Cu Foil or Wire in air offer high inductance(do/di variation flux with variation of current), as are surrounded by air uniformly , compare to, on PCB Trace, and hence the reactance offered(Xl=2.Pie.f.L) would be more in air, or the law current density.
I understood your question differently - aren't you asking about the current carrying capability? Why is it allowed to pass much higher current through a PCB trance than through an solid copper wire?
From my perspective this is mainly related to the ratio of cross-sectional area (that generates heat when current passes thru it) and circumference which helps to get rid of the heat, i.e. dissipate it to ambient. As wire becomes smaller, it is simply that it's capability to dissipate heat is enhanced. Think of the little bond wires, for instance. PCB traces have usually a thickness of 35 µm - and compared to that - a HUGE surface area helping to dissipate the heat resulting from ohmic heating.
Compare to air, as heat dissipation would be faster, hence resistivity offer would be less, and current could rise. Can be considered as second point(Z=Squrt.(RSqr.+XlSqr.)).Yet the effect of Xl would be more considerable at higher frequency.
Higher surface area for a given cross sectional area gives better cooling. Take an example of a round copper wire of diameter d and length L. Then its cross-sectional area is A1 = pi*d*d/4 and its total surface area is S1 = pi*d*L, where pi = 3.1416. If the same wire is flattened out to a thickness x, as in a pcb track, then keeping cross-sectional area and length same, track width will be w2 = pi*d*d/4x and top surface area will be S2 = w2*L = pi*d*d*L/4x = S1(d/4x). Thus, if x
Thank you so much for explaining the concept. I completely got to understand why the Cu in PCB Traces has higher current density as compared to a wire. Your answer has been very helpful for me to know this concept. Thank you so much Sir.
As explained by you, the total surface area of the PCB track now depends on the value of 'x' alone with other dimensional parameters of the wire being same. Now, if x=35microns, then the surface area of the same wire with d=1mm which is flattened out as a PCB track is approximately 7 times bigger than that of a normal wire. So how does this translate into the corresponding Current Density? In other words, in terms of figures, if the surface area increases by 7 times, by how much exactly does the allowable current density increase? I would like to derive this relation. Can you help me out sir.
this is a question that won't get an precise answer. There are simply too many factors to deal with. As you understood, the trace and its surroundings build up a complex thermo-electrical model with heat transfer depending on the temperature differences basing on conduction and convection. You could simulate this for simple set-ups using the material properties of the PCB, air, copper and so on. Perhaps, there are some rule-of-thumb guidelines out there - can anyone help?
look at the IPC-2221 and IPC-2152 design "rules" to get an idea how current (density) and temperature correlate. Those deal simple and very conservative setups. As Michael Meisser said, the reality on an industrial PCB can be much different - usually colder. This is because heat spreading in FR4 and by other copper layers create a bigger heated surface and thus a better dissipation into ambient air. This is the reason why a thin trace with same cross section and current is much colder than the same wire in air. By the way, the current density itself is not a metric for temperature. Same current density can create any temperature.