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How to calculate the line width and current of PCB design

How to calculate the line width and current of PCB design
Regarding the empirical formulas of PCB line width and current, there are many relationship tables and software online. In the actual PCB board design, it is necessary to comprehensively consider the size of the https://www.ipcb.com/https://www.ipcb.com/, and select an appropriate line width through the current.


The relationship between PCB line width and current
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). The cross-sectional area is multiplied by the line width. Note that it is converted to square millimeters.

 There is an empirical value of current density, which is 15-25 A/mm2. Call it the upper cross-sectional area to get the flow capacity.


    K is the correction factor, generally 0.024 for the inner layer of copper clad wire, and 0.048 for the outer layer


    T is the maximum temperature rise, the unit is degrees Celsius (the melting point of copper is 1060°C) A is the cross-sectional area of the copper clad, and the unit is square MIL (not millimeters 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

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 cannot 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 authorities:


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 Capacityper mil std 275

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, the temperature rise is 1060 degrees (that is, the melting point of copper), and the current is 22.8A.