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Short-Circuit Current Calculator — IEC 60909

Short-Circuit Current Calculator — IEC 60909

Compute prospective short-circuit currents (Ik3 max/min, Ik1 Ph-N, peak, thermal loading) at the end of a cable run per IEC 60909. Built-in 100–2500 kVA transformer library, parallel transformers and conductors.

3-phase Isc max (kA) (kA) —
3-phase Isc min (kA) (kA) —
1-phase Isc Ph-N (kA) (kA) —
Peak current ip (kA) (kA) —
Total impedance Zk (mΩ) (mΩ) —
Thermal loading Ik·√t (kA√s) —
Formule 
SOURCE — TRANSFORMER OR UPSTREAM ISC

  Z_trafo  =  u_k × U_LV² / S_r        (Ω, transformer LV impedance)
  Z_source =  c × Un / (√3 × Isc_up)   (manual entry)

  Variables :
    u_k       transformer short-circuit voltage (%)
    S_r       rated power (VA)
    U_LV      LV nominal voltage (V)
    Isc_up    Isc available before the cable (kA, utility data)

  Example : 630 kVA, u_k 4 %, 400 V → 0.04·400²/630000 = 10.16 mΩ.
  The source is split into R/X (inductive-dominant, X/R ≈ 5).
  Parallel transformers divide Z_source by their number.


SHORT-CIRCUIT CURRENTS

  Ik3_max  =  c_max × Un / (√3 × Zk)          c_max = 1.05
  Ik3_min  =  c_min × Un / (√3 × Zk,θ)        c_min = 0.95, hot R
  Ik1      =  c_min × (Un/√3) / Z_loop        Ph-N fault (TN system)

  Rationale : Ik3_max sizes the breaking capacity Icu (Icu ≥ Ik3_max).
  Ik3_min and Ik1 (the weakest: c = 0.95 and hot conductors) verify that
  protection actually trips — the automatic-disconnection condition of
  IEC 60364-4-41.


CABLE · PEAK · THERMAL

  R_c = ρ·L / (S·n)     X_c = X'·L / n       (n conductors per phase)
  ip  = κ × √2 × Ik3_max     κ = 1.02 + 0.98·e^(−3R/X)
  Ith·√t = Ik3_max × √t                      (thermal loading, kA√s)

  Variables :
    ρ    0.0175 (Cu) · 0.029 (Al)  Ω·mm²/m at 20°C
    X'   0.08 mΩ/m  (typical LV cable per-metre reactance)
    t    protection clearing time (s)

  Rationale : ip mechanically stresses busbars (making capacity).
  Ith·√t is compared with the cable withstand k·S (I²t ≤ k²S²).


CHECKS

  Ik3_max  ≤  Icu   (breaker ultimate breaking capacity)
  ip       ≤  Ipk   (admissible peak — busbars, making)
  Ik1_min  ≥  magnetic trip threshold (automatic disconnection)

Reference: IEC 60909-0:2016, IEC 60364-4-41

Short-circuit at end of cable ∼ SourceU_nBusbarIsc_amontCable (R + jX)L, S → RcableIk3→ kAFault Ik3 = c × Un / (√3 × Zk) where Zk = Zs + Rcable Check : Ik3 ≤ Icu of breaker · ip ≤ peak withstand current

The prospective short-circuit current is the fundamental quantity for selecting circuit breakers (Icu / Icw), fuses and busbars. IEC 60909-0 defines the impedance method c·Un/√3·Zk — this calculator implements its simplified form for LV networks.

How to use:

  1. Pick your transformer from the list (rated power, u_k and LV impedance are pre-filled), or select “Manual” to enter the upstream Isc in kA (utility data).
  2. Where relevant, set how many transformers run in parallel.
  3. Fill in the cable: material, cross-section, length, and conductors per phase (parallel cables).
  4. Set the protection clearing time for the thermal loading.

What you get: Ik3 max (sizes Icu) and Ik3 min (c = 0.95, hot conductors), the phase-neutral fault Ik1 (disconnection in a TN system), the peak current ip, the total impedance and the thermal loading Ik·√t to compare with the cable withstand k·S.

Assumptions: the source is split into an inductive-dominant R/X (X/R ≈ 5); cable reactance is taken at 0.08 mΩ/m; the upstream HV network is neglected unless folded into the upstream Isc. For HV faults or a fine selectivity study, use the full complex IEC 60909 method.

Standards

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