AS/NZS 3008 (2025) Cable Sizing Guide: Example Calculations

Learn How to Size Active, Neutral, and Earth Cables Using AS/NZS 3008.1

How to Calculate Cable Size Using AS/NZS 3008

Updated for AS/NZS 3008.1.1:2025 This article has been updated to reflect the changes introduced in the 2025 edition of AS/NZS 3008.1.1. The calculation methodology remains the same, but several reference tables and values have changed.

This comprehensive guide provides step-by-step instructions for sizing electrical cables in accordance with Australian Standard AS/NZS 3008.1.1:2025. Whether you’re an electrical engineer, contractor, or student, this resource will help you master the essential calculations for selecting the correct cable sizes in various applications. Key topics covered:

Current-carrying capacity calculations
Voltage drop requirements
Short-circuit performance
Neutral and earth cable sizing

We’ll walk through two detailed examples, demonstrating how to:

Calculate load currents
Select initial cable sizes
Determine cable operating temperatures
Check voltage drop compliance
Verify short-circuit performance
Choose appropriate active, neutral, and earth cable sizes

By following this guide, you’ll learn how to accurately size cables for different electrical installations while ensuring compliance with Australian Standards. Whether you’re working on residential, commercial, or industrial projects, these calculations are crucial for safe and efficient electrical systems. Use this article as your go-to reference for cable sizing calculations, and feel confident in your ability to design electrical systems that meet all necessary requirements and safety standards.

If you prefer to perform these AS/NZS 3008.1 cable sizing calculations automatically, you can use our free AS/NZS 3008.1 cable sizing calculator. It calculates current-carrying capacity, voltage drop, and short-circuit performance, and instantly selects compliant active, neutral, and earth conductor sizes.

Example 1 - Three-Phase 400 V System: Cable Sizing for High Current Load Over a Long Run

Problem

Three-phase \(400\ V\) system. Load current = \(1200\ A\). Route length = \(260\ m\). The cable type is single-core copper XLPE 90 degrees. Buried in separate conduits. The voltage drop limit is \(3.2\%\). The fault level is \(15\ kA\), and the clearing time is 1 second. Based on AS/NZS 3008.1.1:2025, calculate the smallest active conductor, neutral, and earth cable sizes.

Answer

Step 1: Minimum size to meet current-carrying capacity

To determine which cable can carry the \(1200\ A\) load, an initial cable size assumption must be made. A practical assumption would be to use 2 sets of \(400\ mm^2\) cables, which would equate to \(1252\ A\). Based on this assumption, we would proceed to the subsequent steps to determine whether it could provide sufficient current-carrying capacity, adequate voltage drop, and short-circuit performance. The correct current rating table must be selected from AS/NZS 3008.1.1:2025 based on the cable type and phases to make a valid assumption. Once the correct reference table has been identified, the appropriate column for the installation method and conductor type must be selected. The appropriate table and column for this problem are Table 3.13, Column 19.

Step 2: Minimum size to meet voltage drop requirement

For the voltage drop calculation, we will use 3 sets of \(630\ mm2\) cables.

The voltage drop for 3 sets of \(630\ mm^2\)cables will be calculated with a load current of \(1200\ A\). The rated current of the assumed cable is: \(I_R = 2460\ A\) (the value from Table 3.13, Column 19 is multiplied by 3, which is the number of runs chosen)

Determine cable operating temperature

To determine the Voltage Drop for a cable selected, the operating temperature \((\theta_0)\) must first be calculated using the equation:

\(\left(\frac{I_0}{I_R}\right)^2 = \frac{\theta_0 – \theta_A}{\theta_R – \theta_A}\)

Where:

\(I_0\) = operating current, in Amperes

\(I_R\) = rated current given in Tables 3.9 to 3.32 (AS/NZS 3008.1.1:2025), in amperes

\(\theta_R\) = operating temperature of the cable when carrying \(I_R\), in degrees Celsius

\(\theta_A\) = ambient air or soil temperature, in degrees Celsius

The values for these are as follows:

\(I_0 = 1200\,A\)

\(\theta_R = 90°C\) (temperature rating for the XLPE 90 degrees cable)

\(\theta_A = 25°C\) (standard ambient temperature for soil)

\(I_R = 2460\ A\)

From these, the calculated operating temperature is:

\(\theta_0 = 40.47°C\)

\(\theta_0\) is then raised to the nearest temperature of \(45°C\), \(60°C\), \(75°C\), \(80°C\), \(90°C\) or \(110°C\). For the calculated operating temperature, the nearest would be \(45°C\).

Determine cable resistance based on cable operating temperature

This rounded-up operating temperature would be used to get the a.c. resistance \(R_c\) of the chosen cable.

\(R_c = 0.0389\ \Omega/km\) (from Table 4.5(A) Column 3 of AS/NZS 3008.1.1:2025, single core cables)

Determine cable reactance

The cable reactance \(X_c\) can also be determined from the cable type (XLPE)

\(X_c = 0.0787\ \Omega/km\) (from Table 4.1(A) Column 3 of AS/NZS 3008.1.1:2025, the appropriate table for this cable type)

The \(R_c\) and \(X_c\) values will be used to calculate the voltage drop on the cables, considering the load dictated by the problem. Take note that these would need to be converted to \(Ω/m\).

Calculate voltage drop based on power factor and cable operating temperature

The following equation is used to calculate the voltage drop on the cables:

\(V_{d3\phi} = IL\left[\sqrt{3}(R_c \cos \theta + X_c \sin \theta)\right]\)

\(I = 1260\ A\) (Load Current) \(L=260\, m\) (Route length) A standard power factor of \(0.9\) can be used to calculate the power factor angle: \(\theta = 25.84°\) From these values, the voltage drop on the cable can be calculated: \(V_{d3\phi} = 12.49\ V\)

The \(\%VD\) is as follows:

\(\%VD = (12.49\ V/400\ V)*100\%\)

\(\%VD = 3.12\%\)

The calculated voltage drop is within the \(3.2\%\) limit specified in the problem. Therefore, the chosen cable (3 runs of \(630 \ mm^2\)) is appropriate. The calculated percentage voltage drop is almost at the limit. Any cable smaller than this would result in a voltage drop exceeding the specified limit. It can be said that the voltage drop is the determining criterion for the smallest cable size.

Step 3: Minimum size to handle the prospective short-circuit current

The fault level dictated by the impedances must be calculated to evaluate the short-circuit performance.

Calculate the total impedance of the cable \(Z_t\) from the values of \(R_c\) and \(X_c\)

\(Z_t=\sqrt{R_c^2 + X_c^2}=0.0878\ \Omega/km\)

Multiply this by the length ( \(260\ m\) ) to get the cable impedance:

\(Z_c=0.0228\ \Omega\)

The impedance of the network \(Z_n\) can be calculated by Ohm’s law using the supply voltage and the fault level at the supply end of the cable:

\(Z_n=400\ V/15000\ A=0.027\ \Omega\)

The voltage at the end of the cable can also be calculated by subtracting the calculated voltage drop from the supply voltage.

\(V_{\text{end}}=400\ V-12.49\ V=387.51\ V\)

The fault level can be calculated using the equation:

\(I=\frac{V_{\text{end}}}{Z_{c} + Z_{n}}=\frac{387.51}{0.0228+0.027}\)

\(I=7829.78\ A\)

This is the fault level at the end of the cable, which must be considered in calculating the minimum cross-sectional area.

Now, the minimum cross-sectional area of the cable can be calculated:

\(I^2 t = K^2 S^2\)

Where:

\(I=7829.78\ A\)

\(t=1s\)

The initial and final temperatures are needed to determine the \(K\) constant. The \(45°C\) operating temperature calculated earlier would be the initial temperature. The final temperature would depend on the insulating materials in contact with the conductors. As per Table 5.2 of AS/NZS 3008.1.1:2025, for XLPE 90, the final temperature would be \(250°C\).

Using these temperatures, the value of \(K\) can be determined from Table 5.1 of AS/NZS 3008.1.1:2025.

\(K=167.4\)

From the values of \(I\), \(t\), and \(K\), the minimum cross-sectional area of the conductor can be calculated.

\(S=46.77\ mm^2\)

The result suggests that the chosen cable (3 runs of \(630\ mm^2\)) can also handle the specified fault level.

Active Cable Size = 3 runs of \(630\ mm^2\)

Step 4: Select the neutral cable size

Neutral Cable Size = 3 runs of \(630\ mm^2\)

As per Section 3.5.2 of AS/NZS 3000, the neutral cable will be the same size as the smallest active conductor and meet the load current requirements.

Step 5: Select the earthing cable size

As per Table 5.1 of AS/NZS 3000, for cables greater than \(630\ mm^2\), the nominal size of Copper earthing conductors shall be \(25\%\) of the active size. The cumulative cross-sectional area of the active cables calculated above is \(1890\ mm^2\). \(25\%\) of this is \(472.5\ mm^2\), which can be rounded up to the standard cable size of \(500\ mm^2\).

However, the minimum earth cable size capable of handling a phase-to-earth fault must also be checked. The equation for the minimum conductor cross-section used in Step 5 will also be used. The only difference is that the phase-to-earth short circuit current will be used.

\(I_{\text{1ph}} = \frac{I_{\text{3ph}}}{\sqrt{3}} = \frac{7829.78\ A}{\sqrt{3}}\)

\(I_{\text{1ph}}=4520.53\ A\)

The same values of \(t\), and \(K\) in Step 3 will be used to calculate the minimum conductor cross-section \(S\). \(S=27.00\ mm^2\) Therefore, the earth cable size selected from Table 5.1 is acceptable. Earth Cable Size = \(500\ mm^2\)