Calculators
Determine the protective conductor cross-section
There are two routes to a protective conductor, calculation or allocation. Allocation by line conductor size is quick but not always on 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.
Result
- Governing cross-sectionThe larger of the two routes below.
- —mm²
- Calculated from heatingThe thermal proof using 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
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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.
The grey values are an example. Simply enter your own.
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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.
Method and questions
Protective conductor size
Two routes to the size, via the thermal proof or the simplified allocation.
Discuss a projectWorked 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 and standards
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.
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.
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