Adiabatic equation calculator
A free adiabatic equation calculator for BS 7671 CPC sizing. Punch in the fault current, disconnection time and k factor, and get the minimum earth conductor size before you've found your pen.
- Next standard size–
Guidance only. k values here are the common copper cases from Tables 54.2 to 54.4 of BS 7671. Aluminium, mineral cables and 90°C insulation have different k values, and for very fast MCB clearance times the manufacturer's I²t let-through data gives a sharper answer.
The adiabatic equation, explained properly
When a fault hits, the CPC has one job: carry a very large current for a very short time without cooking itself before the protective device disconnects. The adiabatic equation from Regulation 543.1.3 is how you prove it can:
S = √(I² × t) ÷ k
S is the minimum cross-sectional area in mm², I is the fault current, t is the disconnection time and k is a factor covering the conductor material and its insulation. One thing worth being clear on: this equation validates a CPC you've already chosen. You pick the conductor, run the sums, and check the answer comes out at or below the size you've got. It's the calculated alternative to simply selecting a CPC from Table 54.7.
The no-maths route: Table 54.7
If you'd rather not calculate at all, Table 54.7 lets you select a CPC directly from the line conductor size:
| Line conductor | Minimum CPC (Table 54.7) |
|---|---|
| Up to 16 mm² | Same size as the line conductor |
| 16 to 35 mm² | 16 mm² |
| Above 35 mm² | Half the line conductor size |
Simple, but generous. The adiabatic route usually justifies a smaller CPC than Table 54.7 demands, and that's exactly why flat twin and earth gets away with a reduced earth: 2.5 mm² T&E carries a 1.5 mm² CPC, and the adiabatic check is what proves it's fine.
Where the inputs come from
The fault current I comes from your earth fault loop impedance. From a measured Zs, use I = 230 × 0.95 ÷ Zs, which is the same maths behind my maximum Zs calculator. The disconnection time t comes from the device's time/current curve at that fault current, and for an MCB operating magnetically it's 0.1 s or less. If you want the sharpest answer, the manufacturer's I²t energy let-through data beats reading a curve every time.
The five-step check on site
- Get your Zs: measure it live at the far end of the circuit, or calculate it from dead tests as Ze + (R1 + R2).
- Convert it to fault current: I = 230 × 0.95 ÷ Zs.
- Find the disconnection time at that current from the device curve. An MCB operating magnetically clears in 0.1 s or less.
- Pick the k factor for how the CPC is installed: 115 inside a cable or bunched, 143 separate and insulated, 159 bare copper.
- Run S = √(I²t) ÷ k and compare. If the answer comes out at or below the CPC you have actually installed, it passes.
A worked example
Say a fault current of 800 A clears in 0.1 s, with k = 115 for a copper CPC inside a cable. Then S = √(800² × 0.1) ÷ 115 = √64000 ÷ 115 = 253 ÷ 115 = 2.2 mm². The 2.5 mm² CPC in 4 mm² T&E passes, with a little in hand.
Picking the right k factor
k comes from Tables 54.2 to 54.4 of BS 7671. These are the common copper cases this calculator covers:
| k | Situation (copper) |
|---|---|
| 115 | Insulated CPC in a cable, or bunched with cables, 70°C PVC |
| 143 | Separate insulated CPC, not bunched, 70°C PVC |
| 159 | Bare copper, visible and in restricted areas |
Higher-temperature insulation and aluminium conductors change k, so if you're outside these cases, go back to the tables rather than guessing.
- The CPC in flat T&E is one size down from the line conductor. That's not a shortcut gone wrong, it's precisely the situation this check exists to verify.
- Longer disconnection times punish you hard. t sits under a square root, but I² dominates the sum, so a sluggish device at high fault current eats CPC headroom fast.
- If the sums fail, you've got three levers: a bigger CPC, a different protective device, or a lower Zs. And while you're checking the circuit, run it through my cable size calculator and voltage drop calculator too, the three checks live or die together.
Adiabatic equation FAQs
What is the adiabatic equation?
S = √(I²t) ÷ k, from Regulation 543.1.3. It tells you the minimum conductor cross-section that can soak up the energy of a fault without damage before the device disconnects. In practice it's the calculation you use to prove a CPC, usually a reduced one, is big enough for the fault energy it might have to carry.
What k value should I use?
For copper with 70°C PVC insulation: 115 when the CPC is inside a cable or bunched with cables, which covers the twin and earth case, 143 for a separate insulated CPC that isn't bunched, and 159 for bare copper that's visible and in a restricted area. Step outside those cases, aluminium, mineral cable, 90°C insulation, and the k value changes, so check Tables 54.2 to 54.4 rather than assuming.
Is the 1.5 mm² CPC in 2.5 mm² twin and earth acceptable?
Almost always, yes. It's the classic adiabatic pass, and it's the whole reason flat T&E is allowed a reduced CPC in the first place. That said, "almost always" isn't "always", so verify it with your real measured Zs and the actual device data instead of taking it on faith. That's what this calculator is for.
What disconnection time do I use?
Read it off the device's time/current curve at your fault current. When an MCB trips magnetically, the clearance time is 0.1 s or less, which is why healthy circuits pass so comfortably. For the most accurate answer, use the manufacturer's I²t let-through data, it captures the real energy the device lets past and usually gives a smaller minimum CPC than a curve reading.
When can I skip the calculation?
Whenever you're happy to size the CPC from Table 54.7 instead: same size as the line conductor up to 16 mm², then 16 mm² for lines between 16 and 35 mm², then half the line size above that. No maths, no device data, no arguments. The trade-off is copper: the table route almost always lands you a bigger CPC than the adiabatic check would justify.
Why is it called the adiabatic equation?
"Adiabatic" means no heat gets in or out. The equation assumes the fault is over so fast that every joule of heat stays in the conductor, none of it has time to escape into the insulation or the air around it. Real faults do shed a little heat, so the assumption sits on the safe side: the conductor the maths demands is at least as big as the one physics does.
Does the adiabatic equation only apply to CPCs?
No. BS 7671 applies the same energy check, I²t against what the conductor can absorb, to live conductors under fault conditions too. The CPC case in Regulation 543.1.3 is just where electricians meet it most, because reduced CPCs like the one in twin and earth need proving. This calculator handles the CPC case; for live conductors, work from the fault current section of the regs and the device's energy let-through data.
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