Substation

IEC 60909 Short-Circuit Current Calculator

Three-phase initial symmetrical (I"k), peak (ip), breaking (Ib), and steady-state (Ik) short-circuit currents. Grid infeed, transformer, cable, and motor contributions.

IEC 60909-0:2016 IEC 60076-1 IEC 60947-2 EN 50160
Presets
Grid Feed (Network Impedance)

c_max = 1.10 for max Ik (IEC Table 1)

Transformer

Leave pkr blank to derive R_T from ukr% and X/R ratio

Cable / Feeder Contribution
Include Cable Contribution
Local Motor Load
Include Motor Contribution
Equipment Duty Check
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Total Fault Impedance (Ω)
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I"k — Initial Symmetrical (kA)
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ip — Peak Short-Circuit (kA)
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Ib — Breaking Current (kA)
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Ik — Steady-State (kA)
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κ — Peak Factor
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Bus X/R Ratio
Impedance Breakdown
Network R (Ω) —
Network X (Ω) —
Transformer R (Ω) —
Transformer X (Ω) —
Cable R (Ω) —
Cable X (Ω) —
Total R (Ω) —
Total X (Ω) —
Impedance Diagram (Ω)
Export
Network Impedance ▸
Zq = c · Un² / Sk" (IEC 60909-0 §4.2.2 Eq 6)

The grid infeed impedance Zq represents the source strength. For a local substation with strong grid infeed (high Sk"), Zq is small and the fault level is high.

Transformer Impedance ▸
ZT = ukr% · Un² / SrT (IEC §4.2.4 Eq 12) R_T = pkr / SrT · Un² / SrT (from copper losses) X_T = √(ZT² − R_T²)

Transformer impedance is defined by ukr% at rated current. Copper losses pkr (kW) allow accurate split of Z into R and X. Vector group affects zero-sequence network for unbalanced faults.

Peak Short-Circuit Current ▸
ip = κ · √2 · I"k (IEC §4.3.1 Eq 74) κ = 1.02 + 0.98 · e^(−3 R/X) (IEC §4.3.1)

ip is the first peak of the short-circuit current, important for circuit breaker making capacity. The peak factor κ depends on the R/X ratio at the fault location.

Equipment Duty Check ▸
Ib ≤ Icw (symmetric rating) ip ≤ 2.5 · Icw (LV equipment, IEC 60947-2)

Breakers are rated for symmetric interrupting current (Icw) and peak making current. LV breakers (≤1 kV) typically have a peak-to-symmetric ratio of 2.5. HV breakers use κ factor directly.

Standards & References ▸
• IEC 60909-0:2016 — Short-circuit currents in three-phase AC systems — Part 0: Calculation of currents
• IEC 60076-1:2011 — Power transformers — General
• IEC 60947-2:2016 — Low-voltage switchgear — Circuit-breakers
• EN 50160:2010 — Voltage characteristics at PoC
Frequently Asked Questions
What does the voltage factor c represent? +
The voltage factor c accounts for the deviation of the system voltage from nominal during a fault. IEC 60909-0 Table 1 defines c_max = 1.10 for HV (>1 kV) / 1.05 for LV (≤1 kV) for maximum short-circuit current, and c_min = 1.00 / 0.95 for minimum. Using c_max gives conservative (higher) fault levels for equipment rating; c_min is used for checking minimum fault conditions.
How is the peak factor κ calculated? +
IEC 60909-0 §4.3.1 gives κ = 1.02 + 0.98 · e^(−3R/X) where R/X is evaluated at the fault location. For high X/R (large reactors), κ approaches 2.0 (full asymmetry). For low X/R (close to transformers), κ drops toward 1.02. The formula is derived from the decaying DC component: id.c.(t) = √2·I"k · e^(−2πf·t·R/X).
When should cable contribution be included? +
Cable contribution is important for faults at the end of long feeders, where the cable impedance dominates the total fault impedance. For substation bus faults, the cable section is usually excluded (or included in parallel with the transformer if the fault is on the outgoing feeder). Enable the cable section when calculating faults beyond the first protection zone.
What is the difference between Ib and Ik? +
Ib (symmetrical breaking current) is the current at contact parting time of the circuit breaker (typically 30–100 ms for HV, 10–20 ms for LV). Ik (steady-state short-circuit current) is the current after all transient components have decayed (0.5–5 s). For purely network feeds without generators, Ib = I"k and Ik = I"k (μ = λ = 1). With generators, Ib < I"k due to the decrement factor μ.
What is the 2.5× rule for LV breaker peak rating? +
IEC 60947-2 Table H.1 defines a peak factor of 2.5 for low-voltage circuit breakers at 50 Hz: the making (peak) current rating must be at least 2.5× the symmetric interrupting rating. This is because the first peak of a near-zero-power-factor fault can reach 2.5× the symmetric RMS value. For HV breakers, the peak making current is related to κ directly.
How do motors contribute to short-circuit current? +
Asynchronous motors inject short-circuit current due to their subtransient reactance Xm (typically 15–25% on their own rating). IEC 60909-0 §4.2.6 gives Zm = Xm% × Un² / S_mot. A motor contribution decays in 1–3 cycles as the motor decelerates. Large motor loads in industrial substations can add 20–40% to the fault level.