IEC 60909 Fault Calculations: Short-Circuit Current Formulae

IEC 60909 gives a method for calculating short-circuit currents in three-phase a.c. systems. It is commonly used to estimate maximum and minimum fault levels for switchgear rating, protection settings and cable withstand checks.

This introduction covers faults far from generators in three-phase networks, including unbalanced line-to-line and earth faults. It sits under the wider Fault Current Calculations topic. Standalone single-phase supplies and d.c. systems are outside the scope of this article.

How the IEC 60909 method works

The method places an equivalent voltage source cU_n/√3 at the fault location and represents the supply network and equipment by their impedances. Here, U_n is the nominal line-to-line voltage and c is the voltage factor. All impedances must be referred to the same voltage level, with the applicable IEC 60909 impedance correction factors included.

A fault is “far from generators” when the symmetrical a.c. component of fault current remains essentially constant during the fault. This describes electrical behaviour, rather than simply physical distance from a generator.

For unbalanced faults, symmetrical components represent the network using three sequence impedances, as seen from the fault location:

• Z₁: positive-sequence impedance, associated with balanced currents in the normal phase order.
• Z₂: negative-sequence impedance, associated with balanced currents in the reverse phase order.
• Z₀: zero-sequence impedance, associated with three in-phase currents and their common return path. It depends on the earthing arrangement, transformer connections and available return paths.

The formulae below give initial symmetrical r.m.s. currents, I″. They assume negligible fault impedance. Earth-fault formulae here concern solidly or impedance-earthed neutral systems; isolated and resonant-earthed systems require separate consideration. Peak current, breaking duty and thermal withstand require additional checks; the initial r.m.s. value alone does not describe every short-circuit duty.

Short-circuit current formulae

For the far-from-generator calculations below, take Z₂ = Z₁. This equality applies to non-rotating components such as cables and transformers. Each Z is a complex impedance, R + jX; vertical bars, |Z|, denote its magnitude. Combine complex impedances before taking their magnitude.

Fault type	Initial symmetrical r.m.s. current
Three-phase short circuit	$$I_k^{″}=\frac{c{U}_n}{\sqrt{3}\left|Z_1\right|}$$
Line-to-line short circuit	$$I_{\mathrm{k2}}^{″}=\frac{c{U}_n}{2\left|Z_1\right|}=\frac{\sqrt{3}}{2}{I}_k^{″}$$
Line-to-line-earth short circuit: earth-return current	$$I_{\mathrm{kE2E}}^{″}=\frac{\sqrt{3}c{U}_n}{|Z_1+2Z_0|}$$
Line-to-earth short circuit	$$I_{\mathrm{k1}}^{″}=\frac{\sqrt{3}c{U}_n}{|2Z_1+Z_0|}$$

I″_kE2E is the total current flowing through the earth return for a line-to-line-earth fault. It is not the current in either faulted phase conductor; those conductor currents require separate calculation.

For earth-fault applications in cable calculations, see Earth Fault Loop Impedance.

Combining series impedances

For a three-phase fault supplied through a series equivalent network, add the positive-sequence resistances and reactances separately. Do not add impedance magnitudes. With all values referred to the voltage level at the fault:

$$R_k=\sum _i{R}_i$$

$$X_k=\sum _i{X}_i$$

$$I_k^{″}=\frac{c{U}_n}{\sqrt{3}\sqrt{{R_k}^2+{X_k}^2}}$$

This is the three-phase fault formula. Line-to-line and earth faults use the sequence-network combinations in the table above. Networks with parallel paths or multiple sources require the appropriate network reduction.

Source impedance

If the initial symmetrical three-phase fault current at the supply connection is known, the magnitude of the external network’s positive-sequence impedance can be calculated as:

$$\left|Z_Q\right|=\frac{c{U}_{\mathrm{nQ}}}{\sqrt{3}{I}_{\mathrm{kQ}}^{″}}$$

Here, U_nQ and I″_kQ refer to the supply connection. Use the voltage factor consistent with the basis of the supplied fault-current value. The source R/X ratio is also needed to split this magnitude into resistance and reactance before combining it with downstream impedances.

Maximum and minimum fault levels

Both maximum and minimum fault levels are normally required. Maximum fault current is used for equipment short-circuit ratings. Minimum fault current is used to confirm that protective devices operate correctly under the least favourable fault condition.

• Use c_max or c_min as appropriate.
• Choose the supply contributions and network configuration that give the maximum or minimum current at the fault location. Use the corresponding external network impedance.
• For maximum current, use line and cable resistances at 20 °C and include motor contributions where required.
• For minimum current, use line and cable resistances at the conductor temperature at the end of the short-circuit duration and neglect motor contributions.

Voltage factor

The voltage factor allows for departures from nominal voltage and other assumptions in the equivalent-source method. Select the applicable value for the system and any relevant national requirements.

Nominal voltage Un	cmax	cmin
Low voltage, 100–1000 V, with +6% voltage tolerance	1.05	0.95
Low voltage, 100–1000 V, with +10% voltage tolerance	1.10	0.95
Medium voltage, above 1 kV up to and including 35 kV	1.10	1.00
High voltage, above 35 kV	1.10	1.00

c_max is used for maximum current and c_min for minimum current. The product c_maxU_n should not exceed the highest voltage for the equipment.

Worked example: three-phase fault

Consider a 400 V three-phase network with +10% voltage tolerance. For a maximum-current calculation, use c_max = 1.10. Assume that the equivalent positive-sequence network at the fault has total resistance R_k = 0.030 Ω and reactance X_k = 0.040 Ω, after referring all impedances to 400 V and applying the relevant corrections. The line and cable resistances are at 20 °C.

$$\left|Z_1\right|=\sqrt{{0.030}^2+{0.040}^2}=0.050\text{Ω}$$

$$I_k^{″}=\frac{1.10\times 400}{\sqrt{3}\times 0.050}\approx 5081\text{A}=5.08\text{kA}$$

The initial symmetrical three-phase fault current is therefore approximately 5.08 kA r.m.s. This is not the instantaneous peak current. A minimum-current calculation must use the appropriate minimum supply conditions, voltage factor and conductor resistance, rather than changing the voltage factor alone.

Symbols

Symbol	Meaning
c	Voltage factor
I″k	Initial symmetrical three-phase short-circuit current, r.m.s., A
I″k1	Initial symmetrical line-to-earth short-circuit current, r.m.s., A
I″k2	Initial symmetrical line-to-line short-circuit current, r.m.s., A
I″kE2E	Initial symmetrical earth-return current for a line-to-line-earth short circuit, r.m.s., A
Un	Nominal system voltage, line-to-line r.m.s., V
Z1, Z2, Z0	Complex positive-, negative- and zero-sequence short-circuit impedances at the fault location, Ω
ZQ	Complex positive-sequence impedance of the external network, Ω
UnQ, I″kQ	Nominal line-to-line voltage and initial symmetrical three-phase fault current at the supply connection, V and A
Rk, Xk	Total positive-sequence resistance and reactance at the fault location, Ω
Ri, Xi	Resistance and reactance of each series element, referred to the same voltage level, Ω
j	Imaginary unit, √(−1)
|Z|	Magnitude of complex impedance, √(R² + X²), Ω

For converting upstream fault level data into source impedance, see Network Fault Level.

For estimating a transformer secondary source fault level before IEC 60909 network calculations, see Transformer Secondary Fault Level.

For the cable sequence impedance inputs used by IEC 60909, see Cable Impedance.

See the myCableEngineering software overview for fault-calculation capabilities and how the supporting results are retained with each cable project.