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Fault current at the end of a cable calculator
Work out the maximum three-phase and the minimum line-to-earth fault current at the far end of a low-voltage cable fed from a transformer, with the IEC 60909 method. Free, no signup.
- Maximum and minimum fault current
- Results update as you type
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Maximum three-phase fault current at the cable end
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Minimum line-to-earth: —
Check the inputs: positive values, and a temperature between 20 and 400 °C.
- Peak current ip at the cable end
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- Three-phase at the transformer terminals
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- Minimum line-to-earth at the cable end
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- Earth-fault loop impedance Zs
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- Cable impedance per phase at 20 °C
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- R/X at the fault (maximum case)
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Maximum: IEC 60909 with cmax and the cable resistance at 20 °C. Minimum line-to-earth: TN system, Dyn transformer, protective conductor the same size as the phase conductors, cmin and the resistance at the end-of-fault temperature. Motor contributions are not included.
How the calculation works
The calculator adds the impedances of the upstream network, the transformer and the cable, using the IEC 60909 method. It gives the maximum three-phase current for equipment ratings and the minimum line-to-earth current for checking the protection.
- ImpedancesZT = (uk/100) · Un² / SrT, ZQ = c · Un² / S″kQ, ZL = (RL + jXL) · L / nXQ = 0.995 ZQ and RQ = 0.1 XQ; RL from IEC 60228 for the conductor size.
- Maximum three-phase currentI″k3 = cmax · Un / (√3 · |ZQ + ZT + ZL|)With the cable resistance at 20 °C. Peak current ip = κ · √2 · I″k3.
- Minimum line-to-earth currentI″k1 = √3 · cmin · Un / |2 Z(1) + Z(0)|Z(1) = ZQ + ZT + ZL; Z(0) = ZT + 4 ZL for a Dyn transformer and a protective conductor the same size as the phases.
- Cable resistance for the minimum currentRL = RL20 · (1 + 0.004 · (θe − 20))θe is the conductor temperature at the end of the fault.
- Earth-fault loop impedanceZs = |2 Z(1) + Z(0)| / 3, I″k1 = cmin · U0 / ZsU0 = Un / √3 is the line-to-earth voltage.
Worked example
A 1000 kVA, 0.4 kV transformer (uk = 6 %, X/R = 6) on a 500 MVA network, feeding 100 m of 95 mm² copper cable (X = 0.08 Ω/km), with cmax = 1.10, cmin = 0.95 and θe = 160 °C:
| Three-phase at the transformer terminals | 25.53 kA |
|---|---|
| Three-phase at the cable end | 9.25 kA |
| Peak current at the cable end | 13.72 kA |
| Minimum line-to-earth at the cable end | 3.28 kA |
| Earth-fault loop impedance | 66.9 mΩ |
Using the results
- The maximum current sets the breaking capacity of the devices at the cable end and the short-circuit withstand of the cable; see the cable sizing calculator.
- The minimum current must make the protective device disconnect in time. IEC 60364-4-41 asks for 0.4 s on final circuits up to 32 A at 230 V in TN systems, and 5 s on distribution circuits.
- Not included: motor contributions, arc resistance, TT and IT systems, protective conductors smaller than the phases, and IEC 60909 correction factors.
Get these formulas and more in a free two-page PDF cheat sheet
Frequently asked questions
Why calculate both a maximum and a minimum fault current?
The maximum current, a bolted three-phase fault with cmax, checks that switchgear can break the fault and that cables withstand it. The minimum current, usually a line-to-earth fault at the far end of the cable, checks that the protective device still trips fast enough.
Why is the cable resistance higher for the minimum current?
IEC 60909 calculates minimum currents with the conductor resistance at the temperature it reaches at the end of the fault, for example 160 °C for PVC or 250 °C for XLPE insulation. The higher resistance gives a lower, conservative current.
What if the protective conductor is smaller than the phase conductors?
Then the loop impedance is higher and the real minimum current is lower than the calculator shows, because it assumes a protective conductor the same size as the phases. Check those circuits with the actual conductor or with a network study.
How do I check that a circuit breaker will trip in time?
Compare the minimum line-to-earth current with the current that trips the device within the required time. For miniature circuit breakers that is the instantaneous range: up to 5 × In for type B, 10 × In for type C and 20 × In for type D. For fuses, read the time-current curve.
Does the result include motor contribution?
No. Running motors feed the fault for the first few cycles and raise the maximum current. A network study with the motors modelled, such as an IEC 60909 study in Electrisim, includes them.