After doing more research I found this paper:
http://www.intalek.com/Index/Projects/Research/Permeability.PDF
Figure 1.17a on page 3 shows permeability peaking about at the middle of the BH curve slope.
I measured the current and voltage through the secondary with an adjustable resistor in series with the transformer. I adjusted the resistor to have a voltage drop equal to the voltage drop across the transformer, and used this to calculate the inductance.
The peak inductance should coincide with the peak permeability. No?
If so, then the peak permeability occurs with an input voltage of 11.3V with a current of 78mA for an inductance of 0.477H.
From this and the number of turns I can calculate Mu.
Is this the proper point at which to calculate Mu? Or does this only tell me the peak flux density at saturation?
http://www.intalek.com/Index/Projects/Research/Permeability.PDF
Figure 1.17a on page 3 shows permeability peaking about at the middle of the BH curve slope.
I measured the current and voltage through the secondary with an adjustable resistor in series with the transformer. I adjusted the resistor to have a voltage drop equal to the voltage drop across the transformer, and used this to calculate the inductance.
The peak inductance should coincide with the peak permeability. No?
If so, then the peak permeability occurs with an input voltage of 11.3V with a current of 78mA for an inductance of 0.477H.
From this and the number of turns I can calculate Mu.
Is this the proper point at which to calculate Mu? Or does this only tell me the peak flux density at saturation?
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
Assuming Ac = 5.68 cm^2, Idc = 10mA, Iac = 4mA, L = 44H, and Bm = 1.2T, gives
N = 774 turns.
Next would be to evaluate the max wire size that would fit so many turns in your winding area, or window area Wa.
The wire are in cm^2 is given by
Aw <= Ku Wa / N
Ku = window utilization factor (constant), which for laminates like you have is 0.48.
Hence my question for your core Wa.
Next, calculate the gap length, in cm:
lg = (0.4*pi*N*N*Ac*0.00000001)/L
In your case lg = 0.001198 cm, which is equivalent to just butting the ends together with not paper between them, roughly. One sheet of fish paper is about 7 mils, which is about 0.01778 cm.
We should probably take into account the fringing flux
Frf = 1+(lg/SQRT(Ac))*LN(2*D/lg)
where D is the window height as seen in the image I attached previously.
Knowing F you can recalculate the number of turns:
N' = =SQRT((lg*L)/(0.4*pi*Ac*Frf*0.00000001))
There's more, I'm working on a spreadsheet too.
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
Assuming Ac = 5.68 cm^2, Idc = 10mA, Iac = 4mA, L = 44H, and Bm = 1.2T, gives
N = 774 turns.
Next would be to evaluate the max wire size that would fit so many turns in your winding area, or window area Wa.
The wire are in cm^2 is given by
Aw <= Ku Wa / N
Ku = window utilization factor (constant), which for laminates like you have is 0.48.
Hence my question for your core Wa.
Next, calculate the gap length, in cm:
lg = (0.4*pi*N*N*Ac*0.00000001)/L
In your case lg = 0.001198 cm, which is equivalent to just butting the ends together with not paper between them, roughly. One sheet of fish paper is about 7 mils, which is about 0.01778 cm.
We should probably take into account the fringing flux
Frf = 1+(lg/SQRT(Ac))*LN(2*D/lg)
where D is the window height as seen in the image I attached previously.
Knowing F you can recalculate the number of turns:
N' = =SQRT((lg*L)/(0.4*pi*Ac*Frf*0.00000001))
There's more, I'm working on a spreadsheet too.
Hammond 126C and 126B , to me sound superb for a £30 a piece lump of iron . These bifilar 1:1 IT are a bargain , probably the biggest bang for buck out there
316A
thanks for your input, did you try to measure frequency response? i thought that winding biffilar increases winding capacitance to affect hf?
and here are some typical FR for some bifilar ITs from back in the late 90s with -3dB beyond 120KHz
http://www.sjselectroacoustics.co.uk/images/IT102.GIF
and
http://www.sjselectroacoustics.co.uk/images/IT103.GIF
http://www.sjselectroacoustics.co.uk/images/IT102.GIF
and
http://www.sjselectroacoustics.co.uk/images/IT103.GIF
I've been interested in winding transformers for many years but never seemed to have the time to study it sufficiently to understand what I'm doing.
ikoflexer, window measurements are 1.1cm X 3.33cm for an area of 3.663 sq-cm
The calculations for the 1:1 interstage transformer should be very much like those for an inductor.
If Bi-Filar wound, construction is the same, except two wires are feed at the same time.
Eric, this is an ope forum so every one is welcome to participate or just follow along as they wish.
Tony, bi-filar does increase coupling capacitance, apparently this is not always bad.
sjs, thanks for the links.
I still wish to measure the characteristics of these cores so I can use those values in my calculations. So if anyone can give me some guidance I would appreciate it.
ikoflexer, window measurements are 1.1cm X 3.33cm for an area of 3.663 sq-cm
The calculations for the 1:1 interstage transformer should be very much like those for an inductor.
If Bi-Filar wound, construction is the same, except two wires are feed at the same time.
Eric, this is an ope forum so every one is welcome to participate or just follow along as they wish.
Tony, bi-filar does increase coupling capacitance, apparently this is not always bad.
sjs, thanks for the links.
I still wish to measure the characteristics of these cores so I can use those values in my calculations. So if anyone can give me some guidance I would appreciate it.
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...
This is what I get for your core. This calculates one winding, assuming it occupies half the total winding area. It also assumes standard industry silicon steel core and a target working Bmax of 1.2 tesla. The method used is called the area product method (temperature rise) from the book Transformer and inductor design handbook by Col. Wm. T. McLyman.
Attachments
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.
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.
Attachments
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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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.
Attachments
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 ???
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.
Attachments
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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.
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
http://www.diyaudio.com/forums/tubes-valves/181072-designing-transformers-j-c-maxwell.html
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