FISEKON GmbH – Fischer Elektrokonstruktion

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13 — Calculators

Calculate short-circuit current at the end of a cable

At the start of a cable the short-circuit current is large; at the far end it is small. That is exactly where it is decided whether the protective device still operates under fault — and exactly where nobody recalculates, because the cable "has always run like that".

Input

For the minimum short-circuit current — that is, for the disconnection condition — a value below 1 is used. The standard you apply governs.

Network and transformer up to the start of this cable. From the network impedance calculator or from a measurement.

The current at which the device operates within the required disconnection time. From the characteristic in the data sheet, not the rated current.

Result

Short-circuit current line–protective conductorThe smallest fault current — the disconnection condition turns on it.
—A
Disconnection condition
—
Three-phase short-circuit current
—A
Total loop impedance
—Ω
Ratio of I_k1 to I_a
—

A guide, not a design to standard. We give no warranty for the correctness of the results, the applicable standards and case-by-case verification govern.

Worked example

The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.

Inputs

System voltage (line to line)
400 V
Voltage factor c
0.95
Upstream impedance
0.05 Ω
One-way cable length
60 m
Line conductor cross-section
16 mm²
Protective conductor cross-section
16 mm²
Conductor material
Copper (κ = 56 m/Ω·mm²)
Conductor temperature under fault
80 °C
Operating current of the protective device I_a
160 A

Result

Short-circuit current line–protective conductor
1,018 A
Disconnection condition
Met. The fault current is well above the operating current.
Three-phase short-circuit current
1,653 A
Total loop impedance
0.216 Ω
Ratio of I_k1 to I_a
6.36

Formula

  • Z_loop = Z_upstream + R_L + R_PE
  • I_k1 = c · U₀ / Z_loop with U₀ = U / √3
  • I_k3 = c · U / (√3 · Z_three-phase)
  • Disconnection condition: I_k1 ≥ I_a

Assumptions and standards

  • The upstream impedance is given as a magnitude; its angle is unknown. It is therefore added arithmetically. That slightly overstates the loop and yields the smaller fault current — the safe side.
  • The calculation uses a heated conductor, because the loop impedance is then highest and the fault current smallest.
  • The operating current is not the rated current. It comes from the device characteristic and belongs to the required disconnection time.

Frequently asked

Why is the rated current of the device not enough?

Because a miniature circuit breaker only trips magnetically at a multiple of its rated current. With a type C characteristic that is around ten times. A 16 A device therefore needs roughly 160 A of fault current to disconnect within the required time — not 16 A.

What does "tight" mean in the assessment?

That the condition is met on paper but without meaningful margin. The inputs are rarely exact: the upstream impedance varies with the state of the network, the conductor temperature with the load. Below 1.25 times the operating current I would measure rather than rely on the calculation.

What if the condition is not met?

Four routes: a larger cross-section, a shorter cable, a device with a lower operating current or a different characteristic — or a residual current device, which disconnects independently of the short-circuit current. Which route is right depends on the installation.

Calculating is the easy part.

A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.

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