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my dc dc converter maximum dc current rating is 10A, operating frequency is 20KHZ or 40KHZ, how to choose copper wire dimensions can you please help me
You should calculate the RMS inductor current based on the waveform. Having the RMS value, to get the copper area you can use the formula A = I / J, where A is the conductor's cross section area, I is the RMS current and J is the current density. As a rule thumb, J should not exceed 4,5 A/mm². As lower J goes, fewer are copper losses and the inductor will dissipate less heat.
Another thing to take into account is the skin effect, which reduces the effective copper area due to high frequency. The maximum wire diameter can be calculated as:
Dmax = 2 x 76 / sqrt(fsw)
Where D is the maximum diameter in mm and fsw is the switching frequency in Hertz.
If the required area is greater than the area of a single conductor with diameter Dmax, you should use paralleled condcutors of less diameter.
There is a book called "Inductors and Transformers for Power Electronics" by Vencislav Cekov Valchev and Alex Van den Bossche, it has the complete instruction set for designing the inductor, the method is somewhat related to what is posted by "ysba" but takes into account the effect the maximum permissible copper loss"
you calculate the power handling capability of the inductor that is the sum of product of V and I in all the windings ,primary and secondary, then you divide this power equally in the copper loss and eddy current loss, using the copper loss you get the resistance, which can help you get the cross sectional area assuming that you selected the proper core for the inductor , you get the length by using the mean lenght per turn of the core data sheet. once you have got the area, you get the radius, and hence the wire gauge to be used
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