======================TRANSFORMERS======================
__ Primary __
 __ __ 
__   __
*() () 
\_/\_/\_/ Np
M _ _ _
/*\/ \/ \ NS
__  () ()  __
 __ __ 
__ __
Secondary

Lp_Leak Lsec_Leak
_ _ _ _ _ _ RL*(Np/Ns)^2
__ Rp / \/ \/ \ / \/ \/ \ Rs __
 _ /\_ () () ____ () () _ /\___ _
__ \/    _ _ _ \/  __ 
__ Rc /  / \/ \/ \ __ /
Cp ___ core \ _ () ()  Cs ___ \
__  loss / K*Lprime   __ /
 ________________________________ _
__ __
L
Ip*Np Is*Ns
Vp/Np Vs/Ns
Rp winding resistance primary
Lp winding reactance primary
Rs winding resistance secondary
Ls winding reactance secondary
Rc core heating
LK*Lp magnetic current

FERRITE CHARACTERISTICS.
Saturation
Moment Saturation
Saturation Curie in Bohr FirstOrder Magneto
Moment Temp Magnetons Xray Lattice Anisotropy striction
Ferrite in Gauss in ~C n_B Density Constant Constant K1 Lambda_eX106
NiFe2O4 3400 585 2.3 5.38 8.34 Ñ0.06 Ñ22
Ni0.8Zn0.2Fe2O4 4600 460 3.5    Ñ Ñ18.5
Ni0.5Zn0.5Fe2O4 5800 360 4.8 Ñ Ñ Ñ15.0
Ni0.5Zn0.5Fe2O4 5500 290 5.0 Ñ  Ñ Ñ8.3
Ni0.3Zn0.5Fe2O4 2600 85 4.0 Ñ Ñ0.004 Ñ1.0
MnFe2O4 5200 300 0 5.00 8.50 Ñ0.04 Ñ14
Mn0.5Zn0,
Fe204 100 6.0 Ñ Ñ Ñ0.004 Ñ
FeFe2O4 6000 585 4.1 5.24 8.39 Ñ0.135 +41
CoFe2O4 5000 520 3.8 5.20 8.38 Ñ2000 Ñ250
CuFeoO4 1700 455 1.3 5 35 8.24 Ñ Ñ
8.68
Li0.5Fe2.5O4 3900 670 2.6 4 75 8.33 Ñ 
MgFe204 1400 440 1.1 4.52 8.36 Ñ 0.05 Ñ
MgA1FeO4 Ñ Ñ 0.3 Ñ Ñ
NiAl0.25Fe1.75O4 1300 506 1.30 Ñ 8.31 Ñ Ñ
NiAl_0.45Fe_1.55O4 900 465 0.61 Ñ 8.28 Ñ Ñ
NiAl_0.62Fe_1.38O4 0 360 0 Ñ 8.25 Ñ Ñ
NiAlFeO4 900 198 0.64 5.00 8.20 Ñ Ñ

hysteresis of core Ph = Kh*f*Bm^x*V hysteresis loss
Kh is a constant which depends on the chemical
analysis of the
material and the heat treatment and mechanical
treatment to which
it has been subjected
f is frequency in Hz
Bm is maximum flux density in webers / m^2 (Teslas)
V is the volume of the material in cubic meters
Pe = Ke*f^2*c^2*Bm^2*V eddy current loss
Ke is a constant which depends on the resistivity of
the material
f is frequency in Hz
c is lamination thickness in meters
Bm is maximum flus density in webers / m^2 (Teslas)
V is volume of the material in cubic inches.

eddy current in laminations and another comes fromhysteresis of core.
Ph = Kh*f*Bm^x*V hysteresis loss
Kh is a constant which depends on the chemical
analysis of material and heat treatment and mechanical
treatment to which it has been subjected
f is frequency in Hz
Bm is maximum flux density in webers / m^2 (Teslas)
V is the volume of the material in cubic meters
Pe = Ke*f^2*c^2*Bm^2*V eddy current loss
Ke is a constant which depends on the resistivity of
the material
f is frequency in Hz
c is lamination thickness in meters
Bm is maximum flus density in webers / m^2 (Teslas)
V is volume of the material in cubic inches.
Let's first guess at the Bm....
Assume inductance of core is constant as of
frequency (poor guess for iron but a start)
Bm will inversely proportional to frequency
and proportional to voltage.
So we can change the above equations to
Ph = Kh'*f hysteresis loss
Pe = Ke'*f^2 eddy current loss
since everything else is constant.
Pc = Ph*135/60 + Pe*(135/60)^2 total core loss
If *ALL* the power were in the hysteresis, you'd get a
loss of 3.6 W,
Now let's *GUESS* that half power went into hysteresis
and half intoeddy current losses, then you would have....
Pc = .8*135/60 + .8*(135/60)^2 total core loss
= 5.9 W !!!

* Trace Elliot 15W output transformer, 8k primary, 16/8 ohm sec.
* Part no. 73 TRAN 15W OP
*
* [1] Red . 
* )  . Green (16 ohm) [4]
* )  (
* )  (
* [2] White .  . Yellow (8 ohm) [5]
* )  (
* )  (
* )  . Brown [6]
* [3] Blue . 
*
.SUBCKT trace15 1 2 3 4 5 6
Lleak1 1 20 2mH
Lpri1 20 21 150H
Rpri1 21 2 186
Cpri1 1 2 120p
Lleak2 2 22 2mH
Lpri2 22 23 150H
Rpri2 23 3 201
Cpri2 2 3 120p
* Secondary
Lleak3 5 24 10uH
Lsec1 24 25 0.6H
Rsec1 25 6 0.8
Lleak4 4 27 10uH
Lsec2 27 28 0.3H
Rsec2 28 5 0.4
Kcoup Lpri1 LPri2 Lsec1 Lsec2 1.0
.ENDS

K
'`
MUTUAL INDUCTANCE V1 ___ o o____ V2
\ /
L1 _   _ L2
/ \/ \/ \
\_/\ /\_/
_   _
/ \/ \/ \
\_/\ /\_/ V2=V1*N2/N1
_   _
/ \/ \/ \ I1=I2*N2/N1
\_/\ /\_/
 
____/ \____
V1= L1*di1/dt + M*di2/dt
V2= L2*di2/dt + M*di1/dt
M= K*sqrt( L1+L2)

_ _ _ _ _ _
MODEL WITH / \/ \/ \ / \/ \/ \ LE 1:N
PERFECT TRANSFORMER _ () () _______ () () ______ o o_______ IF K > 1
 \ /
V1 LE ________ _   _ V2 L1 = LM
 / \/ \/ \
_ _ _  \_/\ /\_/ L2 = N^2*L1
/ \/ \/ \  _   _
__ () () _ / \/ \/ \ 1 K = LE/LM
 \_/\ /\_/
 LM _   _
 / \/ \/ \
 \_/\ /\_/
  
_______________________________/ \________

MODEL CORE LOSS _ _ _ _ _ _
___ R1 / \/ \/ \ L1 / \/ \/ \ L2 R2 ___
 _/\ /\ /\__ () () _______ () () __/\ /\ /\__ 
___ \/ \/  LE \/ \/ ___
_ _ _ 
/ \/ \/ \  R3
__ () () ___/\ /\ /\__
___  LM \/ \/  ___
 __________________________________________________ 
___ ___

_ _ _ _ _ _ 1
MODEL WITH ___ / \/ \/ \ L1 / \/ \/ \ L2 1:N ___
PERFECT TRANSFORMER  _/\ /\ /\__ () () _______ () () __/\ /\ /\_______ o o____ 
___ \/ \/  \/ \/ R2 \ / ___
R1 _ _ _  LE _   _
/ \/ \/ \  R3 / \/ \/ \
__ () () ____/\ /\ /\__ \_/\ /\_/
 LM \/ \/  _   _
  / \/ \/ \
  \_/\ /\_/
___     ___
 ________________________________________________________/ \____ 
___ ___

MAGNETOSTRICTION static strain delta_l/l produced by a directcurrent polarizing flux density B_0 is given by
delta_l/l = c*B_0^2
c being a material constant expressed in m^4/weber^2.
Curie temperature temperature at which a material loses all of its magnetic properties.
Conventional stripwound cores and
powder cores generally have such high Curie temperatures (>450 degrees C)
Manganesezinc ferrites, Curie temperatures (120 to 250 degrees C) of ferrites.
bifilar winding? Two strands of wire, usually twisted together.
The dual wire is then wound on the core
or bobbin to produce two equal and parallel
windings which take the place of one large single
strand.
relative costs of tdifferent magnetic materials?
For powder cores, iron powder ranges from 1x  3x
KOOL Mµ  " " 4x  5x
Hig Flux approx. 10x
MPP approx. 12x
For ferrites,F,P,R,J materials,roughlequivalent (1y)
W material1.251.75y
H mate1.502.00y
Ferrite cost is also a function of geometry:
Toroids least
E coresmid
Other shapesmost
BH loop? (or Hysteresis)
defines flux density of material, coercive force,
amount of drive level required to saturate the core,
and permeability
(ability to change magnetic lines of force).
BH loop changes with frequency and drive level.
Why air gap into cores "tilts", or "shears" the BH loop,
making it possible to use core at higher H levels,
thus preventing early saturation of core.
Once a core saturates, permeability reduced,
magnetostriction? When magnetic material is magnetized,
small change in dimension occurs.
in order of several parts per million,
called "magnetostriction". For
applications like ultrasonic generators,
mechanical motion produced by magnetic excitation
through magnetostriction is used to good advantage.
operating in audiblefrequency range,
an annoying audible hum is observed.
For this reason, low magnetostrictive
materials such as Permalloy 80, METGLAS 2714A, KOOL
Mµ and MPP powder cores may be used
disaccommodation occurring in ferrites,
reduction of permeability with time
after a core is demagnetized.
demagnetization can be caused by heating
above Curie point, by applying AC of diminishing
amplitude, or by mechanically shocking the core.
In this phenomenon, permeability increases
towards its original value, then starts to decrease
exponentially. If no extreme conditions are
expected in the application, permeability changes
will be small because most of the change has
occurred during the first few months after
manufacture of the core. High temperature
accelerates the decrease in permeability.
Disaccommodation is repeatable with each successive
demagnetization; thus, it is not the same as aging.
inductance ferrite toroid decrease after winding and potting?
Ferrite materials are susceptible to mechanical stress
both from winding the core and from encapsulation.
High permeability materials are affected. remedies:
(1)after winding, bake and/or temperaturecycle,
(2)thin out the epoxy used for encapsulation or dope
with an inert material such as sand or ground mica,
(3) cushion with tape,(4) silicone (RTV) dip
wound cores prior to potting.
actual core losses in gappedstructures larger than calculated?
When calculating the core losses, it is
assumed that structure is homogeneous. In
reality, when core halves are mated, there is
leakage flux (fringing flux) at mating
surfaces, and gap losses contribute to
total losses.
difference ferrites between nickelzinc and manganesezinc ferrites?
MnZn materials have a high permeability,
NiZn ferrites have low permeability.
Manganesezinc ferrites used less than 5MHz.
Nickelzinc ferrites used at frequencies
from 2MHz to several hundred megahertz.
The exception to this rule of
thumb is common mode inductors where the impedance
of MnZn materials makes it the best choice up to
70MHz and NiZn is recommended from 70MHz to
several hundred GHz.
permeability important in powermaterials?
Permeability is flux density, B,
divided by drive level, H.
Power materials are generally used
for high frequency transformer applications.
Generally, the important characteristics are high
flux density and low core losses. Permeability is
of less importance because of its variability over
an operating flux range.
Why only min AL listed in catalog?
Permeability (and AL) varies with drive level.
For power applications, no need to limit max AL.
minimum ALtranslates into maximum excitation current.
How know ferrite hardware willfit on the core?
Cores are manufactured to standards t
Tolerancesassigned to critical dimensions.
Generally, hardware fit should not be a problem.
tighter dime tolerances in ferrites?
ANSWER: During the sintering operation, parts
shrink to their final dimensions. Different
material and processing techniques result in
variance in this linear shrinkage which can range
from 10 to 200f the pressed dimensions (in
finished parts, this could range from 14%). Some
dimensions cannot be held to a tighter tolerance.
Dimensions that can be machined after firing can
certainly be held to tighter tolerances.
QUESTION: Can I get a custom ferrite part?
ANSWER: It is possible to get a custom part.
Volumes of less than 500 pieces can be readily
machined. Quantities over 20,000 are generally
pressed from custombuilt tools. Adjusting the
heights of existing parts is a practical way to
minimize machining and tooling costs.
QUESTION: What is the proper clamping pressure in
ferrites?
ANSWER: Generally, a recommended figure is about
70,000 kg/m² (100 lbs./sq. in.) of mating surface.
For specific recommended pressures for RM, PQ, EP
and pot cores, consult MAGNETICS' Ferrite Catalog,
FC601.
QUESTION: What is the best core shape?
ANSWER: There is no "best shape." It depends on
the application, space constraints, temperature
limitations, winding capabilities, assembly, and a
number of other factors; this means that
compromises must be made. For additional
information on this subject, consult the MAGNETICS
Ferrite Cores Catalog where geometry considerations
are covered in more detail. Brochure PS01 also
covers this subject.
QUESTION: Why do manufacturers flatgrind ferrite
cores?
ANSWER: Cores are flatground because of the uneven
surface produced during the firing process. It is
important for mating surfaces to mate with a
minimum air gap to keep the gap losses to a minimum
and to achieve an optimum inductance.
QUESTION: Why do cores get lapped? What is the
surface finish?
ANSWER: Lapping is an additional production process
used to decrease the effects of an air gap on mated
surfaces. It is typically done on mated cores with
material permeabilities of 5000 and greater in
order to achieve the maximum AL value for a given
material. A mirrorlike finish is the result. The
typical surface finish for normally flatground
surfaces is 25 microinches (.635 microns) and for
lapped surfaces is 5 microinches (.127 microns).
Proper surface finish is not measured as a rule,
but is maintained by monitoring the AL.
QUESTION: Why is the ferrite gapped tolerance not
always ±3%?
ANSWER: Due to limitations of the machine
performing the gapping, the smaller the gap, the
harder it is to hold tight tolerances. As the AL
value increases, the gap gets smaller, hence the
tolerance gets larger. As the gap gets smaller,
the mechanical tolerance becomes proportionately
larger, plus the influence of variation in the
material permeability becomes greater. Thus, a gap
specified by its AL value yields a tighter
tolerance than a gap specified by its physical
dimensions.
QUESTION: How do you glue ferrite cores?
ANSWER: Gluing should be done with thermosetting
resin adhesives, in particular the epoxy resins.
The available range is very large. Important
factors in the choice are the required temperature
and viscosity. The curing temperature must not be
above the maximum temperature to which the assembly
may be safely raised. As far as viscosity is
concerned, if it is too high, application is
difficult; if it is too low, the resin may run out
of a poorlyfitted joint or may be absorbed by the
porosity of the ferrite. Follow the manufacturer's
instructions for a particular resin. Take care not
to thermally shock ferrites; raising or lowering
the core temperature too rapidly is dangerous.
Ferrites will crack if changes in temperature
exceed 510 degrees C/min. In addition, care must
be taken to match the adhesives' coefficient of
thermal expansion (CTE) to that of the ferrite
material. Otherwise, the resin may expand or
contract more quickly than the bulk ferrite; the
result can be cracks that will degrade the core.
QUESTION: Why are ferrite toroid AL tolerances wide
and powder cores narrow?
ANSWER: Magnetic materials naturally have wide
variations in permeability. Putting an air gap in
the structure can have the effect of not only
reducing permeability but also dramatically
reducing this variation. Powder cores have a
distributed air gap; this results in a narrower
inductance tolerance. Ferrite toroids do not have a
distributed air gap and are thus subject to
variations caused by normal processing. For a
complete description of the ferrite manufacturing
process, consult the MMPA Soft Ferrite User's
Guide, Publication number MMPA SFG. Magnetic
Materials Producer's Association, 600 South Federal
Street, Suite 400, Chicago, IL 60605.
QUESTION: Can you tighten electrical tolerances on
ferrite toroids?
ANSWER: (see also question 33). While a production
batch of ferrite toroids may have a wide tolerance,
when required in rare instances, the cores can be
graded into narrower inductance bands at a premium.
Due to equipment limitations, this is not possible
on all sizes. Check the factory for specific
information and costs.
QUESTION: What is the MAGNETICS specification for
out of roundness for a ferrite toroid ?
ANSWER: Out of roundness is controlled by
mandating that cores meet overall dimensional
tolerances for OD and ID while keeping enough cross
section to meet the specified AL. Refer to the
MAGNETICS Ferrite Catalog FC601 for toroid
physical dimension tolerances.
QUESTION: What is the difference between nylon and
polyester coatings for ferrite toroids?
ANSWER: They are similar. Nylon is thicker, and can
stand temperatures up to 155 degrees C. Polyester
is good to about 200 degrees C. Nylon finishes are
generally applied to cores ranging in OD from 9 mm
to 29 mm. Very large and very small cores are
coated with a polyester finish. Voltage breakdown
guarantee of nylon and polyester coatings is 500
volts. Nylon cushions better and is more resistant
to solvents. Both finishes are held to the same
electrical and mechanical specifications.
QUESTION: What about availability of any other core
coatings for ferrites?
ANSWER: Black lacquer is an inexpensive coating put
on merely for the purpose of providing a smooth
winding surface. It does not have any voltage
breakdown guarantee. Size range is 7.6 mm. to 15.8
mm. in outside diameter. Parylene C is a vacuum
deposited coating providing good resistance to
moisture and organic solvents. Electrical
characteristics are superior to other coatings. The
size range is economically limited to outside
diameters of 14mm.or less.
QUESTION: How do you determine the proper core
size?
ANSWER: Two elements are useful in determining core
size: core window (winding area) and core
crosssectional area. The product of these two
elements (area product, or WaAc) relates to the
power handling capability of a core. The larger the
WaAc, the higher the power able to be handled. As
operating frequency increases, the area product can
be reduced, thus reducing the core size. MAGNETICS
publishes the area products of all cores as a
useful design tool.
QUESTION: Can MAGNETICS press powder cores and
ferrite toroids in different heights?
ANSWER: Many cores can be pressed to different
heights. Dies are made so that the cavities can
accommodate these different heights. Each core size
is different, however. Consult the factory for
specific questions on the sizes of interest,
minimum quantities and price. One advantage this
offers is the ability to produce other core sizes
without the expense of additional tooling.
QUESTION: In powder cores, why is actual inductance
different from calculated?
ANSWER: MAGNETICS measures inductance in a Kelsall
Permeameter Cup. Actual wound inductance outside a
Kelsall Cup is greater than the value calculated
due to leakage flux and flux developed in the
winding. The difference depends on core size,
permeability, core finish thickness, wire size and
number of turns, in addition to the way windings
are put on the core. The difference is negligible
for turns greater than 500 and permeabilities 125µ
and higher. The following table is a guide to the
differences that one might experience:
No. of Turns Actual L No. of Turns
Actual L
1000 0%
100 +3.0%
500 +0.5%
50 +5.0%
300 +1.0%
25 +8.5%
The following formula can be used to approximate
the leakage flux to add to the expected inductance.
This formula was developed from historical data of
cores tested at MAGNETICS. Be aware that this will
only give an approximation based on evenly spaced
windings. You might expect as much as ±50%
deviation from this result.
where LLK= leakage inductance (mH)
N= number of turns
Ae= core crosssection (cm2)
le= core magnetic path (cm)
CONVERSION FACTORS:
MULTIPLY BY TO
OBTAIN
Oersteds
2.0213 ampereturns/inch
Oersteds
0.79577 ampereturns/cm
Oersteds
79.577 ampereturns/m
Ampereturns/cm
1.2566 oersteds
Gausses
104 teslas
Microinches
0.0254 microns
ePanorama.net  Ground Loops

Coils
Coil equations Equation for calculating inductance of a homemade coil:
L=(D*N^2)/(l/D+0,43)
Where
* D is diameter in cm
* l i length in cm
* L is inductance in uH
* N is nuber of turns
Air core coils L = (r^2 * n^2)/(9r + 10l)
For air core coils you can come close with:
* L = ind. in uH
* r = radius of coil in inches
* n = number of turns
* l = coil length in inches
ref: Bauchbaum's Complete Handbook of Practical Electronic
Air cores typically range from .1 to 2000 uH. Bigger
indictances usually
make the coil too bulky and the above formula is not
accurate enough because
inner and outer radii of your windings may vary too
widely.
Another coil equation L = (a^2*n^2)/(9*a+10*b)
where
* a = radius in inches
* b = length in inches
* n = number of turns
this was claimed to be accurate within a few percent.
Esingle layer coils L= N^2*A*u*u0/l
Where:
* L=inductance desired
* N=Number of turns
* A=cross sectional area of core in square centimeters
* u(Mu)=permeability of core (Air=1; Iron~1000)
* u0 (mu subzero)= Absolute permeability of air
(1.26*10^12)
* l=length of coil in centimeters
formulas turn numbers of air and iron core
coils
Air core: N=(SR[A*u*u0/l])*20Pi
Iron core: N=(SR[A*u*u0/1*.5Pi])
Where:
* N=number of turns needed for coil
* SR=Square Root (of bracketed equation)
* A=cross sectional area of core
* u=Permeability of core
* u0=Absolute permeability of air
* l=length of coil
* 20Pi=20 times Pi or about 63
* .5Pi=half of Pi or 1.57
The equations "sort of" govern the inductance. Testing
with inductance
meters will ascertain desired inductance.
Program for calculating coil inductance
Here is a simple basic program for calculating coils in GW

^
/_\ B FIELD
AIR GAP 
 # # # # *# *
 # *
 *
 *
 # *
 *
 *
 # *
 *
 * # NORMAL
 # * # AIR GAP
 *
 *
 *
 *
# *
 *
 *
 * H FIELD
#_____________________________________\
/

^
FREQUENCY AND CORE /_\ B FIELD B_SAT

* #######################
* # # * *
* # # *
* #  # *
* #  # *
* #  # * H FIELD
______________#________#________________\
* #  # * /
* #  # *
* #  # *
* # # * 6000HZ
* # # *
* # # *
###################### *



^
/_\ B FIELD

 # # # # *# *
 # *
 *
 *
 # *
 *
 *
 # *
 *
 * # NORMAL
 # * # AIR GAP
 *
 *
 *
 *
# *
 *
 *
 * H FIELD
#_____________________________________\
/
airgap is useful It can be. The shape of the airgap can be important as well.
Some devices are built with a 'stepped airgap'. This causes the magnetic field in the vicinity of the
gap to concentrate near the narrower portion of the gap (which has a lower reluctance).
This portion of the device will saturate before that near the longer gap. The
device will exhibit an initial higher inductance until a portion of the material saturates,
when the device will exhibit a 'step' to a lower value. Thus for low values of
DC excitation the device will exhibit an inductance higher than for larger values of DC,
creating what is called a 'swinging choke', useful in some filter applications.
One can use a 'wedge' shaped air gap to create a more smoothly varying inductance with DC excitation.
But of course these changes will also occur with instantaneous excitation as well.
Why is there a gap or 'hole' in the middle of a typical BH curve?
It represents a loss component, however. Most magnetic materials will retain some
magnetization after being 'magnetized' and the excitation is removed.
There will be some residual flux present. This must be overcome to return to a benign, zero flux state.

^
/_\ B FIELD B_SAT

 ######################
# N #
# ...... N. #
#. N. #
#. *****N. #
#. * N. #
#. * *. #
COERCIVITY #. *  N. # H FIELD
_______________#._*___N_*_.#_____________\
#. *  *. # /
#. * *. #
#. * *. #
#. ****** . #
#........ # N NORMAL MAGENITZATION
# # * MINIMUM
# # . MIDDLE
##################  # MAXIMUM


AUDIO TRANSFORMERS 
Audio transformers are significantly more difficult to design and manufacture than
power transformers, primarily because they must have an extremely wide bandwidth.
Bandwidth at 3 dB must extend from about 0.7 Hz or less to at least 25
kHz for a Bessel function rolloff, or to as high as 70 kHz for other types of rolloff,
to assure accurate reproduction in the time domain. (1), (2)
To satisfy the low frequency response requirement, primary inductance must be
enormously high, often requiring thousands of turns of wire smaller than a human hair.
Primary inductances over 1000 Henries are not uncommon.
all those thousands of turns of wire have
capacitance, which tends to reduce high frequency response
To reduce the possibility of picking up stray magnetic fields, high performance audio
transformers are totally enclosed in a magnetic shielding can, which is then filled
severe distortion occurs as "saturation"
is approached.
"hysteresis" distortion and is caused by
irregularities in the crystalline structure of the core's metal alloy
Distortion is almost entirely due to the core material. Power transformers use
silicon steel alloys because it is cheap and has good magnetic saturation
properties, but its hysteresis distortion properties are horrible.
Here is the "t" equivalent circuit for a 1:1 audio isolation transformer
(designed for a 300 ohm load):
R1L1+L2R2
Primary  Secondary
Side Lm Side

+
* R1,R2 = primary and secondary winding (copper) resistence. Typically
about 50 ohms. Not necessarily equal.
* L1,L2 = primary and secondary leakage inductances. About 5 mH. Not
necessarily equal.
* Lm = mutual inductance, about 2H.
_ _ _ _ _ _
___ R1 / \/ \/ \ L1 / \/ \/ \ L2 R2 ___
 _/\ /\ /\__ () () _______ () () __/\ /\ /\__ 
___ \/ \/  \/ \/ ___
_ _ _  LE
/ \/ \/ \ 
__ () () _
___  LM ___
 ___________________________________________________ 
___ ___
I called Lm the mutual inductance
same as self inductance or
shunt inductance or
magnetizing inductance or what ever
For simplification you can wind up combining both leakage inductances into a
single inductance on either side of Lm.
Description of model operation
Well lets assume that there is 1.25Vrms at 1KHz on the primary and no load.
The full 1.25V appears on the mutual inductance so that there is about 0.1
ma through the mutual inductance. This is the current that gives rise to the
core flux. There is 0.1ma (.995 ma) through the leakage inductance and
primary resistance too. In short through the primary winding lumped circuit.
N
basic AC transformer formula, V = k f N Ac Bm,
which tells us how much flux is present for any voltage and frequency ?
This is the formula
used to find Bmax, so we can be sure the transformer core isn't too close to
saturation, which would introduce even more losses.
The field in transformer core goes goes actuaally DOWN a bit when the
transformer is loaded. due to the effective primary voltage being
reduced by (primary current * resistance of primary winding):
Vs = IpRp + BA[omega]Np
where:
* B is the r.m.s (not peak) induction
* A is the area of crosssection of the core
* [omega] is 2[pi]f, of course
* Np is the number of turns.
What about the isolation ?
Real transformer provides isolation between the input and output. The model
above does not show the isolatio but is sufficent for most of the analysis.
Where the isolation is needed in model you can pretend that there is an
ideal transformer between the "t" and the load like in the picure below:
1:N ideal transformer
R1L1+L2R2o o
Primary  0 Secondary
Side Lm 00 Side
 0
+o o
One model for ideal transformer wiht isolation
This model displays transformers intuitively, the way we most often think of
them:
> Ip R1L1+, ,L2R2 Is >
Primary  O / Secondary
Side, Vp Lm OO Side, Vs
 O \
+' '
magnetizing perfectratio
inductance transformer
_ _ _ _ _ _
___ / \/ \/ \ L1 / \/ \/ \ L2 1:N ___
 _/\ /\ /\__ () () _______ () () __/\ /\ /\________o o____ 
___ \/ \/  \/ \/ R2 \ / ___
R1 _ _ _  LE _   _
/ \/ \/ \  / \/ \/ \
__ () () _ \_/\ /\_/
 LM _   _
 / \/ \/ \
 \_/\ /\_/
___    ___
 _________________________________________________________/ \____ 
___ ___
Lm is the required magnetizing inductance. The perfect transformer converts
Vp to Vs by the ratio of the turns independant of frequency, and draws a
primary current Ip, related to the secondary current Is by the inverse ratio
of the turns. Should you desire to group the series resistances Rx and the
leakage inductances Lx, all on one side, just move the values from the other
side, translated by the square of the turns ratio. This model also directly
matches the simple bench measurements we can take to characterize a
transformer, measuring both lead resistances, and primary inductance with
the secondary both open (magnetizing inductance) and shorted (leakage
inductance L1 + L2*N^2).

Measuring BH curve R2 senses the current in the primary (The magnetizing force) 
R1 and C1 act as a crude integrator,
1:N 100K ___
______________________ o o_______/\ /\ /\____ Y 
__ \ / \/ \/ _ ___
/_ \ _   _ 
// \ \ / \/ \/ \ MEASURE B __
\ \// \_/\ /\_/ ___
\___/ _   _ 
__ / \/ \/ \  3uF
/// MEASURE H \_/\ /\_/ 
___   
__/\ /\ /\_ X _____/ \___________________
__ \/ \/ ___ __
/// 5 ///
_ _ _ _ _ _
___ / \/ \/ \ L1 / \/ \/ \ L2 1:N ___
 _/\ /\ /\__ () () _______ () () __/\ /\ /\________o o____ 
___ \/ \/  \/ \/ R2 \ / ___
R1 _ _ _  LE _   _
/ \/ \/ \  / \/ \/ \
__ () () _ \_/\ /\_/
 LM _   _
 / \/ \/ \
 \_/\ /\_/
___    ___
 _________________________________________________________/ \____ 
___ ___
Lm is the required magnetizing inductance.
measuring both lead resistances, and primary inductance with
the secondary both open (magnetizing inductance) and shorted (leakage
inductance L1 + L2*N^2).

Transformers "t" equivalen circuit
Here is the "t" equivalent circuit for a 1:1 audio isolation transformer
(designed for a 300 ohm load):
R1L1+L2R2
Primary  Secondary
Side Lm Side

+
* R1,R2 = primary and secondary winding (copper) resistence. Typically
about 50 ohms. Not necessarily equal.
* L1,L2 = primary and secondary leakage inductances. About 5 mH. Not
necessarily equal.
* Lm = mutual inductance, about 2H.
_ _ _ _ _ _
___ R1 / \/ \/ \ L1 / \/ \/ \ L2 R2 ___
 _/\ /\ /\__ () () _______ () () __/\ /\ /\__ 
___ \/ \/  \/ \/ ___
_ _ _  LE
/ \/ \/ \ 
__ () () _
___  LM ___
 ___________________________________________________ 
___ ___
Lm mutual inductance
same as self inductance
or shunt inductance
or magnetizing inductance or what ever
it is best called.
1.25Vrms at 1KHz on the primary and no load. about 0.1 ma through the mutual inductance.
current that gives rise to the core flux.
300 ohm load. voltage on the mutual inductance is decreased very little
a short on secondary, the mutual current is decreased only by factor of about two.
current goes through primary winding is 0.1ma + 4.2ma.
basic AC transformer formula, V = k f N Ac Bm
tells how much flux is present for any voltage and frequency to find Bmax,
Only leakage inductance limits the current during a short.
current through primary is limited by winding resistance and leakage resistance when secondary is shorted.
Secondary voltage drop
field in transformer core loaded by (primary current * resistance of primary winding):
Vs = IpRp + BA[omega]Np
where:
* B is the r.m.s (not peak) induction
* A is the area of crosssection of the core
* [omega] is 2[pi]f, of course
* Np is the number of turns.
Other models for transformers
One model for ideal transformer wiht isolation
This model displays transformers intuitively, the way we most often think of
them:
> Ip R1L1+, ,L2R2 Is >
Primary  O / Secondary
Side, Vp Lm OO Side, Vs
 O \
+' '
magnetizing perfectratio
inductance transformer
_ _ _ _ _ _
___ / \/ \/ \ L1 / \/ \/ \ L2 1:N ___
 _/\ /\ /\__ () () _______ () () __/\ /\ /\________o o____ 
___ \/ \/  \/ \/ R2 \ / ___
R1 _ _ _  LE _   _
/ \/ \/ \  / \/ \/ \
__ () () _ \_/\ /\_/
 LM _   _
 / \/ \/ \
 \_/\ /\_/
___    ___
 _________________________________________________________/ \____ 
___ ___
Lm is the required magnetizing inductance.
The following assumes an ideal transformer:
Vs=Vp*(Ns/Np)
Ip=Is*(Ns/Np)
Pin=Pout

Measuring BH curve
You can easily display the BH curve of a transformer on a scope that can do
XY display with just a couple of components.
R2 senses the current in the primary (The magnetizing force)  it should be
selected to give a couple of volts for the Xaxis of the display  a few
ohms.
R1 and C1 act as a crude integrator, since the voltage across the secondary
of the transformer is proportional to the rate of change of the magnetic
field rather than the field itself. Select R1 to give negligible loading on
the transformer (it could be 100s of K) and C1 so that the voltage across it
is less than 5% of voltage on the secondary of the transformer.
1:N 100K ___
______________________ o o_______/\ /\ /\____ Y 
__ \ / \/ \/ _ ___
/_ \ _   _ 
// \ \ / \/ \/ \ MEASURE B __
\ \// \_/\ /\_/ ___
\___/ _   _ 
__ / \/ \/ \  3uF
/// MEASURE H \_/\ /\_/ 
___   
__/\ /\ /\_ X _____/ \___________________
__ \/ \/ ___ __
/// 5 ///
R1
 /\/\/\ Scope Y input
)( 
)( 240/120 = C1
6.3v )( 
)(_______________ Scope Ground

_________ Scope X input

\
/
\ R2
/

 Scope Ground
You cna fo example use 100 kohms resistor and 3 uF capacitor for this
circuit.
Leakeage inductance measurement
First method
Secondary equivalent circuit:
_ _ _
___ / \/ \/ \ ___
 _/\ /\ /\__ () () ___ 
___ \/ \/ ___
Rs+Rp/N^2 Ls+Lp/N^2
___ ___
 _________________________ 
___ ___
O(Rs+Rp/N^2)(Ls+Lp/N^2)+



E(t) Rl


O+
Open circuit voltages on my transformer show Ns/Np = 8.5. Other measured
values:
Rp = 144.5 ohm
Rs = 2.13 ohm
E(t) = 14.17 V
Rl = 25.20 ohm
Voltage across Rl = 12.20 V
Secondary impedance Zs = 14.17 V / (12.20 V / 25.2 ohm) = 29.3 ohm
(Ls+Lp/N^2) = Sqrt( Zs^2  ( (Rs+Rp/N^2) + Rl )^2 ) = 0 , effectively.
The quantity (Rs+Rp/N^2) + Rl = 29.33 ohm which is not significantly
different from Zs .
It appears that within the limits of experimental error, I cannot resolve
any total leakage inductance in this experiment. Thus the leakage inductance
must be very negligible.
Second approach
In the second approach, I shorted the primary winding, and applied voltage
to the primary.
Secondary equivalent circuit with the primary shorted:
O(Rs+Rp/N^2)+


E(t) 


O(Ls+Lp/N^2)+
This time,
E(t) = 1.463 V
Is = 0.342 A
Zs = 4.28 ohm
(Ls+Lp/N^2) = Sqrt( Zs^2  (Rs+Rp/N^2)^2 ) = 3 mH
Here the quantity Rs+Rp/N^2 = 4.13 which is different from Zs.
In this case Rp/N^2 = 2, which is consistent with Rs = 2.13 for a
welldesigned transformer. The primary should have just a little more
winding area than the secondary.
The measured leakage reactance (3 mH) is a bit on the high side, but not
unreasonable for a laminated transformer. It's too high for a welldesigned
toroid. Anyway in measurements like theat the accuracy of the measurements
needs to taken account.
The current waveforms should be reasonably nearly sinusoidal in both tests,
unlike the noload primary current.
Transformer design and selection for applications
Transformer core type selection
TOROIDS vs. ECORES ADVANTAGES
Toroids:
* More compact than Ecore design
* Materials cost is lower due to single component
* Tighter magnetic coupling  lower stray flux leakage
Ecores:
* Easier to automate winding process
* Can be mounted by pins on the bobbins
* Easier to isolate electrically multiple windings
* Core can be easily gapped to extend energy storage capability
Power transformer designing principles
most power transformerson the ragged edge of saturation,
equations for 50 Hz power transformer using laminated iron transformer Ecore:
primary turns = 45 * primary voltage / core area
secondady turns = 48 * secondary voltage / core area
core area = 1.1 * sqrt ( P )
* core area = cross square ares of the core going through the coil in
square centimeters
* primary voltage = AC voltage fed to prmary in volts
* secondary voltage = AC voltage wanted on secondary in volts
* P = transformer power
secondary needs a little bit more turns per voltage
because there are always some losses inside the transformer
As a rule of thumb do not try to push more
then 2.5 amperes of current per square millimeter of the
wire in coils inside transformer.
size of the transformer core must be determined based on the transformer total power.
The area of the core (as used in equation above) should at
least have the value accoring the following equation (can be larger):
core area = sqrt ( transformer power in watts )
different currents suitable for power transformers:
Current Wire diameter
(mA) (mm)
10 0,05
25 0,13
50 0,17
100 0,25
300 0,37
500 0,48
1000 0,7
3000 1,2
5000 1,54
10000 2,24
Low frequency transformers
determine the minimum impedance for a certain transformer coil using the following formula:
L = Z / (2 * pi * f)
* L = primary coil inductance (secondary open circuit)
* Z = circuit impedance
* pi = 3.14159
* f = lowest frequency the transformer must work
approximare inductance formula (for coils with cores) is
L = N * N * a
* L = inductance
* N = number of turns
* a = a constant value (determine value from coil core data or measure it
with test coil)
iron core and need to transfer some power you can determine
the needed core size using formula:
Afe = sqrt ( P / (Bmax * S * f) )
* Afe = core area (cm^2)
* P = maximum transmitted power
* Bmax = maximum magnetic flux in core (Vs/m^2) (usually 4000 G = 0.4
Vs/m^2)
* S = Current density (A/mm^2) (usually 0.5 A/mm^2)
* f = lowest frequency transformer needs to operate (Hz)
Transformers without air gap
N1 = sqrt ( (10^8 * L1 * l) / (u * Afe) )
* N1 = number of turns in primary coil
* Afe = core area (cm^2)
* L1 = primary coil inductance (H)
* l = average length of magnet flow force lines (cm) (lenght of line
around coil going through inside the core)
* u = relative permiability of magnetic material (around 500 for typical
transformer iron)
number of turns on secondary coil using the following
formula (expects transformer effiency of 90%):
N2 = 1.1 * U2 / U1 = 1.1 * sqrt (Z2 / Z1) =
* N1 = number of turns in primary coil
* N2 = number of turns in secondary coil
* U1 = primary voltage
* U2 = secondary voltage
* Z1 = primary impedance
* Z2 = secondary impedance
you should use as thick wire as you can.
to leave 3050% of the coil volum to the insulation.
Transformers with air gap
If DC current flowing on transformer primary, primary inductance is reduced.
To compensate core should have a small air gap in the core.
air gap should be selected to be around 1/1000 of the length of magnetic lines in the core.
In this case the following equation can be used to determine number of turns needed for primary coil:
N1 = sqrt ( (L1 * li) / (Afe * 10^8) )
* N1 = number of turns in primary coil
* Afe = core area (cm^2)
* L1 = primary coil inductance (H)
* li = size of the air gap (mm)
gives much larger nu,ber of turns for primary coil
Pulse transformers
Impedance matching transformer selection
Matching is required to ensure maximum power transfer from source to the load.
A matched condition exists when:
N = N2 / N1 = sqrt (Zl / Zs)
* N = turns ratio between primary and secondary
* N1 = number of turns in primary
* N2 = number of turns in secondary
* Zs = signal source impedance
* Zl = transformer load impedance
matching transformer will present its own shunt impedance to the source.
should be large comparef with the souce impedance.
A safety factor of 5 should be sufficent f
Lp = 5 * Zs / (2 * pi * fmin)
* Lp = prmary inductance
* Zs = source impedance
* fmin = minimum frequency needed to be transferred through transformer
* pi = 3.14159
too high a primary inductance reduce the highfrequency
Selection procedures for pulse matching transformers
It is necessary to check for pulse distortion when selecting the
transformer.
Et constant maximum area of pulse which a given transformer can transmit. T
may be estimated from the known pulse shape
Et = Vp * tpw
Lp = R * tpw / Ln (I  D)
D = delta / Vp = 1  exp (R * tt / Lp)
0 < tt < tpw
* tpw = the worstcase (maximum) pulse width to be transmitted
* Vp = pulse voltage (voltage from top to bottom)
* delta = how much pulse top is allowed to drop
* tt = time the pulse top is active (tpw  start and end slopes)
* D = droop (usually 10& can be tolerated)
* R = parallel combination of the source and reflected load impedance
(for a matched case this is half of the source impedance)
if no upper limit to pulse length (tpw) transformers don't work with DC.
too high Et constant full pulse width will not be transmitted
and the transformer will cause excessive loading to the due to saturation.
too high an Et will bring high parasitic capacitances and inductances
which will cause poor signal rise times.
droop in i relation to the pulse time, primary inductance and system impedances.
droop of 10% can usually be tolerated.
Power transformer designing principles
Based on [1]
It has been my suspocion that to save iron and weight, most power
transformers are designed to operate right on the ragged edge of saturation,
hence all hell can break loose (at least transformer hears more) when you
take a product designed for 60Hz service and power it with 50Hz.
Designing a power transformer is quite careful thng if an optimized design
is needed. To get a general view of the design of a power transformer I geve
here you some approzimate design equations for 50 Hz power transformer using
laminated iron transformer Ecore:
primary turns = 45 * primary voltage / core area
secondady turns = 48 * secondary voltage / core area
core area = 1.1 * sqrt ( P )
Where:
* core area = cross square ares of the core going through the coil in
square centimeters
* primary voltage = AC voltage fed to prmary in volts
* secondary voltage = AC voltage wanted on secondary in volts
* P = transformer power
Low frequency transformers
Based on [1]
General formulas
For low power low frequency transformers you can generally determine that
the turns ratio determines the voltage transfer ratio. For given impedance
circuti you need to determine the minimum impedance for a certain
transformer coil using the following formula:
L = Z / (2 * pi * f)
Where:
* L = primary coil inductance (secondary open circuit)
* Z = circuit impedance
* pi = 3.14159
* f = lowest frequency the transformer must work
This is the recommended value for the impedance. The impedance of the coil
can be higher than the value determined by thie equations. Using too high
inductance would not generally have much problems, but generally it is not
good idea because many practical reasons (longer primary coil, more
resistance, more capacitance, propably for those reasons poorer high
frequency response etc.).
The actual number of turns needed to get the necessary inductance depends on
the tranformer core model and magnetic material used it. Consult the
datasheet of the coil material you are using for more details or it. Other
option is to first wire one test coul and measure it. Using the measurement
results you can determine how many turns are needed for a specific
inductance. General approximare inductance formula (for coils with cores) is
useful for this:
L = N * N * a
Where:
* L = inductance
* N = number of turns
* a = a constant value (determine value from coil core data or measure it
with test coil)
If you are using iron core and need to transfer some power you can determine
the needed core size using formula:
Afe = sqrt ( P / (Bmax * S * f) )
Where:
* Afe = core area (cm^2)
* P = maximum transmitted power
* Bmax = maximum magnetic flux in core (Vs/m^2) (usually 4000 G = 0.4
Vs/m^2)
* S = Current density (A/mm^2) (usually 0.5 A/mm^2)
* f = lowest frequency transformer needs to operate (Hz)
Transformers without air gap
And when you know core area you can calculate the number of turns for
transformer primary for transformer without air gap in core using the
following formula:
N1 = sqrt ( (10^8 * L1 * l) / (u * Afe) )
Where:
* N1 = number of turns in primary coil
* Afe = core area (cm^2)
* L1 = primary coil inductance (H)
* l = average length of magnet flow force lines (cm) (lenght of line
around coil going through inside the core)
* u = relative permiability of magnetic material (around 500 for typical
transformer iron)
You can determine the number of turns on secondary coil using the following
formula (expects transformer effiency of 90%):
N2 = 1.1 * U2 / U1 = 1.1 * sqrt (Z2 / Z1) =
Where:
* N1 = number of turns in primary coil
* N2 = number of turns in secondary coil
* U1 = primary voltage
* U2 = secondary voltage
* Z1 = primary impedance
* Z2 = secondary impedance
For optimum tranformer performance the resistance of the coils should be
kept as low as possible. This means that you should use as thick wire as you
can. When selecting wire size, remember to leave 3050% of the coil volume
to the insulation.
Transformers with air gap
If there is any DC current flowing on transformer primary, the primary
inductance is reduced. To compensate the effect of this (in circuits where
this is a problem) the core should have a small air gap in the core. In
practice the air gap should be selected to be around 1/1000 of the length of
the magnetic lines in the core. In this case the following equation can be
used to determine the number of turns needed for primary coil:
N1 = sqrt ( (L1 * li) / (Afe * 10^8) )
Where:
* N1 = number of turns in primary coil
* Afe = core area (cm^2)
* L1 = primary coil inductance (H)
* li = size of the air gap (mm)
Note that this formula gives much larger nu,ber of turns for primary coil
than the equation for transformer without air gap. Other calculations for
transformers are made as with the transformer without air gap.
Pulse transformers
Based on [4]
Impedance matching transformer selection
Matching is required to ensure maximum power transfer from source to the
load. A matched condition exists when:
N = N2 / N1 = sqrt (Zl / Zs)
Where:
* N = turns ratio between primary and secondary
* N1 = number of turns in primary
* N2 = number of turns in secondary
* Zs = signal source impedance
* Zl = transformer load impedance
In real world the matching transformer will present its own shunt impedance
to the source. The magnitude of this impedance will depend on the primary
inductance and the frequency of operation. This should be large comparef
with the souce impedance. A safety factor of 5 should be sufficent for most
of the applications. So a suitable value for primary coil inductance can be
calculated using the follwing formula:
Lp = 5 * Zs / (2 * pi * fmin)
Where:
* Lp = prmary inductance
* Zs = source impedance
* fmin = minimum frequency needed to be transferred through transformer
* pi = 3.14159
If too high a primary inductance is chosen, the parasitic components (shunt
capacitance, leake inductance etc.) conspire to reduce the highfrequency
performance of the circuit.
Selection procedures for pulse matching transformers
It is necessary to check for pulse distortion when selecting the
transformer. There is a maximum area of pulse which a given transformer can
transmit. This is known as the Et constant. The following formulas will
describe how this may be estimated from the known pulse shape
Et = Vp * tpw
Lp = R * tpw / Ln (I  D)
D = delta / Vp = 1  exp (R * tt / Lp)
0 < tt < tpw
Where:
* tpw = the worstcase (maximum) pulse width to be transmitted
* Vp = pulse voltage (voltage from top to bottom)
* delta = how much pulse top is allowed to drop
* tt = time the pulse top is active (tpw  start and end slopes)
* D = droop (usually 10& can be tolerated)
* R = parallel combination of the source and reflected load impedance
(for a matched case this is half of the source impedance)
It is worth noting that if no upper limit can be put to pulse length (tpw)
then it will not be possible to use a transformer in this application
because transformers don't work with DC. If too high Et constant is chosen
then the full pulse width will not be transmitted and the transformer will
cause excessive loading to the due to saturation. Conversely, too high an Et
constant will bring attendant high parasitic capacitances and inductances
which will cause poor signal rise times.
^
/\ B FIELD B_SAT

 ##########N##N########
# N #
# N #
# N #
#  ****N #
# * N #
# * * #
COERCIVITY # *  N # H FIELD
_____________#__*__N_*__#______________\
# *  * # /
# * * #
# * * #
# *****  #
# # N NORMAL MAGENITZATION
# # # MIDDLE
# ## # MAXIMUM
################## 




*
* Trace Elliot 15W output transformer, 8k primary, 16/8 ohm sec.
* Part no. 73 TRAN 15W OP
*
* [1] Red . 
* )  . Green (16 ohm) [4]
* )  (
* )  (
* [2] White .  . Yellow (8 ohm) [5]
* )  (
* )  (
* )  . Brown [6]
* [3] Blue . 
*
.SUBCKT trace15 1 2 3 4 5 6
*
* Primary
*
Lleak1 1 20 2mH
Lpri1 20 21 150H
Rpri1 21 2 186
Cpri1 1 2 120p
Lleak2 2 22 2mH
Lpri2 22 23 150H
Rpri2 23 3 201
Cpri2 2 3 120p
*
* Secondary
*
Lleak3 5 24 10uH
Lsec1 24 25 0.6H
Rsec1 25 6 0.8
Lleak4 4 27 10uH
Lsec2 27 28 0.3H
Rsec2 28 5 0.4
Kcoup Lpri1 LPri2 Lsec1 Lsec2 1.0
.ENDS
K
'`
V1 ___ o o____ V2
\ /
L1 _   _ L2
/ \/ \/ \
\_/\ /\_/
_   _
/ \/ \/ \
\_/\ /\_/ V2=V1*N2/N1
_   _
/ \/ \/ \ I1=I2*N2/N1
\_/\ /\_/
 
____/ \____
V1= L1*di1/dt + M*di2/dt
V2= L2*di2/dt + M*di1/dt
M= K*sqrt( L1+L2)
_ _ _ _ _ _
/ \/ \/ \ / \/ \/ \ LE 1:N
_ () () _______ () () ______ o o_______ IF K => 1
 \ /
V1 LE ________ _   _ V2 L1 = LM
 / \/ \/ \
_ _ _  \_/\ /\_/ L2 = N^2*L1
/ \/ \/ \  _   _
__ () () _ / \/ \/ \ 1 K = LE/LM
 \_/\ /\_/
 LM _   _
 / \/ \/ \
 \_/\ /\_/
  
_______________________________/ \________

Typical Transformer __ P =10 + VL1b  ___ __
 __ ____0 0___/ \___ 
__     +\___/ __
*() ()  *() ()  (P/S)*VL2b
L1a\_/\_/\_/ L1b\_/\_/\_/
L1 = L1a+L1b L2 = L2a+L2b
Turn Ratio =P:S
L2a _ _ _ L2b _ _ _ K=L1b/L1=L2b/L2=99%
/*\/ \/ \ /*\/ \/ \ (S/P)*VL1b
__  () ()   () ()  + ___  __
 __ ____ ___/ \___ 
__ S=100 0 0 \___/ __
+ VL2b  M= K*sqrt(L1*L2)
VL1 = (dI1/dt)*L1 + (dI2/dt)*M*(P/S)
K = coupling coeffecient
VL2 = (dI2/dt)*L2 + (dI1/dt)*M*(S/P)
VL1/VL2 = P/S
hysteresis of core Ph = Kh*f*Bm^x*V hysteresis loss
Kh is a constant which depends on the chemical
analysis of the
material and the heat treatment and mechanical
treatment to which
it has been subjected
f is frequency in Hz
Bm is maximum flux density in webers / m^2 (Teslas)
V is the volume of the material in cubic meters
Pe = Ke*f^2*c^2*Bm^2*V eddy current loss
Ke is a constant which depends on the resistivity of
the material
f is frequency in Hz
c is lamination thickness in meters
Bm is maximum flus density in webers / m^2 (Teslas)
V is volume of the material in cubic inches.
Switching Power Supply
: V = k [FNAB]
: where V is the maximum input voltage
: k is a constant that depends on the type of
excitation
: (e.g. square wave, sine wave, etc.)
: F is the frequency of excitation
: N is the switching frequency
: A is the effective core area
: B is the core flux density
In fact, the basic equation is:
V = k [FNAB]
where V is maximum input voltage
k is a constant that depends on type of excitation
(e.g. square wave, sine wave, etc.)
F is the frequency of excitation
N is the number of primary turns
A is the effective core area
B is the core flux density
formula is:
AL = L/N^2
DC saturation is an important criteria as well. For any
nductor wound
on a permeable core (that is not an air core), there is a
maximum field
strength beyond which the inductor will magnetically
saturate, and act
as an ordinary piece of wire. Magnetic field strength can
be determined
according to:
H = (I x N)/Ie
where: H is the DC field strength in Amps/Meter
I is the DC current in the windings
N is the number of turns
Ie is the effective magnetic length if the
core (from the
core data book.
eddy current in laminations and another comes fromhysteresis of core.
Ph = Kh*f*Bm^x*V hysteresis loss
Kh is a constant which depends on the chemical
analysis of material and heat treatment and mechanical
treatment to which it has been subjected
f is frequency in Hz
Bm is maximum flux density in webers / m^2 (Teslas)
V is the volume of the material in cubic meters
Pe = Ke*f^2*c^2*Bm^2*V eddy current loss
Ke is a constant which depends on the resistivity of
the material
f is frequency in Hz
c is lamination thickness in meters
Bm is maximum flus density in webers / m^2 (Teslas)
V is volume of the material in cubic inches.
Let's first guess at the Bm....
Assume inductance of core is constant as of
frequency (poor guess for iron but a start)
Bm will inversely proportional to frequency
and proportional to voltage.
So we can change the above equations to
Ph = Kh'*f hysteresis loss
Pe = Ke'*f^2 eddy current loss
since everything else is constant.
Pc = Ph*135/60 + Pe*(135/60)^2 total core loss
If *ALL* the power were in the hysteresis, you'd get a
loss of 3.6 W,
Now let's *GUESS* that half power went into hysteresis
and half intoeddy current losses, then you would have....
Pc = .8*135/60 + .8*(135/60)^2 total core loss
= 5.9 W !!!

Coils
Coil equations
General formula
Equation for calculating inductance of a homemade coil: L=(D*N^2)/(l/D+0,43)
Where
* D is diameter in cm
* l i length in cm
* L is inductance in uH
* N is nuber of turns
Air core coils
L = (r^2 * n^2)/(9r + 10l)
For air core coils you can come close with:
* L = ind. in uH
* r = radius of coil in inches
* n = number of turns
* l = coil length in inches
ref: Bauchbaum's Complete Handbook of Practical Electronic Ref. Data
Air cores typically range from .1 to 2000 uH. Bigger indictances usually
make the coil too bulky and the above formula is not accurate enough because
inner and outer radii of your windings may vary too widely.
Another coil equation
L = (a^2*n^2)/(9*a+10*b)
where
* a = radius in inches
* b = length in inches
* n = number of turns
this was claimed to be accurate within a few percent.
Equation for single layer coils
L= N^2*A*u*u0/l
Where:
* L=inductance desired
* N=Number of turns
* A=cross sectional area of core in square centimeters
* u(Mu)=permeability of core (Air=1; Iron~1000)
* u0 (mu subzero)= Absolute permeability of air (1.26*10^12)
* l=length of coil in centimeters
Simple formulas for turn numbers of air and iron core coils
Air core: N=(SR[A*u*u0/l])*20Pi
Iron core: N=(SR[A*u*u0/1*.5Pi])
Where:
* N=number of turns needed for coil
* SR=Square Root (of bracketed equation)
* A=cross sectional area of core
* u=Permeability of core
* u0=Absolute permeability of air
* l=length of coil
* 20Pi=20 times Pi or about 63
* .5Pi=half of Pi or 1.57
The equations "sort of" govern the inductance. Testing with inductance
meters will ascertain desired inductance.
Program for calculating coil inductance
TABLE 23ÑFERRITE CHARACTERISTICS.
Saturation
Moment Saturation
Saturation Curie in Bohr FirstOrder Magneto
Moment Temp Magnetons Xray Lattice Anisotropy striction
Ferrite in Gauss in ~C n_B Density Constant Constant K1 Lambda_eX106
NiFe2O4 3400 585 2.3 5.38 8.34 Ñ0.06 Ñ22
Ni0.8Zn0.2Fe2O4 4600 460 3.5    Ñ Ñ18.5
Ni0.5Zn0.5Fe2O4 5800 360 4.8 Ñ Ñ Ñ15.0
Ni0.5Zn0.5Fe2O4 5500 290 5.0 Ñ  Ñ Ñ8.3
Ni0.3Zn0.5Fe2O4 2600 85 4.0 Ñ Ñ0.004 Ñ1.0
MnFe2O4 5200 300 0 5.00 8.50 Ñ0.04 Ñ14
Mn0.5Zn0,
Fe204 100 6.0 Ñ Ñ Ñ0.004 Ñ
FeFe2O4 6000 585 4.1 5.24 8.39 Ñ0.135 +41
CoFe2O4 5000 520 3.8 5.20 8.38 Ñ2000 Ñ250
CuFeoO4 1700 455 1.3 5 35 8.24 Ñ Ñ
8.68
Li0.5Fe2.5O4 3900 670 2.6 4 75 8.33 Ñ 
MgFe204 1400 440 1.1 4.52 8.36 Ñ 0.05 Ñ
MgA1FeO4 Ñ Ñ 0.3 Ñ Ñ
NiAl0.25Fe1.75O4 1300 506 1.30 Ñ 8.31 Ñ Ñ
NiAl_0.45Fe_1.55O4 900 465 0.61 Ñ 8.28 Ñ Ñ
NiAl_0.62Fe_1.38O4 0 360 0 Ñ 8.25 Ñ Ñ
NiAlFeO4 900 198 0.64 5.00 8.20 Ñ Ñ