Hong Life CalculatorsUnderstanding the wire size calculator — concept and principle
Choosing a cable that is too thin wastes energy and leaves equipment short of voltage. This calculator works backwards from an allowed voltage drop to the minimum conductor cross-section, using the IEC 60364-5-52 method, then rounds up to the next standard metric size (IEC 60228) and the nearest American Wire Gauge, and shows the drop you would actually get.
How it is calculated
S = b × ρ₁ × L × cos φ × I ÷ (u_allowed − b × λ × L × sin φ × I)
u_allowed = limit × U₀
- b = 2 for DC and single-phase, 1 for balanced three-phase; U₀ is the phase-to-neutral voltage.
- ρ₁ = 0.0225 Ω·mm²/m for copper and 0.036 Ω·mm²/m for aluminium (at service temperature); λ = 0.08 mΩ/m reactance.
- L one-way length (m), I current (A), cos φ power factor.
- The result is rounded up to the next IEC 60228 size and the smallest AWG with at least that area.
The principle
The voltage drop in a cable is current times resistance, and resistance is ρL ÷ S. If you fix the largest drop you will accept, you can solve for the area S that just meets it. With AC loads that are not purely resistive, the cable's reactance adds a part that does not depend on S at all, which is why it is subtracted from the allowed drop first. A thicker conductor lowers resistance in direct proportion, so doubling the area halves the resistive part of the drop.
Worked example
Default inputs: single-phase 230 V, 20 A, 40 m, copper, cos φ = 1, limit 3 %.
- Allowed drop = 0.03 × 230 = 6.9 V; with cos φ = 1, the reactance term is zero.
- S = 2 × 0.0225 × 40 × 1 × 20 ÷ 6.9 = 36 ÷ 6.9 = 5.217 mm².
- Next standard size = 6 mm²; nearest AWG with at least this area = 10 AWG (about 5.26 mm²).
- Drop with 6 mm²: 2 × 0.0225 × 40 ÷ 6 × 20 = 6.0 V, or 2.61 %.
- Check next: current rating & breaker size.
| Supply voltage (V; line-to-line for three-phase) | Minimum cross-section by voltage drop (mm²) |
|---|---|
| 57.5 | 20.87 |
| 115 | 10.43 |
| 172 | 6.957 |
| 230 | 5.217 |
| 288 | 4.174 |
| 345 | 3.478 |
| 460 | 2.609 |
How to read the result
The size shown is the minimum for voltage drop only. The cable must also carry the current without overheating for its installation method, ambient temperature and grouping, and it must be coordinated with the protective device. The final choice is the larger of the sizes required by all these checks. Typical voltage-drop limits are 3 % for lighting and 5 % for other uses from a public low-voltage supply.
Common mistakes
- Treating the voltage-drop size as the final cable size without checking current rating.
- Entering round-trip length instead of one-way length.
- Assuming AWG and mm² sizes match exactly; they do not.
Frequently asked questions
Why round up to a standard size?
Cables are only made in standard cross-sections, so you choose the next one that meets the requirement.
Is 10 AWG the same as 6 mm²?
No. 10 AWG is about 5.26 mm², slightly smaller than 6 mm².
Does aluminium need a larger cable?
Yes, about 1.6 times the copper area for the same voltage drop, because its resistivity is higher.
Sources: IEC 60364-5-52:2009, Annex G (voltage-drop formula) · IEC 60228, Conductors of insulated cables (standard cross-sections); ASTM B258 (AWG diameters)