Calculate PCB trace width for target current based on copper thickness and temperature rise.
This tool sizes a PCB copper trace using the IPC-2221 current-capacity equation, which relates trace cross-sectional area to allowable temperature rise above ambient. You enter target current, copper weight (oz/ft²), allowed temperature rise, and whether the trace is on an outer or inner layer, and it solves for the required width in mils and mm. The detail people miss most: internal traces need roughly 2.6 times the width of external traces for the same current, not merely double, because the formula raises area to the 0.725 power, which amplifies the 2x gap between the internal and external k constants; internal copper also cannot shed heat by convection to open air the way outer-layer copper can.
A concrete input and expected output from the current implementation.
Input
External layer, 1 oz/ft2 copper, 2 A target current, 20 C allowed temperature rise
Expected output
Required trace width approx 20.2 mils (0.51 mm), based on a required cross-sectional area of about 27.8 sq mils at 1.378 mils copper thickness
Using I = 0.048 x deltaT^0.44 x A^0.725 for external traces, solving for A at I=2A and deltaT=20C gives about 27.8 sq mils; dividing by 1oz copper thickness (1.378 mils) gives the width.
The calculator uses the IPC-2221 fitted curve I = k x deltaT^0.44 x A^0.725, where I is current in amps, deltaT is temperature rise in degrees C, A is cross-sectional area in square mils, and k is 0.048 for external traces or 0.024 for internal traces. It solves this for A given your inputs, then divides A by the copper thickness implied by your chosen ounce weight to get width.
Selecting an ounce weight (0.5, 1, 2, 3 oz/ft2, etc.) fixes trace thickness at roughly 0.7, 1.378, 2.756, 4.134 mils respectively, since ounce weight is a measure of copper mass per square foot, not a direct thickness. The tool converts your chosen weight to thickness internally before dividing the solved area by it to output width.
No, this tool only sizes the flat trace width. Via current capacity depends on barrel plating thickness and diameter and needs a separate via-current calculation; a trace sized correctly here can still bottleneck at an undersized via.
Because the internal k constant (0.024) is half the external one (0.048) in the IPC-2221 formula, but the formula raises area to the 0.725 power. Halving k multiplies the required area by 2^(1/0.725), about 2.6, not 2. Physically this reflects that inner-layer copper is sandwiched in FR-4 and loses heat mainly by conduction through the board rather than convection to air, so it needs more cross-sectional area to hit the same temperature rise.
The example and behavioral notes were checked against the browser implementation. Standards and primary references below define the relevant format, formula, or platform behavior.
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