In my quest to learn about ferrite materials Re: Relative Permeability - Page 1
Forum section: electronics / rf-microwave

author: Qmavam
author_rank: Regular Contributor
timestamp: August 27, 2019, 09:25:22 pm
content: I recently learned that when making an inductor and looking at the graph of relative permeability of your material choice, the point where µ' and µ" cross the Q of any inductor will be 1 or less. Using relative permeability graph, if you pick a frequency 10kHz and say µ" is 1000 and µ" is 4000, that the highest Q inductor you can make is 4000/1000=4.but it will be less because of wire and other losses. See page 9 second graph for an example.https://assets.lairdtech.com/home/brandworld/files/SIP-CAT-TOROID-BALUN%20CORES%200113.pdfMy question how does the Core Q affect a transformer?I have a circuit that used a specific core at frequencies where the Core Q has dropped way below 1.Is that a pore core choice?I think it's not, because of the designer, but I don't know why it wouldn't be.Thanks, Mikek

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 27, 2019, 11:44:05 pm
content: The Q of a transformer is made irrelevant by using a large enough magnetizing impedance (i.e., enough turns).Fortunately this works, otherwise pulse transformers, baluns and so on would be uselessTim

author: Qmavam
author_rank: Regular Contributor
timestamp: August 28, 2019, 02:19:44 am
content: Thanks much, Is the 4x rule altered, if you have a core with a Q under 1 vs say Q of 100?Mikek

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 28, 2019, 05:19:35 am
content: What rule of 4x?The ratio of mu's is correct, for a core of material as given. For power application (not necessarily what you're doing), the core is gapped with air, which has mu' = 1 and mu'' ~= 0*, and so the mu_eff is the weighted average of the two. Typically |mu_eff| in the 20-60 range, so effectively mu'' is reduced by the same factor as mu' is.*Hm, I suppose strictly speaking it isn't zero, but in a purely magnetostatic problem (assuming near fields, infinite speed of light, no propagation) it would be. Anyway, at 100s of kHz, we're hard pressed to notice, compared to losses of materials.A saturable reactor effectively turns some of the core into airgap, so the losses drop with the inductance. This is visible in some ferrite bead data, when Z(F) is shown for different I_DC bias (see Laird ferrite bead catalog for instance). The curve gets lower (L is down significantly), and peak |Z| drops some, but the peak width becomes much narrower. (If it were constant-loss, we would expect peak |Z| to drop proportionally with L.)It is not, however, obvious if your core should have a Q of, say, greater or lesser than 1, at I_DC = 0, when you intend to operate at some arbitrary level of bias and frequency. Higher initial Q should certainly help, and if you expect to operate near zero bias, you won't get as much increase in Q as you would while operating at high bias.Tim

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: August 28, 2019, 05:40:31 am
content: In my world, ferrites don't exist, so I can't really comment here. But is the \$\mu '\$ and \$\mu '' \$ anything like \$\epsilon'\$ and \$\epsilon''\$? Can loss also be related to the angle of this?I would assume so, since duality is a thing.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 28, 2019, 05:49:30 am
content: Yeah, it's the magnetic form, mu = mu' + j mu''. So there's angle involved too; the network in question has an impedance somewhere between an inductor and resistor.Tim

author: Qmavam
author_rank: Regular Contributor
timestamp: August 28, 2019, 02:48:46 pm
content: QuoteI was taught when winding a transformer to make the winding have a reactance 4x the impedance being transformed at the lowest frequency of interest.Example: 50ohm to 450 ohm transformer, the 50 ohm primary should have 200 ohms XL at the lowest frequency of interest.I had no understanding until I found this, I don't know if there was text with it, I'm not finding it in my computer.Mikek

author: Qmavam
author_rank: Regular Contributor
timestamp: August 28, 2019, 09:46:15 pm
content: Just poking around elsewhere, I saw this,Quoteand a response,QuoteI don't agree that the 10x does anything good, and the extra interwinding capacitance can cause problems on a wide band transformer.Anyone have an opinion on this, Mikek

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 29, 2019, 01:07:41 am
content: Ahyeah, that's what I thought. So it follows, if you have some impedance Z being transformed (say it's transformed 1:1 so we don't have to do any ratios here), and you have a transformer magnetizing impedance that's all resistance (lossy) but four times |Z|, then you get a maximum Q of about 4, which is when Z is all reactive, and the parallel combination is now\[ \frac{(4|Z|)(j|Z|) }{ (4|Z|) + (j|Z|) } \]\[ = |Z| \frac{ j 4 }{ (4 + j } \frac{ 4 - j }{ 4 - j } \]\[ = |Z| \frac{ j 16 - j^2 4 }{ 17 } \]\[ = |Z| \frac{ 4 + j 16 }{ 17 } \]and the Q is Im/Re =\[ Q = \frac{ \frac{16}{17} }{ \frac{4}{17} } = 4 \]...or exactly 4, heh.So if you want to transform a load of, say, Q = 100, you need a transformer with Rmag > 100*Xload.Tim

author: Qmavam
author_rank: Regular Contributor
timestamp: August 30, 2019, 02:55:26 pm
content: Quote from: T3sl4co1l on August 29, 2019, 01:07:41 amSo it follows, if you have some impedance Z being transformed (say it's transformed 1:1 so we don't have to do any ratios here), and you have a transformer magnetizing impedance that's all resistance (lossy) but four times |Z|, then you get a maximum Q of about 4, which is when Z is all reactive, and the parallel combination is now(4|Z|)(j|Z|)(4|Z|)+(j|Z|)=|Z|j4(4+j4−j4−j=|Z|j16−j2417=|Z|4+j1617and the Q is Im/Re = Q=1617417=4...or exactly 4, heh.So if you want to transform a load of, say, Q = 100, you need a transformer with Rmag > 100*Xload.TimI appreciate a the math but it mostly over my head.I think you started use the 4 times number I mentioned.did the math to show that to show the formula works with a Q up to about 4.Then you conclude with, when transforming a source impedance with a high Q, you need a transformerthat has a much high ratio than the factor of 4 mentioned.Or, said another way reactive loads need a high ratio than resistive loads.EDIT: Ahh1 I probably have that wrong, at resonance impedance is resistive, so, don't know.Did I get any understanding from your math?However, in your conclusion you use Rmag, is Rmag = my factor (where I use 4).I never saw it defined as Rmag.After seeing the factor of 4 in many sources, I got curiousto see why a factor of four. I did a search and all I came up with is the graph I posted.Thanks for your help, Mikek