What fault current does a 630 kVA transformer at 5 per cent impedance give?

About 16.8 kA at its LV terminals. The base impedance is 433² / 630,000 = 297.6 mΩ, and 5 per cent of that is 14.9 mΩ, so the terminal fault is 433 / (√3 × 0.0149). The shortcut is that a transformer pushes roughly 100 divided by its impedance percentage times its own full load current into a bolted fault — 20 times full load at 5 per cent.

Why is the line-to-neutral fault current so much lower than the three-phase figure?

Two reasons that compound. The loop is twice the cable length, because the current goes out on a phase and back on the neutral, while the driving voltage is only V/√3 rather than V. Less voltage across more impedance puts it typically at a third to a half of the three-phase figure at the same point. A 3.5-core cable makes it worse still, because the reduced neutral has higher resistance than this calculation assumes.

Which figure do I use — maximum or minimum fault?

Both, for opposite purposes. The maximum — a bolted three-phase fault at the origin — is what switchgear has to break and what busbar bracing has to survive, so bigger is worse. The minimum — line to neutral at the far end of the longest circuit — has to be large enough to drive the protective device into its magnetic region, so smaller is worse. A circuit can pass the first comfortably and fail the second, and when it does, a genuine short circuit is cleared by the thermal element seconds later with the cable carrying fault current throughout.

Where does the 2.5 times peak factor come from?

From the DC offset in the first cycle, and it is not always 2.5. The peak is κ × √2 × Isc, where κ = 1.02 + 0.98 e^(−3R/X). A stiff, reactive supply gives κ near 1.8 and a peak around 2.5 times the rms figure, which is the conservative number used for switchgear at the transformer terminals. Add a long cable run and the circuit turns resistive, κ falls towards 1.0, and the peak falls with it — which is why this tool computes κ from the actual X/R rather than assuming it.

What does this calculation leave out?

Four things, three of them in the safe direction. The impedance of the 11 kV network behind the transformer is ignored, which overstates the fault slightly — for an approval submission use the DISCOM's declared fault level. Arc resistance is ignored, which also overstates it. Busbar and joint impedance is small but not zero. The one that goes the other way is motor contribution: running motors feed a fault for the first few cycles, so on a motor-heavy plant add roughly four times the connected motor full load current to the peak.

Short Circuit Current Calculator

Two fault currents, pulling opposite ways

Every point in an installation has two fault figures that matter, and they are used for opposite purposes:

Maximum fault — a bolted three-phase fault at the origin. This is what switchgear must break, and what busbar bracing must survive. Bigger is worse.
Minimum fault — a line-to-neutral fault at the far end of the longest circuit. This must still be large enough to operate the protective device quickly. Smaller is worse.

A circuit can pass the first comfortably and fail the second. When it does, a genuine short circuit is not seen by the magnetic element at all and is cleared — eventually — by the thermal element, seconds later, with the cable carrying fault current the whole time.

Building the impedance

Start at the transformer. Its per-unit impedance scales the base impedance of the system:

Z_base = V² / S
Z_tx   = Z_base × (Z% / 100)

For a 630 kVA transformer at 433 V and 5 %: Z_base = 433² / 630,000 = 297.6 mΩ, so Z_tx = 14.9 mΩ, and the terminal fault is 433 / (√3 × 0.0149) = 16.8 kA.

That impedance is mostly reactance. Splitting it needs the X/R ratio, typically around 6 for an LV distribution transformer — which puts R at about a sixth of X.

Then add the cable:

R_cable = ρ × L / (A × runs)      ρ = 18.51 mΩ·mm²/m copper
X_cable ≈ 0.08 mΩ/m               near enough constant

Resistance is taken at 20 °C. That is deliberate: a cold conductor has lower resistance, so it gives the highest fault current, which is the conservative case for rating switchgear. IEC 60909 calls this the maximum-current condition.

For the minimum fault the standard does the opposite and uses the conductor hot, because that gives the lowest current. If a circuit is marginal on the minimum-fault check, do it again with resistance raised by about 25 % for a 70 °C conductor before deciding it passes.

Why the line-to-neutral figure is so much lower

A three-phase fault sees the phase impedance once. A line-to-neutral fault sees the phase conductor and the neutral — the loop is twice the cable length — while the driving voltage is only V/√3. Less voltage across more impedance, so the current is typically a third to a half of the three-phase figure at the same point.

Two things make it worse in practice. A 3.5-core cable has a reduced neutral, so the return path has higher resistance than this calculation assumes. And on a long run the cable dominates, so the fault current becomes almost independent of the transformer — doubling the transformer size barely moves it.

Peak, and where 2.5 comes from

κ  = 1.02 + 0.98 e^(−3R/X)
ip = κ × √2 × I_sc

The first cycle of a fault is offset by a DC component, so the instantaneous peak is well above the rms value. κ depends on the X/R ratio at the fault point: a stiff, reactive supply gives κ near 1.8 and a peak of 2.5 × the rms figure, while a long cable run makes the circuit resistive, κ falls towards 1.0, and the peak drops with it.

The transformer calculator uses a fixed 2.5 for a quick terminal estimate, which is the conservative switchgear figure. This tool computes κ from the actual X/R once the cable is included, which is the more accurate number for a point downstream.

The withstand check

The last panel checks the cable itself against the fault it has to hold:

S ≥ √(I² t) / k

with k = 115 for PVC-insulated copper and 143 for XLPE copper — the phase conductor values, which are lower than the protective conductor values because a phase conductor is already at its 70 °C operating temperature when the fault starts. Aluminium is 76 and 94.

Cross-section scales with the square root of time, so a device six times slower needs a conductor about two and a half times larger. Every discrimination delay you set upstream is paid for in copper downstream. The reasoning is worked through in short-circuit withstand.

What this does not include

Source impedance. The 11 kV network behind the transformer is ignored, which overstates the fault current slightly — the safe direction. For a design being submitted for approval, use the DISCOM's declared fault level at the point of supply.
Motor contribution. Running motors feed a fault for the first few cycles. On a motor-heavy plant, add roughly four times the connected motor full load current to the peak.
Arc resistance. Real faults are rarely bolted. Arc resistance lowers the current, which matters for the minimum-fault check — another reason to leave margin there rather than accept a marginal pass.
Busbar and joint impedance, which is small but not zero on a long busbar chamber.

Questions people ask

What fault current does a 630 kVA transformer at 5 per cent impedance give?: About 16.8 kA at its LV terminals. The base impedance is 433² / 630,000 = 297.6 mΩ, and 5 per cent of that is 14.9 mΩ, so the terminal fault is 433 / (√3 × 0.0149). The shortcut is that a transformer pushes roughly 100 divided by its impedance percentage times its own full load current into a bolted fault — 20 times full load at 5 per cent.
Why is the line-to-neutral fault current so much lower than the three-phase figure?: Two reasons that compound. The loop is twice the cable length, because the current goes out on a phase and back on the neutral, while the driving voltage is only V/√3 rather than V. Less voltage across more impedance puts it typically at a third to a half of the three-phase figure at the same point. A 3.5-core cable makes it worse still, because the reduced neutral has higher resistance than this calculation assumes.
Which figure do I use — maximum or minimum fault?: Both, for opposite purposes. The maximum — a bolted three-phase fault at the origin — is what switchgear has to break and what busbar bracing has to survive, so bigger is worse. The minimum — line to neutral at the far end of the longest circuit — has to be large enough to drive the protective device into its magnetic region, so smaller is worse. A circuit can pass the first comfortably and fail the second, and when it does, a genuine short circuit is cleared by the thermal element seconds later with the cable carrying fault current throughout.
Where does the 2.5 times peak factor come from?: From the DC offset in the first cycle, and it is not always 2.5. The peak is κ × √2 × Isc, where κ = 1.02 + 0.98 e^(−3R/X). A stiff, reactive supply gives κ near 1.8 and a peak around 2.5 times the rms figure, which is the conservative number used for switchgear at the transformer terminals. Add a long cable run and the circuit turns resistive, κ falls towards 1.0, and the peak falls with it — which is why this tool computes κ from the actual X/R rather than assuming it.
What does this calculation leave out?: Four things, three of them in the safe direction. The impedance of the 11 kV network behind the transformer is ignored, which overstates the fault slightly — for an approval submission use the DISCOM's declared fault level. Arc resistance is ignored, which also overstates it. Busbar and joint impedance is small but not zero. The one that goes the other way is motor contribution: running motors feed a fault for the first few cycles, so on a motor-heavy plant add roughly four times the connected motor full load current to the peak.

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