In short. The IEC 60909 method replaces the network’s sources with one equivalent voltage source at the fault point and calculates the initial symmetrical short-circuit current from the network impedance seen from that point. From it, it derives peak, breaking and thermal values for equipment selection.
The core idea
At the fault location, a voltage source of magnitude c × Un / √3 is applied, where Un is the nominal system voltage and c is a voltage factor that accounts for voltage variation. All other sources are set to zero. The current that flows is the initial symmetrical short-circuit current, I″k.
I″k = c × Un ÷ (√3 × Zk)
Zk is the equivalent impedance at the fault point seen from the equivalent source. The voltage factor takes different values for maximum and minimum calculations and for different voltage levels, as given in the standard.
Quantities calculated
| Quantity | Symbol | Used for |
|---|---|---|
| Initial symmetrical current | I″k | Starting point; protection and rating |
| Peak short-circuit current | ip | Making capacity, mechanical stress on busbars |
| Symmetrical breaking current | Ib | Breaker breaking capacity |
| Steady-state current | Ik | Thermal effects and long-duration faults |
| Thermal equivalent current | Ith | Thermal withstand of equipment |
The peak value includes a factor κ that depends on the ratio of resistance to reactance (R/X) of the network. A low R/X means a longer DC component and a higher peak.
Treatment of elements
- Network feeder: impedance from the supply fault level.
- Transformers: impedance corrected with factors in the standard.
- Synchronous generators: subtransient reactance with correction factors, and decay over time for breaking and steady-state values. See generator contribution.
- Motors: contribute to I″k, ip and Ib, with decay, depending on size and distance.
- Converters: contribution depends on design and is limited. See inverter behaviour.
Maximum and minimum calculations
For maximum currents, use the maximum voltage factor, minimum impedances and include motors. For minimum currents, use the minimum factor, higher conductor temperature resistances and exclude motor contribution. See maximum and minimum studies.
Practical workflow
- Collect impedance data for all elements in per-unit or ohms on a common base.
- Reduce the network to the equivalent impedance at each fault location.
- Apply correction factors and calculate I″k, then ip, Ib, Ik.
- Compare with equipment ratings.
- Feed the results to protection and arc-flash studies.
Illustrative screening example
On a 415 V system with a 1,000 kVA transformer at 5 percent impedance fed from a strong utility, the transformer limits the fault current to about twenty times its rated current. A rapid hand check using this relation helps confirm software output is in the right range.
The IEC 60909 method in outline
IEC 60909 replaces the real network with an equivalent voltage source at the fault point, which allows a consistent calculation without a prior load flow. It applies voltage factors (c factors) to cover voltage variation.
Initial symmetrical short-circuit current I”k = c × Un ÷ (√3 × Zk)
Quantities it provides
| Symbol | Meaning | Use |
|---|---|---|
| I”k | Initial symmetrical current | Starting point |
| ip | Peak current, ip = κ × √2 × I”k | Making capacity, bus bracing |
| Ib | Symmetrical breaking current | Breaker breaking duty |
| Ik | Steady-state current | Thermal and protection settings |
| Ith | Thermal equivalent current | Cable and equipment withstand |
Worked point (illustrative)
At 415 V, c = 1.05 for maximum case, Zk = 15 milliohm: I”k = 1.05 × 415 ÷ (1.732 × 0.015) ≈ 16.8 kA. With X/R of 5, κ is about 1.4, so peak ip ≈ 1.4 × 1.414 × 16.8 ≈ 33 kA.
Generator-specific treatment
The standard applies correction factors to generator impedance and distinguishes generators that feed directly into the network from those connected via a unit transformer. Use the standard’s current edition and confirm its treatment with the study software vendor.
Why X/R matters
High X/R values increase the peak and the DC component, which can exceed a breaker’s asymmetrical rating even when the symmetrical level is within limits.
See study architecture and maximum and minimum studies.
Common mistakes
- Using one voltage factor for all cases.
- Ignoring R/X for the peak current.
- Not applying the standard’s correction factors.
What to send us
Send network data. Send your application details. See also short-circuit study architecture.
Questions engineers ask
What is the voltage factor c?
A factor in the standard that accounts for voltage variation and other effects in the equivalent source.
Why does peak current exceed the symmetrical value?
Because the first cycle includes a DC component, depending on the network’s R/X ratio.
IEC or ANSI?
Use the standard your equipment and region follow, and state it in the report.
Is hand calculation acceptable?
For simple systems and as a check. Complex networks need software.
This guide gives general principles. Calculations should follow the current edition of the standard.