If Juancito says that things are of such of way, I put my signature under his one.
I don't think the site from edcorusa is very reliable but as i say before the difference with for a µrel=480 is not that big.
Here is a site i use: http://www.nssmc.com/product/catalog_download/pdf/D004je.pdf
M6=35z155 (the data is for un-cut cores so EI will be less good)
I use 23zh85 and 30zh105 (i have also other materials)
Here is a site i use: http://www.nssmc.com/product/catalog_download/pdf/D004je.pdf
M6=35z155 (the data is for un-cut cores so EI will be less good)
I use 23zh85 and 30zh105 (i have also other materials)
That would be great, an experimental check.
Calculations I did with μeff ≈ 481, however in datasheet there is not μ, then I took that value from Edcor M6 lamination μ ≈ 47000 @ 16000 Gauss (DC) only to estimate the air gap.
https://www.edcorusa.com/t/Mtr-Core-Steel
μeff = [Bdc(max) 9 l] / [4 π Np i(DC)]
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For UI you need two bobbins and 3UI is for three phase (power)transformers.
UI is very useable for PP transformers
btw for really high quality c-cores, series sm; se; sg; su or s3u, are much better
See also:
http://www.waasner.de/produkte/transformatorenbauteile/transformatorenbauteile/
UI is very useable for PP transformers
btw for really high quality c-cores, series sm; se; sg; su or s3u, are much better
See also:
http://www.waasner.de/produkte/transformatorenbauteile/transformatorenbauteile/
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That "horizontal partitioning" is called "Vertical Sectioning" because original winding is divided into vertical sections, is just semantic anyway.
Nothing to thank, we are here to help each other.
If you have any doubt, please ask me, if I know the answer no problem, if I don't, I will sign to you an "explainenture" remember that I think slowly.
Magnetic materials has a magnetic permeability µ, that is in the manufacturer datasheet, if you add an air gap that magnetic permeability is strongly decreases to the µeff value, I bet that, as is common today, you use mks units, the worst of the worst for electromagnetism, introducing µ, µo, ε, εo, with its own units, this is a madness to confuse people.
Fortunately Edcor guys use cgs units, =Gauss, [µ]=dimensionless.
Dumb people like me, we love cgs units, because all is easier.
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Tip#6 How to evaluate distributed capacitance
Suppose a winding like fig.2, we have N primary layers, at point Mp1
Ce = (2.5/N)² C1
At point Mp2
Ce = (3.5/N)² C2
At point Mp3
Ce = (8.5/N)² C3
At point Mp4
Ce = (9.5/N)² C4
Where Ci is each interwinding capacitance between P ans S, so total distributed capacitace
Cd = (2.5/N)² C1 + (3.5/N)² C2 + (8.5/N)² C3 + (9.5/N)² C4 +...
If C is the average interwinding capacitance, then
Cd ≈ C [(2.5/N)² + (3.5/N)² + (8.5/N)² + (9.5/N)² +...]
Suppose a winding like fig.2, we have N primary layers, at point Mp1
Ce = (2.5/N)² C1
At point Mp2
Ce = (3.5/N)² C2
At point Mp3
Ce = (8.5/N)² C3
At point Mp4
Ce = (9.5/N)² C4
Where Ci is each interwinding capacitance between P ans S, so total distributed capacitace
Cd = (2.5/N)² C1 + (3.5/N)² C2 + (8.5/N)² C3 + (9.5/N)² C4 +...
If C is the average interwinding capacitance, then
Cd ≈ C [(2.5/N)² + (3.5/N)² + (8.5/N)² + (9.5/N)² +...]
Hi Steven
Glad to know that equations works so well for you, indeed, I suffered the same in early days, that's the reason I derived it.
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Tip#8 Vertical Sectioning
Let's consider now a normal winding as in fig.3a, suppose that interwinding capacitance is C and interlayer capacitance is 3C, total distributed capacitance is
Cd ≈ C [(2.5/12)² + (3.5/12)² + (8.5/12)² + (9.5/12)²] ≈ 1.26 C
Total winding capacitance is
(Cw)⁻¹ ≈ (3C)⁻¹ {2 {(4/3x3) [1-(1/3)]}⁻¹+ {(4/3x6) [1-(1/6)]}⁻¹}
Then
Cw ≈ 0.25 C
Then total shunt capacitance
Cs = Cd + Cw ≈ 1.51 C
Let's consider now a split winding as in fig.3a, all areas are halved, so
Cd ≈ (C/2) [(5.5/24)² + (6.5/24)² + (17.5/24)² + (18.5/24)²] ≈ 0.63 C
Total winding capacitance is
(Cw)⁻¹ ≈ (3C/2)⁻¹ 4 {(4/3x6) [1-(1/6)]}⁻¹
Then
Cw ≈ 0.07 C
Then total shunt capacitance
Cs = Cd + Cw ≈ 0.7 C
It works!!!
Let's consider now a normal winding as in fig.3a, suppose that interwinding capacitance is C and interlayer capacitance is 3C, total distributed capacitance is
Cd ≈ C [(2.5/12)² + (3.5/12)² + (8.5/12)² + (9.5/12)²] ≈ 1.26 C
Total winding capacitance is
(Cw)⁻¹ ≈ (3C)⁻¹ {2 {(4/3x3) [1-(1/3)]}⁻¹+ {(4/3x6) [1-(1/6)]}⁻¹}
Then
Cw ≈ 0.25 C
Then total shunt capacitance
Cs = Cd + Cw ≈ 1.51 C
Let's consider now a split winding as in fig.3a, all areas are halved, so
Cd ≈ (C/2) [(5.5/24)² + (6.5/24)² + (17.5/24)² + (18.5/24)²] ≈ 0.63 C
Total winding capacitance is
(Cw)⁻¹ ≈ (3C/2)⁻¹ 4 {(4/3x6) [1-(1/6)]}⁻¹
Then
Cw ≈ 0.07 C
Then total shunt capacitance
Cs = Cd + Cw ≈ 0.7 C
It works!!!
Thanks! But I do not deserve.
That's why my wife calls me Calculín.
You're right looking at shunt capacitance but this kind of sectionizing rises Leakage Inductance to the sky
Yves.
Agree, but the trick is to lower shunt capacitance, nothing more, sometimes can be useful.
I will make calculations for cross connected primaries, I bet leakage inductance will be not so high, but not today...
Maybe for a small SMPS transformer, but with many primary layers is a nightmare to implement, thanks, but no thanks.
Z windings really help downloading capacitance. In a obscure project I'm working on, it helped me too much, as to reduce capacitance almost 3 times less.
Certainly, Popilin, can you explain where a "humble TV repairman" get those complex equations and demonstrations?.
Something doesn't close for me...
Certainly, Popilin, can you explain where a "humble TV repairman" get those complex equations and demonstrations?.
Something doesn't close for me...
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