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.