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Transformer short-circuit current calculator
Estimate the three-phase short-circuit current at a transformer’s secondary terminals with the IEC 60909 method: initial current, peak current and fault level. Free, no signup.
- IEC 60909 voltage and peak factors
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Initial short-circuit current I″k
—
Peak current ip: —
Check the inputs: every value must be a positive number.
- Short-circuit power S″k
- —
- Peak factor κ
- —
- Transformer rated current
- —
- Transformer impedance ZT
- —
- Network impedance ZQ
- —
- Total impedance Zk
- —
- R/X of the fault loop
- —
Estimate for a balanced three-phase fault at the transformer terminals. Cables, motors and IEC 60909 correction factors are not included.
How the calculation works
The calculator follows the IEC 60909-0 method for a balanced three-phase fault at the transformer’s secondary terminals.
- Transformer impedanceZT = (uk / 100) · UrT² / SrTSplit into RT and XT with the X/R ratio.
- Upstream network impedance, referred to the secondary sideZQ = c · UrT² / S″kQWith XQ = 0.995 ZQ and RQ = 0.1 XQ, the IEC 60909 values when the network’s R/X is not known.
- Initial symmetrical short-circuit currentI″k = c · UrT / (√3 · |ZQ + ZT|)
- Peak short-circuit currentip = κ · √2 · I″k, κ = 1.02 + 0.98 · e−3R/X
- Short-circuit powerS″k = √3 · UrT · I″k
Worked example
A 1000 kVA, 20/0.4 kV transformer with uk = 6 % and X/R = 6, fed from a 500 MVA network, with c = 1.10:
| Transformer impedance ZT | 9.6 mΩ |
|---|---|
| Network impedance ZQ | 0.352 mΩ |
| Initial short-circuit current I″k | 25.53 kA |
| Peak factor κ | 1.619 |
| Peak current ip | 58.44 kA |
| Short-circuit power S″k | 17.69 MVA |
With an infinite source and c = 1.00 you get the familiar rule of thumb: rated current divided by uk, 1443 A / 0.06 = 24.06 kA.
What this estimate leaves out
- Cables, busbars and protective devices between the transformer and the fault, which lower the current.
- Motor and generator contributions, which raise the current in the first cycles.
- IEC 60909 impedance correction factors, such as KT for network transformers, which change the result by a few percent.
- Unbalanced faults (line-to-earth, line-to-line), which need zero- and negative-sequence data.
For a complete study, read the short circuit analysis guide and the 33 MW wind farm example.
Frequently asked questions
What is the voltage factor c?
IEC 60909 multiplies the nominal voltage by c to cover voltage variations in the network. For maximum currents use cmax = 1.10 at medium and high voltage, and 1.05 (systems with +6 % tolerance) or 1.10 (+10 % tolerance) at low voltage. For minimum currents use cmin = 1.00 at medium and high voltage, and 0.95 or 0.90 at low voltage.
Why is the result higher than the rated current divided by uk?
Because c = 1.10 raises the driving voltage by 10 %. The upstream network adds some impedance, which lowers the current slightly. With c = 1.00 and an infinite source the calculator gives exactly the rated current divided by uk.
I don’t know the network short-circuit power. What should I enter?
Ask the network operator for the fault level at the point of connection. Leaving the field empty assumes an infinite source, which gives the highest possible current: a safe choice when you check the breaking capacity of switchgear.
What is the peak current ip used for?
Switchgear and busbars must withstand the electromagnetic forces of the first current peak, so ip is compared with their peak withstand and making capacity. The breaking capacity of circuit breakers is checked against I″k or the breaking current.
Does the result include motor contribution?
No. Running motors feed the fault for the first few cycles, so the total current needs a network study with the motors modelled.