Construction of cable cores according to IEC 60228

When we say a conductor has a “stranded core" or a "class 2 core" or "solid class 1", what’s this about? We explain it here. Cable cores are always constructed in accordance with the requirements of the IEC 60228 standard. They are subdivided into 4 classes, each with their own characteristics.

For which materials does IEC 60228 apply ?

  • Copper and tinned copper: class 1, class 2, class 5 and class 6 are possible.
  • Aluminium and aluminium alloys: class 1 and class 2 are possible. Classes 5 and 6 are not defined in IEC 60228.

Which are the classes in IEC 60228?

  • Class 1: solid conductor: the core is formed by a single (solid) wire.
    The nominal section is determined by the maximum DC resistance of the conductor at 20°C.
  • Class 2: stranded conductor: the core is formed by several wires that are stranded together into a "cord".
    The nominal section is determined by a minimum number of wires and the maximum DC resistance at 20°C.
  • Class 5: flexible conductor: the core is formed by a large number of thin wires.
    The nominal section is determined by the maximum diameter of the wires and the maximum DC resistance at 20°C. Note that class 5 has a higher resistance than class 2 cores with identical cross-section.
    YMvK ss Dca-s2, YMz1K ss Cca, and YMz1K ss B2ca contain Class 5 conductors. In the standards, it was decided to increase the number of wires, so that these Class 5 conductors still have the same resistance as Class 2 conductors.
     
  • Class 6: highly flexible conductor: the core is formed by even more and still finer wires.
    The nominal cross-section is determined by the maximum diameter of the wires and the maximum DC resistance at 20°C (Ω/km).

Classes 3 and 4 are not described in IEC 60228.

Related document
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Classification of conductors according to IEC 60228

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Nominal cross-section

From this follows that a nominal cross-section is in fact a codified way of indicating the resistance of the conductor.
If you’d directly measure on a particular conductor its diameter and then calculate its cross-section (diameter² / 4 x π), you’d notice that the result would slightly vary from the nominal cross-section.

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