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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

https://www.tdk-electronics.tdk.com/inf/80/db/fer/e_70_33_32.pdf

Manjesh Gowda
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    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](https://www.tdk-electronics.tdk.com/download/528882/71e02c7b9384de1331b3f625ce4b2123/pdf-n87.pdf). 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. – Andy aka Feb 09 '21 at 17:44
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    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). – Andy aka Feb 09 '21 at 17:50
  • Related: https://electronics.stackexchange.com/questions/547321/core-loss-of-an-inductor – winny Feb 09 '21 at 17:54
  • 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? – Manjesh Gowda Feb 09 '21 at 18:04
  • 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. – Andy aka Feb 09 '21 at 18:43
  • Airflow from a dedicated cooling fan can help, but as Andy hints to, you'll want a bigger core and lower loss for reliable operation. – rdtsc Feb 09 '21 at 19:01
  • I got 8kW core loss. But it is mentioned in the datasheet that "/set" Does it mean I have to double the core loss which is 16kW or is it 8kW itself? Sorry I feel kinda dumb – Manjesh Gowda Feb 09 '21 at 19:12
  • https://www.tdk-electronics.tdk.com/download/ 531596/133c4190b4d8aac6ea08cc21352bf2d8/pdf-application.pdf On the page number 31 they have mentioned the Rth as 5.5K/W. Is it correct? – Manjesh Gowda Feb 09 '21 at 19:53
  • @Andyaka How much temperature did u consider as ambient temperature for the core? – Manjesh Gowda Feb 10 '21 at 07:45
  • @manesh my top comment alludes to a temperature increase of 100 degC when 9.5 watts is being lost in the core. So, you decide on what starting ambient is suitable for your application but, remember, that the local ambient around the core will warm as the core (and copper) throw out their heat unless it is adequately dissipated. – Andy aka Feb 10 '21 at 08:25
  • So now, it depends on the copper loss. Is it possible to have a theoretical calculation of copper losses? I used signal generator to calculate Ac resistance. I would like to know if i can calculate theoretically also. Btw i am using litz wire for winding – Manjesh Gowda Feb 10 '21 at 08:58
  • @Andyaka I have mentioned the losses. Is it possible to rougly tell the temperature rise? – Manjesh Gowda Feb 13 '21 at 09:12
  • Read my very first comment. – Andy aka Feb 13 '21 at 09:39
  • @Andyaka I have edited the question with providing copper losses. Could you give me any literature or any links which gives proper explanation for temperature rise. I guess you are not considering the copper loss. I am considering ambient temperature as 40degreeC – Manjesh Gowda Feb 13 '21 at 12:14
  • There is no direct relationship between temperature rise of a volume of mass and the power it is dissipating. It entirely depends on where the heat flows to and, at some point, when the device warms to a certain higher temperature, the heat removed by external factors will match the heat generated by power losses. Your question does not mention these external factors but, if you dig into the data provided by ferrite vendors then you should be able to rationalize my opening comment in terms of what ambient conditions were. The DS says 100 degC rise due to 9.5 watts. Look up what that means. – Andy aka Feb 13 '21 at 12:21
  • In effect, you need to find the thermal resistance of the core and windings together. That thermal resistance tells you how high the temperature will rise above ambient for a certain heat power that needs dissipating. If local ambient rises due to the heat then, the temperature of the core and windings rise accordingly. There is no direct relationship between power dissipated in a certain volume or shape and how warm it will get. It's a complex calculation but my opening comment contains some good clues. – Andy aka Feb 13 '21 at 12:23
  • Could you share any links or literature for the calculation of thermal resistance because I am new to magnetics and dont know much about these things. Would be really great if u could share about the thermal resistance. – Manjesh Gowda Feb 13 '21 at 17:58

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