Cable size calculator: mm² and AWG, protection, short circuit and medium voltage

What this calculator does

It is four tools on one page. The low voltage tab returns the minimum conductor cross-section in the two sizing systems that coexist on the market — the metric cross-sections of IEC 60228 and the American AWG and kcmil sizes — for copper and aluminium with thermoplastic (PVC, 70 °C) or thermosetting (XLPE or EPR, 90 °C) insulation. The medium voltage tab computes the current rating of 3.6/6 kV to 18/30 kV single-core cables with the IEC 60287 method. The short-circuit withstand tab tells you how much fault current a conductor can take, or what size a given fault requires. The IEEE methods tab adjusts the IEEE 835 ampacity tables to the actual installation (IEEE 399) and calculates voltage drop, short circuit and pulling tension per IEEE 525.

At low voltage the size is decided by several conditions that the calculator checks at once, and the one demanding the largest size governs:

1. heating in service (current-carrying capacity, with temperature, grouping and soil);
2. coordination with the overload protection;
3. voltage drop along the circuit;
4. short-circuit thermal stress;
5. the harmonics that load the neutral.

Conditions 2 to 5 are optional: each is enabled with its own checkbox. For the chosen size the calculator also returns the neutral, the protective conductor and the IEC 60228 conductor data.

First condition: current-carrying capacity

A cable heats up by Joule effect, and its insulation has a maximum temperature it must not exceed in continuous service: 70 °C for PVC and 90 °C for cross-linked polymers. The current that brings it exactly to that temperature is its current-carrying capacity Iz, and it depends on how the cable sheds heat: a cable in free air is not the same as one buried in an insulating wall, nor is a single cable the same as one in a bundle of twelve.

IEC 60364-5-52 tabulates Iz for every size, material and insulation as a function of the reference installation method, identified by a letter:

• A1 and A2: insulated conductors or a multicore cable in conduit in a thermally insulated wall. The worst case.
• B1 and B2: the same, with the conduit on the wall or in trunking.
• C: multicore cable clipped direct to the wall or on an unperforated tray.
• D1 and D2: cable in a buried duct or buried direct. The reference temperature here is that of the ground, 20 °C.
• E: multicore cable in free air or on a perforated tray, with air circulating around it.
• F: single-core cables touching, in free air or on a perforated tray, in trefoil or flat (Tables B.52.10 to B.52.13, from 25 to 630 mm²).
• G: single-core cables spaced at least one diameter apart, flat horizontal or vertical. This method allows the highest current.

The table values are given at 30 °C ambient (20 °C buried), for a single circuit and, when buried, for a soil of 2.5 K·m/W. When reality differs, correction factors apply:

• Temperature (Tables B.52.14 and B.52.15): from 1.22 at 10 °C down to 0.50 at 60 °C for PVC.
• Soil thermal resistivity (Table B.52.16), for D1 and D2 only: a moist soil of 1 K·m/W raises the current by 18% in duct and 50% buried direct; a dry soil of 3 K·m/W lowers it by 4% and 10%.
• Grouping, with the table matching the actual layout: B.52.17 for bunched cables (row 1: 0.80 with two circuits, 0.70 with three, 0.50 with nine) or a single layer on a wall, ceiling, tray or ladder (rows 2 to 5, more favourable); B.52.18 for cables buried direct according to their spacing; B.52.19 for buried ducts; B.52.20 and B.52.21 for multicore cables or single-core circuits on several stacked trays.

The condition checked is Ib ≤ Iz × kT × kG × kR: the design current may not exceed the corrected capacity. When the grouping falls outside what the standard tabulates — more than six circuits in buried ducts, for example — the calculator says so instead of extrapolating.

NEC 310.16 does the same with American sizes and three temperature columns (60, 75 and 90 °C) according to the insulation type (TW; THW/THWN/XHHW; THHN/XHHW-2). Ambient temperature correction uses the formula of 310.15(B)(1) and the adjustment for more than three current-carrying conductors in the same raceway uses Table 310.15(C)(1). For 14, 12 and 10 AWG the overcurrent protection is capped at 15, 20 and 30 A even if the table allows more (240.4(D)): if a larger protective device is entered, the calculator increases the size.

Second condition: overload protection

A cable that meets the current-carrying capacity still has to be properly protected. IEC 60364-4-43 (433.1) sets two conditions for the device protecting against overload:

• Ib ≤ In ≤ Iz: the rated current (or setting) of the protection lies between the design current and the cable’s capacity.
• I2 ≤ 1.45 × Iz: the current that ensures tripping within the conventional time, I2, does not exceed 1.45 times the capacity.

I2 depends on the device: 1.45 In for IEC 60898 miniature circuit-breakers, 1.30 Ir for IEC 60947-2 industrial breakers and, for IEC 60269 gG fuses, 2.1 In up to 4 A, 1.9 In from 4 to 16 A and 1.6 In from 16 A. With a miniature circuit-breaker the second condition is met automatically; with a gG fuse it requires the cable’s corrected Iz to reach 1.1 In.

For example, 19 A three-phase in conduit on a wall (method B1), copper and PVC. With a 20 A circuit-breaker 2.5 mm² is enough (Iz = 21 A). With a 20 A gG fuse, I2 = 32 A requires Iz ≥ 22.1 A and 4 mm² is needed. The calculator picks the next standard In above Ib, or uses the one entered, and finds the size that meets both conditions. With parallel conductors, Iz is the sum of all of them (433.4.1). When protection is enabled, the NEC side also returns the equipment grounding conductor from Table 250.122.

Third condition: voltage drop

A long conductor can run cool and still deliver less voltage to the load than it needs. The drop is computed from the conductor impedance at its operating temperature:

ΔU = k × Ib × L × (R·cos φ + X·sin φ)

with k = 2 for single-phase and DC and k = √3 for three-phase; R is the conductor resistance per metre — the maximum resistance fixed by IEC 60228 for that size, material and class, brought to 70 °C or 90 °C — and X the reactance, which in low-voltage cables is around 0.08 Ω/km and only matters for large sizes at low power factor. The calculator finds the smallest size in the IEC 60228 series that keeps ΔU below the permitted percentage of the nominal voltage.

Fourth condition: short circuit

During a short circuit the current is so large and so brief that the conductor has no time to shed heat: all the energy stays inside and heats it. IEC 60364-4-43 (434.5.2) requires the protection to clear the fault before the conductor reaches the limit temperature of its insulation, which translates into the adiabatic condition:

k² × S² ≥ I² × t

with S the cross-section in mm², I the short-circuit current and t the disconnection time. The constant k depends on the metal and on the initial and final temperatures (Table 43A): 115 for copper with PVC (from 70 °C to 160 °C), 143 for copper with XLPE or EPR (from 90 °C to 250 °C), 76 and 94 for aluminium, and 103 and 68 for PVC above 300 mm², whose limit drops to 140 °C. The calculator does not use the table but the formula behind it, that of IEC 60949:

k = √( Qc (β + 20) / ρ20 × ln((β + θf) / (β + θi)) )

which reduces to 226 √ln(…) for copper and 148 √ln(…) for aluminium. This allows any initial temperature, for example 75 °C for the NEC side or 30 °C for a separate protective conductor.

An example: 20 A on a line with 10 kA prospective short-circuit current and 0.1 s disconnection time. Heating alone would allow 2.5 mm², but the fault requires √(10 000² × 0.1) / 115 = 27.5 mm² with PVC: 35 mm². With XLPE (k = 143), 22.1 mm²: 25 mm². The formula is valid from 0.1 s to 5 s; below 0.1 s the current asymmetry matters and the I²t let through by the protection, from the manufacturer, must be used — it can be entered directly. The American sizes use the same equation, which is that of ICEA P-32-382 and IEEE 525, with a final temperature of 150 °C for thermoplastics and 250 °C for XLPE and EPR.

Neutral, harmonics and protective conductor

In a balanced three-phase circuit without harmonics the neutral carries no current. But switch-mode power supplies, drives and electronic lighting generate third harmonic, which does not cancel in the neutral: it adds up. With 33% third harmonic in the phase, the neutral carries as much current as each phase, and above that, more. Annex E of IEC 60364-5-52 gives the reduction factor for 4- and 5-core cables with a full-size neutral:

The standard’s example: 39 A in a four-core cable on a wall, PVC. Without harmonics 6 mm² is enough; with 20%, 39 / 0.86 = 45 A and 10 mm²; with 40%, the neutral carries 46.8 A and is sized for 54.4 A: 10 mm²; with 50%, 58.5 A and 16 mm².

The neutral has the same size as the phase in single-phase circuits and in three-phase circuits up to 16 mm² copper or 25 mm² aluminium (524.2). Above that it may be reduced, but only if the maximum expected neutral current, harmonics included, fits in the reduced size, if the neutral is protected against overload (431.2) and if it is not below 16 mm² Cu / 25 mm² Al (524.3). With more than 10% harmonics it is not reduced.

The protective conductor is chosen from Table 54.2 of IEC 60364-5-54 — equal to the phase up to 16 mm², 16 mm² up to 35 mm² and half the phase above — or calculated with the same adiabatic short-circuit equation (543.1.2), which may give a smaller size. The k differs depending on whether the conductor is a core of the cable (starting at operating temperature: 115 or 143 for copper) or a separate insulated conductor (starting at 30 °C: 143 with PVC and 176 with XLPE for copper).

The IEC 60228 series: sizes, classes and resistance

The European standard that defines the conductor is not the installation standard but IEC 60228 (BS EN 60228 in the UK): it fixes the series of nominal cross-sections, from 0.5 to 2500 mm², and for each one the maximum DC resistance at 20 °C. The nominal size is defined by that resistance, not by its millimetres: two “4 mm²” conductors from different makers may have different geometric areas, but neither may exceed 4.61 Ω/km. That is why the calculator uses that resistance, rather than the generic resistivity of the metal, for the voltage drop.

The standard distinguishes four classes by flexibility: class 1 is a single solid wire (up to 35 mm² in copper), class 2 is the stranded conductor for fixed installation (the usual one in power cables, with a minimum number of wires from 7 to 91), and classes 5 and 6 are the flexible and very flexible conductors of flexible cords and machine cables, which exist only in copper and have slightly higher resistance for the same size. The calculator asks for the class and returns, next to the recommended size, its maximum resistance at 20 °C and at operating temperature, and the minimum number of wires or the maximum wire diameter according to the class.

Class 2 values (plain copper and aluminium) with the American size of equal or greater area:

The current-carrying tables of IEC 60364-5-52 stop at 300 mm² for methods A to E and at 630 mm² for methods F and G. Above that, the calculator keeps sizing for voltage drop with the IEC 60228 series up to 2500 mm², but warns that heating has not been checked and that the usual practice is to split the current over several conductors per phase.

Metric versus AWG

The two systems are not interchangeable one for one. In IEC 60228 the nominal cross-sections follow a progression from 1.5 to 300 mm² (and up to 1000 mm² for large conductors). The American Wire Gauge defines diameters in a geometric progression in which the area doubles every three sizes: 10 AWG is 5.26 mm², so 7 AWG would be 10.5 mm². Above 4/0 AWG (107 mm²) the size is expressed in kcmil, thousands of circular mils, where 1 kcmil = 0.5067 mm².

What matters in practice: the calculator solves the two tables separately and, next to each result, shows the equivalent in the other system. That a metric size “equals” an AWG size only means its area is equal or greater; the current-carrying capacity remains the one in the relevant table.

Worked example

Three-phase motor at 400 V with a design current of 63 A and cos φ = 0.85. Copper cable with XLPE insulation, clipped direct to the wall (method C), 30 °C, no other circuits nearby. Length 120 m, maximum permitted drop 3%.

• By current-carrying capacity: Table B.52.3 gives 71 A for 10 mm² with three loaded conductors, and 71 ≥ 63. Minimum size 10 mm². In the NEC 90 °C column, 6 AWG carries 75 A.
• By voltage drop: IEC 60228 sets a maximum resistance of 0.727 Ω/km at 20 °C for 25 mm² class 2 copper, which is 0.927 Ω/km at 90 °C. With 10 mm² (1.83 Ω/km at 20 °C), ΔU = √3 × 63 × 120 × (0.002333 × 0.85 + 0.00008 × 0.527) = 26.5 V, or 6.6%: far too much. With 16 mm² it falls to 4.22%; with 25 mm², to 2.72% (10.9 V).
• Recommended size: 25 mm² copper XLPE, equivalent to 3 AWG. Voltage drop governs, not heating, as is usual on circuits longer than 50 or 60 m.

Were the same circuit in aluminium, heating would call for 16 mm² (Table B.52.5, method C: 76 A) and voltage drop for 50 mm²: 35 mm² (0.868 Ω/km) lands at 3.23%, over the limit.

Several conductors in parallel per phase

When the current will not fit in a single conductor — or the one in stock is smaller than the calculation asks for — the usual answer is several identical conductors in parallel per phase. The calculator handles it two ways: fix the number of conductors and it sizes each one for Ib/n; or fix the available size and it tells you how many are needed. In both cases the grouping factor is recomputed with n times as many circuits, which is why two conductors in parallel do not carry exactly twice as much as one, and the voltage drop is checked with the current of each branch.

The conditions for the current to really share are set by IEC 60364-5-52 (523.7) and match those of the NEC (310.10(G), which additionally allows it only from 1/0 AWG): same material, same size, same length and same route, no intermediate branches, and one protective device for the set. With single-core conductors they should also be laid symmetrically (trefoil or alternating phases) so each branch has the same impedance; otherwise one branch takes more current than the others and the calculation no longer holds. In copper a single conductor is usually cheaper than two smaller ones up to 240 mm²; beyond that, and always for large aluminium sizes, parallel is the norm.

Medium voltage: the IEC 60287 calculation

For medium-voltage cables there is no installation table that fits every case: the current rating is calculated with IEC 60287, the same method manufacturers use to build their catalogues. The idea is a thermal circuit: the heat generated in the conductor (I²R) and in the screen flows in series through the insulation, the oversheath and the soil or air, and the sum of those thermal resistances sets how far the conductor heats above ambient. For three single-core cables without armour:

I = √( Δθ / ( R·T1 + R·(1 + λ1)·(T3 + T4) ) )

• Δθ is the difference between the maximum conductor temperature (90 °C for XLPE and EPR) and that of the soil or air.
• R is the AC resistance at 90 °C: the IEC 60228 value plus skin and proximity effects.
• T1 and T3 are the thermal resistances of the insulation and the oversheath, derived from the cable geometry: the fictitious conductor diameter, the IEC 60502-2 insulation thickness (5.5 mm at 12/20 kV, 8.0 mm at 18/30 kV) and the semiconducting layers.
• T4 is the external thermal resistance: of the soil, with its resistivity and depth; of the duct and the air inside it, if ducted; or of dissipation to air by convection and radiation, which is solved iteratively.
• λ1 is the screen loss as a fraction of the conductor loss. With screens earthed at both ends, induced currents circulate; with single-point bonding or cross-bonding only eddy currents remain.

An example: 240 mm² aluminium single-core cable, 12/20 kV, XLPE, 16 mm² screen earthed at both ends, in trefoil, buried direct at 1 m in soil of 1.5 K·m/W and 25 °C. T1 = 0.327, T3 = 0.068 and T4 = 2.907 K·m/W; R = 0.161 Ω/km and λ1 = 0.013. Result: 347 A. With the three cables in a 160 mm duct, 316 A; in free air at 40 °C, 449 A; with two circuits in the same trench 20 cm apart, 282 A each. Manufacturers’ catalogues give 345, 320 and 455 A for that same cable under the same conditions. Soil resistivity is the input that weighs most: at 1 K·m/W the same cable carries 413 A, and at 2.5 K·m/W, 276 A.

The calculator finds the smallest size that carries the current, checks the conductor short circuit (k = 94 for aluminium and 143 for copper, from 90 °C to 250 °C) and, if the length is given, the voltage drop with the actual reactance of the arrangement. It also shows the current rating of every size under those conditions.

IEEE methods: ampacity adjustment, voltage drop, short circuit and cable pulling

The fourth tab brings together the calculations from the IEEE guides used in North American substations and industrial plants, and in international projects that take them as their reference.

Ampacity adjustment (IEEE 399, Chapter 13). The IEEE 835 tables — more than 3000 of them, calculated with the Neher-McGrath method — and those of NEC 310.60 give a cable’s ampacity under fixed conditions: typically 90 °C at the conductor, 20 °C earth ambient and a thermal resistivity (RHO) of 90 °C·cm/W, or 40 °C in air. IEEE 399 carries that base ampacity over to the actual installation with three factors, I’ = I × Ft × Fth × Fg:

• Ft, for conductor and ambient temperature: Ft = √( (Tc’ − Ta’) / (Tc − Ta) × (K0 + Tc) / (K0 + Tc’) ), with K0 = 234.5 for copper and 228.1 for aluminum.
• Fth, for soil thermal resistivity, from Tables 13-5 to 13-7 (60 to 250 °C·cm/W, depending on conductor size and number of circuits).
• Fg, for grouping, from Tables 13-8 and 13-9 for duct banks of up to 4 × 15 ducts at 5 in to 7.5 in center-to-center, and Tables 13-10 and 13-11 for directly buried cables.

The standard’s example: 15 kV three-conductor 350 kcmil cables, with a base ampacity of 375 A, in a 3 × 5 duct bank, 120 °C·cm/W soil at 30 °C and the conductor limited to 75 °C. Ft = 0.82, Fth = 0.90 and Fg = 0.479: 133 A. The calculator reproduces that result. The 5 to 35 kV grouping tables contain several obvious misprints (values that do not follow the progression of the neighboring sizes); they have been corrected using those neighbors, always on the safe side.

Voltage drop (IEEE 525, Annex C). Instead of a typical reactance, IEEE 525 calculates the cable impedance: the DC resistance with the stranding (1.02) and lay (1.04) factors, the AC resistance with skin and proximity effects, circulating currents in shields grounded at both ends and losses in steel conduit, and the reactance from the geometric mean spacing of the arrangement: X = 2πf (0.4606 log10(S’/rc) + 0.0502) µΩ/m. The drop is given both by the approximate formula ΔV = I (R cos θ + X sin θ) and by the exact one, VS² = (VL cos θ + IR)² + (VL sin θ + IX)², which the standard recommends for power factors below 0.7, as in motor starting. With the standard’s example — 6 AWG copper at 75 °C, 36 A and 2 × 38 m at 240 V — the result is 4.6 V, or 1.9%.

Short circuit (IEEE 525, C.4). This is the ICEA P-32-382 adiabatic equation in circular mils: (I/A)² t = K log10((T2 + K0) / (T1 + K0)), with K = 0.0297 for copper and 0.0125 for aluminum, and the maximum temperatures from Table C.8: 250 °C for XLPE and EPR, 200 °C for paper and rubber, 150 °C for PE and PVC. In addition to the withstand current and the minimum conductor size, the calculator gives the temperature the conductor reaches with the expected fault. Beware of the 2007 edition: its Equation C.15b and its example use 0.0125 — the aluminum constant — for a copper conductor, which oversizes the conductor by more than 50%. With 0.0297 the result matches the k = 143 of IEC 60949 for copper and XLPE.

Pulling through conduit (IEEE 525, Annex J). A cable that is correctly sized electrically can still be damaged during installation. The calculator follows the run section by section: in straight sections the tension grows as T = L·w·f·c (length, weight per meter, coefficient of friction and weight correction factor), on slopes the weight component is added or subtracted, and in bends it is multiplied by e^(c·f·θ). The factor c accounts for the additional friction of several cables in the conduit: 1 + 4/3 (d/(D − d))² for cradled, 1/√(1 − (d/(D − d))²) for triangular. At the exit of each bend it calculates the sidewall bearing pressure on the cable (7300 N/m maximum for power cables, 4380 N/m for control cables) and compares the tension with the allowable value: 70 N/mm² of copper or hard aluminum conductor, twice the single-conductor value with two or three, and no more than 26.7 kN with a pulling eye or 4.45 kN with a basket grip. It does this in both directions, because pulling from one end or the other changes the result considerably, and it checks conduit fill and the jam ratio D/d for three cables, which between 2.8 and 3.0 makes them likely to jam in a bend.

With the standard’s Example J.4 — three 750 kcmil aluminum cables in triplex, 8 kg/m in total, 5 in conduit, f = 0.5 and a 152 m run, 45° bend, 30 m, 45° bend and 60 m — the final tension is 21.1 kN and the maximum sidewall bearing pressure 2.74 kN/m, within limits; pulling from the other end, with the 90° vertical bend at the pole, the standard reaches 37.2 kN and advises against that direction.

Fire performance

Size is not the only choice: in public buildings, tunnels or long vertical tray runs, how the cable behaves in a fire also matters. IEC 60332-1-2 tests a single vertical insulated cable with a 1 kW burner: the flame must not spread beyond a set limit. IEC 60332-3 tests a bunch of cables on a 3.5 m vertical ladder and distinguishes four categories by the amount of non-metallic material in the bunch and the flame application time: A (7 litres per metre, 40 min), B (3.5 l/m, 40 min), C (1.5 l/m, 20 min) and D (0.5 l/m, 20 min). Halogen-free cables add smoke (IEC 61034) and acid gas (IEC 60754) tests. In the European Union the Construction Products Regulation classifies cables as Aca, B1ca, B2ca, Cca, Dca, Eca and Fca. None of this changes the current-carrying capacity, but it determines which cable may be installed.

What this calculator does not cover

It does not account for solar radiation or for cyclic or emergency load ratings, which IEC 60287 and IEC 60853 treat separately. At low voltage, groups are assumed to be of identical, equally loaded cables; for groups of very different sizes the standard refers to specific calculations. At medium voltage it solves a single circuit in free air (groups in air are calculated with IEC 60287-2-2), does not check the screen short circuit, which depends on the earth fault, and does not include dielectric losses, negligible below 127 kV for XLPE. The NEC side uses Table 310.16 for cables in free air too, where 310.17 allows more. The table values correspond to the 2009 edition of IEC 60364-5-52 and to NEC Table 310.16; national rules — in Spain, the REBT with UNE-HD 60364-5-52 at low voltage and ITC-LAT 06 of the RLAT at medium voltage — may impose their own conditions and values. In the IEEE tab the base ampacity is entered by hand — the more than 3000 IEEE 835 tables are not reproduced — and the grouping factors apply only to the IEEE 399 spacings.

What checks it

The actual current in a circuit in service is measured by power and power-quality analysers, which also log the power factor and harmonics the calculator takes for granted. Loop impedance and conductor continuity of the installed cable are verified with multifunction installation testers, and a conductor running above its temperature gives itself away on a thermal imaging camera long before the insulation fails.