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 IPC2221 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 rearranging this formula to
determine the crosssectional 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 IPC2221 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 IPC2221 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 = 1.7E6 ohmcm
Copper Temp_Co = 3.9E3 ohm/ohm/C
Voltage Drop is Current * Resistance
Power Loss is Current^2 * Resistance