Part 1- Fundamentals
WARNING!
If you don't like math, skip this post to the next with abbreviated equations and practical advices.
I would have liked to write a little more, but the forum server does not let me…
Just a few extra details to clarify (or obscure) some concepts.
Real Transformers -The Core Nightmare
In the attachment we have almost all the math to design transformers, but we need a core to transport energy from primary to secondary.
We need a ferromagnetic material, but in ferromagnetic materials, the relation between B and H exhibits both nonlinearity and hysteresis, B is not a single valued function of H.
Houston, we have a problem…
Magnetic anisotropy is a prerequisite for hysteresis in ferromagnetic materials, so μ is not a scalar, but a tensor.
We can still using previously derived equations in the attachment, on condition to renouncing the correct description of influence of the material medium, at others; they are an astonishingly accurate approximation of reality.
How transformers work
Following analysis, for simplicity and better comprehension, supposes no copper losses, our wire is an ideal conductor.
-No load
If we put an AC voltage, Up on the primary, it produces a magnetic field B with B given by eq (23) and it induces a voltage Us on the secondary, furthermore a current im flows on the primary, this current is related with magnetic field H with H given by eq (43) and is called magnetizing current.
However is not a sinusoidal current, even when Up and B both were sinusoidal.
Both fields B and H are interrelated by the magnetic hysteresis curve B=f(H), the area inside this loop represents core losses, and like magnetizing current, remains the same on transformer operation regardless of load.
-Full load
If on the secondary flows a current is , it needs a current ip on the primary to maintain the relation Np ip = Ns is, clearly responsible for this is the field H, however both fluxes, primary and secondary, are canceled and the field B remains the same, and fortunately, when Up is sinusoidal, currents are also sinusoidal, except magnetizing current which is usually negligible.
The magic to transfer energy from primary to secondary with the same field B is provided by field H, the energy transport depends on B.H
If we consider copper losses, the major difference is that B reach a maximum at no load when Up reaches its maximum, so core losses reach its maximum too.
The idea is counter intuitive, it even verges on not making sense, but is a fact, when exists primary resistance Rp it reduces primary voltage Up proportionally to primary current, so B is reduced with load.
Note1: Name the fields B and H (or E and D) as you want, is just semantic, but be in mind that they have different properties unless they were in vacuum.
Note2: AFAIK, all equations in the attachment are correct and double checked, maybe I screwed up something to derive eq (80), (81) for leakage inductance, its value differs in excess of about 15% from accepted value eq (82)
Also I apologize for drawings, made with my old DOS program to design PCBs.
If anyone finds any error, please let me notify. Thanks.
WARNING!
If you don't like math, skip this post to the next with abbreviated equations and practical advices.
I would have liked to write a little more, but the forum server does not let me…
Just a few extra details to clarify (or obscure) some concepts.
Real Transformers -The Core Nightmare
In the attachment we have almost all the math to design transformers, but we need a core to transport energy from primary to secondary.
We need a ferromagnetic material, but in ferromagnetic materials, the relation between B and H exhibits both nonlinearity and hysteresis, B is not a single valued function of H.
B = μ H
Houston, we have a problem…
Magnetic anisotropy is a prerequisite for hysteresis in ferromagnetic materials, so μ is not a scalar, but a tensor.
We can still using previously derived equations in the attachment, on condition to renouncing the correct description of influence of the material medium, at others; they are an astonishingly accurate approximation of reality.
How transformers work
Following analysis, for simplicity and better comprehension, supposes no copper losses, our wire is an ideal conductor.
-No load
If we put an AC voltage, Up on the primary, it produces a magnetic field B with B given by eq (23) and it induces a voltage Us on the secondary, furthermore a current im flows on the primary, this current is related with magnetic field H with H given by eq (43) and is called magnetizing current.
However is not a sinusoidal current, even when Up and B both were sinusoidal.
Both fields B and H are interrelated by the magnetic hysteresis curve B=f(H), the area inside this loop represents core losses, and like magnetizing current, remains the same on transformer operation regardless of load.
-Full load
If on the secondary flows a current is , it needs a current ip on the primary to maintain the relation Np ip = Ns is, clearly responsible for this is the field H, however both fluxes, primary and secondary, are canceled and the field B remains the same, and fortunately, when Up is sinusoidal, currents are also sinusoidal, except magnetizing current which is usually negligible.
The magic to transfer energy from primary to secondary with the same field B is provided by field H, the energy transport depends on B.H
If we consider copper losses, the major difference is that B reach a maximum at no load when Up reaches its maximum, so core losses reach its maximum too.
The idea is counter intuitive, it even verges on not making sense, but is a fact, when exists primary resistance Rp it reduces primary voltage Up proportionally to primary current, so B is reduced with load.
Note1: Name the fields B and H (or E and D) as you want, is just semantic, but be in mind that they have different properties unless they were in vacuum.
Note2: AFAIK, all equations in the attachment are correct and double checked, maybe I screwed up something to derive eq (80), (81) for leakage inductance, its value differs in excess of about 15% from accepted value eq (82)
Also I apologize for drawings, made with my old DOS program to design PCBs.
If anyone finds any error, please let me notify. Thanks.
Part 2- Abbreviated equations
Core area
Actually, needed core area for a given power is pretty small, but for practical reasons there are plenty rules of thumb; for PT or PP OPT
For SE OPT
Where
P = Power [W]
fo = Lowest frequency at full power [Hz]
Bmax = Maximum B [Gauss]
Number of primary turns
With
Where
Uac = Primary RMS voltage [V]
Zp = Primary impedance [Ω]
A = Apparent core area [cm²]
ζ = Stacking factor, between 0.9 and 0.95 (thinner lamination the smaller value)
Number of secondary turns
Where
η = Efficiency factor ∼ 0.9 to 0.95
Zs = Secondary impedance [Ω]
AC magnetic field
DC magnetic field
Where
i(DC) = Primary DC current [A]
l = Magnetic circuit length [cm]
Primary Inductance
Air Gap
Where
l = Magnetic circuit length [cm]
μeff = Magnetic permeability for preceding equations work [cgs] dimensionless
μ = Magnetic permeability from lamination datasheet [cgs] dimensionless
Copper Losses
Where
ρ = 0.017 (Ω mm2) / m , copper resistivity.
lm = average turn length around bobbin [m]
N = number of turns
r = wire radius [mm]
Or
Core Losses
According to Steinmetz empirical equation
Hysteresis loss
Eddy current loss
Where
[Wh] = [We] = W/cm³
f = Frequency [Hz]
d = Lamination thickness [mm]
η , ξ = Coefficients for the magnetic material
Leakage Inductance
Where
Np = number of primary turns
lm = average turn length around bobbin [mm]
n = number of dielectrics between windings
c = thickness of dielectric between windings [mm]
a = winding height [mm]
b = winding traverse [mm]
Capacitance
-Between primary and secondary windings
Where
ε = Dielectric constant of insulator [cgs] dimensionless
A = Average winding area [cm²]
d = Insulator thickness [cm]
[C] = cm, 1cm ≈ 1.113 pF (I love cgs units)
-Between layers
-Total winding capacitance Cw
Where
n = Number of layers from +B to a layer midpoint (after/before insulation interphace)
NP = Total number of primary windings
-Total distributed capacitance
Where
n = Number of layers from +B to a layer midpoint (after/before insulation interphace)
N = Total number of layers in each primary winding
m = Number of insulation interphases between primaries and secondaries
-Shunt Capacitance
Insulation
Valve amps work with high voltages, we must very cautious with insulation, as my own rule of dumb I use
Distortion
THD for silicon steel has been calculated by Dr.N.Partridge (see RDH4, p.215)
Vh/Vf = [(Shx10⁹ l Ra)/(8 π² Np² S f)] [1 - (Ra/4 Zf)]
Where
Vh = Harmonic voltage across the primary
Vf = Fundamental voltage across the primary
Sh = Distortion coefficient of the magnetic material
l = Lenght of the magnetic path
Np = Number of primary turns
Ra = Resistance (or equivalent resistance) in series with the primary
S = Cross-sectional area of the core
f = Fundamental frequency in Hz
Zf = Primary impedance at fundamental frequency ≈ 2 π f L
L = Primary inductance at chosen field B
Core area
Actually, needed core area for a given power is pretty small, but for practical reasons there are plenty rules of thumb; for PT or PP OPT
S = 1000 √(P / fo Bmax) [cm²]
S = 1500 √(P / fo Bmax) [cm²]
P = Power [W]
fo = Lowest frequency at full power [Hz]
Bmax = Maximum B [Gauss]
Number of primary turns
Np = (Uac x 10⁸) / [√2 π fo S Bac(max)]
Uac = √(P Zp) For OPT
S = ζA
S = ζA
Uac = Primary RMS voltage [V]
Zp = Primary impedance [Ω]
A = Apparent core area [cm²]
ζ = Stacking factor, between 0.9 and 0.95 (thinner lamination the smaller value)
Number of secondary turns
Ns = (Np Us) / (η Up) For PT
Ns = Np √(η Zs / Zp) For OPT
Ns = Np √(η Zs / Zp) For OPT
η = Efficiency factor ∼ 0.9 to 0.95
Zs = Secondary impedance [Ω]
AC magnetic field
Bac(max) = (Uac x 10⁸) / (√2 π fo S Np)
Bdc(max) = [4 π μ Np i(DC)] / (9 l)
i(DC) = Primary DC current [A]
l = Magnetic circuit length [cm]
Primary Inductance
Lp = (4 π μ S Np²) / (9 l x 10⁸)
Air Gap
lG = l (μ - μeff) / (μ μeff)
l = Magnetic circuit length [cm]
μeff = Magnetic permeability for preceding equations work [cgs] dimensionless
μ = Magnetic permeability from lamination datasheet [cgs] dimensionless
Copper Losses
R = ρ N lm / (π r²)
ρ = 0.017 (Ω mm2) / m , copper resistivity.
lm = average turn length around bobbin [m]
N = number of turns
r = wire radius [mm]
Ploss [%] = 100xRp/(Zp+Rp)
Sloss [%] = 100xRs/(Zs+Rs)
Sloss [%] = 100xRs/(Zs+Rs)
InsertionLoss = 10 log {1 – [Rp/(Zp+Rp)] – [Rs/(Zs+Rs)]} dB
Core Losses
According to Steinmetz empirical equation
Hysteresis loss
Wh ≈ η f (Bmax)¹˙⁶ x 10⁻⁸
We ≈ ξ d² f² (Bmax)² x 10⁻¹¹
[Wh] = [We] = W/cm³
f = Frequency [Hz]
d = Lamination thickness [mm]
η , ξ = Coefficients for the magnetic material
Leakage Inductance
Ls = [0.417 Np² lm (2 n c + a)] / (10⁹ n² b)
Np = number of primary turns
lm = average turn length around bobbin [mm]
n = number of dielectrics between windings
c = thickness of dielectric between windings [mm]
a = winding height [mm]
b = winding traverse [mm]
Capacitance
-Between primary and secondary windings
C ≈ (ε A) / (4πd)
ε = Dielectric constant of insulator [cgs] dimensionless
A = Average winding area [cm²]
d = Insulator thickness [cm]
[C] = cm, 1cm ≈ 1.113 pF (I love cgs units)
-Between layers
Cl ≈ (ε A) / (4πd)
(Cw)⁻¹ ≈ (Cl)⁻¹ ∑ {(4/3nı) [1 - (1/nı)]}⁻¹ ı=1,...,NP
n = Number of layers from +B to a layer midpoint (after/before insulation interphace)
NP = Total number of primary windings
-Total distributed capacitance
Cd ≈ C ∑ (nı/N)² ı=1,...,m
n = Number of layers from +B to a layer midpoint (after/before insulation interphace)
N = Total number of layers in each primary winding
m = Number of insulation interphases between primaries and secondaries
-Shunt Capacitance
Cs = Cw + Cd
Insulation
Valve amps work with high voltages, we must very cautious with insulation, as my own rule of dumb I use
Vi = 3 (Up₋p + Udc)
Distortion
THD for silicon steel has been calculated by Dr.N.Partridge (see RDH4, p.215)
Vh/Vf = [(Shx10⁹ l Ra)/(8 π² Np² S f)] [1 - (Ra/4 Zf)]
Where
Vh = Harmonic voltage across the primary
Vf = Fundamental voltage across the primary
Sh = Distortion coefficient of the magnetic material
l = Lenght of the magnetic path
Np = Number of primary turns
Ra = Resistance (or equivalent resistance) in series with the primary
S = Cross-sectional area of the core
f = Fundamental frequency in Hz
Zf = Primary impedance at fundamental frequency ≈ 2 π f L
L = Primary inductance at chosen field B
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Transformers can be highly complex beasts.
I usually just buy off the shelf parts depending on the job I want it for.
I leave an experts job to the experts.
They also have the right kit for building the transformers quickly and accurately.
I have built a few for SMPS and learned a lot from that.
For parallel windings it is important to get them exactly the same or high currents can flow and cause extreme heating.
I usually just buy off the shelf parts depending on the job I want it for.
I leave an experts job to the experts.
They also have the right kit for building the transformers quickly and accurately.
I have built a few for SMPS and learned a lot from that.
For parallel windings it is important to get them exactly the same or high currents can flow and cause extreme heating.
To improve accuracy I will collaborate with the following:
Stacking factor (depends on type of assembly as well as lamination thickness):
1:1 0.88
4:4 0.92
Butt joint 0.95
(Values for typical 0.35mm lamination, adjust accordingly)
Equivalent air gap of assembly types (cm):
1:1 0.012
4:4 0.025
8:8 0.05
16:16 0.1
Butt joint 0.12
(Values for typical 0.35mm lamination, adjust accordingly)
Source: Electronic designers handbook Mc-Graw Hill 1957
Stacking factor (depends on type of assembly as well as lamination thickness):
1:1 0.88
4:4 0.92
Butt joint 0.95
(Values for typical 0.35mm lamination, adjust accordingly)
Equivalent air gap of assembly types (cm):
1:1 0.012
4:4 0.025
8:8 0.05
16:16 0.1
Butt joint 0.12
(Values for typical 0.35mm lamination, adjust accordingly)
Source: Electronic designers handbook Mc-Graw Hill 1957
I can vouch for that!
No need to be an expert to build excellent transformers, just patience and perseverance.
When I need various secondaries paralleled, with an auxiliary core and an auxiliary winding, I measure carefully each secondary winding to mV tolerance, I do not want smoke.
Gracias por tu colaboración !
Thanks for your reply !
And for the air gap, the idea is to adjust magnetic permeability to desired primary inductance in order to not exceed Bdc(max) in SE OPTs.
I applaud the above dissertation, but for practical purposes designers would rather use tables and design charts taking care of much of the spadework, not to mention committed cad programmes available these days. I still use the Crowhurst charts published in RDH-4 (as mentioned) I know of at least one cad program derived from that and including properties of steels, which is even more convenient.
Respectfully, I would disagree with that, Poplin. There are practical factors like winding curvature, end-of-layer, core assembly etc. which cannot just be rushed into. Initial experience is necessary. Thus, yes; one can build an excellent transformer - but don't expect it to come out right after the first try. Unless you are singularly fortunate that it did ... in which case you would not have learnt much.
I do not fully understand that; do you in the middle of winding move the lot to a different core and first measure each winding?? In my experience it is quite sufficient to begin and end similar windings on the same imaginary cross-line. That is easily accomplished to within 1 mm; most commercial transformers are made that way.
No need to be an expert to build excellent transformers, just patience and perseverance.
Respectfully, I would disagree with that, Poplin. There are practical factors like winding curvature, end-of-layer, core assembly etc. which cannot just be rushed into. Initial experience is necessary. Thus, yes; one can build an excellent transformer - but don't expect it to come out right after the first try. Unless you are singularly fortunate that it did ... in which case you would not have learnt much.
When I need various secondaries paralleled, with an auxiliary core and an auxiliary winding, I measure carefully each secondary winding to mV tolerance, I do not want smoke.
I do not fully understand that; do you in the middle of winding move the lot to a different core and first measure each winding?? In my experience it is quite sufficient to begin and end similar windings on the same imaginary cross-line. That is easily accomplished to within 1 mm; most commercial transformers are made that way.
Yes, agreed, but everyone has their own method, there are lots over there, I like mine.
Not a dissertation, but my humble collaboration to share at least the equations, the correct ones, I hate to read 4.44, instead of √2 π
Maybe I'm a rara avis, my first set of transformers worked wonderfully from the first time, maybe I had luck, but with my winding method there is only one chance, a silver bullet.
That I meant is about counting turns, with a cheap calculator, a reed switch and a magnet, everytime that finish a winding I must check if all was OK.
Turner is a good source for practical transformer design and it is good in that respect. This thread (as I see it) was to put together every formula someone could need to build a transformer, which I personally (as other people) find useful. If you think it´s useless don´t comment and let the thread slide by.
With respect to the thread if someone could add the RDH4 leakage inductance and capacitance charts by crowhurst it would add a lot.
Bien Marcelo, un poco de buena onda al thread!
All who make transformers had visited sometime Patrick Turner's page, very useful and detailed.
That's precisely the idea, unfortunately words are not my thing and I must write equations, with bigger characters because of super indexes, my eyes are not as before...
BTW, RDH4 here on Pete Millett's page
Technical books online
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Tip#1 The winding lathe
Almost anything that spins, a drill with variable speed, or the mom's old sewing machine.
You have to put a screw and tightened with nuts and washers two pieces of wood to hold the bobbin.
To count the turns of wire, place a small magnet on the part that rotates, on the fix part, a reed switch connected with the "=" key on a cheap calculator, type "1" and "+" and each time the magnet pass over the reed switch it be counted a turn.
Almost anything that spins, a drill with variable speed, or the mom's old sewing machine.
You have to put a screw and tightened with nuts and washers two pieces of wood to hold the bobbin.
To count the turns of wire, place a small magnet on the part that rotates, on the fix part, a reed switch connected with the "=" key on a cheap calculator, type "1" and "+" and each time the magnet pass over the reed switch it be counted a turn.
Hi Martin
Good to see you around here.
Of course, if you already have a winding machine, things are a lot easier.
Thank you very much !
Good to see you around here.
I prefer my Bobifil winder
Of course, if you already have a winding machine, things are a lot easier.
Thank you very much !
Font size of first two posts reduced due to complaints about the size.
Tip#2 How to start a winding
For me it is the hardest part, we start by straightening the wire making it a little run through your fingers without exerting too much pressure.
Then we wrap the wire with tape type "Scotch" as shown in fig 1, and then put the wire over the tape so it is snug, and we can continue winding.
Every few more turns, with a soft plastic spatula, we press the wire so that they are as together as possible.
For me it is the hardest part, we start by straightening the wire making it a little run through your fingers without exerting too much pressure.
Then we wrap the wire with tape type "Scotch" as shown in fig 1, and then put the wire over the tape so it is snug, and we can continue winding.
Every few more turns, with a soft plastic spatula, we press the wire so that they are as together as possible.
Tip#3 Interlayer insulation
Modern magnet wire has a breakdown voltage of more than 4 KV, so no need a great insulator, unless you make a very high voltage transformer.
However there are two more reasons for use insulation between layers
1- Decrease mechanical stress between layers
2- Decrease interlayer capacitance
One or two layers of 0.05mm polyester would be enough.
Modern magnet wire has a breakdown voltage of more than 4 KV, so no need a great insulator, unless you make a very high voltage transformer.
However there are two more reasons for use insulation between layers
1- Decrease mechanical stress between layers
2- Decrease interlayer capacitance
One or two layers of 0.05mm polyester would be enough.
Small correction, I am not a professional. I do this for a hobby and i have just a normal job at daylight
Hey, that's not fair, you are a professional, and your transformers are amazing!
Here we have a saying "It is counting money in front of the poor"
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