Installation design
Protective-device let-through energy
- where
- device let-through energy (A²s)
- material/temperature factor
- conductor area (mm²)
Explain this equation
Checks whether a conductor can withstand the energy a protective device lets through during a fault. Use it with current-limiting fuses and circuit-breakers, where clearing times are very short.
I²t is the device's let-through energy in A²s, from manufacturer data at the relevant fault level. k is the material and insulation factor. S is the conductor area in mm². k²S² is the most energy the conductor can absorb without its insulation overheating. The calculator gives k²S² and the margin left after subtracting I²t.
Heating in a very short fault depends on I²t, and a conductor's ability to absorb it depends on k² times its area squared. If the device lets through less than the conductor can take, the conductor is protected.
2.5 mm² copper with k = 115 can withstand about 82,600 A²s, so a device letting through 50,000 A²s leaves a positive margin. Doubling the area quadruples the withstand.
Watch out: a negative margin means the conductor is not protected. Use the manufacturer's figure, not a guess. This is a thermal check only; breaking capacity and disconnection times are separate.
A current-limiting device changes the current waveform during interruption. Its published let-through energy can be compared directly with conductor withstand.
Use manufacturer data for the actual device, voltage and prospective fault current. The simple constant-current I² × t estimate is not a substitute for those data.
Enter your values to see the result.
Common mistake
Comparing an A²s energy figure with a current in A.
Limits
This is a thermal comparison only; device breaking capacity, selectivity and disconnection requirements remain separate.
Sources
The tools calculate the displayed equation. They do not automatically choose cable sizes, approve installations, supply current regulatory tables, certify equipment or replace inspection and testing.