Ac

Once you determined Ac, you can find the required trace width for a given copper weight: width=Ac/thickness, where minimum thickness(mil)=oz/1.3. The above equation provides reasonably accurate approximation of the generic Fig.5-2 charts. For example, for i=10A and ∆T=20

For those who work with metric units, here is a quick conversion reference: 1 mm=0.03937", 1 mil^2=0.000645 mm^2, 1 oz/ft^2 copper is 0.033 mm thick minimum.

- All the results here are obtained by interpolation of the IPC plots, so there is always some inaccuracy.
- For simplicity, our calculations don't include the effect of the board material (FR4 is worse than polyimide just by 2%).
- The analysis is valid for traces spaced apart by more than 1" (which often may not be practical). If parallel tracks are spaced closer, their temperatures will increase. In this case, you need to use combined current to determine their combined cross-sectional area. The presence of heat dissipating components may also raise the temperature.
- The tests that formed the basis of the new standard were conducted for electric currents up to 30 ampere and ∆T up to 100 degree C.
- The output data here do not include any derating. It is always recommended to add some
**safety margin**.

The information and the widget are provided here with

IPC-2152 is © Copyright 2009 IPC, Bannockburn, Illinois, USA.

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