K_fe: Core Loss Coefficient of an Inductor - Page 1
Forum section: electronics / projects

author: TimNJ
author_rank: Super Contributor
timestamp: March 18, 2019, 10:10:04 pm
content: Hi all,I'm working through the design of a resonant AC inductor for an LLC power supply. This is basically my first time doing this.I'm trying out the design procedure outlined in Ch 15. of 'Fundamentals of Power Electronics' by Erickson and Maksimovic. There is a parameter K_fe, "core loss coefficient" which is necessary to determine the optimal AC flux density for lowest total losses. The book, and countless white-papers I've found online, say "K_fe can be determined from the manufacturer's data" but none of them explain how to do this. (Someone on EDABoard asked this same exact question without getting a good answer:https://www.edaboard.com/showthread.php?379779-How-to-calculate-core-loss-coefficient).I can determine Beta, the "Core Loss Exponent" by finding the slope of the log-log core loss plot. Does anyone know how to determine K_fe, which has the following units?IthinkK_fe depends on B, flux density, so I'm a little confused on how this equation (with output AC peak flux density) also would have a flux density input.Thank you.

author: TimNJ
author_rank: Super Contributor
timestamp: March 19, 2019, 10:46:48 pm
content: I decided to re-familiarize myself with some math concepts that I haven't touched in 10 years or so, since high school. Here's what I came up with.First off, I believe the "core loss coefficient" to be k in the Steinmetz equation.In a log-log plot, like those typically used to describe a magnetic material's core loss, you can determine the exponent of the power function by determining the slope of the log-log plot. You can do this by evaluating b = log(F1/F0) / log(X1/X0), where (X0, F0) and (X1, F1) are a set of points that exist on the plot.For the 100KHz, 100C plot, I found b to be 4.16.Here's where my confusion remains...I'm taking a flux density of 200mT as a nominal operating value, 100KHz switching frequency. If we want to determine the coefficient k, we can use the following (In this case m is equal to b above):I believe this allows you to calculate the coefficient. However, I've tried finding the coefficient at different points on the plot, and the numbers seem to vary.For example, on the 100Khz/100C plot...At (300mT, 8.1W/cm3), k = 1212W/cm3*TbAt (200mT, 0.90W/cm3), k = 728W/cm3*TbAt (100mT, 0.04W/cm3), k = 578W/cm3*TbMath and I aren't best friends, but I don't understand why the coefficient would change...surely that can't be correct?Anyone know what I'm missing here?Thanks!

author: floobydust
author_rank: Super Contributor
timestamp: March 20, 2019, 03:37:23 am
content: <crickets>I'm not fluent in going down the rabbit hole for core loss calulations, they are super extremely complicated. I find almost always the copper losses are higher.I found one reference in "Modern Ferrite Technology" to a book byColonel W.T. McLyman(pic related) first published 1982.Colonel W.T. McLyman, "Magnetic Core Selection for Transformers and Inductors"which I would hunt down and see his empirical equation discussed.He also has: "Transformer and Inductor Design Handbook"This is about all the help I can give.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 20, 2019, 03:57:32 am
content: Right, you can derive the exponent from the log-log plot, although when the exponent itself varies say with frequency, that's a bit awkward. You may have to iterate an approximation in that case; but that's alright, it will converge quickly, and getting within 10% of the true answer is better than the real material will likely do anyway.Anyway, theoffsetof the log-log line is in units of loss (W/cm^3 or whatever) per flux density (tesla). The slope is beta.Um, give or take a +1 to beta I think?It may help to understand core loss in terms of the Q factor of the core itself. Constant-Q curves on the loss diagram are diagonal lines of slope = 1. A lot of materials have an exponent near there, so that Q essentially depends on frequency alone. I've some powder core datasheets that I've doodled such notes on; typical figures for a Sendust mu=60 core are:FreqQ1k1005k8325k75100k52250k37500k22This is a little trickier for gapped ferrite, because you can control the gap. The gap itself is lossless* (air), so as you make more of the core's effective length into airgap, the Q rises proportionally. (A terrible Q < 1 core, with air gap, can still be used quite effectively, if you need to.)Mind, you aren't varying Bpk by varying gap -- what you are varying is the amount of magnetization required to reach Bpk. Which means for the same inductance, you need more turns, and that's where Bpk and therefore losses go down with increasing gap.*Not counting possible eddy currents in wires near the gap.Tim

author: TimNJ
author_rank: Super Contributor
timestamp: March 20, 2019, 04:59:00 am
content: floobydust and Tim, thank you for your answers. Being a complicated subject, I'm going to sleep on your responses and see if I can come up with meaningful answers!For now, I'd like to share my progress in determining K_fe, some problems I ran into, and maybe tips for others attempting to do the same. (Props to my dad, the math teacher, for helping.)-----1.) First mistake I made was reading the values off the core loss plot incorrectly. Notice on the x-axis (flux density), each tick is displayed at 100, 200, 300, etc. On the y-axis (core loss), the ticks are 101, 102, 103, etc. In between 102and 103are minor tick marks corresponding to 2*102, 3*102, 4*102, etc.I however, initially read these as 202, 302. 402, etc. which is totally not the same thing! Don't do that!-----2.) I didn't really understand what the "core loss exponent" and "core loss coefficient" were, but now it is quite obvious. Core loss fits the power function model, that is y = a*xb. In order to determine the optimum flux density that yields the lowest total power loss, that is Pcu + Pfe (copper + core loss), we need to know exactly what the core loss function looks like. In "Fundamentals of Power Electronics", Kfecorresponds to a, and beta corresponds to b. The optimum flux density is found where Ptot = Pcu + Pfe is minimized, that is the derivative = 0, i.e. at the bottom of a parabolic function.-----3.) While b, the exponent, can be found by determining the slope between two points on the core loss curve, I found this method to be rather inaccurate. Since core loss is on a log-log plot, small errors tend to be amplified, especially where the tick markets get compressed. Calculating by hand, using two points, I got lots of different exponent values, between 2 and 4. This error/variance means the calculated coefficient will also be off. tl;dr Doing this entirely by eye/hand does not seem very accurate.A more accurate method is to select a number of points on the curve (4 to 6, maybe), and fit an equation to it, using Excel. In this case, I chose 4 points, and determined the exponent to be 2.83, and the coefficient to be 27.42W/cm3*Tb. I'm much more comfortable with these values because they are quite similar to the "typical" values noted in the book. (Note, before I was getting coefficient values on the order of 500W/cm3*Tbor more.)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 20, 2019, 07:41:08 am
content: FWIW, the way to reproduce the plotted curve is:Set C2 "=LOG(A2)"Set D2 "=LOG(B2)"Repeat/drag/copy to rows 3-5 as wellSelect E2:F2, enter "=LINEST(D2:D5, C2:C5)", and CTRL+SHIFT+ENTER (this sets the pair as an array, and the contents will gain curly braces). This gives E2 = slope of the log-log plot (the exponent), and F2 = the offset (the factor k).Uh, assuming Excel of course. YMMV.Using formulas and test values in this way, and you can also set the background of the plot as a screenshot of the curves you're fitting -- you can recreate the published values and see that they overlap.Should be alright to do by hand, even just eyeballing it. The slope is the exponent -- consider for example, the PC44 curve at 100kHz, 100C: it crosses 60mT, 10 kW/m^3. The x axis spans not quite a full decade, but we can do half a decade, which is a factor of 3* so we look at 3 * 60 = 180mT, which is basically 200mT which crosses at 300 kW/m^3. This is 30 times more, or a decade and a half. We started with half a decade, so the exponent is 1.5 / 0.5 = 3.*Rough eyeballing guess. It's closer to 3.16, or exactly 10^(0.5). 3.2 is pretty close too, and can be easier to multiply and divide in your head.You should never get an exponent of 2 or 4, at least on this curve. If you get other numbers for other curves, that are obviously sloping differently, well, that would be a good thing, not a bad thing.(The 60C curves are decidedly flatter, so we expect a lower, 2.something exponent there. Doesn't look like anything is pushing 4 here.)Tim

author: TimNJ
author_rank: Super Contributor
timestamp: March 20, 2019, 03:53:36 pm
content: Thanks Tim.I may have over-stated the range of slope values I was getting. It was more like 2.5 to 3.5.As I cleaned up some of my errors, i became fairly convinced the exponent was between 3 and 3.2, as you suggest. But, the problem I found was that the calculated coefficient (out front) would come out differently, for different points, again using this equation:For example, if using b = 3.2At (60mT, 9KW/m3), Kfe = 73.1 (W/cm3*T)At (100mT, 40KW/m3), Kfe = 63.4 (W/cm3*T)At (200mT, 290KW/m3), Kfe = 50.0 (W/cm3*T)But if we use b = 2.83 (what I found in Excel),At (60mT, 9KW/m3), Kfe = 25.83 (W/cm3*T)At (100mT, 40KW/m3), Kfe = 27.04 (W/cm3*T)At (200mT, 290KW/m3), Kfe = 27.57 (W/cm3*T)Notice the variance in Kfe values is much less. I'm not sure how much the accuracy really matters in the end, but just a thought.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 20, 2019, 06:53:53 pm
content: Ah yep, that'll do it. If the constant isn't constant, well...Tim

author: Kleinstein
author_rank: Super Contributor
timestamp: March 20, 2019, 07:44:52 pm
content: The formula for the power loss is to a large part just an empirical one. The actual curve can deviate from this. So the curve in the log-log scale may be slightly nonlinear and thus the slope different if different ranges are used.There can also be a temperature effect: at high power loss the core can heat up quite a bit and thus the temperature can change with the magnetization use. This gives possibly different values for a slow (equilibrium) and fast measurement.

author: TimNJ
author_rank: Super Contributor
timestamp: March 21, 2019, 05:36:35 pm
content: Kleinstein,Right. A close look at the log-log graphs show that the core loss curves do not follow a perfect power relationship, though they are very close.floobydust,Colonel W.T. McLyman seems likethe manwith respect to understanding of magnetic materials. I really should get that book.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: March 21, 2019, 07:01:12 pm
content: I picked up McLyman some years ago when Amazon was running a special on it. It's a great practical book written for people that actually have to build something. Watch out though, as you'll find the occasional error has slipped in.