I'm having trouble calculating the core loss coefficient Kfe for ferrite materials. I'm referencing Chapter 15 of "Fundamentals of Power Electronics" by Robert W. Erickson and Dragan Maksimović.

The textbook defines Kfe in units of W cm-3 T, where β is the core loss exponent. For modern ferrite materials, β is between 2.6 and 2.8. In the Cuk converter example (15.3.1), a ferrite pot core with Kfe = 24.7 W cm-3 T is used, while in example 15.3.2, Kfe = 7.6 W cm-3 Tβ.

I attempted to calculate Kfe using Steinmetz's equation with datasheet parameters from Ferroxcube's "Soft Ferrites and Accessories" for materials 3C91 and 3C93:

For 3C91 (100 kHz, β = 2.7, Bpeak = 200 mT, Pv = 300 kW m-3):

$$K_{fe} = (0.3\ \mathrm{W\,cm^{-3}}) / ((0.2T)^{2.7}) \approx 23.13\ \mathrm{W\,cm^{-3}\,T^{-\beta}}$$

For 3C93 (100 kHz, β = 2.7, Bpeak = 100 mT, Pv = 50 kW m-3):

$$K_{fe} = (0.05\ \mathrm{W\,cm^{-3}}) / ((0.1T)^{2.7}) \approx 25.06\ \mathrm{W\,cm^{-3}\,T^{-\beta}}$$


  1. Are my calculations correct, or am I missing something?
  2. Is there a better way to compute the core loss coefficient (Kfe) for ferrite materials?

1 Answer 1


You can do it that way, roughly, but it's only a one-point calibration. β will be only an assumption. Better data requires more effort. As tabulated data are not provided, at this point I just break out the spreadsheet and fit curves by hand:

enter image description here

With Microsoft Excel (and almost certainly everything else spreadsheets these days?), the plot area background can be filled with a screenshot from the ferrite material datasheet, lined up with axes, and curves plotted on top of it.

The table is filled with formula e.g. C21 = =$B$12*($B21^$B$13+$B$15*$B21^$B$14)*C$19^$B$16 repeated by row and col, and B6 = =B3*B12*((B4*1000)^B13+B15*(B4*1000)^B14)*B5^B16/1000000*1000^B16. Basically, the expected formula, accounting for unit multipliers, and applied across rows/cols.

I used a modified formula, effectively to vary the frequency exponent α, or Kfe generally, with frequency, to better fit the particular curve of this material (N97, 3C97, etc.; most major manufacturers have an equivalent). As long as the Kfe is calculated properly, this has no effect on their optimization formulas (incorporating β and Bpeak). Or you can simply remove the extra terms.

  • \$\begingroup\$ Where are you getting the frequency exponents, α and α1, from and why are there two of them? What is Cm and why are there two? \$\endgroup\$
    – Hector
    Commented May 19 at 2:36
  • \$\begingroup\$ Cm is K_fe under a different name. Cm, alphas and beta are fitting parameters. As the Steinmetz formula always has been; it's an empirical curve fit that does well enough for most things (but, as you can see, even when modified, isn't a perfect fit). \$\endgroup\$ Commented May 19 at 4:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.