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 3rd March 2013, 05:56 PM #31 diyAudio Member     Join Date: Aug 2009 Location: Johnson City, TN That is the work I referenced earlier where it shows the peak permeability part way up the BH curve. I have not read the full text yet.
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Check chapter 5 of RDH4

http://www.paleoelectronics.com/RDH4/CHAPTR05.PDF

Figure 5.20 shows how the effective permeability changes for various DC bias, for one type of core. The problem is that for most cores I have I won't have this data. If you indeed want to measure the permeability of your core, it'd be better to do it with DC bias applied.

I personally will avoid formulas using permeability because I don't intend to measure it. For me it's good enough to assume a safe Bmax somewhere between 1.0 and 1.2 tesla.
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 dcpermeability.PNG (205.0 KB, 487 views)

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Quote:
 Originally Posted by ikoflexer These are the formulas I found for the design of an inductor with DC, and using these the relative permeability is not needed. N = (L*Ipk*10^4)/(Ac*Bm) where N = the number of turns L = desired inductance Ipk = peak current = Idc + Iac/2 Ac = effective core area Bm = desired max flux density, usually 1.2T
This is inconsistent with Electromagnetism, there is no magnetic field for the term Iac/2.

I will use cgs units, because Maxwell's equations are written differently in different systems of units.

The magnetic field B(DC) only depends on current i(DC)

B(DC) = 4 π μ N i(DC) / (9 l)

The magnetic field B(AC) only depends on voltage U(AC)

B(AC) = U(AC) x 10^8 / (√2 π f S N)

Only by making the assumption

B = B(AC) = B(DC) = Bt/2

We can write

N = L x 10^8 i(DC) / (S B)

Also, μ is implicit

L = 4 π μ S N^2 / (9 l x 10^8)
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Last edited by popilin; 3rd March 2013 at 07:14 PM. Reason: Alz someone

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Quote:
 Originally Posted by popilin This is inconsistent with Electromagnetism, there is no magnetic field for the term Iac/2.
Sorry, I don't understand what you mean. There is DC current through the primary. This DC current flow will induce flux in the core. In the same time there is an AC current component, which, if we assume a sin wave shape, will vary between Idc - Iac/2 and Idc + Iac/2. The AC current variation produces a change in the flux. The peak current Ipk = Idc + Iac/2 will change the flux towards saturation of the material.

Surely you don't claim that AC current in the wire does not have an effect on induction.

See the attached image for an illustration. My Iac is the delta(I) in the image.
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Quote:
 Originally Posted by ikoflexer Sorry, I don't understand what you mean. There is DC current through the primary. This DC current flow will induce flux in the core. In the same time there is an AC current component, which, if we assume a sin wave shape, will vary between Idc - Iac/2 and Idc + Iac/2. The AC current variation produces a change in the flux. The peak current Ipk = Idc + Iac/2 will change the flux towards saturation of the material. Surely you don't claim that AC current in the wire does not have an effect on induction.
The AC current variation produces nothing, but losses in the wire.

Think a little, in a transformer, if you draw a different current, the magnetic field B(AC) remains almost unchanged, even more, major core losses occur when there is no load, so U(AC) reach a maximum, so B(AC) reach a maximum.

The confusion is that a voltage U(AC) is associated with a current I(AC), which is incorrect.

Ohm's law is in the collective unconscious, but imagine a transformer with a winding ideal conductor R = 0 ==> U = 0 ???
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Quote:
 Originally Posted by 6L6 Over in the Pass Labs forum we are using a 1:1 to 1:1 quadra-filar interstage transformer as an interstage transformer/phase inverter for the F6 amp, the capacitance of the windings is not an issue. I also don't think it's avoidable, so no worries...
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Quote:
 Originally Posted by ikoflexer See the attached image for an illustration. My Iac is the delta(I) in the image.
Here we go again, lets consider Maxwell's equations

∇ . D = 4 π ρ (1)

∇ . B = 0 (2)

∇ x E + (1/c) ∂ B / ∂ t = 0 (3)

∇ x H – (1/c) ∂ D / ∂ t = (4 π/c) J (4)

And constitutive relations

D = ε E (5)

B = μ H (6)

If the medium is homogeneous and isotropic, ε and μ are constants.

In ferromagnetic materials

B = f(H)

The curve B vs H is called the magnetic hysteresis curve.

The area within the magnetic hysteresis curve is proportional to the energy dissipated as heat in the irreversible process of magnetization and demagnetization.

From eq.(4) it follows that

B(DC) = 4 π μ N i(DC) / (9 l) (7)

Therefore, equation (5-7) in the attachment is correct.

From eq.(3)

B(AC) = U(AC) x 10^8 / (√2 π f S N) (8)

Therefore, equation (5-8) in the attachment is a joke, and eq.(5-9) is clearly incorrect.

Fig.(5-1) in the attachment is also incorrect, the author confuses ∆H with ∆Iac, and/or Iac with ∆Idc.

This can be clearly seen if we consider a transformer with a given nucleus which corresponds to it a certain magnetic hysteresis curve.
If, eg. apply a voltage U(AC) in the primary, we have a magnetic field given by (8), regardless of the current i(AC).
Even more, as has been mentioned, is a consequence of Maxwell's equations, that the greater area inside the magnetic hysteresis curve occurs when there is no load, so i(AC) reach a minimum, in open contradiction with fig.(5-1)

This is also a fact observable and measurable.
Take a power transformer and do the experiment, and do not believe everything the books say.
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 IMG_1664.JPG (393.8 KB, 361 views)
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Last edited by popilin; 4th March 2013 at 12:22 AM.

 4th March 2013, 01:42 AM #38 diyAudio Moderator     Join Date: May 2003 Location: Palatiw, Pasig City so poppilin if you have the spreadsheets, can you make the corrections and post them here please.....i take it that you want to help your fellow diy'ers eh... OT. btw, are you still selling tubes? __________________ planet10 needs your help: Let's help Ruth and Dave...http://www.diyaudio.com/forums/plane...ml#post5010547[B
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Quote:
 Originally Posted by Tony so poppilin if you have the spreadsheets, can you make the corrections and post them here please.....i take it that you want to help your fellow diy'ers eh...
Hi Tony
Nice to see you around here !

Look, I do what I can, I don't know what is a spreadsheet.

I always work with my old calculator.

Quote:
 Originally Posted by Tony OT. btw, are you still selling tubes?
What are you talking about ? Here valves are scarce and expensive, I have just a couple of projects for.

Once, I was tempted to get rid of my ECC82s, do you remember ?

After the overwhelming success of my last preamp and two class A monoblocks, I don't sell nor drunk.

BTW. All is here

Designing Transformers with J.C.Maxwell
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He has the most who is most content with the least. - Diogenes of Sinope

Last edited by popilin; 4th March 2013 at 02:10 AM.

 4th March 2013, 02:35 AM #40 diyAudio Member     Join Date: Aug 2009 Location: Johnson City, TN What! No sliderules?

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