Cable Cross-Section Selection Guide: From Load Current to mm²

Step 1: the design current is not the nameplate total

Start from the load, and be honest about what the load is. A distribution board with 120 kW of connected load does not draw 120 kW, because lighting, small power and machine tools are never all on at their rating at the same moment. The demand factor, sometimes called the diversity factor, converts the installed load into the load the feeder has to carry. Mixed commercial loads commonly land between 0.6 and 0.8; a workshop where several motors may run together is nearer 0.7 to 0.9. If the project has a design basis or a national annex with a factor, use that and record it.

For a single motor, do the opposite and take the nameplate: the full-load current printed on the motor, which already includes the efficiency and the power factor. Do not derive it from kW with an assumed efficiency, because the nameplate is right and the assumption may not be. For heating loads, the load is continuous and no diversity applies at all; a 30 kW heater draws 30 kW for as long as the thermostat is closed.

The result of this step is the design current Ib, in amperes. For a balanced three-phase load it is Ib = P / (√3 × U × cosφ), with P in watts and U the line-to-line voltage. Every later step compares against this number, so an error here propagates through the whole calculation.

• Single motor: take the nameplate full-load current, not a kW conversion
• Mixed small power and lighting: apply the demand factor from the project design basis, typically 0.6 to 0.8
• Continuous loads such as heating: no diversity, size for the full connected load
• Three-phase: Ib = P / (√3 × U × cosφ); single-phase: Ib = P / (U × cosφ)
• Write the power factor down; do not assume 0.9 and move on

Step 2: base ampacity and the correction factors

The base current-carrying capacity Iz comes from a table in the standard you are working to, at a stated set of reference conditions. For IEC 60364-5-52 those conditions are 30 °C ambient air, a single circuit, in free air (installation method E), with the conductor at its maximum operating temperature. Real installations are never at the reference conditions, so the tabulated value is then multiplied by correction factors.

The factors are multiplicative, never additive. Three circuits grouped in a 45 °C plant room means Iz × 0.70 × 0.87, not Iz × (1 − 0.30 − 0.13). And each factor has to match the insulation: the ambient factors for a 90 °C XLPE cable are kinder than those for a 70 °C PVC cable, because a hotter-rated cable has more temperature rise in hand. The full table of those factors is in the ampacity chart article in this series.

Step 3: voltage drop, the criterion that usually governs

Ampacity is a thermal question: how hot does the conductor get. Voltage drop is a completely different question, and on any run longer than a few tens of metres it is the one that decides the size. The reason is that ampacity depends on the current alone, while voltage drop depends on the current multiplied by the length. Double the length and the drop doubles, but the heating does not change at all.

The three-phase formula, with the length in metres and the resistance and reactance in ohms per kilometre, is:

ΔU% = (√3 × I × L × (R × cosφ + X × sinφ)) / (10 × U)

The term R × cosφ is the resistive part and X × sinφ the reactive part. For cables up to about 50 mm² the resistive part dominates and the reactance can often be neglected on a first pass; from 95 mm² upward the reactance term is a significant share of the total, and ignoring it will underestimate the drop. Cable tables usually publish the combined result as mV/A/m, which is √3 × (R × cosφ + X × sinφ), so the arithmetic becomes ΔU = I × L × (mV/A/m) / 1000 in volts.

The commonly applied limits are 3 % for lighting circuits and 5 % for other uses, measured from the origin of the installation to the furthest point. If there is a long feeder upstream feeding the board, part of that budget is already spent before the circuit you are sizing begins.

Worked example: an 80 kW feeder at 150 m and at 350 m

Take a three-phase 400 V load of 80 kW at a power factor of 0.85, fed from a distribution board over a distance L, installed as a 4-core XLPE copper cable on a ladder rack in a plant room at 40 °C ambient, sharing the route with three circuits in total.

First the design current: Ib = 80000 / (1.732 × 400 × 0.85) = 80000 / 588.9 = 135.8 A. Call it 136 A. Next the correction factors: ambient 40 °C gives 0.91 for a 90 °C cable, and three circuits grouped gives 0.70, so the combined factor is 0.91 × 0.70 = 0.637. Try 50 mm²: its base rating is 192 A, and 192 × 0.637 = 122.3 A, which is below 136 A, so 50 mm² fails. Try 70 mm²: 246 × 0.637 = 156.7 A, which clears 136 A. On the thermal criterion alone the answer is 70 mm².

Now the voltage drop at 150 m. Use the formula with R = 0.320 Ω/km and X = 0.080 Ω/km at cosφ 0.85, so sinφ = 0.527: R × cosφ = 0.272 and X × sinφ = 0.0422, giving 0.3142 Ω/km. Then ΔU% = (1.732 × 136 × 150 × 0.3142) / (10 × 400) = 11103 / 4000 = 2.78 %. That is inside a 5 % budget, so at 150 m the answer stands at 70 mm².

Change only the length, to 350 m, and rerun it. The thermal answer does not move: still 70 mm². The drop scales with length: 2.78 % × 350 / 150 = 6.48 %, well over 5 %, so 70 mm² fails. Try 95 mm², R = 0.236, X = 0.075: R × cosφ = 0.2006, X × sinφ = 0.0395, total 0.2401. ΔU% = (1.732 × 136 × 350 × 0.2401) / 4000 = 19795 / 4000 = 4.95 %. That is technically inside 5 % but leaves no margin at all for the tabulated tolerances, so move up to 120 mm²: R × cosφ = 0.159, X × sinφ = 0.0395, total 0.1985, and ΔU% = (1.732 × 136 × 350 × 0.1985) / 4000 = 16364 / 4000 = 4.09 %. The selected size for 350 m is 120 mm², even though the load only needed 70 mm².

Step 4: the short-circuit thermal check

The fourth check asks a question the other three ignore: if a fault occurs at the far end, will the conductor survive the few hundred milliseconds before the device clears, without the insulation being damaged? The adiabatic expression in IEC 60364-5-54 assumes all the heat stays in the conductor, which is conservative and slightly overestimates the temperature rise. It is:

S ≥ I × √t / K

where I is the rms short-circuit current in amperes, t the clearing time in seconds, and K a constant: 143 for copper with XLPE insulation, 115 for copper with PVC. Suppose the prospective fault current at the board is 10 kA and the protective device clears it in 0.5 s. Then S ≥ 10000 × √0.5 / 143 = 10000 × 0.7071 / 143 = 49.4 mm², so the minimum is 50 mm². In the 150 m case that is smaller than the 70 mm² already chosen, so it does not govern. In the 350 m case it is far below 120 mm² and does not govern either.

Sensitivity matters here. If the same circuit were connected to a board with a 25 kA prospective fault, the requirement becomes 25000 × 0.7071 / 143 = 123.6 mm², and 150 mm² becomes the minimum for both lengths. The short-circuit current and the clearing time are properties of the network and the device, not of the cable, so get them from the switchgear supplier or the fault study before you finalise the schedule.

• K = 143 for copper with XLPE insulation, 115 for copper with PVC (IEC 60364-5-54)
• Use the device clearing time, not an assumed 1 s; fast current-limiting breakers clear in under 10 ms
• A larger cable is not automatically safer on this criterion: it also lowers the loop impedance, which raises the fault current
• For aluminium conductors the K values are different: 94 for XLPE and 76 for PVC

Which criterion governs, and what it costs

The three electrical criteria behave differently as the installation changes, and knowing which one governs tells you what to change. When ampacity governs, the fix is usually cheaper than a bigger cable: separate the circuits, ventilate the route, or move the run out of the sun. When voltage drop governs, there is no such escape: the only real fixes are a larger cross-section, a shorter route, or raising the system voltage.

That is also the reason long rural and large-site feeders are often economical at a higher voltage. The same 80 kW at 350 m needs 4 × 120 mm² at 400 V. Move the feeder to 11 kV and the current falls to about 5 A, at which point a 3-core 35 mm² medium voltage cable comfortably handles both the current and the drop, and a transformer at the far end does the rest. The cable cost falls by a wide margin even after paying for the transformer and the switchgear.

Fitting the protective device into the picture

Once the size is chosen, the protective device has to sit inside it. The rule in IEC 60364-4-43 is Ib ≤ In ≤ Iz, and I2 ≤ 1.45 × Iz, where In is the device rating and I2 the current that causes it to operate within the conventional time. In words: the load current must not exceed the device rating, the device rating must not exceed the derated cable rating, and the device must actually trip before the cable is damaged.

In the 150 m example, the derated rating of the 70 mm² cable is 156.7 A and the load is 136 A, so a 160 A device would violate In ≤ Iz and a 125 A device would violate Ib ≤ In. A 160 A moulded-case breaker is therefore not usable on that circuit; the options are a 160 A cable size or a device with a 140 A or 150 A setting if the range allows it. This is the step where a size chosen on paper comes back for another look, and it is worth doing before the schedule is issued rather than after the cable is delivered.

What to put on the calculation sheet

A cable schedule without its assumptions is impossible to check. Record the following for every circuit, and the next engineer to touch the job will be able to verify it in five minutes instead of starting again.

Every entry on this list is also a question the supplier will ask if it is missing. A schedule that carries them arrives as a specification rather than an enquiry, and it comes back as a price that can be checked instead of one that has to be trusted.

• Load, demand factor, power factor and the resulting design current Ib in amperes
• The standard and the table the base ampacity came from, plus the installation method letter
• Each correction factor separately, with its justification: ambient, grouping, soil, depth, and the combined product
• The derated rating Iz and the required minimum S ≥ I√t/K with its fault level and clearing time
• The voltage drop in per cent, with the length, the R and X used, and the limit applied
• The selected size, the protective device type and rating, and the spare capacity left in the cable