What trace width do I need?
Enter the current, the temperature rise you allow and the copper weight, and pick an outer or inner layer. The trace width calculator gives the minimum width in mils and millimetres from the IPC-2221 formula.
- Minimum width (mil)
- 11.83
Use a trace at least 11.83 mils (0.3004 mm) wide.
- Minimum width (mm)
- 0.3004
- Cross-section (mil²)
- 16.3
- Copper thickness (mil)
- 1.378
Minimum width (mil): 11.83. Use a trace at least 11.83 mils (0.3004 mm) wide.
Minimum width (mil) by current
How to calculate
Works out the minimum width of a printed circuit board trace for a current, a temperature rise and a copper thickness, on an outer or inner layer, from the IPC-2221 formula.
Example with the default inputs (Current 1 A, Temperature rise (°C) 10, Copper weight (oz/ft²) 1, Layer Outer (external)): Use a trace at least 11.83 mils (0.3004 mm) wide.
Method: A (mil²) = (I ÷ (k × ΔT^0.44))^(1 ÷ 0.725), k = 0.048 (outer layer) or 0.024 (inner layer). Thickness (mil) = oz × 1.378. Width = A ÷ thickness; mm = mil × 0.0254.
- IPC-2221 is a curve fit to old measurements of bare traces in still air; it is a starting point, often conservative for inner layers. Planes, nearby copper and airflow change the real temperature.
- The charts behind the fit cover currents up to 35 A, widths up to 400 mil, rises of 10 to 100 °C and 0.5 to 3 oz copper; outside that range treat the answer with care.
Worked examples
Each example is checked against the calculator on every build.
- Current 1 A, Temperature rise (°C) 10, Copper weight (oz/ft²) 1, Layer Outer (external) gives Minimum width (mil) 11.825836, Minimum width (mm) 0.300376, Cross-section (mil²) 16.296001.Source: Salitronic, "PCB trace width calculator and guidelines" (IPC-2221: I = k × ΔT^0.44 × A^0.725, k = 0.048 external, 0.024 internal, A in mil²; 1 oz copper = 35 µm), https://salitronic.com/kb/pcb-trace-width/ (retrieved 2026-10-05); Advanced Circuits, "PCB trace width calculator" (width = area ÷ (oz × 1.378 mil); the IPC-2221 charts cover up to 35 A, 400 mil, 10 to 100 °C rise, 0.5 to 3 oz), https://www.advancedpcb.com/en-us/tools/trace-width-calculator/ (retrieved 2026-10-05)
- Current 5 A, Temperature rise (°C) 20, Copper weight (oz/ft²) 2, Layer Inner (internal) gives Minimum width (mil) 92.986118, Cross-section (mil²) 256.269741.Source: Salitronic, "PCB trace width calculator and guidelines" (IPC-2221: I = k × ΔT^0.44 × A^0.725, k = 0.048 external, 0.024 internal, A in mil²; 1 oz copper = 35 µm), https://salitronic.com/kb/pcb-trace-width/ (retrieved 2026-10-05)
- Current 10 A, Temperature rise (°C) 10, Copper weight (oz/ft²) 1, Layer Outer (external) gives Minimum width (mil) 283.231909, Minimum width (mm) 7.19409.Source: Salitronic, "PCB trace width calculator and guidelines" (IPC-2221: I = k × ΔT^0.44 × A^0.725, k = 0.048 external, 0.024 internal, A in mil²; 1 oz copper = 35 µm), https://salitronic.com/kb/pcb-trace-width/ (retrieved 2026-10-05)
How it works
IPC-2221 gives the current a trace carries for a temperature rise:
I = k × ΔT^0.44 × A^0.725
with I in amperes, ΔT in °C, A the copper cross-section in square mils, and k = 0.048 for an outer (external) layer or 0.024 for an inner (internal) layer. Solved for the area:
A = (I ÷ (k × ΔT^0.44))^(1 ÷ 0.725)
The copper thickness is t = copper weight (oz/ft²) × 1.378 mil, and the minimum width is
width (mil) = A ÷ t, width (mm) = width (mil) × 0.0254
Rules
- Current from 1 mA to 100 A; temperature rise from 1 to 100 °C; copper weight from 0.25 to 10 oz/ft².
- The page computes outside the charts’ range (35 A, 400 mil, 10 to 100 °C, 0.5 to 3 oz) too; such answers are extrapolations.
- The powers make every value a float; values are shown to 4 significant figures.
Worked examples by hand
1 A, 10 °C, 1 oz, outer layer. 10^0.44 = 2.75423. 0.048 × 2.75423 = 0.132203. 1 ÷ 0.132203 = 7.56413. 7.56413^(1/0.725) = 7.56413^1.37931 = 16.296 mil². Width = 16.296 ÷ 1.378 = 11.83 mil = 0.3004 mm.
5 A, 20 °C, 2 oz, inner layer. 20^0.44 = 3.73640. 0.024 × 3.73640 = 0.0896736. 5 ÷ 0.0896736 = 55.7578. 55.7578^1.37931 = 256.27 mil². Thickness = 2 × 1.378 = 2.756 mil. Width = 256.27 ÷ 2.756 = 92.99 mil.
10 A, 10 °C, 1 oz, outer layer. 10 ÷ 0.132203 = 75.6412; 75.6412^1.37931 = 390.29 mil². Width = 390.29 ÷ 1.378 = 283.2 mil = 7.194 mm.
Other questions people ask
How do I calculate PCB trace width?
IPC-2221 relates current I (A), temperature rise ΔT (°C) and cross-section A (square mils): I = k × ΔT^0.44 × A^0.725, with k = 0.048 on outer layers and 0.024 on inner layers. Solve for A, then divide by the copper thickness: 1 oz copper is 1.378 mil. For 1 A, 10 °C and 1 oz on an outer layer the width is about 11.8 mil (0.30 mm).
Why do inner layers need wider traces?
An inner trace is buried in the board material, which conducts heat poorly, so it cannot shed heat into the air. IPC-2221 halves the constant k for inner layers; for the same current and rise, the trace needs about 2.6 times the cross-section.
What temperature rise should I use?
10 °C above the surroundings is a common, cautious choice; 20 °C is often used where space is tight. A larger rise allows a narrower trace but runs the copper hotter, which matters near heat-sensitive parts and in hot enclosures.
What does copper weight in ounces mean?
It is the weight of copper spread over one square foot. 1 oz of copper on a square foot is about 35 µm (1.378 mil) thick; 2 oz is twice that. Thicker copper carries the same current in a narrower trace.
How accurate is the IPC-2221 formula?
It is a curve fit to old charts of bare traces in still air, covering up to 35 A, widths up to 400 mil, rises of 10 to 100 °C and 0.5 to 3 oz copper. Planes, nearby copper, vias and airflow change the real temperature. Use it as a starting point and leave a margin.
How do I convert mils to millimetres?
A mil is a thousandth of an inch, so 1 mil = 0.0254 mm. A 10 mil trace is 0.254 mm wide, and 0.1 mm is about 3.94 mil.