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ThermalJuly 20, 20267 min read

Why Your PCB Trace Is Hotter Than You Think (How to ...

Calculate PCB trace temperature rise from current, width, and copper weight. Includes worked examples, IPC-2152 context, and common thermal design mistakes.

Contents

The Problem Nobody Talks About Until Something Melts

Most engineers size PCB traces for current capacity and call it done. The trace can handle 3 A, you're running 2.5 A, everything's fine. Except it's not, because that 2.5 A is heating your trace to 40°C above ambient, and now your nearby temperature-sensitive analog circuitry is drifting out of spec. Or worse, your solder joints are slowly degrading because they're sitting at 85°C instead of the 60°C you assumed.

Trace temperature rise is one of those things that seems straightforward until you actually work through the numbers. The relationship between current, trace geometry, and temperature is nonlinear, the thermal environment matters enormously, and the IPC standards everyone cites have limitations that aren't always obvious.

What the Calculator Actually Does

The PCB Trace Temperature Rise calculator takes your trace parameters—current, width, copper weight, ambient temperature, and trace length—and gives you the thermal picture: how hot the trace gets, how much power it dissipates, and what the trace resistance looks like.

The outputs include:

  • Temperature Rise (ΔT): How many degrees above ambient your trace will run
  • Maximum Trace Temperature: The actual operating temperature (ambient + rise)
  • Trace Resistance: The DC resistance of your trace at operating temperature
  • Power Dissipated: The I²R losses heating everything up
  • Current Capacity at ΔT=10°C: A sanity check showing what current would give you a 10°C rise

That last one is useful for quick comparisons. If you're running 4 A and the calculator says your ΔT=10°C capacity is 2.1 A, you know you're pushing the trace hard.

The Physics (Briefly)

Trace heating comes from resistive losses. Current flowing through the trace's finite resistance dissipates power:

P=I2RP = I^2 R

The trace resistance depends on geometry and temperature. For a rectangular trace:

R=ρcdotLWcdottR = \frac{\rho \\cdot L}{W \\cdot t}

where ρ\rho is copper resistivity (about 1.7×1081.7 \times 10^{-8} Ω·m at 20°C), LL is length, WW is width, and tt is thickness.

Copper weight translates to thickness: 1 oz copper is approximately 35 μm (1.4 mils), 2 oz is 70 μm, and ½ oz is 17.5 μm. These are nominal values—your fab house might be ±10% on any given board.

Here's where it gets interesting. Copper resistivity increases with temperature at about 0.39% per °C. So a trace that's 50°C hotter than room temperature has roughly 20% higher resistance than you'd calculate from room-temperature values. This creates a feedback loop: higher resistance means more power dissipation, which means higher temperature, which means higher resistance. The system finds an equilibrium, but it's not at the temperature you'd get from a naive calculation.

A Real Worked Example

Let's say you're designing a motor driver board. The motor draws 3 A continuous, and you've got a 12-inch (305 mm) trace running from the connector to the H-bridge. The board uses 1 oz copper, and you've allocated 50 mils (1.27 mm) for the trace width. Ambient temperature in the enclosure is 45°C.

First, trace cross-sectional area:

  • Width: 1.27 mm
  • Thickness: 35 μm (1 oz copper)
  • Area: 1.27×0.035=0.04451.27 \times 0.035 = 0.0445 mm² = 4.45×1084.45 \times 10^{-8}

Trace resistance at 20°C:

R20=1.7×108cdot0.3054.45×108=0.116 ΩR_{20} = \frac{1.7 \times 10^{-8} \\cdot 0.305}{4.45 \times 10^{-8}} = 0.116 \text{ Ω}

Power dissipation (first approximation):

P=32×0.116=1.05 WP = 3^2 \times 0.116 = 1.05 \text{ W}

That's over a watt dissipated in a narrow copper strip. The temperature rise depends on how well that heat can escape—whether the trace is on an internal layer (poor), an outer layer with solder mask (okay), or an outer layer with airflow (better).

Using IPC-2152-derived curves for an external trace, 3 A through a 50 mil trace in 1 oz copper gives roughly 35°C temperature rise. So your trace is running at:

Tmax=45+35=80°CT_{max} = 45 + 35 = 80°C

Now we need to correct the resistance for temperature:

R80=R20×[1+0.0039×(8020)]=0.116×1.234=0.143 ΩR_{80} = R_{20} \times [1 + 0.0039 \times (80 - 20)] = 0.116 \times 1.234 = 0.143 \text{ Ω}

Actual power dissipation:

P=32×0.143=1.29 WP = 3^2 \times 0.143 = 1.29 \text{ W}

That's 23% more power than our initial estimate. In practice, the thermal curves already account for some of this, but the point stands: temperature-dependent resistance matters.

Open the PCB Trace Temperature Rise calculator and plug in these values. You'll see the temperature rise, the corrected resistance, and whether you need to rethink your layout.

What If We Made the Trace Wider?

Doubling the trace to 100 mils cuts resistance in half and dramatically improves thermal performance (more copper area to spread and radiate heat). The temperature rise drops to around 12°C, putting the trace at 57°C instead of 80°C. That's a much more comfortable margin.

Alternatively, going to 2 oz copper with the original 50 mil width gets you similar results—roughly 15°C rise. The choice depends on your stackup constraints and cost tolerance.

The IPC Standards: What They Do and Don't Tell You

IPC-2152 replaced the older IPC-2221 charts that everyone used for decades. The old charts were based on data from the 1950s—single-sided boards, no solder mask, no planes nearby. They were conservative for modern multilayer boards in some cases and dangerously optimistic in others.

IPC-2152 is based on actual thermal testing of modern PCBs, which is good. But the standard still makes assumptions:

  • Still air (no forced convection)
  • Trace in isolation (no nearby heat sources)
  • Steady-state conditions (no pulsed currents)
  • Specific board constructions

Your actual board might have a ground plane 8 mils below the trace acting as a heat spreader, or it might have three other power traces running parallel and adding their heat to the mix. The calculator gives you a reasonable baseline, but you need to apply engineering judgment.

Common Mistakes and Gotchas

Ignoring Vias in the Current Path

A single via has significant resistance compared to a wide trace. A 12 mil via in 1 oz copper has maybe 0.5-1 mΩ of resistance—not much in isolation, but if you've got 3 A flowing through it, that's 4.5-9 mW per via. String ten vias in series (which happens in poorly-planned layouts), and you've got a localized hot spot that won't show up in trace calculations.

Forgetting About Thermal Coupling

Two 2 A traces running parallel for 3 inches will each run hotter than a single 2 A trace in isolation. The heat from one trace raises the local ambient for the other. I've seen boards where the calculated temperature rise was 25°C but the actual rise was 40°C because of adjacent traces and components.

Using Nominal Copper Weight

Your "1 oz" copper might be 0.9 oz after etching, especially for fine-pitch traces where the etch factor eats into the sidewalls. For critical current paths, assume 10-15% less copper than nominal.

Treating Internal and External Layers the Same

Internal traces run significantly hotter than external traces at the same current. They're surrounded by FR-4, which is a decent thermal insulator. An internal trace might see 50% higher temperature rise than an external trace with the same geometry and current. The calculator typically assumes external traces—if you're routing power on internal layers, add margin.

Ignoring Transient Behavior

The steady-state temperature rise assumes continuous current. If your load is pulsed—say, 5 A for 100 ms every second—the trace won't reach steady-state temperature. It'll cycle between some lower and higher temperature. This can actually be worse for reliability than steady heating because of thermal fatigue in solder joints. But it also means you might be able to use a narrower trace than steady-state calculations suggest, if you understand the thermal time constants involved.

The Solder Mask Myth

Some engineers think solder mask insulates the trace and makes it run hotter. In reality, solder mask has minimal thermal impact—it's thin, and while it's not a great conductor, it doesn't trap heat significantly. What matters more is whether you have exposed copper for convective cooling or a ground plane underneath for conductive spreading.

When to Worry (And When Not To)

A 10°C temperature rise on a power trace is generally fine. 20°C is worth paying attention to. 40°C or more means you need to redesign or add thermal management.

But context matters. A 30°C rise that puts your trace at 55°C (in a 25°C ambient) is probably fine. The same 30°C rise in a 70°C ambient enclosure puts you at 100°C, which is getting close to solder reflow temperatures and will accelerate electromigration in your copper.

Also consider what's nearby. A hot trace next to a precision voltage reference is a problem even if the trace itself is within spec. A hot trace in the middle of a power section surrounded by other power components? Probably fine.

Quick Rules of Thumb

For external traces in still air with 1 oz copper, these rough guidelines work for a 10°C temperature rise:

  • 1 A: ~10-15 mils wide
  • 3 A: ~50-60 mils wide
  • 5 A: ~120-150 mils wide
  • 10 A: ~400+ mils wide (you probably want a polygon pour)

Double the width for internal traces. Halve it if you're using 2 oz copper. These are ballpark numbers—always verify with actual calculations for anything that matters.

Try It

Grab your schematic, find your highest-current trace, and open the PCB Trace Temperature Rise calculator. Plug in the real numbers—actual trace width, actual copper weight, realistic ambient temperature. If the temperature rise surprises you, you're not alone. Most engineers underestimate trace heating until they measure it on a real board with a thermal camera.

Better to find out in the calculator than in production.

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