What limits the current in wires

[mag_therm]

Hi Mr Kokoriantz,
To get an approximation for a cylindrical conductor,
First read about the  power loss as a function of  current and resistance to correct your above calculation.

To get an approximate geometric progression of current rating versus  diameter,you will need to do some math.
Hints, take surface area into account, and do some math on the exponent of the ratio of the diameter increases.
You will get a progression based on constant Power / Surface_area
It is not so easy.

That progression might help your understanding, but in the practical tables, other factors are taken into account including empirical tests.

[Kokoriantz]

The resistance is inverse the square of the thickness, so for same amount of calories, the current should be 25 times, as it is cooled by 5 times more surface, then, a wire of 1mm should carry 125 times the current of 0.2mm. If so, the limit given on the table of 0.5A for 0.2mm should be 62.5A for 1mm. It is shows 14A.

The last 3 days I was traveling  from Thailand to central Vietnam and I wrote it tired at 3 am local time. Sorry for that.
If the wire is thicker by N, than the same calories to dissipate needs a current N times (I²R). As the surface to dissipate is N times higher then it needs √ N more current to get the same temperature. If the wire is 0.2mm given to have 0.51A max chassis current, then the current should be 5×√5×0.51 =5.7A  for 1mm, it is still not 15A.
All this doesn't explain why my classA amplifier couldn't pass 1A with a lead  given to pass 6A max in chassis but limited 0.8A in power transmission and why it can pass high currents if the same lead is paired by an opposite phase carrying current.
If it is a thermal threshold that the copper becomes nonlinear resistance, what is that precise temperature above which the wire cannot pass more current?  The curve bellow shows linear increase up to 600K° or 330C° of 0.4%/C°. There is no any singular limiting temperature.

 

[mag_therm]

I take liberty to partially quote:
"If the wire is 0.2mm given to have 0.51A max chassis current, then the current should be 5×√5×0.51 =5.7A  for 1mm, "

I agree, that you are now correct. For constant (Watt_per_surface m^2) the exponent on (diameter_ratio) is 1.5 .

There are many other factors in design.
a) For example adiabatic heating calculations for short time current until a protective device opens.
b) For example a transformer, inductor winding if impregnated  will usually allow currents in excess of  {0.2 mm, 0.51 Amp} based on temperature class.

[Kokoriantz]

As IanB did well remarked, the max current is slightly more than the square of the diameter. This means if you tie a cable of 25 brines of 0.2mm wires passing through each the maximum 0.51A then you get 25 times the power to dissipate but with a surface of only 5 times of each. This results the 1mm cable with 12.75A  get 5 times hotter than 0.51A through 0.2mm, with 15A, it will be 7 times hotter. This proves that the current limit is independent from temperature.
Thank you IanB, thank you very much.

[Kokoriantz]

This is the size of TPA3255 produces 255W/ch @4ohm, that is 8A peak from a single lead. The size of the lead is 0.27×0.17. This is equivalent to 0.22mm diameter. The table shows max. current of 0.7A. How do you think it does work? The output lead is sandwiched between its own +/- supplies.   

[Kokoriantz]

I found this video that shows the energy flow doesn't obey lump ohms law but rather electric/magnetic fields. I will try to put my hand on the simulation software to see what is going on in the wire that is limiting the current.

[tooki]

I found this video that shows the energy flow doesn't obey lump ohms law but rather electric/magnetic fields. I will try to put my hand on the simulation software to see what is going on in the wire that is limiting the current.

Bear in mind that that video and its claims have been discussed at length in other threads here back when it, and the controversial predecessor video (below) came out.

https://youtu.be/bHIhgxav9LY