Using this PCB Trace Width Calculator the external tace width for 11A (default settings) is suggested as 8.21mm.
Is this acceptable? I was expecting something bigger, and just wanted to check that there isn't a bug. Would this include a margin of safety or should I add to this figure?
There's an underlying question here that you want to ask. The width it recommends is for the specified temperature rise on the trace (default 10 C with the linked calculator) above the ambient temperature. In other words, the trace width recommended is the minimum width required to have the temperature of the trace rise no more than 10 C while carrying your 11A current. It's not a binary question of the trace working or not.
With that in mind, the question you have to ask yourself is: What temperature rise is acceptable on the trace for your application?
Be careful with those calculators. Use them as first quick calculation, never as an accurate, definitive tool.
When current passes through a PCB trace there is an equilibrium between the generated heat and the heat going away, basically, by NATURAL CONVECTION of the surrounding air.
Heat transfer by natural convection is highly empirical and depends on many factors, like geometry (horizonta, vertical, top face, bottom face) as well as non-idealities (holes in a plate, air flow perturbations, conducting/ non conducting objects, etc.)
When designing, most of the times there is a maximum environment temperature (like 60 C or 80 C) plus a maximum temperature in your trace (depends on the PCB quality as well as the mean life you want to give to your product, it is around 150 - 200 C), so your temperature rise in a standard application would be the difference.
Keep in mind if there are other components or traces that could add local additional heat to your trace 'under test'.
Also, final thickness of copper traces are different from 'nominal' values (35 microns for 1 ounce)
I suggest one of these three ways:
Go and use a FE/CFD tool (Finite Elements/ Computational Fluid Dynamics) if you like theories and simulations and have money for computers and not for prototypes.
Go and build a prototype and measure the actual temperature. Correct your model if necessary.
Use the calculator and be very conservative (choose a thicker trace)
This sounds about right. According to MiscEl I get the same result for temperature rise of 12°C. If it's an inner layer though, the current rating is less than half, at 5A (as the heat cannot dissipate as easily.
IPC-D-275 is the standard associated with the above calculations.
Also bear in mind the voltage drop and power dissipation. According to MiscEl for a 10cm trace of 8.2mm width, the resistance is ~6mΩ. So if your trace is quite long you could drop a fair amount.
For a 20cm example:
11A * 12mΩ = 132mV drop
11A^2 * 12mΩ = 1.452W
