IEC 60909 — Short-Circuit Current Calculation
IEC 60909 is the standard used to calculate short-circuit currents in three-phase a.c. systems. It defines an equivalent voltage source at the short-circuit location and gives the impedance of every element that feeds the fault, so the result does not depend on the load flow before the fault. This page follows the calculation clause by clause, with the formula number for each step.
What is IEC 60909?
IEC 60909 is the IEC series for calculating short-circuit currents in three-phase a.c. systems. Part 0 gives the calculation procedure: the equivalent voltage source c·Un/√3 is applied at the fault location, every other source is replaced by its internal impedance, and the currents follow from the resulting network impedance.
The series is split by purpose: 60909-0 is the calculation, 60909-1 holds the factors behind it, TR 60909-2 collects typical equipment data, and TR 60909-4 publishes worked examples against which an implementation can be checked — which is exactly what the calculators on this site are checked against.
When is IEC 60909 used?
It is the reference for any calculation where the size of a fault current decides equipment or settings:
- switchgear selection — the breaking capacity a circuit-breaker must have, and the peak current the busbars and contacts must survive;
- cable and busbar verification — the thermal duty of the fault through the protection's clearing time;
- protection design — the maximum current for grading, and the minimum current that decides whether a device operates at all;
- earthing and touch-voltage studies, which need the earth-fault current rather than the three-phase one.
The scope of Part 0 covers low-voltage systems from 100 V and high-voltage systems up to 550 kV, at 50 Hz or 60 Hz. Because the low-voltage and medium-voltage rules differ in practice, the calculators here are split the same way: up to 1 kV and 1 kV to 33 kV.
Which currents does it calculate?
| Symbol | Current | What it decides | Clause |
|---|---|---|---|
| I″k | Initial symmetrical short-circuit current | Breaking capacity, thermal duty, protection grading | 7.2.1, Formula (33) |
| ip | Peak short-circuit current | Electrodynamic withstand of switchgear, busbars, supports | 8.1.1, Formulas (56) and (57) |
| Ib | Symmetrical breaking current | Duty at the moment the breaker parts its contacts | Clause 9 |
| Ik | Steady-state short-circuit current | Sustained fault, generator behaviour | Clause 10 |
| Ith | Thermal equivalent short-circuit current | Cable and equipment thermal withstand, I²t | Clause 14, Formulas (108) and (109) |
Each of those exists in a three-phase, line-to-line, line-to-line with earth connection and line-to-earth version. Which one is largest depends on the network: where Z(0) is smaller than Z(1), the line-to-earth current exceeds the three-phase one, and clause 7.5 says so explicitly.
The method: an equivalent voltage source at the fault
IEC 60909 does not simulate the pre-fault load flow. It places a single voltage source c·Un/√3 at the short-circuit location, replaces every generator, motor and feeder by its impedance, and computes the current from the network impedance seen from that point (clause 5.3, Figure 4).
That is why the calculation is reproducible: two engineers with the same network data get the same answer, independent of assumptions about loading. The price is that the method is deliberately conservative — the factor c exists precisely to cover operating voltage above nominal, transformer taps and the tolerance of the data.
The short-circuit impedance itself can be found either by network reduction or, for anything with a parallel path, from the diagonal element Zii of the nodal impedance matrix — the standard points at Annex B for that, and it is what the calculators here solve.
The voltage factor c (Table 1)
c is the factor applied to the nominal voltage to form the equivalent voltage source. It has one value for maximum-current calculations and another for minimum-current calculations, and for low voltage it depends on the tolerance of the system voltage.
| Nominal system voltage | Tolerance | c for maximum current | c for minimum current |
|---|---|---|---|
| 100 V to 1 000 V | ±6 % | 1,05 | 0,95 |
| 100 V to 1 000 V | ±10 % | 1,10 | 0,90 |
| above 1 kV to 230 kV | — | 1,10 | 1,00 |
| above 230 kV | — | 1,10 | 1,00 |
Table 1 also carries the constraint that cmaxUn should not exceed the highest voltage Um for equipment, and notes that above Um = 420 kV the factors are not defined by the standard.
Element impedances, clause by clause
Clause 6 gives the impedance of each element. Where the standard permits an assumption in the absence of data, it states the assumption — and a calculation that uses one should say so:
| Element | Clause / formula | Notes and permitted assumptions |
|---|---|---|
| Network feeder | 6.2, Formulas (4) to (6) | ZQ from I″kQ at the connection point; referred to the low-voltage side by 1/tr². Without data on the feeder resistance: RQ = 0,1 XQ with XQ = 0,995 ZQ. |
| Two-winding transformer | 6.3.1, Formulas (7) to (9) | ZT from ukr, RT from the load loss PkrT, XT from the difference. |
| Transformer correction | 6.3.3, Formula (12a) | KT = 0,95 cmax / (1 + 0,6 xT), applied to maximum currents only, and to the negative- and zero-sequence impedances as well. |
| Overhead lines and cables | 6.4 | From the conductor data; the zero-sequence impedance comes from measurement or the manufacturer. |
| Asynchronous motors | 6.10, Formulas (30) and (31) | ZM from the locked-rotor ratio. Permitted ratios: R/X = 0,10 with X = 0,995 Z for HV motors ≥ 1 MW per pole pair; 0,15 with 0,989 Z below 1 MW; 0,42 with X = 0,922 Z for LV motor groups with their connection cables. |
| Static converter drives | 6.11 | Treated as a motor with ILR/IrM = 3 and R/X = 0,10; all other static converters are disregarded. |
| Generators, for ip | 8.1.1 | Fictitious resistances RGf = 0,05 X″d (UrG > 1 kV, SrG ≥ 100 MVA), 0,07 X″d (below 100 MVA), 0,15 X″d for UrG ≤ 1 000 V. |
| Capacitors, non-rotating loads | 6.12 | Not included; the discharge current of shunt capacitors may be neglected for ip. |
Initial symmetrical short-circuit current I″k
I″k is the r.m.s. value of the symmetrical current at the instant the fault appears, and it is the current every other short-circuit quantity is derived from.
where c is the voltage factor of Table 1, Un the nominal system voltage and Zk = Rk + jXk the short-circuit impedance seen from the fault location.
The unbalanced faults
The line-to-earth current needs the zero-sequence impedance, which is where most of the uncertainty in an earth-fault calculation lives: for a cable it is manufacturer data, and for a transformer it depends on the winding connection and the star-point earthing.
Peak short-circuit current i_p
ip is the largest instantaneous value the current reaches, and it is what the electrodynamic withstand of switchgear and busbars is checked against. It follows from I″k through the factor κ, which depends on R/X.
For a single-fed fault the R/X of that one branch is used. For a multiple-fed or meshed network clause 8.1.2 offers three ways, and they do not give the same answer:
| Method | Basis | What the standard requires |
|---|---|---|
| a) | Uniform ratio R/X | The smallest R/X of all branches carrying fault current is used for the whole network. |
| b) | R/X at the fault location | κ from Rk/Xk, then multiplied by 1,15 to cover inaccuracy. The factor is not needed while R/X stays below 0,3 in every branch, and the product 1,15κ need not exceed 1,8 in low-voltage networks or 2,0 in high-voltage networks. |
| c) | Equivalent frequency | The network impedance is evaluated at fc = 20 Hz for a 50 Hz system or 24 Hz for 60 Hz, and R/X is scaled by fc/f per Formula (62). This is the method the standard recommends. |
In practice method c) usually gives the lower and more realistic peak, and method b) the conservative one. A calculation that reports only one of them is hiding a decision; the LV calculator prints both.
Thermal equivalent current I_th and the Joule integral
Ith is the constant current that would heat the equipment as much as the real, decaying fault current does over the fault duration Tk. It is what a cable's or busbar's thermal withstand is compared against.
where m covers the heat effect of the d.c. component and n that of the decaying a.c. component. The formula for m is given in Annex A; for a far-from-generator short circuit I″k/Ik = 1 and therefore n = 1.
Maximum and minimum short-circuit current — two different calculations
They are not the same calculation with different numbers — clause 7.1.2 changes the rules. The maximum current sizes the equipment; the minimum current decides whether protection operates at all, and it must be calculated with its own set of assumptions.
| Aspect | Maximum current | Minimum current |
|---|---|---|
| Voltage factor | cmax | cmin |
| System configuration | The arrangement giving the largest contribution | The arrangement giving the smallest contribution |
| Impedance correction factors | Applied (KT, KG, KS) | Not applied |
| Motors | Included where relevant, per 6.10 | Neglected (item 6) |
| Photovoltaic station units | Included | Neglected (item 5) |
| Conductor resistance | At 20 °C (item f) | At the conductor temperature at the end of the fault, θe, by Formula (32) with α = 0,004/K (item 7) |
Raising the conductor resistance and lowering the voltage factor is what makes the minimum current genuinely small — and it is the number to compare against the operating current of a protective device, as IEC 60364-4-41 clause 411.4.4 requires for disconnection.
Calculation workflow
- Fix the nominal voltage and pick c from Table 1, for the regime you are calculating.
- Take the impedance of every element from clause 6, referring everything to the voltage side where the fault sits.
- Apply the impedance correction factors — for maximum currents only (6.3.3, 6.6, 6.7).
- Reduce the network, or invert the nodal admittance matrix, to get Zk at the fault busbar.
- Compute I″k by Formula (33), then the unbalanced faults by (45), (46) and (54) — the last needs Z(0).
- Compute ip by (56) and (57), choosing the method of 8.1.2 for a meshed network.
- Compute Ith and the Joule integral for the actual clearing time (Clause 14).
- Repeat the whole thing for the minimum regime under 7.1.2, and use that result for protection sensitivity.
Worked example and reference case
IEC TR 60909-4:2000 clause 3 publishes a complete 400 V example: a 20 kV feeder with I″kQ = 10 kA, a 630 kVA and a 400 kVA transformer in parallel, four cables and three fault locations. Its printed answers are I″k3 = 34,62 kA, 34,12 kA and 6,95 kA, with a peak current of 81,36 kA by method b and 70,85 kA by method c at the first location.
That case is the reference the LV short-circuit calculator is pinned to — the comparison runs inside the tool, and the summary for every engine on this site is on the validation page. Two internal inconsistencies in that report are recorded there rather than smoothed over.
IECCalc calculators for IEC 60909
- #012 Short circuit — LV up to 1 kV One network table from feeder to fault; I″k3, I″k2, I″k1, peak current by methods b and c, I_th, minimum currents and the protective-device check.
- #002 Short circuit — MV 1 to 33 kV Sequence networks, several sources in parallel per busbar, fault level at every busbar, earthing-arrangement selector.
- #010 Transformer sizing and selection The short-circuit contribution a transformer brings to its low-voltage side, with cmax and the impedance band of IEC 60076-5.