Trace fusing current calculator

The current at which a trace stops being a conductor and becomes smoke — Onderdonk's adiabatic estimate of the melt boundary, run either way: the current that opens a known trace in a given time, or the width that rides out a known fault.

Fig 1 — the melt boundary of a 10 mil trace: 2.57 A opens it in 1.00 s; longer faults need less.

Solve for

Trace width (mil)

Copper weight (oz)

Fault duration (s)

Starting temperature (°C)

Fusing current: 2.57 A opens the trace in 1.00 s
Cross-section: 13.8 mil² · 17.5 cmil
At other durations: 25.7 A at 10 ms · 8.12 A at 100 ms

This is the melting estimate, not a rating: the continuous current for a sane temperature rise is far lower — size that with the trace width tool, and use this number only against faults.

How this is calculated

Standard: Onderdonk's fusing equation — an adiabatic estimate; the continuous rating belongs to IPC-2152 (see the trace width tool)

$I = A_{cmil} \sqrt{ \frac{ \log_{10}\!\left( \frac{T_m - T_a}{234 + T_a} + 1 \right) }{ 33 \, t } }$: Area in circular mils (1 cmil = π/4 mil²), t in seconds, T_m = 1083 °C for copper. The log argument is (T_m+234)/(T_a+234): resistivity rising toward copper’s −234.5 °C inferred zero.
$I^2 t \cdot \rho(T) = c_v \, A^2 \, dT$: What the equation is underneath: adiabatic Joule heating. Integrating with copper’s heat capacity and resistivity reproduces the 1/33 constant within 2 % — the unit tests perform that derivation so a mistyped constant cannot ship.
$A_{mil^2} = w \cdot 1.378 \cdot oz$: 1 oz/ft² of copper is 1.378 mil (34.79 µm) thick; the width solve inverts this.

Assumptions

Adiabatic: no heat leaves the copper during the fault. Fair below a few seconds; after that the laminate heat-sinks the trace and the estimate turns conservative.
The uniform, narrowest cross-section fuses. Neck-downs, thermal-relief spokes and via barrels fail first and are not modelled here.
An estimate of melting, not a design rating — Saturn PCB’s toolkit ships the same equation with the same caveat. Operate nowhere near it.
Solder mask, adjacent copper and stacked plating all move the real number; the equation carries none of them.
T_m is copper’s melting point. The board is on fire figuratively well before the trace is literally.

What current melts a trace, and how fast

Two different questions hide under "how much current can this trace carry". The everyday one — how much before it runs unacceptably hot — is the trace width tool's job, sized for a deliberate rise of a few tens of degrees. This page answers the other one: how much current before the copper reaches 1083 °C and the trace opens like the fuse it briefly is. That number matters exactly once per fault — when a short circuit has to blow the upstream fuse, trip the hot-swap controller, or ride out the supply's current limit without the board losing a trace first.

The model is Onderdonk's equation: dump energy into the copper faster than it can leave, and the temperature rides the I²t up to the melting point. It is an estimate, and every serious user of it says so — the page's job is to say so too. The constant everyone copies, the 33, is not copied here: the unit tests re-derive it from copper's volumetric heat capacity and its resistivity-temperature law, and it lands within 2 %.

Trace fusing current chart

The current that melts a trace, from Onderdonk's equation at 25 °C, computed by the calculator above for the widths a layout uses and the times a protective device takes to act. Read it with the fault duration in mind: the same 10 mil trace survives ten times the current for a hundredth of the time, which is the whole reason a fast fuse and a thin trace can coexist.

Width	1 oz, 10 ms	1 oz, 100 ms	1 oz, 1 s	1 oz, 5 s	2 oz, 1 s
5 mil (0.13 mm)	12.8 A	4.06 A	1.28 A	574 mA	2.57 A
8 mil (0.20 mm)	20.5 A	6.49 A	2.05 A	918 mA	4.11 A
10 mil (0.25 mm)	25.7 A	8.12 A	2.57 A	1.15 A	5.13 A
15 mil (0.38 mm)	38.5 A	12.2 A	3.85 A	1.72 A	7.70 A
20 mil (0.51 mm)	51.3 A	16.2 A	5.13 A	2.30 A	10.3 A
30 mil (0.76 mm)	77.0 A	24.4 A	7.70 A	3.44 A	15.4 A
50 mil (1.27 mm)	128 A	40.6 A	12.8 A	5.74 A	25.7 A
100 mil (2.54 mm)	257 A	81.2 A	25.7 A	11.5 A	51.3 A

None of these is a rating. The continuous current a trace carries is set by its temperature rise, which is thetrace width calculator's question and is an order of magnitude lower; this chart is the other end, the current at which the copper is gone.

Worked example: a 10 mil 1 oz trace in a one-second fault

The defaults: a 10 mil, 1 oz trace, a 1 second fault, 25 °C board.

A       = 10 × 1.378                    = 13.78 mil²  =  17.5 cmil
log arg = (1083 + 234) / (25 + 234)     = 5.085   →  log₁₀ = 0.706
I_fuse  = 17.5 × √(0.706 / 33 × 1)      = 2.57 A

same trace, faster interruption:
  100 ms → 8.12 A        10 ms → 25.7 A

Read the second line the way protection engineers do: the faster the upstream device clears, the more current the trace tolerates on the way. A 10 mil trace and a fuse that clears a dead short in 10 ms coexist with a 20 A fault; the same trace with a sleepy polyfuse taking a full second does not survive 3 A of overload. The coordination question is the figure's curve against the protective device's time-current curve.

Where Onderdonk's equation stops being valid

The equation is adiabatic, so time is its weak axis. Under about a second it is a fair model of a trace in a fault; by ten seconds the laminate is visibly helping and the real trace survives more than predicted — the error is at least in the survivable direction. For sustained overload there is no fusing calculation worth doing: that is a temperature-rise problem, and the continuous rating owns it.

It also assumes the trace is the uniform weakest link, which on a real board it often is not. The neck-down into a pad, a thermal-relief spoke, or a single via carrying the fault will open first at a lower I²t. If the fault path includes vias, check them with thevia tool — one 12 mil barrel is a far smaller cross-section than a 10 mil trace.

Nothing here is a safety approval. Where a trace is the deliberate sacrificial element, or a creepage/clearance question rides on the fault, the relevant standard and testing own the answer — a century-old estimating equation does not.

Common fusing current mistakes

Using the fusing current as a rating. The 2.57 A that melts the default trace in a second is only two or three times its comfortable continuous current; the gap between "works" and "melts" is the entire design margin, not usable headroom.
Sizing the trace for the fault and forgetting the fuse, or the fuse and forgetting the trace. The trace must sit above the protective device's clearing I²t with margin — either alone proves nothing.
Ignoring the starting temperature. A power trace already at 105 °C in an enclosure has meaningfully less absorbing capacity than the 25 °C the calculation was done at on a desk in January.
Trusting artwork width at the failure point. Acid traps, neck-downs and teardrops all move the melt point to somewhere narrower than the number entered above.