Skip to content
RFrftools.io
PCB

PCB Trace Fusing Current Calculator

Calculate the current that will fuse a PCB trace using Onderdonk's equation. Includes etch-factor trapezoid correction, time-to-fuse at your operating current, and safety margin.

Loading calculator...

Formula

I=Alog⁡10 ⁣[Tm−Ta234+Ta+1]33 SI = A\sqrt{\frac{\log_{10}\!\left[\frac{T_m - T_a}{234 + T_a} + 1\right]}{33\,S}}

Reference: I. M. Onderdonk, as published in E. R. Stauffacher, "Short-time Current Carrying Capacity of Copper Wire", General Electric Review, 1928; IPC-2152 Appendix A

I— Fusing current (A)
A— Cross-sectional area (circular mils)
S— Fault duration (s)
T_m— Copper melting point, 1083 (°C)
T_a— Ambient temperature (°C)

How It Works

There are two completely different questions about how much current a trace can carry, and confusing them destroys hardware. IPC-2152 answers how much current can this trace carry forever without getting too hot. Onderdonk's equation answers how much current will melt this trace in a given time. The second number is far larger, and it is a destruction limit, not a rating.

Onderdonk assumed the fault is fast enough that no heat escapes the conductor — every joule of I2RI^2R goes into raising the copper's own temperature until it reaches the melting point. Integrating the temperature-dependent resistivity of copper across that adiabatic rise gives:

I=Alog⁡10 ⁣[Tm−Ta234+Ta+1]33 SI = A\sqrt{\frac{\log_{10}\!\left[\frac{T_m - T_a}{234 + T_a} + 1\right]}{33\,S}}

with AA in circular mils, SS in seconds, and temperatures in °C. The 234 is copper's inferred zero-resistance temperature; the 33 collects its density, specific heat, and resistivity for that unit system. At 25 °C ambient and one second, the whole square-root term evaluates to about 0.1463 A per circular mil — the same figure IEEE Std 80 uses for copper grounding conductors.

The inverse-square-root scaling

Fusing current scales as the inverse square root of duration. A trace that survives 10 A for one second survives about 32 A for ten milliseconds. That is why a short-circuit event and a slow overload need entirely different analyses.

Etch factor

Etched traces are not rectangular. The etchant attacks sideways as well as down, leaving a trapezoid narrower at the top than at the base. With a sidewall slope expressed as vertical rise per horizontal run, each side loses t/EFt/EF of width, and the cross-section uses the mean of the top and bottom widths. On 2 oz copper with a 1:1 slope, that removes a meaningful fraction of the conductor.

Where the model stops being conservative

Beyond about 1.5 s, the adiabatic assumption breaks down. A trace is bonded to laminate and often adjacent to copper planes, all of which absorb heat, so the real fusing current at long durations is higher than Onderdonk predicts. Brooks and Adam measured PCB traces taking 1.5 to 6 times longer to fuse than the equation predicts, which corresponds to surviving roughly 1.2 to 2.5 times more current. The error is always in the safe direction, but treat long-duration results as a lower bound and measure if the design depends on it.

Worked Example

Problem: A 0.5 mm wide, 1 oz trace carries 5 A. A downstream short could apply that fault for up to one second before the upstream protection clears. Will the trace survive?

Step 1 - Cross-sectional area. Convert to mils: W = 0.5 mm = 19.685 mil t = 35 um = 1.378 mil A = 19.685 * 1.378 = 27.13 sq mil

Step 2 - Convert to circular mils (1 circular mil = pi/4 sq mil): A = 27.13 * 4/pi = 34.54 circular mil

Step 3 - Temperature term at 25 C ambient: log10[(1083 - 25)/(234 + 25) + 1] = log10(4.085 + 1) = log10(5.085) = 0.7062

Step 4 - Fusing current for S = 1 s: I = 34.54 sqrt(0.7062 / 33) = 34.54 0.14629 = 5.05 A

Step 5 - Verdict. The trace fuses at 5.05 A in one second and the fault is 5 A. Margin is 1.01 times — effectively none. This trace will fail.

Step 6 - Fix by widening. Scaling the width by four gives A = 138 circular mil and I_fuse = 20.2 A, a four times margin over the 5 A fault.

Step 7 - Cross-check against continuous rating. IPC-2152 gives roughly 1.4 A continuous for the original 0.5 mm 1 oz trace at a 10 C rise. Note how far apart the two numbers are: 1.4 A forever versus 5 A for one second. Never confuse them.

Step 8 - Faster clearing. If the protection clears in 100 ms instead, the same 0.5 mm trace fuses at 5.05 times the square root of ten, or 16.0 A, and the 5 A fault now has a 3.2 times margin.

Practical Tips

  • ✓Aim for at least a four times margin between the fusing current and the worst-case fault current the protection will allow.
  • ✓Get the actual clearing time from your fuse, breaker, or current-limit circuit — the duration drives the answer more strongly than anything else.
  • ✓Include the etch factor for 2 oz copper and heavier, where the sidewall slope removes a real fraction of the cross-section.
  • ✓Prefer a real fuse over a fuse trace anywhere safety, repairability, or reproducibility matters.
  • ✓Check the continuous rating with IPC-2152 and the fault rating with Onderdonk. Both must pass; they are different failures.
  • ✓Remember that a hotter ambient lowers the fusing current, so run the calculation at the maximum internal temperature of the enclosure, not at room temperature.

Common Mistakes

  • ✗Using the fusing current as a continuous rating. It is the current that destroys the trace, typically three to four times the IPC-2152 continuous figure. Size normal operating current with IPC-2152 instead.
  • ✗Ignoring the fault duration. Fusing current scales as one over the square root of time, so the same trace tolerates three times more current for a tenth of the duration.
  • ✗Applying the adiabatic model to slow overloads. Past about 1.5 s the board absorbs significant heat and the equation becomes pessimistic by a factor of 1.2 to 2.5 — useful as a lower bound, not as a prediction.
  • ✗Designing a deliberate fuse trace without accounting for etch tolerance. Width, copper weight, and etch factor all vary, so the actual fusing point can move by tens of percent.
  • ✗Forgetting that a blown fuse trace vents molten copper onto the board and often damages the laminate and neighbouring nets.

Frequently Asked Questions

An adiabatic model for the current that melts a copper conductor in a given time. It assumes no heat leaves the conductor, so all resistive heating goes into raising its own temperature to the melting point. It is the same relation IEEE Std 80 uses for short-time ratings of grounding conductors.
No. Onderdonk gives a destruction limit, not a rating. Continuous current is set by temperature rise, which is what IPC-2152 addresses. The two numbers commonly differ by a factor of three or four.
Because the energy delivered scales with current squared times time, the fusing current scales as one over the square root of duration. Ten times shorter means about 3.2 times more current tolerated.
Occasionally, in cost-sensitive consumer products. The drawbacks are real though: poor tolerance because width, copper weight, and etch all vary; molten copper deposited on the board when it blows; and no way to reset or service it. Use a real fuse where any of that matters.
Good for short faults, conservative for long ones. Below about 1.5 s the adiabatic assumption holds well, and the coefficient agrees with IEEE Std 80's copper short-time withstand figure to under one percent. Beyond that, the laminate and adjacent copper absorb heat the model ignores: Brooks and Adam measured real traces taking 1.5 to 6 times longer to fuse, so they survive roughly 1.2 to 2.5 times more current than predicted. The error is always in the safe direction.

Shop Components

As an Amazon Associate we earn from qualifying purchases.

PCB Manufacturing (JLCPCB)

Affordable PCB fabrication with controlled impedance options

JLCPCB →

FR4 Copper Clad Laminate

FR4 laminate sheets for custom PCB prototyping

Amazon →

Thermal Paste

Thermal interface material for component heat management

Amazon →

Related Calculators

PCB

Trace Width

Calculate minimum PCB trace width for current capacity per IPC-2221 and IPC-2152. Get resistance, voltage drop, and power dissipation. Free, instant results.

PCB

Trace Resistance

Calculate PCB copper trace DC resistance from width, length, thickness, and temperature. Get sheet resistance and temp coefficient. Free, instant results.

Thermal

Trace Temp

Calculate PCB copper trace temperature rise under load current using IPC-2152 and IPC-2221 standards. Enter trace width, thickness, and current to get ΔT above ambient. Essential for thermal PCB design and reliability engineering. Free, instant results.

Power

Inrush Current Limiter

Calculate NTC thermistor requirements for inrush current limiting, including cold resistance, peak inrush current, time constant, and energy absorbed