IEC 60909 peak short-circuit current — the kappa formula κ = 1.02 + 0.98·e^(−3R/X), explained and worked out

The answer: ip = κ·√2·Ik″ and κ = 1.02 + 0.98·e^(−3R/X)

IEC 60909-0 computes the peak short-circuit current ip — the highest instantaneous value of the fault current, reached within the first half-cycle — from the initial symmetrical short-circuit current Ik″ and a dimensionless peak factor κ (kappa):

ip = κ · √2 · Ik″
κ  = 1.02 + 0.98 · e^(−3·R/X)

In plain ASCII, as it is usually typed into a search box or a spreadsheet:

kappa = 1.02 + 0.98*exp(-3*R/X)
ip    = kappa*sqrt(2)*Ik''

R/X is the resistance-to-reactance ratio of the equivalent short-circuit impedance at the fault point — the sum of everything between the source and the fault, referred to the fault voltage. κ depends on nothing else: not on the current magnitude, not on the voltage level.

Why the peak matters more than people think

Ik″ is an RMS value. It rates the breaking duty of a circuit-breaker (Icu, Ics) and the thermal stress of conductors. The peak ip rates something different: the mechanical forces, which are proportional to the square of the instantaneous current. Three checks depend on it:

• Making capacity of circuit-breakers (Icm). A breaker closing onto a fault must survive the first peak. IEC 60947-2 ties Icm to Icu through a ratio n that depends on the current level — for 20 kA < Icu ≤ 50 kA, n = 2.2 at a test power factor of 0.2.
• Peak withstand of busbars and assemblies (Ipk). IEC 61439-1 requires the assembly to declare a rated peak withstand current; it has to cover ip at the busbar.
• Supports, insulators and bus duct joints, whose electrodynamic loading is set by ip, not by Ik″.

A board can pass the Icu check and still fail the peak check when the network X/R is higher than the X/R at which the breaker was tested. That is the case in the worked example below.

Where κ comes from: the DC component

When a fault occurs on an inductive circuit, the current cannot jump instantaneously. The fault current therefore starts with a DC component that decays with the circuit time constant τ = L/R = X/(ωR). The first peak adds the AC crest (√2·Ik″) to whatever is left of that DC offset after about half a cycle.

• With R/X = 0 (pure reactance), the DC component never decays and the peak reaches twice the AC crest: κ = 2.0.
• With a purely resistive circuit there is no offset: κ tends to 1.02 (the formula’s floor).
• Everything in between is captured by the exponential term e^(−3R/X), which is the IEC 60909 fit to the exact transient curve for a series R-L circuit.

Higher X/R means a slower decay, a larger first peak and a larger κ.

κ for typical R/X ratios

Values below are computed directly from κ = 1.02 + 0.98·e^(−3R/X):

κ = 1.8 corresponds to R/X ≈ 0.076 (X/R ≈ 13.1). κ = 2.0 is the theoretical ceiling.

Worked example: 500 MVA grid, 1000 kVA transformer, 440 V board

The inputs are those of case A in the LOGOS verification suite against pandapower:

• Utility source: S″kQ = 500 MVA at 13.8 kV, X/R = 10.
• Transformer: Sr = 1000 kVA, uk = 6 %, X/R = 6, secondary 440 V.
• Fault: three-phase, at the 440 V busbar, maximum case (c = 1.05 for a LV system with +6 % tolerance).

All impedances are referred to U = 440 V.

1 — Utility impedance. The source sits at medium voltage, so it uses cQ = 1.10:

Z_Q = cQ · U² / S″kQ = 1.10 · 440² / 500 MVA = 0.4259 mΩ
R_Q = 0.0424 mΩ   X_Q = 0.4238 mΩ   (X/R = 10)

2 — Transformer impedance and K_T.

Z_T = uk · U² / Sr = 0.06 · 440² / 1000 kVA = 11.616 mΩ
R_T = 1.9097 mΩ   X_T = 11.4580 mΩ   (X/R = 6)
x_T = uk · X/Z = 0.06 · 6/√37 = 0.0592 pu
K_T = 0.95 · cmax / (1 + 0.6 · x_T) = 0.95 · 1.05 / (1 + 0.6 · 0.0592) = 0.9633
Z_TK = K_T · Z_T → R = 1.8396 mΩ, X = 11.0374 mΩ

3 — Equivalent impedance and R/X.

R_k = 0.0424 + 1.8396 = 1.8819 mΩ
X_k = 0.4238 + 11.0374 = 11.4612 mΩ
Z_k = 11.6147 mΩ   →   R/X = 0.1642   (X/R = 6.09)

4 — Ik″, κ and ip.

Ik″ = c · U / (√3 · Z_k) = 1.05 · 440 / (√3 · 11.6147 mΩ) = 22.97 kA
κ   = 1.02 + 0.98 · e^(−3 · 0.1642) = 1.619
ip  = 1.619 · √2 · 22.97 = 52.58 kA

The minimum current of the same bus (cmin = 0.95, no K_T, source with cQmin = 1.00) is Ik″min = 20.11 kA.

These numbers were produced by the same IEC 60909 module that runs the LOGOS short-circuit calculation in production. For this case, pandapower 3.5.4 (calc_sc, IEC 60909, ip=True) returns Ik″ = 22.965 kA and ip = 52.576 kA — the same values to the third decimal. The comparison, with four more cases, is documented on the validation page.

What the result means for the equipment. The board needs a breaking capacity above 22.97 kA, so the next standard rating is Icu = 25 kA. With n = 2.2, that breaker has Icm = 55 kA, against an ip of 52.58 kA: it passes, with a 4.6 % margin. Note the reason the margin is thin: the breaker was tested at power factor 0.2 (X/R ≈ 4.9), and this bus has X/R = 6.09. The busbar must declare Ipk ≥ 52.6 kA.

Add 3,000 A of locked-rotor current from running motors and the picture changes: I″kM = 1.05 · 3.0 kA = 3.15 kA, the motor peak uses its own κM = 1.3, and ip rises to 58.37 kA — above the 55 kA of making capacity, with Ik″ (26.12 kA) also above the 25 kA Icu.

Meshed networks: methods A, B and C

The single-loop approach above is exact for a radial network, where every element between source and fault is in series. When the fault is fed through a meshed network, the equivalent impedance comes from a complex network reduction, and its R/X no longer describes the transient of each branch. IEC 60909-0 offers three ways to get κ in that case:

• Method A — uniform ratio. κ is taken from the smallest R/X (largest X/R) among the relevant branches: those carrying partial short-circuit currents at the fault voltage level and the transformers adjacent to the fault. Simple and usually conservative.
• Method B — ratio at the fault location. κ_b is computed from the R/X of the equivalent impedance Z_k at the fault, and multiplied by 1.15 to cover the error of that reduction: κ = 1.15·κ_b. The product need not exceed 1.8 in LV or 2.0 in MV/HV, and the 1.15 factor is not needed if R/X stays below 0.3 in all branches.
• Method C — equivalent frequency. The network is solved again with reactances at an equivalent frequency fc = 20 Hz for 50 Hz systems (24 Hz for 60 Hz). With Z_c = R_c + jX_c from that run, R/X = (R_c/X_c)·(fc/f), and κ follows from the usual formula. Method C is the most accurate of the three and is what network analysis programs normally implement.

Applied to the worked case for illustration, method B would give 1.15 · 1.619 = 1.862, capped to 1.8 at LV — ip = 58.46 kA instead of 52.58 kA. In a radial circuit that 11 % is pure margin, which is why the standard keeps method B for meshed networks.

Common mistakes

1. Using the transformer X/R instead of the equivalent R/X. At the transformer terminals the difference is small here (κ = 1.614 vs 1.619), because the transformer dominates. Behind a cable it is not: with 30 m of 3 × (1 × 240 mm²) the equivalent X/R falls to 4.65, κ to 1.534 and ip to 46.15 kA; using the transformer’s X/R = 6 gives 48.56 kA (+5 %). After 100 m, X/R = 3.21, κ = 1.405, ip = 35.83 kA against 41.16 kA with the transformer value (+15 %).
2. Forgetting K_T. IEC 60909-0:2016 (6.3.3) applies the correction factor K_T to network transformers in the maximum case. Without it, the example gives Ik″ = 22.15 kA instead of 22.97 kA — 3.5 % low, on the unsafe side. K_T does not apply to the minimum case.
3. Confusing ip with √2·Ik″. √2·Ik″ is the crest of the symmetrical current: 32.48 kA in the example. The true peak is 52.58 kA. Dropping κ underestimates the mechanical stress by 38 %.
4. Using κ = 1.8 (or 2.0) as a fixed value. Fine as a quick upper bound in many LV boards, but it oversizes Icm and Ipk where X/R is moderate (+11 % in the example) and can be unconservative where X/R is above 13 — MV busbars near large transformers or generators.
5. Typing X/R in the exponent. The formula uses R/X. With X/R = 6.09 in the exponent, κ ≈ 1.02 and ip drops to 33.1 kA — 37 % below the real peak, on the unsafe side.
6. Applying the network κ to motor contributions. Motors have their own impedance ratio: IEC 60909-0 gives R_M/X_M = 0.42 (κ_M = 1.3) for LV motor groups and 0.10 (κ_M = 1.75) or 0.15 (κ_M = 1.65) for MV motors, depending on power per pole pair. In a radial network the partial peaks add.

What the LOGOS short-circuit calculation does, and what it does not

The LOGOS short-circuit calculator builds the impedance path from the selected bus up to the utility source — utility Skq, transformers with K_T, cables referred to the bus voltage — and returns Ik″ (maximum and minimum), ip with κ from the equivalent R/X, two-phase and phase-to-earth currents. When motor contribution is enabled, LOGOS adds I″kM = c·ΣI_LR of the motors fed from that bus, with κ_M = 1.3 at LV and 1.75 at MV, and sums the partial peaks. The physics runs in a versioned server-side function; the calculation report is exported as a Word (.docx) document.

What LOGOS does not do:

• Meshed networks. The path is radial; methods A, B and C are not implemented. For ring mains, parallel transformers with interconnected secondaries or meshed MV, use a network analysis program.
• Near-to-generator faults. The source is modelled as a utility infeed (far-from-generator). Local generators and their decaying AC component are not modelled.
• Decaying currents for breaking duty. The motor contribution is added at its initial value; the decay factors μ and q for the breaking current Ib, the DC component i_DC at a given time and the thermal equivalent current I_th are not computed. This is conservative for breaking capacity, but it is not a full time-domain study.
• Motor cables. The impedance between the bus and motors fed from intermediate panels is neglected, which overstates their contribution.

The pandapower comparison covers the network side (Ik″, ip, two-phase and minimum currents); the motor contribution follows the standard’s simplified expression and is not part of that benchmark. As with any calculation tool, the result supports the engineer’s judgment; the responsibility for the design stays with the engineer who signs it.