PCB Trace Width Calculator (IPC 2221)
Find the minimum PCB track width for a given current and allowed temperature rise, using the IPC 2221 formula.
- A = (I ÷ (k × ΔT0.44))1/0.725 with k = 0.048 → 42.39 mil²
- W = A ÷ thickness = 42.39 ÷ 1.38 = 30.8 mil (0.781 mm)
IPC 2221 Trace Width Formula
A (mil²) = (I ÷ (k × ΔT0.44))1/0.725
W = A ÷ (1.378 × copper oz)
k = 0.048 for external layers and 0.024 for internal layers. 1 oz copper is 1.378 mil (35 µm) thick. IPC 2152 is the newer standard and is less conservative for internal layers.
Worked example
2 A on a 1 oz outer layer with 10 °C rise needs A = 42.4 mil², so W = 30.8 mil (0.78 mm). On an inner layer the same current needs about 2.6 times the width.
Frequently Asked Questions
How wide should a PCB trace be for 1 A?
About 0.3 mm (12 mil) on a 1 oz outer layer for a 10 °C rise.
Why are internal traces wider for the same current?
They are surrounded by FR4, which conducts heat poorly, so IPC 2221 halves the current capacity.
What is the difference between IPC 2221 and IPC 2152?
IPC 2152 is based on newer test data and accounts for board thickness and planes; IPC 2221 is simpler and more conservative.
Learn more on the blog
PCB design fundamentals
A practical, field focused guide on instrumentationblog.in that explains the theory behind this calculator and how engineers apply it on real plants.
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