The cross-section of the protective conductor (PE) is determined according to the phase cross-section: S ≤ 16 → PE = S; 16 < S ≤ 35 → PE = 16; S > 35 → PE = S/2. The main equipotential bonding conductor must be ≥ half the PE, with a minimum of 6 mm² (Cu) and an upper limit of 25 mm². For example, for a 25 mm² phase, the PE would be 16 mm² and the main busbar 10 mm².

PE: S≤16→S · 1635→S/2 · Main busbar ≥ PE/2 (min 6, max 25 mm² Cu)
Example calculation: Largest phase 25 mm² → PE 16 mm² (16<25≤35); main equipotential busbar ≥ 16/2 = 8 → 10 mm² (min 6, max 25 Cu).

Main Grounding / Equipotential Busbar Cross-Section

View the cross-section of the protective conductor (PE) and the main equipotential bonding conductor from the cross-section of the largest phase conductor (TS EN 60364-5-54).

Protective conductor (PE)
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Main equipotential bus
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Notes: PE rule (TS EN 60364-5-54, 543.1.2): phase S ≤ 16 → PE = S; 16 < S ≤ 35 → PE = 16; S > 35 → PE = S/2. Main equipotential bonding conductor ≥ PE/2, minimum 6 mm² (Cu), upper limit 25 mm² (Cu equivalent). Earthing conductor min 25 mm² Cu (unshielded buried) or 6 mm² (shielded). Copper based; for aluminum/steel, the equivalent is considered.
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How is it calculated?

  1. Select the largest phase conductor cross-section (mm²).
  2. Press Calculate; the PE and main equipotential busbar cross-sections will be displayed.

Frequently Asked Questions

What should the area of the main grounding conductor be in mm²?
The protective conductor (PE) is determined according to the phase cross-section (543.1.2); the main equipotential bonding conductor must be half the PE, and at least 6 mm² of copper.
What is the cross-section of a grounding conductor (electrode)?
If unshielded, at least 25 mm² of copper; if mechanically shielded, at least 6 mm² of copper (TS EN 60364-5-54).
Why is the busbar cross-section dependent on the phase cross-section?
To safely carry the fault current, the protective and connecting conductors are selected in proportion to the cross-section of the phase conductor.

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