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2nd April 2014, 03:30 AM  #231  
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Location: Traslasierra

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µeff is not a capricious value, neither a random number, it depends on previous choice of Bdc(max) μeff = Bdc(max) (9 l) / [4 π Np i(DC)] No need to put random numbers here, e.g. 330350. And is the other way around, with this calculated µeff, now you can calculate the air gap lG = l (μ  μeff) / (μ μeff) As GOSS have been improved over the years in terms of saturation and losses, its very high permeability is quite similar, so if you have the core datasheet better, but if not, an astoundingly good estimate can be made with almost any similar standard datasheet. In the particular case of transformer of post#71 I used the data from Edcor M6 GOSS lamination, as permeability has a maximum in the region between 10 KGauss12 KGauss, and the curve μ=f(B) is almost symetrical in that region, I took the value for B=16 KGauss because I needed μ at 6500 Gauss. So with that value, μ=47000 @ 16000 Gauss, I obtained lG≈0.35 mm Let suppose now a "Standard Unspecified M6" with μ=80000 @ unspecified B value, I would obtained lG≈0.36 mm, not too much difference, a so experienced winder would know it. Quote:
As have been perfectly cleared on the first post, magnetic anisotropy makes extremely difficult to obtain the values of magnetic permeability for a given hysteresis curve, so we must work with DC values, such curves are in datasheets, like this, made with data from here http://www.nssmc.com/product/catalog...pdf/D004je.pdf If we add an air gap, things are completely different, we have not μ anymore, but μeff, that is between one to two orders of magnitude lower, and above certain value of B, it is quite constant almost up to saturation, like this, which is constant at about 1% from 100 Gauss up to saturation!!! Linearity comes from line, if the medium were homogeneous and isotropic, the relation between B and H would be linear too B = μ H Unfortunately, the core nightmare is a reality, but not all is lost, if we have a superposition of two fields, DC and AC, we can apply linear superposition principle, then we have something like this With a careful design, you can make hysteresis loss very small, so the area into the curve B=f(H) will be also very small, as μ(AC) is the slope of the curve, it can be almost constant, that's the goal for linearity. Then, for properly designed SingleEnded OPTs, μ(AC) is quite constant under all its normal operation. Quote:
BTW. For calculation purposes, always you can consider the gap as single valued, except you need another strange property for your transformer. Quote:
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To lower distortion, your goal is to increase primary inductance Lp = (4 π μ S Np²) / (9 l x 10⁸) Three ways to do that 1 Increasing μ 2 Increasing Np 3 Increasing S Has been shown that you almost can not increase μ, because of your obsession to obtain astoundingly low copper losses, you resign to use many turns, so Np is kept low, as it is squared, you must increase sky high S, because of l, you must increase S even more! Then enormous cores are needed, even when you work with lower values of B, hysteresis losses deppends mostly on how big is your core, mind you that hysteresis losses are expressed in W/Kg and EI cores has higher losses, then the B=f(H) curve has a bigger area inside it, and with lower B the curve is even thicker. Bingo! You did it! You increased μ, but now it is not quite constant anymore, you ruined linearity!!! Is a valid approach increase Lp in order to decrease distortion, but must be do in the right way, your particular way to maximize Lp is the recipe to disaster in terms of linearity. Not necessarily Bac(max) = (Uac x 10⁸) / (√2 π fo S Np) Hysteresis loss Wh ≈ η f (Bmax)¹˙⁶ x 10⁻⁸ Eddy current loss We ≈ ξ d² f² (Bmax)² x 10⁻¹¹ B is a linear function on its parameters, core losses doesn't. If you reduces B increasing Np, no problem, but if you increases S, the relation of S with the volume (weight) of the core is highly nonlinear, as you use enormous cores, you would achieve the paradox of reducing B with more core losses.
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2nd April 2014, 08:00 AM  #232 
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2nd April 2014, 07:49 PM  #233 
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Late at night, tired and dizzy after struggling hours with mks units (I hate them) I can write almost anything.
B is not a linear function on the sense B = a x + b I meant that its parameters varies linearly, that is!
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3rd April 2014, 03:18 AM  #234 
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Join Date: Oct 2010
Location: Traslasierra

Let's consider again the transformer of post#71, why not use a bigger core?
OK, let's go for an EI 150 ! Uac = √(P Zp) = √(10 x 3000) ≈ 173 VRMS For the same lamination, stacking factor is 0.95, so S = 5 x 5 x 0.95 ≈ 23.75 cm² l = 30 cm Supposing now Bac=Bdc=4000 Gauss Np = (Uac x 10⁸) / [√2 π fo S Bac(max)] ≈ 2049 (rounded to 2048) Ns = Np √(η Zs / Zp) ≈ 101 (rounded to 102) We have i(DC) = 80 mA, so μeff = Bdc(max) (9 l) / [4 π Np i(DC)] ≈ 525 Lp = (4 π μ S Np²) / (9 l x 10⁸) ≈ 24 H Supposing μ≈47000 we can estimate the air gap lG = l (μ  μeff) / (μ μeff) ≈ 0.56 mm Mean turn length, with correction for rounded corners is about lm ≈ 278 mm Then, for Φp=0.45 mm Φs=0.60 mm (trifilar) Rp = ρ Np lm / (π rp²) ≈ 61 Ω Rs = ρ Ns lm / (π rs²) ≈ 0.19 Ω So Ploss [%] = 100xRp/(Zp+Rp) ≈ 2 % Sloss [%] = 100xRs/(Zs+Rs) ≈ 2.3 % Insertion Loss ≈ 0.18 dB !!! Suggested winding pattern (two primaries and three paralleded secondaries) S8PS8PS S = 102 turns P = 128 turns With 0.1 mm (NOMEX) between primary layrers, 0.56 mm (seven layers 0.08 mm NOMEX) between each primary and secondary windings, and another 0.56 mm to finish the bobbin 16x0.50+9x0.65+20x0.1+5x0.56 = 18.65 mm Coil former has about 22 mm, so you must winding to 85% window, easy. Leakage inductance Ls = [0.417 Np² lm (2 n c + a)] / (10⁹ n² b) ≈ 10 mH Leakage inductance increased a bit, but not to throw out the hairs. Interwinding capacitance C ≈ (ε A) / (4πd) ≈ 413 cm ≈ 460 pF Interlayer capacitance Cl ≈ (ε A) / (4πd) ≈ 2581 cm ≈ 2873 pF Total winding capacitance (Cw)⁻¹ ≈ (Cl)⁻¹ ∑ {(4/3nı) [1  (1/nı)]}⁻¹ ı=1,...,NP Cw ≈ 209 pF Total distributed capacitance Cd ≈ C ∑ (nı/N)² ı=1,...,m Cd ≈ 663 pF So shunt capacitance is Cs = Cw + Cd ≈ 872 pF Shunt capacitance was almost doubled, even increasing insulators, so winding pattern could be changed. Hysteresis loss comparison {[B(EI 150)] / [B(EI 84)]}¹˙⁶ [V(EI 150) / V(EI 84)] ≈ 0.39 * 5.23 ≈ 2 Then an EI 150 core has double hysteresis loss than an EI 84 core, even when B(max) was almost halved ! This is the most optimistic case, because modern lamination is better than on times of Steinmetz, and the exponent 1.6 would be reduced to at least 1.4 As I said before this could happen, even when Bac(max) was halved, bad news is that now, because double hysteresis loss with halved Bac(max), hysteresis loop is very thick, we don't have a quite constant μ(AC) anymore, we have destroyed linearity!!! The moral is: Do not use big cores unnecessarily, if you really need a big core, E I cores are not the best option, go for a C core, unless you have the money for a more exotic material with lower losses.
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I do not suffer from insanity, I enjoy every minute of it.  Edgar Allan Poe I hate people who don't face up.  Anonymous 
3rd April 2014, 04:22 PM  #235  
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Join Date: Oct 2010
Location: Traslasierra

As now elementary math has abandoned us, due to magnetic anisotropy, we only can do a graphical description.
At each point on the B=f(H) curve, must be B = μ H Then μ(AC) is the "slope" of the curve at each point. Bac1 and μ(AC)1 corresponds to transformer with EI 150 core, Bac2 and μ(AC)2 corresponds to humble transformer with EI 84 core. A properly designed SE OPT, must have μ(AC) quite constant, so Lp must have the same behavior. Huge variations on Lp, says us that linearity was destroyed. However that has nothing to do with this variation Quote:
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Totally agree !!!
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I do not suffer from insanity, I enjoy every minute of it.  Edgar Allan Poe I hate people who don't face up.  Anonymous Last edited by popilin; 3rd April 2014 at 04:33 PM. 

3rd April 2014, 04:46 PM  #237  
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Location: Traslasierra

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Let's clear that "colifa" means "a bit crazy guy" That are friends!
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I do not suffer from insanity, I enjoy every minute of it.  Edgar Allan Poe I hate people who don't face up.  Anonymous Last edited by popilin; 3rd April 2014 at 04:57 PM. Reason: TarzanEnglish 

4th April 2014, 02:43 AM  #239 
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Join Date: Mar 2003
Location: NYC


4th April 2014, 05:33 AM  #240  
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Join Date: Oct 2010
Location: Traslasierra

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If you has a curve B=f(H), the derivative at each point represents the tangent to the curve at each point, as on each point B = μ H Then we can write μ = dB/dH Maybe you can understand this way. Wow, thank you! Don't help me so much... Quote:
Last option, DC bias plus two minor loops. Quote:
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I do not suffer from insanity, I enjoy every minute of it.  Edgar Allan Poe I hate people who don't face up.  Anonymous 

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