How to use the pcb trace width calculator
- Choose to find the trace width for a current or the maximum current for a width.
- Enter the current or width, the allowed temperature rise and the copper weight.
- Pick external or internal for the layer, and enter a trace length if you want resistance and voltage drop.
- Read the width in mils and millimeters, the cross-section, the resistance and the power lost.
Formula
I = k x dT^0.44 x A^0.725, with I in amps, dT in degrees C, A in square mils
k = 0.048 for outer layers, 0.024 for inner layers
Width = A / thickness; 1 oz copper is 1.378 mil thick
Resistance = resistivity x length / (width x thickness), resistivity 1.724e-8 ohm m at 20 C, +0.393 % per degree
This is the IPC-2221 relation for trace current capacity. The resistance uses the copper temperature, which is the surrounding air plus the allowed rise.
Three cases
To carry 1 A on an outer layer of 1 oz copper with a 10 C rise needs a cross-section of 16.30 square mils, which is a width of 11.83 mil or 0.300 mm. A 50 mm trace of that width has 86.82 milliohms at 35 C, so it drops 86.82 mV and loses 86.82 mW.
To carry 3 A on an inner layer with a 20 C rise needs 126.68 square mils, a width of 91.93 mil or 2.335 mm. In the other direction, a 10 mil outer trace in 1 oz copper with a 10 C rise carries 0.8855 A.
Why the width depends on the rise
A current heats a trace in proportion to its resistance. A wider or thicker trace has less resistance, and more surface to lose heat. The formula links the current to the temperature rise that results, so you choose the rise you can accept. Inner layers lose heat less easily, which is why their constant is half that of outer layers.
Related tools include the wire voltage drop calculator, the Ohm's law calculator, the LED resistor calculator and the op-amp gain calculator.
Limits of the formula
The IPC-2221 curves were fitted to older measurements on certain boards and are extended beyond that range, so treat them as an estimate. Thick boards with planes nearby carry more, and crowded or enclosed boards carry less. Short narrow necks, vias and connectors often decide the real limit. Allow a margin and follow your board maker's rules and any safety standard for your product.
Assumptions and limits
- The relation is IPC-2221, with the temperature rise from 1 to 100 C and currents up to 100 A.
- Copper thickness is 1.378 mil per ounce per square foot, with no plating added.
- Resistance is calculated at the surrounding air temperature plus the rise, with 25 C as an example for the air.
- The result is an estimate for steady direct current, not a rating for pulses or high-frequency current.
Frequently asked questions
How wide should a PCB trace be for 1 amp?
About 12 mil (0.30 mm) on an outer layer of 1 oz copper for a 10 C rise, by the IPC-2221 formula. Many designers use a wider trace for a safety margin.
Why are internal layers wider for the same current?
Inner traces are surrounded by board material and lose heat more slowly, so they run hotter. The formula uses half the constant for them, which calls for a wider trace.
What temperature rise should I choose?
It depends on the board, the other parts and your product. A common starting point is 10 C, with larger values accepted only when the board and nearby parts can tolerate the heat.
How does copper weight change the width?
Thicker copper has more cross-section for the same width. Doubling the weight to 2 oz needs about half the width for the same current and rise.