STEP 01
Transformer Impedance
ZT = uk · Un² / SrT = 0.06 × 400² / 1,000,000 = 9.6 mΩ
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Standard Overview
What Is IEC 60909 and How Is Short Circuit Calculated?
IEC 60909 is the international standard titled Short-circuit currents in three-phase AC systems, published by the International Electrotechnical Commission. Part 0 (IEC 60909-0:2016) sets out the calculation method for both low-voltage (up to 1 kV) and high-voltage three-phase systems at 50 Hz or 60 Hz.
The standard's core idea is the equivalent voltage source method. Instead of modelling the entire network in detail, IEC 60909 lets you replace everything upstream of the fault with a single ideal voltage source behind a single equivalent impedance. The short-circuit current is then simply:
I″k = c · Un / (√3 · Zk)
where c is the voltage factor, Un is the nominal line-to-line voltage, and Zk is the total short-circuit impedance.
This matters because it makes short circuit calculation practical. You don't need a load-flow model or a transient simulation; you need the rated data of your equipment — the transformer's impedance, the generator's subtransient reactance, the cable's resistance and reactance — and the standard does the rest. Two results are always calculated: maximum short-circuit current (using cmax) for equipment rating, and minimum short-circuit current (using cmin) for relay coordination.
Core Method
Short Circuit Current Formula & Voltage Factor c
Every IEC 60909 short circuit calculation starts from the initial symmetrical short-circuit current formula. The voltage factor c deliberately overstates the source voltage so that the maximum current is always conservative for equipment rating.
Initial Symmetrical Short-Circuit Current
The master value from which all other currents are derived:
I″k = (c · Un) / (√3 · Zk)
For a simple radial network fed from a single source, the impedance is a straightforward sum: Zk = Zsource + ZT + Zcable + …
Voltage Factor c
The signature of the IEC 60909 method. It deliberately overstates the source voltage so that the maximum short-circuit current is always conservative.
| Nominal Voltage Un | cmax (max. current) | cmin (min. current) |
|---|---|---|
| Low voltage (100 V – 1 kV) | 1.05 (6%) or 1.10 (10%) | 0.95 |
| Medium voltage (>1 kV – 35 kV) | 1.10 | 1.00 |
| High voltage (>35 kV) | 1.10 | 1.00 |
Use cmax for rating and withstand checks, and cmin for protection-sensitivity checks.
Five Short-Circuit Currents
I″k, ip, Ib, Ik & Ith
The initial symmetrical current I″k is only the start. IEC 60909 derives a family of currents from it, each used to size a different piece of equipment — from breaker interrupting rating to thermal withstand of cables and busbars.
| Symbol | Name | Formula / Basis | What It Sizes |
|---|---|---|---|
| I″k | Initial symmetrical | I″k = c·Un / (√3·Zk) | Master value; basis for all others |
| ip | Peak (make) current | ip = κ·√2·I″k | Mechanical bracing, busbar withstand, breaker close-and-latch |
| Ib | Symmetrical breaking current | Ib = μ·I″k | Breaker interrupting rating |
| Ik | Steady-state current | Ik = I″k × (steady-state factor) | Generator/relay coordination, long-time settings |
| Ith | Thermal equivalent current | Ith = I″k × √(m + n) | Thermal withstand of cables, busbars, transformers |
Peak factor κ:
κ = 1.02 + 0.98 · e(−3·R/X), ranging from about
1.0 (purely resistive) up to nearly
2.0 (highly inductive, close to a generator
or transformer). This determines the mechanical forces
switchgear must withstand.
Step-by-Step Method
How to Calculate Short-Circuit Current (IEC 60909)
Let's work a complete example so the method is concrete. Calculate the maximum short-circuit current at the 400 V busbar of a 1000 kVA transformer (20 kV / 400 V, uk = 6%), using cmax = 1.05.
STEP 02
Apply KT
KT = 0.95·cmax / (1 + 0.6·xT)
= 0.95×1.05 / (1+0.6×0.06) = 0.963
ZTK = 0.963 × 9.6 = 9.24 mΩ
STEP 03
Calculate I″k
I″k = cmax·Un / (√3·Zk) = 1.05×400 / (1.732×9.24×10⁻³) = 26.3 kA
STEP 04
Peak Current ip
Transformer-fed fault, R/X ≈ 0.1, so κ ≈ 1.7.
ip = κ·√2·I″k
= 1.7 × 1.414 × 26.3
≈ 63 kA
STEP 05
Result
A breaker on this busbar must interrupt at least 26.3 kA and withstand a 63 kA peak. Motor contributions would be added if present.
Key Insight
Generator-Fed Systems Are Different
For a generator-fed system, the fault level is set by the generator's subtransient reactance X″d, not by an infinite bus. A 1000 kVA generator with X″d = 15% delivers roughly 1 / 0.15 = 6.7× its rated current — about 9.7 kA at 400 V — far lower than the transformer example above. This is the critical factor when sizing switchgear on generator-fed or islanded systems.
Standard Comparison
IEC 60909 vs ANSI / IEEE C37
North America uses ANSI/IEEE C37; most of the rest of the world uses IEC 60909. Both reduce the network to a voltage source behind an impedance, but they differ in enough details that the results — typically 5–15% apart — can change an equipment rating.
| Parameter | IEC 60909 | ANSI / IEEE C37 |
|---|---|---|
| Base current | Initial symmetrical RMS (I″k) | Symmetrical RMS at contact-parting time |
| Voltage source | Voltage factor c (cmax = 1.05–1.10) | ~1.0 pu (max operating voltage) |
| Peak current | ip = κ·√2·I″k (κ depends on X/R) | ip = 1.6 × Isym (fixed) |
| DC decay | Via time constant τ | Integrated via X/R ratio |
| Impedance corrections | Extensive (KT, KG, KM, KS) | Limited temperature corrections |
| Motor contribution | Included with μ and q decay factors | Separate, with reactance multipliers |
Practical difference: IEC 60909 usually yields a
~10% higher initial symmetrical and peak current
than ANSI for the same network, because of the cmax uplift.
The two methods should never be cross-substituted: a breaker rated
to ANSI interrupting duty may not meet the IEC breaking current,
and vice versa.
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Get Your Short-Circuit Calculation Right
ShanHua Power has manufactured diesel generator sets since 1999, with a range spanning 8 kVA to 4,000 kVA and certifications to ISO9001, CE, and CCC. We specify the subtransient reactance and short-circuit contribution of every machine we build — and help you select switchgear rated to interrupt it.
Frequently Ask Questions
Q: What is IEC 60909?
A: IEC 60909 is the international standard for calculating short-circuit currents in three-phase AC systems at 50 Hz or 60 Hz, covering low-voltage and high-voltage networks. Part 0 (IEC 60909-0:2016) defines the equivalent voltage source method and the voltage factor c.
Q: How do you calculate short-circuit current per IEC 60909?
A: Use the equivalent voltage source method: I″k = c · Un / (√3 · Zk), where c is the voltage factor, Un is the nominal voltage, and Zk is the total short-circuit impedance to the fault. Apply the transformer (KT) and generator (KG) correction factors to the impedances first.
Q: What is the voltage factor c in IEC 60909?
A: The voltage factor c artificially raises the source voltage to guarantee a conservative maximum current. For low voltage it is 1.05–1.10 (cmax) and 0.95 (cmin); for medium and high voltage it is 1.10 (cmax) and 1.00 (cmin).
Q: What are I″k, ip, Ib, and Ik?
A: I″k is the initial symmetrical short-circuit current, ip is the peak current, Ib is the symmetrical breaking current, and Ik is the steady-state current. They size switchgear withstand, breaker interrupting rating, and relay coordination respectively.
Q: How much short-circuit current does a generator contribute?
A: A generator’s contribution is limited by its subtransient reactance X″d, so it typically delivers only 6 to 12 times its rated current — much less than a utility “infinite bus” source. This is the key factor when sizing switchgear on a generator-fed system.
Q: What is the difference between IEC 60909 and ANSI/IEEE C37?
A: IEC 60909 applies a voltage factor c and calculates the initial symmetrical current with extensive correction factors, while ANSI/IEEE C37 uses a ~1.0 pu source and a fixed 1.6× peak multiplier. Results differ by roughly 5–15%, and the two must not be cross-substituted.