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How to calculate the trace width and copper weight

Table of production minimum sizes

 Copper Weight 0.5oz 1oz 2oz 3oz 4oz or above Outer layers Minimum Trace Width 3mil 4mil 5mil 6mil RFQ Minimum Trace Spacing 4mil 5mil 7mil 10mil RFQ Via holes to other copper featuer 7mil 9mil 12mil 16mil RFQ Inner layers Minimum Trace Width 3mil 3.5mil 5mil 6mil RFQ Minimum Trace Spacing 3mil 4mil 6mil 9mil RFQ Via holes to other copper featuer 7mil 8mil 11mil 15mil RFQ

Notes:

The formula for calculating allowable current through a trace is published in the IPC-2221 standard section 6.2 as shown below.

Where I is current, k is a constant, ΔT is Temperature Rise, & A is cross sectional area of trace.

The trace width can then be calculated by re-arranging this formula to determine the cross-sectional area that our desired current can safely pass through.

Area[mils^2] = (Current[Amps]/(k*(Temp_Rise[deg. C])^ 0.44))^(1/0.725)

Then, the Width is calculated from the cross sectional area for a chosen thickness:

Width[mils] = Area[mils^2]/(Thickness[oz]*1.378[mils/oz])

As per IPC-2221 for internal layers k = 0.024 and for external layers: k = 0.048

Disclaimer:

These calculations are industry standard and believed to be correct, but not guaranteed. May not be appropriate for all designs.

Trace Width Calculator FAQs

Q: Is there a limit to the amount of current this tool can calculate a width for?

A: Yes. The IPC-2221 data from which these formulas are derived only covers up to 35 Amps, trace width up to 400 mils, allowable temperature rise from 10 to 100 degrees Celsius, and copper of 0.5 to 3 ounces per square foot. If used outside of these ranges, this calculator will extrapolate thus becoming more inaccurate with higher currents.

Q: Instinctively, I would predict that internal trace widths would need to be less than external traces since the external trace can peal off of the board if too hot. Your calculator gives the opposite result. Why?

A: External layers have better heat transfer than internal layers as air dissipates heat due to convection, while the internal dielectric does not conduct heat as well. Since the goal of the Trace Width Calculator is to prevent an excess temperature rise of the traces, it makes the internal traces wider because they store more heat. In the case of a circuit in vacuum, or in a potted assembly, external layers do not have the benefit of heat convection in the air thus you should use the internal trace width for all traces.

Q: What does temperature rise mean in this context?

A: Temperature rise is the difference between the maximum safe operating temperature of your PCB material and the typical operating temperature of your board. Higher current flow increases the temperature of the copper traces therefore, temperature rise is a design parameter for how much added heat you would like to design for. Based on this limit the formula chooses a width to stay within it. Ten degrees is a safe rule of thumb for most applications. If you need to reduce trace width then you can increase this value if your PCB material and operating temperature allows.

Q: In some cases thermal relief lines called "wagon wheels" or "spokes" are used when connecting a pad to a large copper area for easier soldering. I used the trace width calculator and the width given for these spokes are so wide that it is impractical to use. How should I calculate these?

A: Thermal relief spokes are usually very short. The formula this calculator is based on was determined empirically for reasonably long transmission lines. The purpose of this calculator is to prevent traces form overheating, thus if these spokes are connected to dissipate heat then they do not need to be as wide as this tool predicts. Please consult other PCB design resources for this issue.

Q: What unit of measure is Mils?

A: A Mil is one thousandth (1/1000) of an inch. Its name is derived from Latin mille meaning thousand. In electronics mil is commonly used but in other disciplines it may be referred to as thou and a mil being a millimeter.

The below is the formula for resistance of a trace:

Resistance = Resistivity*Length/Area*(1 + (Temp_Co*(Temp - 25))

Where, Area = Thickness*Width

A copper Thickness of 1 oz/ft^2 = 0.0035 cm

Copper Resistivity = 1.7E-6 ohm-cm

Copper Temp_Co = 3.9E-3 ohm/ohm/C

Voltage Drop is Current * Resistance

Power Loss is Current^2 * Resistance