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Cable sizing calculator
Find the smallest standard cable that keeps the voltage drop within your limit and withstands the short-circuit current, and the current-carrying capacity it needs after derating. Free, no signup.
- Copper and aluminium, PVC and XLPE
- Results update as you type
- Share any calculation as a link
Recommended cable size
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Check the inputs: positive values, disconnection time up to 5 s, ambient below the conductor limit.
- Smallest size for the voltage drop limit
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- Smallest size for the short circuit
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- Voltage drop at the recommended size
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- Load current
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- Correction factor (temperature × grouping)
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- Required tabulated current-carrying capacity
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Check that the cable’s tabulated current-carrying capacity for your installation method (IEC 60364-5-52 or the manufacturer’s datasheet) is at least the required value.
How the calculator works
A cable has to pass three checks. The calculator finds the smallest IEC 60228 size that passes the first two and tells you what the third one needs.
- Voltage dropΔU = √3 · Ib · L · (R cos φ + X sin φ) / nWith R at the maximum conductor temperature (70 °C for PVC, 90 °C for XLPE), which is conservative. Single-phase and DC circuits use 2 instead of √3.
- Short-circuit withstand (adiabatic equation)S ≥ Ik · √t / kWith k = 143 for copper XLPE, 115 for copper PVC (103 above 300 mm²), 94 for aluminium XLPE and 76 for aluminium PVC (68 above 300 mm²), from IEC 60364-4-43. Valid for disconnection times up to 5 s.
- Current-carrying capacityIz,table ≥ Ib / (n · kθ · kg), kθ = √((θmax − θa) / (θmax − θref))kg is the grouping factor for circuits bunched together. Compare the result with the tabulated value for your installation method.
Worked example
A 400 V three-phase feeder, 100 A at cos φ = 0.9, 80 m of copper XLPE cable in air at 30 °C, 10 kA cleared in 0.1 s:
| Smallest size for a 5 % voltage drop | 16 mm² (4.69 %) |
|---|---|
| Short circuit: 10 000 · √0.1 / 143 | 22.1 mm² → 25 mm² |
| Recommended size | 25 mm² (3.01 % drop) |
| Required tabulated current-carrying capacity | ≥ 100 A |
Here the short circuit sets the size. With a longer cable the voltage drop takes over; with several circuits in one tray the current-carrying capacity often does.
What this calculator leaves out
- The current-carrying capacity tables themselves, which depend on the installation method and the soil.
- Harmonic currents in the neutral conductor, which can call for a larger size.
- Protection coordination (Ib ≤ In ≤ Iz), earth fault loop impedance and disconnection times for protection against electric shock.
- Voltage dips when large motors start; see the motor starting guide.
Frequently asked questions
Why doesn’t the calculator pick the size from the current-carrying capacity?
Current-carrying capacity depends on the installation method (clipped direct, in conduit, on trays, in the ground and so on) and comes from the tables in IEC 60364-5-52, the NEC ampacity tables or the cable manufacturer. The calculator gives the value your cable has to reach after correction, so you can check it against the right table.
Where does the k factor come from?
IEC 60364-4-43 gives k for each conductor and insulation. It depends on the conductor material and on the conductor temperature before and at the end of the fault: 90 °C to 250 °C for XLPE, and 70 °C to 160 °C for PVC up to 300 mm².
Which fault current and time should I enter?
The maximum prospective fault current at the start of the cable and the time the protective device takes to clear it, from its time-current curve. For very short times (below about 0.1 s) and current-limiting devices, use the let-through energy instead: the cable needs k²S² ≥ I²t.
How are ambient temperature and grouping taken into account?
The temperature factor kθ = √((θmax − θa) / (θmax − θref)) reproduces the IEC 60364-5-52 temperature tables. The grouping factors are those for circuits bunched together (0.80 for 2 circuits, 0.70 for 3, 0.50 for 9). For single layers, trays or several cables in the ground, use the standard’s tables.
Is a bigger cable always better?
A bigger cable lowers the voltage drop and the losses and withstands more fault current, but it costs more and may not fit the terminals. For large currents, two or more cables in parallel are often easier to install than one very large cable.