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Cable Sizing Calculator

Determine minimum conductor size for current capacity, voltage drop, and fault withstand across 4 international standards.

AS/NZSBSIEC

Circuit Parameters

Voltage

PhasesLoad Current

Load Power

Power FactorCable Length

Cable Parameters

Insulation

Installation Method

Installation Conditions

Ambient Temperature

°C

Number of CircuitsGrouping ArrangementCircuit Type

Earth Fault Loop Impedance

Verify Zs (earth fault loop impedance)

Protective Device

Device Rating

Max Voltage Drop

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Cable sizing is the engineering process of selecting the minimum conductor cross-sectional area that safely carries the design current while satisfying voltage drop limits, short-circuit withstand requirements, and installation derating factors. The procedure is defined in IEC 60364-5-52 Clause 523 and produces a conductor size in square millimetres.

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How to Size a Cable to BS 7671

1
Determine the design current — Calculate the design current Ib from the load power, voltage, and power factor using P = V x I x pf for single-phase or P = 1.732 x V x I x pf for three-phase circuits.[BS 7671 Regulation 433.1]
2
Select the protective device — Choose a protective device with nominal rating In where In is greater than or equal to Ib. The device must comply with the discrimination requirements of the upstream protection.[BS 7671 Regulation 433.1]
3
Apply ambient temperature correction — Look up the ambient temperature correction factor Ca from BS 7671 Table B.52.14 based on the actual ambient temperature and the cable insulation type (PVC 70C or XLPE 90C).[BS 7671 Table B.52.14]
4
Apply grouping correction factor — Determine the grouping factor Cg from Table C.3 based on the number of circuits and the installation arrangement. More circuits grouped together reduce individual cable capacity.[BS 7671 Table C.3]
5
Calculate required current capacity — Compute the tabulated current-carrying capacity It = In / (Ca x Cg x Ci x Cc). This is the minimum value the selected cable must achieve in the ampacity tables.[BS 7671 Appendix 4]
6
Select cable from ampacity table — Choose the smallest cable size whose tabulated current rating equals or exceeds It from the appropriate table for the installation method, conductor material, and insulation type.[BS 7671 Table 4D1A-4J4A]
7
Verify voltage drop and fault — Check voltage drop using Vd = mV/A/m x Ib x L / 1000 against the 5% limit. Verify short-circuit withstand using the adiabatic equation k2S2 >= I2t.[BS 7671 Regulation 525.1]

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How Cable Sizing Works

The cable sizing calculator determines the minimum conductor cross-sectional area that satisfies current carrying capacity, voltage drop limits, and short circuit withstand requirements across four major electrical standards.

The process follows a sequential methodology. First, the design current (Ib) is established from the load parameters — voltage, power, power factor, and circuit configuration (single-phase or three-phase). A protective device is then selected with a nominal rating In where In >= Ib.

Next, derating (correction) factors are applied based on installation conditions. These include factors for installation method, ambient temperature, conductor grouping, and thermal insulation contact. Under AS/NZS 3008.1.1:2017, these come from Tables 3, 22, 24, and 26 respectively. Under BS 7671:2018+A2, the equivalent factors Ca, Cg, Ci, and Cc are drawn from Tables B.52.14, B.52.17-B.52.21, Regulation 523.7, and Table B.52.5. IEC 60364-5-52 uses the same tabulated approach per Clause 523, while NEC/NFPA 70:2023 applies adjustment factors from Article 310.15.

The minimum required current-carrying capacity is calculated as Iz = In / (k1 x k2 x k3 x ...), where each k-factor represents one derating condition. The cable is then selected from the appropriate ampacity table — Table 13/14 (AS/NZS), Appendix B tables (BS 7671), Tables B.52.2-B.52.13 (IEC), or Table 310.16 (NEC).

Two verification checks follow. Voltage drop is confirmed against allowable limits using the formula Vd = (mV/A/m x Ib x L) / 1000. Short circuit withstand is verified using the adiabatic equation k^2 x S^2 >= I^2 x t, where S is conductor area, I is fault current, t is disconnection time, and k is a material constant.

Results include the selected cable size, tabulated and derated current ratings, voltage drop percentage, short circuit withstand capacity, and full clause references for every derating factor applied.

Current-Carrying Capacity — 2.5mm² Copper, PVC, in Conduit (Reference Method B)

Standard	Single-phase (A)	Three-phase (A)	Reference Table
AS/NZS 3008.1.1	23	20	Table 13, Col 4
BS 7671	24	20	Table 4D1A
IEC 60364-5-52	24	20	Table B.52.4
NEC (NFPA 70)	25 (12 AWG)	20 (12 AWG)	Table 310.16

Source: AS/NZS 3008.1.1:2017 Table 13, BS 7671 Table 4D1A, IEC 60364-5-52 Table B.52.4, NEC Table 310.16

Frequently Asked Questions

How do I determine the correct cable size per AS/NZS 3008.1.1:2017?

Under AS/NZS 3008.1.1:2017, cable sizing starts by determining the design current (Ib), then selecting a protective device rating (In) where In >= Ib. You then apply derating factors from Table 3 (installation method), Table 22 (ambient temperature), and Table 24 (grouping) to calculate the minimum current-carrying capacity Iz. The final cable size is selected from Table 13 (thermoplastic) or Table 14 (thermosetting) ensuring the tabulated rating meets or exceeds Iz.

What derating factors are required for cable sizing under BS 7671?

BS 7671:2018+A2 requires application of several correction factors per Regulation 523.1. These include Ca for ambient temperature (Table B.52.14), Cg for grouping (Table B.52.17 to B.52.21), Ci for thermal insulation (Regulation 523.7), and Cc for the type of protective device (Table B.52.5). The effective current-carrying capacity is calculated as It = In / (Ca x Cg x Ci x Cc), and the cable must be selected from the appropriate Appendix B table for the installation method.

What is the difference between thermoplastic (PVC) and thermosetting (XLPE) cable insulation for sizing?

Thermoplastic insulation (PVC/V-90) has a maximum continuous conductor temperature of 70-75 degrees C, while thermosetting insulation (XLPE/X-90) is rated to 90 degrees C per AS/NZS 3008.1.1 Table 1. This higher temperature rating means XLPE cables can carry significantly more current for the same cross-sectional area. For example, a 16mm2 XLPE cable in a typical installation can carry approximately 20-25% more current than the equivalent PVC cable.

How does NEC Article 310.15 differ from IEC cable sizing methods?

NEC Article 310.15 uses ampacity tables (Table 310.16 for 0-2000V conductors) based on AWG/kcmil sizes, whereas IEC 60364-5-52 uses metric mm2 sizes with Table B.52.2 to B.52.13. The NEC applies adjustment factors for ambient temperature (Table 310.15(B)(1)) and conduit fill (Table 310.15(C)(1)), while IEC uses correction factors from Tables B.52.14 and B.52.17. The fundamental difference is that NEC treats the protective device as a ceiling for conductor ampacity, whereas IEC/BS methods calculate minimum conductor capacity from the protective device rating divided by combined correction factors.

Why must cable sizing consider both current carrying capacity and voltage drop?

A cable that satisfies current carrying capacity may still fail voltage drop requirements. AS/NZS 3008.1.1 Clause 4.5 limits voltage drop to 5% from the point of supply to the load. BS 7671 Regulation 525.1 recommends 3% for lighting and 5% for other uses. In long cable runs, voltage drop often governs the cable selection, requiring a larger conductor than what current capacity alone would dictate. Both criteria must be checked independently.

What is the short circuit withstand requirement for cable sizing?

Cables must withstand the prospective fault current for the disconnection time of the protective device. Per AS/NZS 3008.1.1 Clause 5, the adiabatic equation k2S2 >= I2t determines minimum conductor size, where k is a material constant (Table 52 for copper, Table 53 for aluminium), S is the conductor cross-sectional area in mm2, I is the fault current in amperes, and t is the disconnection time in seconds. This often governs cable size in installations close to transformers where fault levels are high.

Can two engineers size the same 63 A three-phase circuit and legitimately get cables two sizes apart just by choosing a different standard?

Yes, and it is more common than most engineers realise. A 63 A three-phase circuit at 40 degrees C ambient, four circuits grouped on a perforated cable tray using XLPE copper conductors gives different results under AS/NZS 3008 and BS 7671. AS/NZS 3008 uses 40 degrees C as its reference ambient so the temperature factor is 1.00, with grouping of 0.77 from Table 21, yielding a required capacity of 81.8 A and a 16 mm2 cable. BS 7671 uses 30 degrees C as its reference, so at 40 degrees C the correction is 0.87, with a harsher grouping factor of 0.65 from Table 4C1, requiring 111.4 A and a 25 mm2 cable. The grouping factor difference (0.77 vs 0.65) drives a full cable size jump because the standards model mutual heating differently based on their respective research programs.

Why does adding thermal insulation on only one side of a cable sometimes trigger the full enclosure derating factor, halving the cable's capacity?

Under AS/NZS 3008.1.1:2017 Clause 3.5.5 and Table 23, a cable in contact with thermal insulation on one side receives a derating factor of 0.75. But if insulation covers more than one surface so heat cannot dissipate freely, the factor drops to 0.5. The standard does not define 'surrounded' with a precise angular coverage, so many inspectors apply 0.5 whenever insulation covers more than half the cable circumference. A 32 A circuit using 4 mm2 XLPE copper rated at 44 A in free air has an effective rating of 33 A with one-side contact (0.75) but only 22 A if embedded in insulation (0.5), making it undersized by 31%. BS 7671 Method 100 and IEC 60364-5-52 Clause 523.7 mirror this. Always document the exact relationship between cable and insulation, because 'touching' versus 'embedded' is the difference between a compliant and non-compliant installation.

In AS/NZS 3008, why does selecting a larger conductor sometimes fail the protective device coordination check even though the smaller conductor passed?

This is a trap in Clause 5.2 of AS/NZS 3008.1.1:2017. When you upsize a cable, you reduce circuit impedance, which increases the prospective fault current at the far end. The protective device (e.g., a 32 A MCB) has a fixed energy let-through (I2t). On very long circuits where the original cable was marginal, upsizing can push the fault current into a higher energy let-through band on the MCB's I2t curve, particularly near the B-curve to C-curve boundary at 5x In versus 10x In. If the fault current crosses from the instantaneous trip region to the short-time delay region, the let-through energy can jump dramatically. In most cases the larger cable wins because S2 grows faster than fault current, but always re-verify the device's I2t against the new cable's k2S2 when upsizing.

How does the 'direct buried' derating stack with soil thermal resistivity, and why does everyone get this wrong in desert climates?

Most engineers apply the standard ambient soil temperature derating from AS/NZS 3008.1.1:2017 Table 22 and stop there, but Clause 3.5.4 also requires a correction for soil thermal resistivity when it differs from the reference value of 1.2 K.m/W (Table 20). In arid or sandy soils the actual resistivity can reach 2.5 to 3.0 K.m/W, giving a derating of approximately 0.80 from Table 20. Stacking ambient soil temperature of 35 degrees C (factor 0.96), soil resistivity of 2.5 K.m/W (factor 0.80), and three circuits in a trench (factor 0.73 from Table 21) gives a combined derating of 0.56. A 200 A cable now carries only 112 A. The IEC equivalent in Table B.52.16 uses 2.5 K.m/W as its reference instead of 1.2. Applying an IEC soil correction to AS/NZS base ratings or vice versa gives the wrong result.

Why does the neutral conductor sometimes need to be larger than the phase conductors in a three-phase circuit, and which standards address this?

In balanced three-phase linear loads the neutral current is zero, but with predominantly non-linear loads such as LED drivers, switch-mode power supplies, and VFDs, the third harmonic currents from each phase add in the neutral rather than cancelling. The neutral current can theoretically reach 1.73 times the phase current. BS 7671:2018 Regulation 523.6 and Table 4Ab provide reduced grouping factors when the neutral carries more than the phase current. When third harmonic content exceeds 33%, the neutral is treated as the loaded conductor for sizing purposes. For example, with 100 A per phase and 45% third harmonic, the neutral current is approximately 135 A. IEC 60364-5-52 Section 523.6 and Table C.3 mirror this. AS/NZS 3008 does not have an explicit harmonic neutral table but Clause 2.4 references the general obligation to account for all current paths. In any modern commercial building with predominantly electronic loads, assume at least 33% THD and size the neutral accordingly.

Standards Reference

AS/NZS 3008.1.1:2017 — Section 3, Tables 3-22
BS 7671:2018+A2 — Tables B.52.2-B.52.20, Appendix 4
IEC 60364-5-52 — Clause 523, Tables B.52.1-B.52.20
NEC/NFPA 70:2023 — Article 310, Table 310.16

Cable Sizing Calculator

How to Size a Cable to BS 7671

How Cable Sizing Works

Current-Carrying Capacity — 2.5mm² Copper, PVC, in Conduit (Reference Method B)

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