======================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 First-Order Magneto- Moment Temp Magnetons X-ray 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 strip-wound cores and powder cores generally have such high Curie temperatures (>450 degrees C) Manganese-zinc 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 material------------1.25-1.75y H mate-----------1.50-2.00y Ferrite cost is also a function of geometry: Toroids -------------least E cores--------------mid Other shapes-------most B-H 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). B-H loop changes with frequency and drive level. Why air gap into cores "tilts", or "shears" the B-H 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 temperature-cycle, (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 nickel-zinc and manganese-zinc ferrites? MnZn materials have a high permeability, NiZn ferrites have low permeability. Manganese-zinc ferrites used less than 5MHz. Nickel-zinc 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 1-4%). 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 custom-built 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, FC-601. 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 PS-01 also covers this subject. QUESTION: Why do manufacturers flat-grind ferrite cores? ANSWER: Cores are flat-ground 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 mirror-like finish is the result. The typical surface finish for normally flat-ground surfaces is 25 micro-inches (.635 microns) and for lapped surfaces is 5 micro-inches (.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 poorly-fitted 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 5-10 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 FC-601 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 cross-sectional 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 cross-section (cm2) le= core magnetic path (cm) CONVERSION FACTORS: MULTIPLY BY TO OBTAIN Oersteds 2.0213 ampere-turns/inch Oersteds 0.79577 ampere-turns/cm Oersteds 79.577 ampere-turns/m Ampere-turns/cm 1.2566 oersteds Gausses 10-4 teslas Micro-inches 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 |#_____________________________________\ / air-gap is useful It can be. The shape of the air-gap can be important as well. Some devices are built with a 'stepped air-gap'. 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 B-H 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): ------R1---L1-----+----L2----R2------ 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 cross-section 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 ------R1---L1-----+----L2----R2-----o o----- Primary | 0|| Secondary Side Lm 0||0 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 -----R1---L1---+---, ,---L2----R2------ Is -> Primary | O|| / Secondary Side, Vp Lm O||O Side, Vs | O|| \ ---------------+---' '----------------- magnetizing perfect-ratio 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 B-H 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): ------R1---L1-----+----L2----R2------ 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 cross-section 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 -----R1---L1---+---, ,---L2----R2------ Is -> Primary | O|| / Secondary Side, Vp Lm O||O Side, Vs | O|| \ ---------------+---' '----------------- magnetizing perfect-ratio 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 B-H curve You can easily display the B-H curve of a transformer on a scope that can do X-Y 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 X-axis 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 well-designed 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 well-designed 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 no-load primary current. Transformer design and selection for applications Transformer core type selection TOROIDS vs. E-CORES ADVANTAGES Toroids: * More compact than E-core design * Materials cost is lower due to single component * Tighter magnetic coupling - lower stray flux leakage E-cores: * 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 E-core: 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 30-50% 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 high-frequency 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 worst-case (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 E-core: 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 30-50% 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 high-frequency 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 worst-case (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 First-Order Magneto- Moment Temp Magnetons X-ray 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