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30 — Calculators

Determine the protective conductor cross-section

There are two routes to a protective conductor: calculate or allocate. Allocation by line conductor size is quick but not always the safe side — with high short-circuit currents and slow protective devices the thermal proof demands more. This calculator shows both and takes the larger.

Input

The short-circuit current returning through the protective conductor under fault.

For the protective conductor, depending on material, insulation and installation. Given in the standard you apply.

Result

Governing cross-sectionThe larger of the two routes below.
—mm²
Calculated from heatingThe thermal proof: A = I · √t / k.
—mm²
From the simplified allocationUp to 16 mm² equal to the line conductor, up to 35 mm² sixteen, above that half.
—mm²
Next standard cross-section
—mm²

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

Line conductor cross-section
35 mm²
Fault current through the protective conductor
6 kA
Disconnection time
0.2 s
Material factor k
143

Result

Governing cross-section
18.8 mm²
Calculated from heating
18.8 mm²
From the simplified allocation
16 mm²
Next standard cross-section
25 mm²

Formula

  • Thermal: A = I · √t / k
  • Simplified: A ≤ 16 mm² → A_PE = A
  • 16 < A ≤ 35 mm² → A_PE = 16 mm²
  • A > 35 mm² → A_PE = A / 2

Assumptions and standards

  • The material factor k applies to the protective conductor and differs from that for the line conductor — it depends on material, insulation and whether the protective conductor is part of the cable or laid separately. It is entered, not looked up.
  • The simplified allocation assumes protective and line conductors are of the same material. With different materials a conversion is required.
  • Protective conductors that are not part of a cable are subject to additional minimum sizes against mechanical damage.

Frequently asked

Why is the simplified allocation not always enough?

Because it does not know the fault current or the disconnection time. On short feeders with high short-circuit current and a slow protective device, heating can demand more than the allocation gives. Conversely, on small installations the allocation is usually generous.

Which disconnection time should I use?

The actual time of the protective device at this fault current, from its characteristic. With current-limiting devices the stress is lower than the formula suggests — then the let-through value from the data sheet applies.

Does this apply to a PEN conductor?

No. A PEN conductor also carries the operating current and is subject to its own minimum sizes. This calculation covers the protective conductor alone.

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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