Earth conductor size calculator (adiabatic equation)
The minimum copper protective conductor area that passes the adiabatic check, from fault current and disconnection time or from a manufacturer's let-through energy.
Every result has a reason.
Enter your values and calculate to see the result, formula and working here.
S = √(I² × t) ÷ kThe adiabatic check of Regulation 543.1.3 for copper protective conductors up to 300 mm², with the k values of Tables 54.2 and 54.3. You supply the fault current and disconnection time (0.1 s to 5 s), or the manufacturer's let-through energy for faster or current-limiting devices. The result is the thermal requirement only, not a conductor selection: BS 7671 also sets minimum sizes for separate protective conductors (Regulation 543.1.1) and has its own rules for earthing and bonding conductors, and none of these are applied.
What the numbers mean.
The adiabatic equation assumes a fault is cleared before heat can escape the conductor, so all the fault energy (I²t) heats the copper. k combines copper's properties with the insulation's starting and maximum temperatures: a conductor inside a cable, or bunched with cables, starts at the cable's operating temperature, and a separate unbunched conductor at 30 °C. What matters is the energy I²t. A larger fault current usually makes the device disconnect faster, so the largest current does not always give the largest I²t.
Formula reference: IET · Earthing and protective conductors (Regulation 543.1.3) ↗A worked example
1,000 A for 0.1 s in twin and earth (k = 115): √(1,000² × 0.1) ÷ 115 ≈ 2.7498067 mm², so the next size up in the list is 4 mm².
Validation, precision and limitations