I have a customized inductor which is of E70 core. I would like to calculate the temperature rise in the inductor. Ac copper loss= 5W, DC copper loss 2.5W, core loss 8W (according to steinmetz). I want to rougly calculate the temperature rise/operating temperature of the inductor. Is there any way to calculate it? Please let me know


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    \$\begingroup\$ Your core (N87) will warm 100 degC when the core loss is 9.5 watts so, personally speaking, I believe that with a core loss of 20 watts, it will get too warm because you are running far too close to the curie temperature see this on pg3. The diagram of relevance is top right and, given that doubling the power will about double the temperature rise, you will be in trouble. Choose a bigger core that has less loss is my advice. \$\endgroup\$
    – Andy aka
    Feb 9, 2021 at 17:44
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    \$\begingroup\$ A bigger core also facilitates a larger winding cross sectional area and fewer turns so your copper losses will also come down (which I believe they need to). \$\endgroup\$
    – Andy aka
    Feb 9, 2021 at 17:50
  • \$\begingroup\$ Related: electronics.stackexchange.com/questions/547321/… \$\endgroup\$
    – winny
    Feb 9, 2021 at 17:54
  • \$\begingroup\$ Yea. Does coper losses nothing to do with the heating of inductor? I am also talking in the point of bobbin material temperature! If I use the bobbin which has glass temperature of 100degree c. Temperature rise is only about the core? \$\endgroup\$ Feb 9, 2021 at 18:04
  • \$\begingroup\$ The copper loss will heat the core and drive it further towards the curie temperature. It doesn't look good to me. The core will heat 100 degC above the local ambient with 9.5 watts loss. Local ambient is going to be mainly dictated by copper losses so, core temperature rise sits on top and looks not-good. Core loss at 20 watts is going to cause this design to fail. \$\endgroup\$
    – Andy aka
    Feb 9, 2021 at 18:43


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