes through the legs and leaves at the other foot. The body's resistance $R_{B}$ and the contact resistances of the two feet with the surface are in series (Figure 3). IEEE Std 80-2013 takes $R_{B}$ as 1000 Ω and the contact resistance of one foot on the surface layer as $3 C_{s} ρ_{s}$ (§7). Two feet in series give the $6 C_{s} ρ_{s}$ of Eq. (4).

Figure 3.In the equivalent circuit, the step voltage drives the body current through the contact resistance of each foot and the body's resistance in series.

3The tolerable step voltage

The tolerable step voltage drives the largest body current that a person of 50 kg tolerates for the duration of the shock. IEEE Std 80-2013 sets that current from the fibrillation threshold of the body. It is inversely proportional to the square root of the shock time.

I_{B} = \frac{0.116}{t _{s}}

(1)

IEEE Std 80-2013 §5.3, Eq. 6

A layer of crushed rock at the surface raises the contact resistance of the feet. The derating factor $C_{s}$ corrects the layer's resistivity for its finite depth over the soil below.

C_{s} = 1 - \frac{0.09 ( 1 - ρ / ρ _{s} )}{2 h _{s} + 0.09}

(2)

IEEE Std 80-2013 §7.4, Eq. 27

E_{step} = (R_{B} + 2 R_{f}) I_{B}

(3)

IEEE Std 80-2013 §8.3

E_{step50} = (1000 + 6 C_{s} ρ_{s}) \frac{0.116}{t _{s}}

(4)

IEEE Std 80-2013 §8.3, Eq. 29

where
$I_{B}$	body current that a person of 50 kg tolerates	A
$t_{s}$	duration of the shock current	s
$ρ$	resistivity of the soil below the surface layer	Ω·m
$ρ_{s}$	resistivity of the surface layer	Ω·m
$h_{s}$	depth of the surface layer	m
$C_{s}$	surface layer derating factor
$R_{B}$	resistance of the body	Ω
$R_{f}$	contact resistance of one foot on the surface layer	Ω
$E_{step}$	tolerable step voltage	V
$E_{step50}$	tolerable step voltage for a person of 50 kg	V

Figure 4 shows the limit against the surface layer's resistivity. Over soil of 150 Ω·m, the tolerable step voltage is 312 V without a surface layer and 2246 V with a layer of 3000 Ω·m and 0.10 m.

Surface layer resistivity2500 Ω·m · hs 0.10 m: 1907 V · hs 0.15 m: 2091 V

Figure 4.The tolerable step voltage for a person of 50 kg by Eq. (4) rises with the surface layer's resistivity, for layers of 0.10 m and 0.15 m over soil of 150 Ω·m. The shock time is 0.5 s. The dashed line is the step voltage of the default 220 kV mat in Example 1. The design service calculated the values.

4The step voltage of an earth mat

For a rectangular mat in uniform soil, IEEE Std 80-2013 gives the largest step voltage from the soil's resistivity, the grid current and the mat's geometry. The effective length $L_{S}$ counts 75 % of the horizontal conductor and 85 % of the electrodes.

E_{s} = \frac{ρ K _{s} K _{i} I _{G}}{L _{S}}

(5)

IEEE Std 80-2013 §16.5.2, Eq. 92

L_{S} = 0.75 L_{C} + 0.85 L_{R}

(6)

IEEE Std 80-2013 §16.5.2, Eq. 93

K_{s} = \frac{1}{π} [\frac{1}{2 h} + \frac{1}{D + h} + \frac{1}{D} (1 - 0. 5^{n - 2})]

(7)

IEEE Std 80-2013 §16.5.2, Eq. 94

K_{i} = 0.644 + 0.148 n

(8)

IEEE Std 80-2013 §16.5.1, Eq. 89

where
$E_{s}$	step voltage of the mat	V
$I_{G}$	maximum grid current	A
$K_{s}$	spacing factor for step voltage
$K_{i}$	irregularity factor
$L_{S}$	effective length of conductor and electrodes for step voltage	m
$L_{C}$	total length of horizontal conductor	m
$L_{R}$	total length of electrodes	m
$D$	spacing between parallel conductors	m
$h$	burial depth of the mat	m
$n$	geometry factor of the mat
$π$	the ratio of a circle's circumference to its diameter

Eq. (7) holds for a burial depth $h$ between 0.25 m and 2.5 m (§16.5.2). The geometry factor $n$ is the one the mesh voltage uses (Eq. 84 to 88 of §16.5.1).

Example 1Step voltage of the default 220 kV mat

Given the earth mat of a new 220 kV project with the default inputs, as the design service calculates it.

Quantity	Value
Soil resistivity, measured × season factor	100 Ω·m × 1.5 = 150 Ω·m
Surface layer, resistivity and depth	3000 Ω·m, 0.10 m
Shock time	0.5 s
Grid current, fault current × split factor × decrement factor	50 kA × 0.6 × 1.062 = 31 853 A
Mat, length × width, conductors	155.0 m × 133.4 m, 40 × 35
Mesh spacing $D$ , burial depth $h$	3.97 m, 0.6 m
Electrodes	96 of 3 m
Conductor length $L_{C}$ , electrode length $L_{R}$	10 761 m, 288 m
Geometry factor $n$ (Eq. 84 to 88)	37.37

Derating factor by Eq. (2):

C_{s} = 1 - \frac{0.09 ( 1 - 150 / 3000 )}{2 \times 0.10 + 0.09} = 0.705

Tolerable step voltage by Eq. (4):

E_{step50} = (1000 + 6 \times 0.705 \times 3000) \frac{0.116}{0.5} = 2246 V

Factors by Eq. (6), Eq. (8) and Eq. (7):

L_{S} = 0.75 \times 10761 + 0.85 \times 288 = 8316 m

K_{i} = 0.644 + 0.148 \times 37.37 = 6.174

K_{s} = \frac{1}{π} [\frac{1}{1.2} + \frac{1}{4.57} + \frac{1}{3.97} (1 - 0. 5^{35.37})] = 0.415

Step voltage by Eq. (5):

E_{s} = \frac{150 \times 0.415 \times 6.174 \times 31 853}{8316} = 1472 V

Result The step voltage of 1472 V is below the tolerable 2246 V. The mat complies with the step voltage limit.

Sources

[1]IEEE Std 80-2013, IEEE Guide for Safety in AC Substation Grounding. §5.3, Eq. 6; §7.4, Eq. 27; §8.3; §8.3, Eq. 29; §16.5.2, Eq. 92; §16.5.2, Eq. 93; §16.5.2, Eq. 94; §16.5.1, Eq. 89.

CitationSubstationLayout Reference, Step voltage, revision 2, 8 Nov 2026, https://substation.solarlayout.app/library/reference/step-voltage

Step voltage

1The surface potential during a faultCopy link to The surface potential during a fault

2The body in the circuitCopy link to The body in the circuit

3The tolerable step voltageCopy link to The tolerable step voltage

4The step voltage of an earth matCopy link to The step voltage of an earth mat

1The surface potential during a fault

2The body in the circuit

3The tolerable step voltage

4The step voltage of an earth mat