The earth fault can be the bigger one: LV short-circuit in calculator #012

On a 400 V main board fed by a Dyn transformer, the single-phase-to-earth fault current can be larger than the three-phase one. In the run below it is 22,89 kA against 22,18 kA — so the number that rates the switchgear is not the number most people calculate. Move to the end of a 20 m final circuit and the order reverses, hard: 12,68 kA against 15,83 kA. Same installation, same afternoon, and the answer to "which fault governs" is different at each busbar. This guide walks the LV short-circuit tool through a real network. The worked example is a real run — every figure below was taken from the calculator, not typed in by hand.

What the tool is for

Three-phase a.c. systems from 100 V to 1 000 V, per clause 1 of IEC 60909-0:2016, calculated the way that standard calculates: with the equivalent voltage source c·U_n/√3 applied at the short-circuit location.

The whole network is one table. Each row is an element of Clause 6 between two named busbars, and GND is the reference of the equivalent circuit — so a network feeder, a motor group, a generator or a converter-fed unit connects GND to its busbar, while transformers, cables, busways and reactors sit between two busbars. Mark any busbar as the fault and the nodal admittance matrix of the positive- and zero-sequence networks is built and inverted there; Z_k is the diagonal element Z_ii that clause 7.2.1 asks for. Two transformers onto one board is two rows — parallel and meshed arrangements are entered as they stand, not reduced by hand first.

Every step names its clause

A worked network

A 20 kV feeder with I″kQ 10 kA maximum and 8 kA minimum, R_Q/X_Q 0,1, c_Q 1,1; a 630 kVA 20/0,41 kV transformer with u_kr 4 % and P_krT 6,5 kW; two parallel 10 m cables at 0,077 + j0,079 Ω/km to a distribution board; then a 20 m final circuit at 0,271 + j0,087 Ω/km. System 400 V, ±6 %, 50 Hz, T_k 0,2 s, conductors at θ_e 160 °C for the minimum currents.

Three things in that table are worth stopping on.

The earth fault is the larger one at the board. I″k1 comes out of Z(1), Z(2) and Z(0) by Formula (54), and here the transformer's zero-sequence impedance is smaller than its positive-sequence impedance — X(0)/X = 0,95 for this unit, which is ordinary for a Dyn distribution transformer. Smaller Z(0) means more current. Twenty metres of cable later the picture inverts, because the cable's R(0)/R is 3 and its X(0)/X is 4,46, and the zero-sequence impedance of the route now dominates everything. If you rate a main board on the three-phase current alone, you have used the second-largest number on the page.

The minimum earth-fault current is a different order of magnitude from the maximum. 9,41 kA against 15,83 kA at the same point, because clause 7.1.2 asks for a genuinely different calculation: c_min instead of c_max, K_T = 1 instead of the correction factor, no motor and no converter contribution, and the conductor resistance recomputed at θ_e — 160 °C here — by Formula (32) with α = 0,004/K. That is the current that has to operate a protective device, and using the maximum for it is how a circuit ends up protected on paper only.

The two peak-current methods do not agree, and the standard has a preference. Clause 8.1.2 offers method a) (the smallest R/X of any branch), method b) (R/X at the fault location, with the κ then multiplied by 1,15 to cover inaccuracies), and method c) (the same network re-evaluated at an equivalent frequency of 20 Hz for a 50 Hz system, 24 Hz for 60 Hz) — and states plainly: "Method c) is recommended." At the end of the final circuit method b) gives 29,62 kA and method c) gives 25,76 kA, 15 % apart, and the tool prints both so the difference is visible rather than a matter of which spreadsheet you inherited.

The exemption in 8.1.2 b) that is easy to get wrong

Method b)'s 1,15 factor has a condition, and the wording of it matters:

As long as R/X remains smaller than 0,3 in all branches which carry a short-circuit current, it is not necessary to use the factor 1,15. It is also not necessary for the product 1,15·κ to exceed 1,8 in low-voltage networks or to exceed 2,0 in high-voltage networks.

Which carry a short-circuit current. A branch hanging below the fault — the final circuit, when the fault is at the board — carries none, and its R/X must not drag the factor in. In the run above, the fault at the main board is fed only by the feeder and the transformer, both well under 0,3, so no 1,15 applies and methods b) and c) land on the same 46,15 kA. Move the fault to the end of the final circuit and that circuit now does carry the current, at R/X 3,11, so the factor applies and the two methods separate by 15 %.

That distinction was a real defect in this tool, found while writing this guide: the branch test ran over every row in the table rather than only the rows carrying fault current, which inflated i_p by method b) at the board from 46,15 kA to 53,07 kA — 15 % of over-specified busbar bracing, in the conservative direction, for the wrong reason. It is fixed, the reference validation below still reproduces the case where the 1,15 factor genuinely does apply, and the result panel now states which way the exemption fell and why.

The protective device check

Three currents decide whether the device in front of the fault is the right one, and each goes against its own rating. A 160 A moulded-case device at the end of that final circuit, with I_cu 25 kA, I_cw 5 kA for 1 s and an operating current I_a of 1600 A:

• PASS — breaking capacity. I_cu 25,0 kA ≥ I″k3 15,83 kA (IEC 60947-2, rated ultimate short-circuit breaking capacity)
• FAIL — short-time withstand. I_cw²·t = 25,0 (kA)²·s < the fault's Joule integral of 51,4 (kA)²·s (IEC 60947-2 against IEC 60909-0 Clause 14)
• PASS — disconnection. I″k1 minimum 9409 A ≥ I_a 1600 A (IEC 60364-4-41, 411.4.4 — the fault current has to operate the device within the time of Table 41.1)

The middle one is the check that gets skipped, and it is not the same question as the first. Breaking capacity asks whether the device can interrupt the current; short-time withstand asks whether it can hold it for the clearing time — which is exactly what an upstream device is asked to do while a downstream one clears. Here it cannot: it would need I_cw ≥ 7,17 kA at 1 s for a T_k of 0,2 s.

I_cu, I_cw and I_a are not filled from any library. They are manufacturer data that no IEC standard tabulates, so the tool asks for them and says so. The rated-current list is the IEC 60947-2 and IEC 60898-1 preferred series, which is standardised.

How the implementation is checked

Eighteen cases run in the page on every load, and every one of them is a published number from IEC TR 60909-4:2000, clause 3 — the 400 V system of its Figure 6, with a 20 kV feeder, a 630 kVA and a 400 kVA transformer in parallel and four cables, solved at all three of its fault locations:

All eighteen pass, and the reference network is one click away in the tool so you can step through it yourself.

Two figures in that technical report are internally inconsistent, and the tool records them rather than quietly matching one: Z(1) at F1 is printed as 1,881 + j6,746 mΩ in 3.4.1 and as 1,881 + j6,764 mΩ in 3.5.1, a 0,2 % difference; and Table 4a prints |Z(0)| = 6,421 mΩ where its own components 2,140 + j6,009 give 6,378 mΩ, 0,7 %. Neither changes an engineering conclusion, and pretending they are not there would mean tuning the code to a typo.

Limits worth knowing before you rely on it

• n = 1, far-from-generator. The thermal equivalent current assumes the fault current does not decay. A network with a generator close to the fault needs n from Annex A, and that is not implemented.
• Zero-sequence data is an input, not an assumption. R(0)/R and X(0)/X of a cable, Z(0) of a transformer or a motor — these are manufacturer data. The tool offers ieccalc practice ratios by construction (three-core, single-core trefoil or flat, overhead line, busway) and labels them as practice to be confirmed against the cable data. I″k1 is only ever as good as that ratio.
• Above 1 kV, use the other tool. Calculator #002 covers 1 kV to 33 kV and stops at the boundary; every LV busbar there hands over its equivalent source — Z(1) and Z(0) in milliohms for both regimes — and this tool continues the same network from that busbar, reproducing the published figure exactly so the join is verifiable in one line.

The short version

I″k3 at the incoming busbar is one of five numbers a low-voltage fault study owes you, and on a Dyn-fed board it is not even the largest. The others are the minimum earth-fault current, which decides whether the protection works at all; the peak, which decides mechanical withstand and depends on which of three permitted methods you used; the Joule integral, which decides cable and device withstand; and the thermal equivalent current. They come from one impedance model, and the reason to build that model properly once is that four of the five answers change when you move one busbar down.

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