PCB Trace Width Calculator

How wide should a copper trace be for your current? Find the width, or check the current a trace already has.
Current → Width
Width → Current

Required Trace Width for a Current

A = (I / (k × ΔT0.44))1/0.725  →  Width = A / (copper thickness)
1A, ≤10°C rise, 1oz external
3A, ≤20°C rise, 1oz external
2A, ≤10°C rise, 1oz internal
A
°C
Enter values and press Calculate.
Rule of thumb: a 1 oz external trace roughly needs about 12 mils (0.3 mm) of width per amp for a modest 10 °C rise — use this calculator for the exact figure, especially at higher current or on internal layers where the margin is smaller.

Current-Carrying Capacity vs Trace Width

1 oz, this layer 2 oz, this layer

Current Capacity of an Existing Trace

I = k × ΔT0.44 × A0.725
20mil, ≤10°C, 1oz ext
50mil, ≤20°C, 2oz ext
mils
°C
Enter values and press Calculate.

Everything You Need to Know About Trace Width

Copper is a good conductor, but it is not a perfect one — a trace carrying current always has some resistance, and that resistance turns part of the current's energy into heat (I²R). A trace that is too narrow for its current will heat up, and if it heats up too much it can damage the board's fibreglass substrate, discolour or lift the copper, or in extreme cases open-circuit entirely. The industry-standard way to size a trace safely is the IPC-2221 formula (the same standard used by every professional PCB design tool):

I = k × ΔT0.44 × A0.725 — where I is the current in amps, ΔT is how much hotter the trace is allowed to get above the surrounding board (°C), A is the copper cross-section area (trace width × copper thickness, in square mils), and k is 0.048 for external (top/bottom) layers or 0.024 for internal (buried) layers.

Why two different k values? (external vs internal)

A trace on the outer layer of the board is directly exposed to the surrounding air, so heat can escape by convection straight off its surface — it runs cooler for a given current. A trace on an inner layer is sandwiched between layers of fibreglass (FR4), which is a poor thermal conductor, so the heat has to travel much further to escape. That is why internal traces get the smaller k=0.024 (half of the external k=0.048) and, for the same current and temperature rise, need roughly 2.6× more copper area than an external trace.

Step-by-step: how to size a trace

  1. Know your current. Use the worst-case (maximum) current the trace will ever carry, not the average.
  2. Pick an allowed temperature rise. 10–20 °C above ambient is a common, safe default for general traces.
  3. Decide the copper weight. 1 oz is standard for most consumer boards; 2 oz+ is common for higher-current power boards.
  4. Decide internal vs external. Route high-current traces on outer layers where possible — they need less width for the same current.
  5. Calculate the required width using the tool above, then round up to a clean design-rule value (e.g. the nearest 5 or 10 mil).
QuantityFormula / Value
Required area (given I, ΔT)A = (I/(k×ΔT0.44))1/0.725
Current capacity (given A, ΔT)I = k×ΔT0.44×A0.725
k, external layer0.048
k, internal layer0.024
1 oz copper thickness1.378 mils (35 µm)

If you need to control the exact temperature rise more precisely (e.g. keep a trace under 5 °C rise near a sensitive component), or want to work through the maths interactively, the companion PCB Trace Temperature Rise calculator uses the identical formula in more depth. This page focuses purely on the everyday question: "how wide does my trace need to be?"

Real-World Applications & Fully-Explained Examples

Worked examples — explained in full

1. A 1 A signal/power trace. 1 A, 10 °C rise, 1 oz external. First, area A = (1/(0.048×100.44))1/0.725. 100.44≈2.75, so A=(1/(0.048×2.75))1/0.725=(1/0.132)1.379=7.581.37916.3 mils². Dividing by the 1 oz thickness (1.378 mils) gives a width of 16.3/1.378≈11.8 mils (about 0.30 mm). So a fairly thin trace — well within typical minimum trace-width design rules.
2. A 3 A power trace. 3 A, 20 °C rise, 1 oz external: this needs 35.3 mils (0.90 mm) of width — exactly three times example 1's width, since both the higher current and the higher allowed rise push in the same direction.
3. The same 2 A trace, internal vs external. At 10 °C rise, 1 oz: the external version needs 30.8 mils, but the identical trace buried as an internal layer needs 80.0 mils — 80.0/30.8≈2.6× wider, simply because internal layers cannot shed heat to open air.
4. Same current, thicker copper. Switching example 2's board from 1 oz (35.3 mils) to 2 oz copper for the same 3 A, 20 °C rise drops the required width to 17.7 mils — exactly half — a common way to save board space on high-current designs without adding more layers.
5. Checking an existing trace's capacity. A 20 mil, 1 oz external trace: A=20×1.378=27.56 mils². I=0.048×100.44×27.560.725=0.048×2.75×11.06≈1.46 A before exceeding a 10 °C rise.
6. Tight thermal budget near a sensor. If a design demands only 5 °C rise instead of 10 °C for the same 1 A example-1 trace, the required width grows from 11.8 mils to 18.0 mils — about 53% wider, since ΔT only enters the formula to the power of 0.44 (a diminishing-returns relationship, not linear).

Frequently Asked Questions

How wide should a PCB trace be for a given current?

It depends on the current, your allowed temperature rise, the copper weight, and whether the trace is on an external or internal layer. As a rough starting point, a 1 oz external trace needs about 12 mils (0.3 mm) per amp at a modest 10 °C rise, but always calculate the exact value with the IPC-2221 formula for your specific conditions.

What is the IPC-2221 standard?

IPC-2221 is the industry-standard PCB design guideline (successor to the older MIL-STD-275) that defines the empirical relationship between trace current, cross-sectional area, and temperature rise. It is the formula used by virtually every professional PCB CAD tool's trace-width calculator.

Why do internal traces need to be wider than external ones?

External traces sit exposed to open air on the top or bottom of the board, so they can shed heat easily by convection. Internal traces are buried inside layers of fibreglass, a poor heat conductor, so for the same current they run hotter — needing roughly 2.6× more copper area to hit the same temperature rise.

What temperature rise should I allow?

10 °C is a common conservative choice for general traces; 20 °C is acceptable for less critical or short runs. Traces near heat-sensitive components (crystals, precision sensors) are often designed for 5 °C or less.

What is 1 oz copper and how thick is it?

Copper weight on a PCB is specified in ounces per square foot; 1 oz copper is about 1.378 mils (35 µm) thick. 2 oz is twice as thick (70 µm), and so on — thicker copper carries more current for the same trace width.

Does trace length affect the required width?

The IPC-2221 formula is independent of length — it predicts the steady-state temperature rise from cross-sectional area and current alone. Very long, unbroken high-current traces may benefit from extra margin in practice due to voltage drop (see the Trace Resistance calculator) even if the thermal calculation alone doesn't require it.

Can I use a copper pour instead of a routed trace for high current?

Yes. For currents where the formula demands a very wide trace, a solid copper pour (plane) behaves like an extremely wide trace and can carry much more current at a lower temperature rise — the standard approach for power and ground distribution.

How much current can a 10 mil (0.25 mm) trace carry?

On a standard 1 oz external layer at a 10 °C rise, about 1 A. On 2 oz copper it can carry noticeably more; on an internal layer it can carry meaningfully less for the same temperature budget — use the "Width → Current" tab above for your exact numbers.

Should I round the calculated width up or down?

Always round up, and preferably to a clean design-rule increment (e.g. the nearest 5 or 10 mils), to keep a safety margin over the bare-minimum calculated value and to match common manufacturing capabilities.

Does this formula apply to flex PCBs?

IPC-2221 was developed primarily for rigid FR4 boards. Flex circuits (polyimide substrates) have different thermal behaviour, so while the formula gives a reasonable starting estimate, flex-specific design guidelines should be consulted for critical designs.

What happens if my trace is too narrow?

It will run hotter than intended, potentially discolouring or delaminating the board locally, and in extreme overcurrent situations the trace can act as a fuse and burn open — sometimes desirable (intentional fuse traces) but usually a reliability failure.

Is there a simple rule of thumb instead of doing the maths?

A commonly quoted rough guide is about 12 mils (0.3 mm) of 1 oz external trace per amp for a modest 10 °C temperature rise, but this is only a starting estimate — always verify with the actual formula for anything current-critical, especially on internal layers or with tighter thermal budgets.

Why does doubling the current more than double the required width?

It doesn't quite — area scales as I1/0.725 ≈ I1.38, so doubling current roughly requires about 2.6× the area (and since width is proportional to area, roughly 2.6× the width) for the same temperature rise, slightly more than a straight doubling.

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