IEC & BS 7671 calculators

CPC size calculator (adiabatic equation)

Check that a protective conductor survives the fault energy until the device disconnects, with k worked out from first principles and the simplified table as a second answer.

Calculated with electrician.wiki · https://electrician.wiki/calculators/cpc-size/

Code / standard

Fault current

Zs at the end of the circuit

Disconnection

Protective device

Device rating In

The protective conductor is

Its material

Cable insulation

Line conductor

cpc size to check

Result · BS 7671

0.572 mm² minimum → 1 mm² · 1.5 mm² OK

With 208.1 A for 0.1 s and k = 115, the cpc needs at least 0.572 mm², so the chosen 1.5 mm² is large enough. Without calculating, Table 54.7 would ask for 2.5 mm².

Minimum cpc (adiabatic): 0.572 mm²
Smallest standard size: 1 mm²
Table 54.7 (no calculation): 2.5 mm²
k: 115
Fault current: 208 A
Time used: 0.1 s
Chosen 1.5 mm²: Large enough

Using 0.1 s for a circuit-breaker in its instantaneous region is an upper bound: an MCB usually clears in a few milliseconds and lets through far less energy. For operating times under 0.1 s, BS 7671 and IEC 60364 compare k²S² with the I²t the device lets through, from the maker’s data. That let-through rises with fault current, so also check the highest fault current, at the start of the circuit, with the let-through option.
1.5 mm² is smaller than the simplified table asks for (2.5 mm²), which is normal for twin and earth with its reduced cpc. That is allowed: the adiabatic check above is what shows it complies.

Show the work

1. Earth fault current at the end of the circuit (the minimum)
$I_{f} = \frac{C_{\min} \cdot U_{0}}{Z_{s}} = \frac{0.95 \cdot 230 V}{1.05 Ω} = 208.1 A$
Cmin = 0.95 (BS 7671 Appendix 3).
2. Disconnection time
$t = 0.1 s$
208 A is at least 5 × In = 100 A, the top of the type B instantaneous band, where the breaker opens within 0.1 s.
3. k for a copper conductor from 70 °C to 160 °C
$k = 226 \cdot \sqrt{\ln (1 + \frac{160 - 70}{234.5 + 70})} = 115$
Rounded to 115, the value in BS 7671 Table 54.3 / IEC 60364-5-54 Table A.54.3 (core of a cable; same k as Table 43.1).
4. Minimum cross-sectional area (adiabatic equation)
$S = \frac{\sqrt{I^{2} \cdot t}}{k} = \frac{\sqrt{{(208.1 A)}^{2} \cdot 0.1 s}}{115} = 0.572 mm²$
5. The chosen 1.5 mm² is large enough
$1.5 mm² \geq 0.572 mm²$
It withstands k²S² = 29,760 A²s against 4,330 A²s let through.
6. Simplified selection (BS 7671 Table 54.7)
2.5 mm² line → the same as the line conductor (up to 16 mm²) = 2.5 mm² → 2.5 mm²
No calculation needed if the cpc is at least this size.

References: BS 7671 Regulation 543.1.3 (2018): S = √(I²t)/k · BS 7671 Tables 54.2 to 54.4 (2018): k for protective conductors · BS 7671 Table 54.7 (2018): simplified selection · BS 7671 Appendix 3 (2018): device time–current curves and Cmin

Adiabatic heating: all the fault energy stays in the conductor, which is on the safe side for faults up to 5 s.
k from the material constants and the 70 °C start / 160 °C limit temperatures; the formula reproduces the published tables.
The fault current is the minimum, at the far end of the circuit; for a fuse that is the worst case.

New to this? How TN-S, TN-C-S and TT earthing systems work

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How to use the cpc size calculator

Give the fault current, either directly or from the earth fault loop impedance Zs at the end of the circuit, and say how long it flows: the calculator can derive the time from a circuit-breaker or a fuse, take a time you read from the device’s curve, or use the let-through energy the manufacturer publishes. Then describe the protective conductor (cpc): a core of the cable such as the bare earth in twin and earth, a separate insulated conductor, or the steel armour of an SWA cable, with its material and the cable’s insulation. Add the line conductor size for the simplified table, and the cpc you plan to use if you want a verdict on it.

The adiabatic equation

During an earth fault the whole fault current flows through the cpc until the protective device opens. For a few seconds at most, almost none of the heat escapes: the conductor heats “adiabatically”, and its temperature rise depends only on the energy I²t and its own size. The cpc is big enough if that energy can’t take it past the temperature its insulation, or the cable around it, survives (BS 7671 Regulation 543.1.3; IEC 60364-5-54, 543.1.2):

S \geq \frac{\sqrt{I^{2} \cdot t}}{k}

S is the cross-sectional area in mm², I the fault current in amperes, t the disconnection time in seconds (up to 5 s) and k a factor for the conductor material and its start and limit temperatures. The same equation checks a line conductor’s fault withstand, read the other way: t = k²S²/I².

Where k comes from

k is not a table to memorise: it follows from the conductor’s heat capacity and resistivity. For copper the constants give K = 226 and B = 234.5 °C, and with θi the starting and θf the limit temperature:

k = 226 \cdot \sqrt{\ln (1 + \frac{θ_{f} - θ_{i}}{234.5 + θ_{i}})}

A core inside a PVC cable may already be at 70 °C when the fault starts and must stay under 160 °C, which gives 115.0, published as 115. A separate insulated cpc starts at 30 °C, so it has more headroom: 142.9, published as 143. Our formula reproduces all 103 published k values we cross-checked, from BS 7671 Table 43.1 to the ABB and Schneider tables.

k from the formula above, rounded as the tables round it. Start and limit temperatures follow the arrangement; conductors up to 300 mm².
Protective conductor	Start → limit	Copper	Aluminium
Core of a PVC (70 °C) cable, e.g. twin and earth	70 → 160 °C	115	76
Core of an XLPE (90 °C) cable	90 → 250 °C	143	94
Separate PVC-insulated conductor	30 → 160 °C	143	95
Separate XLPE-insulated conductor	30 → 250 °C	176	116
Steel wire armour of a PVC / XLPE cable (single source)	60 or 80 → 200 °C	51 / 46 (steel)

Fault current and disconnection time

The fault current at the far end of a circuit, through the full loop impedance, is the smallest the cpc will carry. BS 7671 allows for the supply running below its nominal voltage with Cmin = 0.95:

I_{f} = \frac{C_{\min} \cdot U_{0}}{Z_{s}}

For a fuse the smallest current is the worst case, because a fuse blows more slowly as the current falls and the energy grows. The calculator reads the device’s tabulated operating currents for 0.4 s and 5 s: if the fault current reaches the 0.4 s value, the fuse is certain to blow within 0.4 s, and using 0.4 s in the equation can only overstate the energy. A BS 88-3 63 A fuse facing 1,000 A gives t ≤ 0.4 s and a minimum separate cpc of 4.42 mm².

A circuit-breaker behaves differently. Once the fault current reaches the top of its instantaneous band (5 × In for type B, 10 × In for type C, 20 × In for type D), BS EN 60898 has it open within 0.1 s, so 0.1 s is a safe upper bound. If the current is below that threshold, the breaker relies on its slow thermal trip, and the circuit fails its Zs check before the cpc is even in question; the calculator says so rather than guessing a time.

Type B breakers with the fault current at the edge of instantaneous tripping (5 × In) for 0.1 s, copper cpc inside a PVC cable (k = 115).
Breaker	If = 5 × In	I²t (0.1 s)	Minimum cpc
B6	30 A	90 A²s	0.082 mm²
B10	50 A	250 A²s	0.14 mm²
B16	80 A	640 A²s	0.22 mm²
B20	100 A	1,000 A²s	0.27 mm²
B32	160 A	2,560 A²s	0.44 mm²
B40	200 A	4,000 A²s	0.55 mm²

Let-through energy: the better number for breakers

In practice a modern breaker clears a heavy fault in a few milliseconds, well before 0.1 s, and limits the current while it does. For operating times under 0.1 s, BS 7671 and IEC 60364 compare k²S² directly with the I²t the device lets through, which manufacturers publish for each rating and fault level. That number grows with fault current, so for breakers the hardest case is often the highest fault current, at the start of the circuit, not the lowest. Taking 0.1 s there would be wildly pessimistic: 3,000 A for 0.1 s implies 8.25 mm² of copper, while a let-through of 20,000 A²s needs only 1.23 mm². Use the let-through option with the maker’s figure for the prospective fault current at the origin.

The simplified table: no calculation

If you’d rather not calculate, BS 7671 Table 54.7 (IEC 60364-5-54 Table 54.2) sets a size that always works: the same as the line conductor up to 16 mm², 16 mm² for line conductors over 16 and up to 35 mm², and half the line size above that. A cpc of a different material is scaled by k1/k2, the ratio of the line conductor’s k to the cpc’s. The steel armour of a 25 mm² XLPE cable would need 49.7 mm² that way, which is why armour is usually checked by calculation instead.

Twin and earth is the everyday case. Its cpc is reduced, 1.5 mm² with a 2.5 mm² line and 2.5 mm² with 6 mm², so it never meets the simplified rule. It complies by calculation: in the default example the 1.5 mm² cpc needs only 0.572 mm². A protective conductor that isn’t part of a cable must also be at least 2.5 mm² copper when mechanically protected and 4 mm² when not, whatever the calculation gives.

Worked examples

Twin and earth on a B20

A radial in 2.5/1.5 mm² twin and earth with Zs = 1.05 Ω at the far end.

1. Earth fault current at the end of the circuit (the minimum)
$I_{f} = \frac{C_{\min} \cdot U_{0}}{Z_{s}} = \frac{0.95 \cdot 230 V}{1.05 Ω} = 208.1 A$
2. Disconnection time
$t = 0.1 s$
3. k for a copper conductor from 70 °C to 160 °C
$k = 226 \cdot \sqrt{\ln (1 + \frac{160 - 70}{234.5 + 70})} = 115$
4. Minimum cross-sectional area (adiabatic equation)
$S = \frac{\sqrt{I^{2} \cdot t}}{k} = \frac{\sqrt{{(208.1 A)}^{2} \cdot 0.1 s}}{115} = 0.572 mm²$
5. The chosen 1.5 mm² is large enough
$1.5 mm² \geq 0.572 mm²$
6. Simplified selection (BS 7671 Table 54.7)
2.5 mm² line → the same as the line conductor (up to 16 mm²) = 2.5 mm² → 2.5 mm²

With 208.1 A for 0.1 s and k = 115, the cpc needs at least 0.572 mm², so the chosen 1.5 mm² is large enough. Without calculating, Table 54.7 would ask for 2.5 mm².

NICEIC’s fuse example: a 6 mm² cpc that isn’t enough

A 16 mm² cable with a 6 mm² cpc on a BS 88-3 63 A fuse. At the far end the fault current is 320 A, which the fuse takes about 5 s to clear (NICEIC, Professional Electrician, 2018).

1. Disconnection time
$t = 5 s$
2. k for a copper conductor from 70 °C to 160 °C
$k = 226 \cdot \sqrt{\ln (1 + \frac{160 - 70}{234.5 + 70})} = 115$
3. Minimum cross-sectional area (adiabatic equation)
$S = \frac{\sqrt{I^{2} \cdot t}}{k} = \frac{\sqrt{{(320 A)}^{2} \cdot 5 s}}{115} = 6.22 mm²$
4. The chosen 6 mm² is too small
$6 mm² < 6.22 mm²$
5. Simplified selection (BS 7671 Table 54.7)
16 mm² line → the same as the line conductor (up to 16 mm²) = 16 mm² → 16 mm²

With 320 A for 5 s and k = 115, the cpc needs at least 6.22 mm², so the chosen 6 mm² is too small. Without calculating, Table 54.7 would ask for 16 mm².

Using the maker’s let-through energy

The same 2.5/1.5 mm² circuit, with a breaker whose data sheet gives 20,000 A²s at the prospective fault current.

1. Earth fault current at the end of the circuit (the minimum)
$I_{f} = \frac{C_{\min} \cdot U_{0}}{Z_{s}} = \frac{0.95 \cdot 230 V}{1.05 Ω} = 208.1 A$
2. Let-through energy (maker’s data)
$I^{2} \cdot t = 20,000 A²s$
3. k for a copper conductor from 70 °C to 160 °C
$k = 226 \cdot \sqrt{\ln (1 + \frac{160 - 70}{234.5 + 70})} = 115$
4. Minimum cross-sectional area
$S = \frac{\sqrt{I^{2} \cdot t}}{k} = \frac{\sqrt{20,000 A²s}}{115} = 1.23 mm²$
5. The chosen 1.5 mm² is large enough
$1.5 mm² \geq 1.23 mm²$
6. Simplified selection (IEC 60364-5-54 Table 54.2)
2.5 mm² line → the same as the line conductor (up to 16 mm²) = 2.5 mm² → 2.5 mm²

With a let-through of 20,000 A²s and k = 115, the cpc needs at least 1.23 mm², so the chosen 1.5 mm² is large enough. Without calculating, Table 54.2 would ask for 2.5 mm².

BS 7671 and IEC 60364

The equation, the k values and the simplified table are the same in both: BS 7671 numbers the k tables 54.2 to 54.6 and the simplified table 54.7; IEC 60364-5-54 puts k in Annex A and the simplified table at 54.2. The difference is the fault current. BS 7671 gives Cmin = 0.95 for UK supplies; in IEC mode the calculator uses the same 0.95 as IEC 60909-0’s minimum voltage factor for low voltage, and lets you set U0. UK fuse data is not used in IEC mode.

Common mistakes

Using the maximum fault current for a fuse. For fuses the far end, with the lowest current and the longest time, is the hard case.
Using the minimum fault current for a breaker’s let-through. For breakers the let-through grows with current; check the origin too.
Taking k for a separate cpc when it is a core of the cable. A core starts hot: 115, not 143, for copper in PVC.
Treating Table 54.7 as the minimum. It is a shortcut that always works; a smaller cpc can comply by calculation.
Forgetting the cpc’s effect on Zs. A smaller cpc raises R2, which can break the disconnection time even when the adiabatic check passes.

Get Zs first from the earth fault loop impedance calculator, and size the line conductors with the cable size calculator. The fault current at the origin comes from the prospective fault current calculator.

Method and verification

k is computed from the material constants with the rounded K values the published tables use, and the tests confirm all 103 cross-checked values, including 115 and 143. The NICEIC worked example above is a test case: k²S² for a 6 mm² cpc is 476,100 A²s, less than the 512,000 A²s a 320 A fault lets through in 5 s, so the cpc must be 10 mm². Steel armour k values come from one publisher only and are flagged.

Adiabatic method, disconnection up to 5 s, conductors up to 300 mm²; steel conduit and trunking as cpc are not covered. Updated 2026-10-14. Corrections log.

Questions people ask

What is the adiabatic equation?

S = √(I²t)/k: the smallest conductor area that withstands a fault current I for t seconds, where k depends on the conductor material and its start and limit temperatures. It assumes no heat escapes during the fault, which holds for times up to 5 s.

What is the k value for a cpc in twin and earth?

115: copper starting at 70 °C inside PVC and limited to 160 °C. A separate PVC-insulated copper cpc, starting at 30 °C, has k = 143.

Is the 1.5 mm² earth in 2.5 mm² twin and earth big enough?

Usually, yes, by calculation. On a B20 with Zs = 1.05 Ω it needs only 0.572 mm². It doesn’t meet the simplified Table 54.7, which is why the adiabatic check is needed.

What disconnection time should I use for an MCB?

If the fault current reaches the top of the instantaneous band (5, 10 or 20 × In for types B, C and D), 0.1 s is a safe upper bound. For a tighter answer, use the manufacturer’s let-through energy instead of a time.

When can I use Table 54.7 instead of calculating?

Any time: a cpc of the same material at least as large as the table says always complies. Calculation lets you use a smaller one, such as the reduced cpc of twin and earth, or check steel armour.

Does the cpc size affect Zs?

Yes. R2, the cpc’s resistance, is part of the earth fault loop, so a smaller cpc raises Zs and lowers the fault current. Check both the adiabatic equation and the maximum Zs.