1. The calculation method is as follows

Calculate the cross-sectional area of the track first. The copper foil thickness of most PCBs is 35um (if you are not sure, you can ask the PCB manufacturer, 1 ounce is 35um, in fact, it is less than 35um) multiplied by the line width is the cross-sectional area, pay attention to convert to square millimeters.

There is an empirical value of current density, which is 15~25 amperes/mm2. Call it the upper cross-sectional area to get the flow capacity. I=KT0.44A0.75 (K is the correction factor, generally 0.024 for the inner layer of the copper clad wire, and 0.048T for the outer layer as the maximum temperature rise, in degrees Celsius (the melting point of copper is 1060°C) A is the copper clad Cross-sectional area, the unit is square MIL (not millimeter mm, note that it is square mil.) I is the maximum allowable current, the unit is ampere (amp) generally 10mil=0.010inch=0.254 can be 1A, 250MIL=6.35mm, it is 8.3A

2. Data

The calculation of PCB current-carrying capacity has always lacked authoritative technical methods and formulas, and experienced CAD engineers can make more accurate judgments relying on personal experience. But for CAD novices, it can be said that they have encountered a problem. The current carrying capacity of the PCB depends on the following factors: line width, line thickness (copper foil thickness), and allowable temperature rise. Everyone knows that the wider the PCB trace, the greater the current-carrying capacity. Here, please tell me: assuming that under the same conditions, a 10MIL trace can withstand 1A, how much current can a 50MIL trace withstand, is it 5A? The answer is naturally no. Please see the following data from international authoritative organizations: The unit of line width is: Inch (inch = 25.4 millimetres) 1 oz. Copper = 35 microns thick, 2 oz. = 70 microns thick, 1 OZ = 0.035mm 1mil. =10-3inch. Trace Carrying Capacity per mil std 275

Three, experiment

In the experiment, the voltage drop caused by the wire resistance caused by the wire length must also be considered. The tin on the process welding is only to increase the current capacity, but it is difficult to control the volume of the tin. 1 OZ copper, 1mm wide, generally used as a 1-3 A ammeter, depending on your cable length and voltage drop requirements.

The maximum current value should refer to 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 (ie copper melting point), the current is 22.8A.