The calculation method of PCB line width and current is as follows:

First calculate the cross-sectional area of Track. The copper foil thickness of most PCBS is 35um (if you are not sure, you can ask THE PCB manufacturer). The cross-sectional area is multiplied by the width of the line. There is an empirical value for current density ranging from 15 to 25 amperes per square millimeter.

Weigh the cross-sectional area to get the through-flow capacity. I= KT0.44a0.75K is the correction coefficient. Generally, 0.024 is taken in the inner layer of copper clad wire, and 0.048t is taken in the outer layer as the maximum temperature rise, and the unit is Celsius (the melting point of copper is 1060℃). A is the cross-sectional area of copper clad, and the unit is square MIL(not mm mm, I is the maximum allowable current, the unit of amperes (AMP) is generally 10mil=0.010inch=0.254, which can be 1A, 250MIL=6.35mm, and is 8.3A data. The calculation of PCB current-carrying capacity has been lacking of authoritative technical methods and formulas. Experienced CAD engineers rely on personal experience to make more accurate judgments. But to CAD novice, can not be said to meet a difficult problem.

The current carrying capacity of PCB depends on the following factors: line width, line thickness (copper foil thickness), allowable temperature rise. As we all know, the wider the PCB line, the greater the current-carrying capacity. Here, please tell me: assuming that 10MIL can withstand 1A under the same conditions, how much current can 50MIL withstand, is it 5A? The answer, of course, is no. The line width is in the unit of Inch (Inch Inch =25.4 millimetres) 1 oz. Copper =35 micron thick, 2 oz.=70 micron thick, 1 oz =0.035mm 1mil.=10-3inch. Trace Capacityper MIL STD 275

The pressure drop caused by the resistance of the wire length must also be considered in the experiment. The tin on process welds is only used to increase the current capacity, but it is difficult to control the volume of tin. 1 OZ copper, 1mm wide, generally 1-3 A galvanometer, depending on your line length, the pressure drop requirements.

The maximum current value should be the maximum allowable value under the temperature rise limit, and the fuse value is the value at which the temperature rise reaches the melting point of copper. Eg. 50mil 1oz temperature rise 1060 degrees (i.e. copper melting point), current is 22.8A.