Design of Magnetic Components

Design of Magnetic Components William P. Robbins Dept. of Electrical and Computer Engineering University of Minnesota Outline A. Inductor/Transformer Design Relationships B. Magnetic Cores and Materials C. Power Dissipation in Copper Windings D. Thermal Considerations E. Analysis of Specific Inductor Design F. Inductor Design Procedures G. Analysis of Specific Transformer Design H. Eddy Currents J. Transformer Leakage Inductance K. Transformer Design Procedures Magnetics - 1

Magnetic Component Design Responsibility of Circuit Designer Ratings for inductors and transformers in power electronic circuits vary too much for commercial vendors to stock full range of standard parts. Core (double E) Instead only magnetic cores are available in a wide range of sizes, geometries, and materials as standard parts. Circuit designer must design the inductor/transformer for the particular application. Design consists of: 1. Selecting appropriate core material, geometry, and size. Selecting appropriate copper winding parameters: wire type, size, and number of turns. Winding Bobbin Assembled core and winding Magnetics -

Review of Inductor Fundamentals Assumptions No core losses or copper winding losses Linearized B-H curve for core with µ m >> µ o l m >> g and A >> g l = mean path length m i 1 Cross-sectional area of core = A Magnetic circuit approimations (flu uniform over core cross-section, no fringing flu) N 1 Air gap: H g g Starting equations H m l m + H g g = N I (Ampere s Law) Core: H m B m A = B g A = φ (Continuity of flu B assuming no leakage flu) µ m H m = B m (linearized B-H curve) ; µ o H g = B g B s H B H Results B s > B m = B g = LI = Nφ ; L = NI l m /µ m + g/µ o = φ/a A N l m /µ m + g/µ o linear region µ = B B = H H Magnetics - 3

Review of Transformer Fundamentals Assumptions same as for inductor Starting equations H 1 L m = N 1 I 1 ; H L m = N I (Ampere's Law) H m L m = (H 1 - H )L m = N 1 I 1 - N I µ m H m = B m (linearized B-H curve) dφ 1 dφ v 1 = N 1 ; v dt = N dt (Faraday's Law) Net flu φ = φ 1 - φ = µ m H m A = µ m A(N 1 I 1 - N I ) L m l = mean path length m i 1 + + v 1 N v 1 N - - φ 1 Magnetic flu φ B s B Cross-sectional area of core = A φ i Results assuming µ m, i.e. ideal core or ideal transformer approimation. φ µ m = 0 and thus N 1 I 1 = N I d(φ 1 - φ ) = 0 = dt v 1 N 1 - v N ; v 1 N 1 = v N H B linear region µ = B B = H H H Magnetics - 4

Current/Flu Density Versus Core Size Larger electrical ratings require larger current I and larger flu density B. B s B Core losses (hysteresis, eddy currents) increase as B (or greater) Winding (ohmic) losses increase as I and are accentuated at high frequencies (skin effect, proimity effect) Minor hystersis loop H To control component temperature, surface area of component and thus size of component must be increased to reject increased heat to ambient. At constant winding current density J and core flu density B, heat generation increases with volume V but surface area only increases as V /3. core - B s Maimum J and B must be reduced as electrical ratings increase. Flu density B must be < B s g Higher electrical ratings larger total flu larger component size Flu leakage, nonuniform flu distribution complicate design fringing flu Magnetics - 5

Magnetic Component Design Problem Challenge - conversion of component operating specs in converter circuit into component design parameters. Goal - simple, easy-to-use procedure that produces component design specs that result in an acceptable design having a minimum size, weight, and cost. Design procedure outputs. Core geometry and material. Core size (A core, A w ) Number of turns in windings. Conductor type and area A cu. Air gap size (if needed). Inductor electrical (e.g.converter circuit) specifications. Inductance value L Inductor currents rated peak current I, rated rms current I rms, and rated dc current (if any) I dc Operating frequency f. Allowable power dissipation in inductor or equivalently maimum surface temperature of the inductor T s and maimum ambient temperature T a. Three impediments to a simple design procedure.. 1. Dependence of J rms and B on core size.. How to chose a core from a wide range of materials and geometries. Transformer electrical (converter circuit) specifications. Rated rms primary voltage V pri Rated rms primary current I pri Turns ratio N pri /N sec Operating frequency f Allowable power dissipation in transformer or equivalently maimum temperatures T s and T a 3. How to design low loss windings at high operating frequencies. Detailed consideration of core losses, winding losses, high frequency effects (skin and proimity effects), heat transfer mechanisms required for good design procedures. Magnetics - 6

Core Shapes and Sizes Magnetic cores available in a wide variety of sizes and shapes. Ferrite cores available as U, E, and I shapes as well as pot cores and toroids. Laminated (conducting) materials available in E, U, and I shapes as well as tape wound toroids and C-shapes. Open geometries such as E-core make for easier fabrication but more stray flu and hence potentially more severe EMI problems. Closed geometries such as pot cores make for more difficult fabrication but much less stray flu and hence EMI problems. Bobbin or coil former provided with most cores. Dimensions of core are optimized by the manufacturer so that for a given rating (i.e. stored magnetic energy for an inductor or V-I rating for a transformer), the volume or weight of the core plus winding is minimized or the total cost is minimized. insulating layer magnetic steel lamination Larger ratings require larger cores and windings. Optimization requires eperience and computerized optimization algorithm. Vendors usually are in much better position to do the optimization than the core user. Magnetics - 7

Double-E Core Eample d a a a 1.9 a 1.4 a h w h a b a Core Bobbin b w Assembled core and winding Characteristic Relative Size Absolute Size for a = 1 cm Core area A core 1.5 a Winding area A w 1.4 a Area product AP = A w A c.1 a 4 Core volume V core 13.5 a 3 Winding volume V w 1.3a 3 Total surface area of assembled core and winding 1.5 cm 1.4 cm.1 cm 4 13.5 cm 3 1.3 cm 3 59.6 a 59.6 cm Magnetics - 8

Types of Core Materials Iron-based alloys Various compositions Fe-Si (few percent Si) Fe-Cr-Mn METGLASS (Fe-B, Fe-B-Si, plus many other compositions) Important properties Resistivity _ = (10-100) ρ Cu B s = 1-1.8 T (T = tesla = 10 4 oe) METGLASS materials available only as tapes of various widths and thickness. Ferrite cores Various compositions - iron oides, Fe-Ni-Mn oides Important properties Resistivity ρ very large (insulator) - no ohmic losses and hence skin effect problems at high frequencies. B s = 0.3 T (T = tesla = 10 4 oe) Other iron alloys available as laminations of various shapes. Powdered iron can be sintered into various core shapes. Powdered iron cores have larger effective resistivities. Magnetics - 9

Hysteresis Loss in Magnetic Materials B Minor hystersis loop B s 0 B ac t H B ac B dc B(t) - B s 0 t Area encompassed by hysteresis loop equals work done on material during one cycle of applied ac magnetic field. Area times frequency equals power dissipated per unit volume. Typical waveforms of flu density, B(t) versus time, in an inductor. Only B ac contributes to hysteresis loss. Magnetics - 10

Quantitative Description of Core Losses Eddy current loss plus hysteresis loss = core loss. 3F3 core losses in graphical form. Empirical equation - P m,sp = k f a [B ac ] d. f = frequency of applied field. B ac = base-to-peak value of applied ac field. k, a, and d are constants which vary from material to material P m,sp = 1.510-6 f 1.3 [B ac ].5 mw/cm 3 for 3F3 ferrite. (f in khz and B in mt) P m,sp = 3.10-6 f 1.8 [B ac ] mw/cm 3 METGLAS 705M (f in khz and B in mt) Eample: 3F3 ferrite with f = 100 khz and B ac = 100 mt, P m,sp = 60 mw/cm 3 Magnetics - 11

Core Material Performance Factor Volt-amp (V-A) rating of transformers proportional to f B ac Core materials have different allowable values of B ac at a specific frequency. B ac limted by allowable P m,sp. Most desirable material is one with largest B ac. Choosing best material aided by defining an emperical performance factor PF = f B ac. Plots of PF versus frequency for a specified value of P m,sp permit rapid selection of best material for an application. Plot of PF versus frequency at P m,sp = 100 mw/cm 3 for several different ferrites shown below. Magnetics - 1

Eddy Current Losses in Magnetic Cores Eddy current B i (r) B o = ep({r - a}/δ) B (t) o B (r) i B o B (r) i B o δ = skin depth = ωµσ a 0 a r ω = π f, f = frequency µ = magnetic permeability ; µ o for magnetic materials. σ = conductivity of materia AC magnetic fields generate eddy currents in conducting magnetic materials. Eddy currents dissipate power. Shield interior of material from magnetic field. Numerical eample σ = 0.05 σ cu ; µ = 10 3 µ o f = 100 Hz δ = 1 mm Magnetics - 13

Laminated Cores Cores made from conductive magnetic materials must be made of many thin laminations. Lamination thickness < skin depth. Stacking factor k stack = t t + 0.05t 0.5 t t (typically 0.3 mm) Insulator Magnetic steel lamination Magnetics - 14

Eddy Current Losses in Laminated Cores Flu φ(t) intercepted by current loop of area w given by φ(t) = wb(t) Voltage in current loop v(t) = w db(t) dt = wωbcos(ωt) wρ core Current loop resistance r = L d ; w >> d Instantaneous power dissipated in thin loop δp(t) = [v(t)] r Average power P ec dissipated in lamination given by P ec = < w L d 3 ω B δp(t)dv > = 4 ρ core P ec P ec,sp = V = w L d 3 ω B 1 4 ρ core dwl = d ω B 4 ρ core Average power P ec dissipated in lamination given by P ec = < w L d 3 ω B δp(t)dv > = 4 ρ core P ec P ec,sp = V = w L d 3 ω B 1 4 ρ core dwl = d ω B 4 ρ core z y L w d d B sin( ωt) - Eddy current flow path Magnetics - 15

Power Dissipation in Windings Average power per unit volume of copper dissipated in copper winding = P cu,sp = ρ cu (J rms ) where J rms = I rms /A cu and ρ cu = copper resistivity. bobbin airgap Winding conductor Average power dissipated per unit volume of winding = P w,sp = k cu ρ cu (J rms ) ; V cu = k cu V w where V cu = total volume of copper in the winding and V w = total volume of the winding. Copper fill factor k cu = N A cu A w < 1 N = number of turns; A cu = cross-sectional area of copper conductor from which winding is made; A w = b w l w = area of winding window. k cu = 0.3 for Leitz wire; k cu = 0.6 for round conductors; k cu 0.7-0.8 for rectangular conductors. h w A w > g b w Double-E core eample k cu < 1 because: Insulation on wire to avoid shorting out adjacent turns in winding. Geometric restrictions. (e.g. tight-packed circles cannot cover 100% of a square area.) g Magnetics - 16

Eddy Currents Increase Winding Losses AC currents in conductors generate ac magnetic fields which in turn generate eddy currents that cause a nonuniform current density in the conductor. Effective resistance of conductor increased over dc value. P w,sp > k cu ρ cu (J rms ) if conductor dimensions greater than a skin depth. J(r) J = ep({r - a}/δ) o δ = skin depth = ωµσ ω = π f, f = frequency of ac current µ = magnetic permeability of conductor; µ = µ o for nonmagnetic conductors. σ = conductivity of conductor material. Numerical eample using copper at 100 C Frequency 50 Hz Skin 10.6 Depth mm 5 khz 1.06 mm 0 khz 0.53 mm 500 khz 0.106 mm H(t) I(t) I(t) Eddy currents B sin( ωt) - + - + + - + - B sin( ωt) J(t) Mnimize eddy currents using Leitz wire bundle. Each conductor in bundle has a diameter less than a skin depth. a 0 a J(t) Twisting of paralleled wires causes effects of intercepted flu to be canceled out between adjacent twists of the conductors. Hence little if any eddy currents. Magnetics - 17 r

Proimity Effect Further Increases Winding Losses a) d (b) d A A B MMF B Proimity effect - losses due to eddy current generated by the magnetic field eperienced by a particular conductor section but generated by the current flowing in the rest of the winding. 10 Eddy Current Losses 100 30 Design methods for minimizing proimity effect losses discussed later. 3 10 3 1 1 Magnetics - 18

Minimum Winding Loss P w = P dc + P ec ; P ec = eddy current loss. Optimum conductor size P w = { R dc + R ec } [I rms ] = R ac [I rms ] Resistance R dc R ac = F R R dc = [1 + R ec /R dc ] R dc R ec Minimum winding loss at optimum conductor size. P w = 1.5 P dc P ec = 0.5 P dc High frequencies require small conductor sizes minimize loss. d opt δ d = conductor diameter or thickness P dc kept small by putting may small-size conductors in parallel using Litz wire or thin but wide foil conductors. Magnetics - 19

Thermal Considerations in Magnetic Components Losses (winding and core) raise core temperature. Common design practice to limit maimum interior temperature to 100-15 C. Core losses (at constant flu density) increase with temperature increases above 100 C Saturation flu density B s decreases with temp. Increases Nearby components such as power semiconductor devices, integrated circuits, capacitors have similar limits. Temperature limitations in copper windings Copper resistivity increases with temperature increases. Thus losses, at constant current density increase with temperature. Reliability of insulating materials degrade with temperature increases. Surface temperature of component nearly equal to interior temperature. Minimal temperature gradient between interior and eterior surface. Power dissipated uniformly in component volume. Large cross-sectional area and short path lengths to surface of components. Core and winding materials have large thermal conductivity.. Thermal resistance (surface to ambient) of magnetic component determines its temperature. P sp = T s - T a R θsa (V w + V c ) ; R θsa = h A s h = convective heat transfer coefficient = 10 C-m /W A s = surface area of inductor (core + winding). Estimate using core dimensions and simple geometric considerations. Uncertain accuracy in h and other heat transfer parameters do not justify more accurate thermal modeling of inductor. Magnetics - 0

Scaling of Core Flu Density and Winding Current Density Power per unit volume, P sp, dissipated in magnetic component is P sp = k 1 /a ; k 1 = constant and a = core scaling dimension. J rms = P sp k cu r cu = k 1 k cu a ; k = constant P w,sp V w + P m,sp V m = T s - T a R θsa : T a = ambient temperature and R θsa = surface-to-ambient thermal resistance of component. P m,sp = P sp = k f b [B ac ] d ; Hence B ac = d Psp kf b = k 3 d f b a where k 3 = constant For optimal design P w,sp = P c,sp = P sp : Hence P sp = T s - T a R θsa (V w + V c ) R θsa proportional to a and (V w + V c ) Plots of J rms, B ac, and P sp versus core size (scale factor a) for a specific core material, geometry, frequency, and T s - T a value very useful for picking appropriate core size and winding conductor size. proportional to a 3 Magnetics - 1

Eample of Power Density and Current Density Scaling J rms P sp A/mm mw/cm 3 8 400 7 350 Assumptions 6 300 1. Double-E core made from 3F3 ferrite 5 50. T s = 100 C and T a = 40 C. 4 00 3. Winding made with Leitz wire - k cu = 0.3 3 J rms 150 100 P sp 1 50 0 0.5 1 1.5.5 3 3.5 4 4.5 5 0 Core scaling parameter a [cm] Magnetics -

Analysis of a Specific Inductor Design Inductor specifications Maimum current = 4 ams rms at 100 khz Double-E core with a = 1 cm using 3F3 ferrite. Distributed air-gap with four gaps, two in series in each leg; total gap length Σg = 3 mm. Winding - 66 turns of Leitz wire with A cu = 0.64 mm g Inductor surface black with emissivity = 0.9 T a,ma = 40 C Find; inductance L, T s,ma ; effect of a 5% overcurrent on T s fringing flu Power dissipation in winding, P w = V w k cu ρ cu (J rms ) = 3. Watts V w = 1.3 cm 3 (table of core characteristics) k cu = 0.3 (Leitz wire) ρ cu at 100 C (appro. ma. T s ) =.10-8 ohm-m J rms = 4/(.64) = 6.5 A/mm g g Power dissipation in 3F3 ferrite core, P core = V c 1.510-6 f 1.3 (B ac ).5 = 3.3 W d A core B ac A g µ o N I rms = 0.18 mt; assumes H A c Σg g >> H core a A g = (a + g)(d + g) = 1.71 cm ; g = 3mm/4 =.075 mm A c = 1.5 cm (table of core characteristics V c = 13.5 cm 3 (table of core characteristics) f = 100 khz A g Magnetics - 3

Analysis of a Specific Inductor Design (cont.) L = N φ = 310 µh I φ = B ac A c = (0.18 T)(1.510-4 m ) =.610-5 Wb Surface temperature T s = T a + R θsa (P w + P core ) = 104 C R θsa = R θ,rad R θ,conv = 9.8 C/W 60 R θ,rad = (5.1) (0.006) 373 4 - = 0.1 [ C/W] 313 100 4 100 g fringing flu R θ,conv = 1 (1.34)(0.006) 4 0.035 60 = 19.3 [ C/W] g Overcurrent of 5% (I= 5 amp rms) makes T s = 146 C d A core g P w = (3. W)(1.5) = 5 W ; P core = (3.3 W)(1.5).5 = 5.8 W T s = (9.8 C/W)(10.8 W) + 40 C = 146 C a A g Magnetics - 4

Stored Energy Relation - Basis of Inductor Design Input specifications for inductor design Inductance value L. Rated peak current I Rated rms current I rms. Rated dc current (if any) I dc. Operating frequency f. Maimum inductor surface temperature T s and maimum ambient temperature T a. Design consists of the following: Selection of core geometric shape and size Core material Winding conductor geometric shape and size Number of turns in winding Design procedure starting point - stored energy relation [L I] I rms = [N φ] I rms N = k cu A w A cu φ = B A core ; I rms = J rms A cu L I I rms = k cu J rms B A w A core Equation relates input specifications (left-hand side) to needed core and winding parameters (right-hand side) A good design procedure will consists of a systematic, single-pass method of selecting k cu, J rms, B, A w, and A core. Goal: Minimize inductor size, weight, and cost. Magnetics - 5

Core Database - Basic Inductor Design Tool Interactive core database (spreadsheet-based) key to a single pass inductor design procedure. User enters input specifications from converter design requirements. Type of conductor for windings (round wire, Leitz wire, or rectangular wire or foil) must be made so that copper fill factor k cu is known. Spreadsheet calculates capability of all cores in database and displays smallest size core of each type that meets stored energy specification. Also can be designed to calculate (and display as desired) design output parameters including J rms, B, A cu, N, and air-gap length. Multiple iterations of core material and winding conductor choices can be quickly done to aid in selection of most appropriate inductor design. Information on all core types, sizes, and materials must be stored on spreadsheet. Info includes dimensions, A w, A core, surface area of assembled inductor, and loss data for all materials of interest. Pre-stored information combined with user inputs to produce performance data for each core in spreadsheet. Sample of partial output shown below. Core No. Material AP = Aw A core R θ T=60 C P sp @ T=60 C J rms @ T=60 C & Psp B ac @ T=60 C & 100 khz k cu J rms B^ A w A core 8 3F3.1 9.8 C/W 37 cm 4 mw/cm 3 3.3/ k 170 mt cu.015 k cu Magnetics - 6

Details of Interactive Inductor Core Database Calculations User inputs: L, I, I rms, I dc, f, T s, T a, and k cu Stored information (static, independent of converter requirements) Core dimensions, A w, A core, V c, V w, surface area, mean turn length, mean magnetic path length, etc. Quantitative core loss formulas for all materials of interest including approimate temperature dependence. Calculation of core capabilities (stored energy value) 1. Compute converter-required stored energy value: L I I rms.. Compute allowable specific power dissipation P sp = [T s - T a ] /{ R θsa [V c + V w ]}. R θsa = h/a s or calculated interactively using input temperatures and formulas for convective and radiative heat transfer from Heat Sink chapter. 3. Compute allowable flu density P sp = k f b [B ac ] d and current density P sp = k cu ρ cu {J rms }. 4. Compute core capabilities k cu A w A core B J rms Calculation of inductor design parameters. 1. Area of winding conductor A cu = I / J rms.. Calculate skin depth δ in winding. If A cu > δ at the operating frequency, then single round conductor cannot be used for winding. Construct winding using Leitz wire, thin foils, or paralleled small dia. ( δ) round wires. Magnetics - 7

Details of Interactive Core Database Calculations (cont.) 3. Calculate number turns of N in winding: N = k cu A w / A cu. 4. Calculate air-gap length L g. Air-gap length determined on basis that when inductor current equals peak value I, flu density equals peak value B. Formulas for air-gap length different for different core types. Eample for double-e core given in net slide. 5. Calculate maimum inductance L ma that core can support. L ma = N A core B peak / I peak. If L ma > required L value, reduce L ma by removing winding turns. Save on copper costs, weight, and volume. P w can be kept constant by increasing P w,sp Keep flu density B peak constant by adjusting gap length L g. 6. Alternative L ma reduction procedure, increasing the value of L g, keeping everything else constant, is a poor approach. Would not reduce copper weight and volume and thus achieve cost savings. Full capability of core would not be utilized. Magnetics - 8

Setting Double-E Core Air-gap Length Set total airgap length L g so that B peak generated at the peak current I peak. L g = N g g ; N g = number of distributed gaps each of length g. Distributed gaps used to minimize amount of flu fringing into winding and thus causing additional eddy current losses. R m = N I peak A c B peak = R m,core + R m,gap R m,gap = L g µ o A g g L g = N I peak µ o A g A c B peak For a double-e core, A g = (a + L g N g ) (d + L g N g ) fringing flu L g L g A g ad + (a + d) ; << a N g N g Insertion of epression for A g (L g ) into epression for L g (A g ) and solving for L g yields d g A core g L g = a B peak A c d µ o N I peak - a + d d N g a Above epression for L g only valid for double-e core, but similar epressions can be developed for other core shapes. A g Magnetics - 9

Single Pass Inductor Design Procedure Start Enter design inputs into core database Eamine database outputs & select core Yes Neglect skin, proimity effects? No Select wires Iterative selection of conductor type/size. No Estimate L ma. Too large? Yes Remove turns and readjust airgap Finish Magnetics - 30

Inductor Design Eample Assemble design inputs L = 300 microhenries Peak current = 5.6 A, sinewave current, I rms = 4 A Frequency = 100 khz T s = 100 C ; T a = 40 C Stored energy L I I rms = (310-4 )(5.6)(4) = 0.00068 J-m -3 Core material and geometric shape High frequency operation dictates ferrite material. 3F3 material has highest performance factor PF at 100 khz. Double-E core chosen for core shape. Double-E core with a = 1 cm meets requirements. k cu J rms B^ A w A core 0.015 k cu 0.0068 for k cu > 0.3 Database output: R θ = 9.8 C/W and P sp = 37 mw/cm 3 Core flu density B =170 mt from database. No I dc, B peak = 170 mt. Winding parameters. Litz wire used, so k cu = 0.3. J rms = 6 A/mm A cu = (4 A)/(6 A/mm ) = 0.67 mm N = (140 mm )((0.3)/(0.67 mm ) = 63 turns. L ma = (63)(170 mt)(1.510-4 m ) 5.6 A 90 microhenries 10 - L g = (0.17) (1.510-4 ) (1.510 - )(4π10-7 )(63)(5.6) -.510 - (4)(1.510 - ) L g 3 mm L ma L so no adjustment of inductance value is needed. Magnetics - 31

Iterative Inductor Design Procedure Start Assemble design inputs Design winding (k,j, A, N) cu cu Iterative design procedure essentially consists of constructing the core database until a suitable core is found. Compute L I I rms Choose core size using initial values of J and B No Corrected inductancecurrent product greater than L I I rms? Yes Choose core material and shape and conductor type as usual. Use stored energy relation to find an initial area product A w A c and thus an initial core size. Find allowable power dissipation density P sp Select larger core size Use initial values of J rms = -4 A/mm and B ac = 50-100 mt. Find corrected core flu density B ac Find corrected peak core flu density B Find maimum inductance Design airgap length g Set L to design value Use initial core size estimate (value of a in double-e core eample) to find corrected values of J rms and B ac and thus corrected value of k cu J rms B^ A w A core. Compare k cu J rms B^ A w A core with L I I rms and iterate as needed into proper size is found. Magnetics - 3

Simple, Non-optimal Inductor Design Method Start Assemble design inputs and compute required LI I rms Assemble design inputs Choose core geometry and core material based on considerations discussed previously. Compute L I I rms Determine core size using assumed values of J and B Assume J rms = -4 A/mm and B ac = 50-100 mt and use LI I rms = k cu J rms B ac A w A core to find the required area product A w A core and thus the core size. Assumed values of J rms and B ac based on eperience. Design winding (k,j, A, N) cu cu Complete design of inductor as indicated. Set airgap length g to obtain desired inductance L Select larger core size Check power dissipation and surface temperature using assumed values of J rms and B ac. If dissipation or temperature are ecessive, select a larger core size and repeat design steps until dissipation/temperature are acceptable. No Check power dissipation and surface temperature. Ecessive?. Done Yes Procedure is so-called area product method. Useful in situations where only one ore two inductors are to be built and size/weight considerations are secondary to rapid construction and testing.. Magnetics - 33

Analysis of Specific Transformer Design Transformer specifications Wound on double-e core with a = 1 cm using 3F3 ferrite. I pri = 4 A rms, sinusoidal waveform; V pri = 300 V rms. Frequency = 100 khz Turns ratio N pri /N sec = 4 and N pri = 3. Winding window split evenly between primary and secondary and wound with Litz wire. Transformer surface black (E = 0.9) and T a 40 C. Find: core flu density, leakage inductance, and maimum surface temperature T s, and effect of 5% overcurrent on T s. Areas of primary and secondary conductors, A cu,pri and A cu,sec. A w,pri = N pri A cu,pri N sec A cu,sec k ; A cu,pri w,sec = k cu,sec N pri A cu,pri N sec A cu,sec A w,pri + A w,sec = A w = k + cu k cu where k cu,pri = k cu,sec = k cu since we assume primary and secondary are wound with same type of conductor. Equal power dissipation density in primary and secondary gives I pri I sec = A cu,pri A cu,sec = N sec N pri Using above equations yields A cu,pri = A cu,sec = k cu A w N sec k cu A w N pri and Numerical values: A cu,pri = (0.3)(140 mm ) ()(3) = 0.64 mm and A cu,sec = (0.3)(140 mm ) ()(8) =.6 mm Magnetics - 34

Analysis of Specific Transformer Design (cont.) Power dissipation in winding P w = k cu ρ cu (J rms ) V w Top view of bobbin Mean turn length l w radi us = b / w J rms = (4 A)/(0.64 mm ) = (16 A)/(.6 mm ) = 6. A/mm P w = (0.3)(.10-8 ohm-m) (6.10 6 A/m ) (1.310-5 m 3 ) P w = 3.1 watts 1.4 a 1.9 a Flu density and core loss V pri,ma = N pri A c ω B ac = (1.414)(300) = 45 V b = 0.7a w 45 B ac = (3)(1.510-4 m )(π)(10 5 = 0.140 T Hz) P core = (13.5 cm 3 )(1.510-6 )(100 khz) 1.3 (140 mt).5 = 1.9 W l w = ()(1.4a) + ()(1.9a) + π (0.35b w ) = 8 a Surface temperature T s. Assume R θ,sa 9.8 C/W. Same geometry as inductor. T s = (9.8)(3.1 + 1.9) + 40 = 89 C µ o (N pri ) b w l w Leakage inductance L leak = 3 h w l w = 8 a = 8 cm L leak = (4π10-7 )(3) (0.7)(10 - )(810 - ) (3)(10-1 microhenries ) Effect of 5% overcurrent. No change in core flu density. Constant voltage applied to primary keeps flu density constant. P w = (3.1)(1.5) = 4.8 watts ` T s = (9.8)(4.8 + 1.9) + 40 = 106 C Magnetics - 35

Sectioning of Transformer Windings to Reduce Winding Losses Primary Secondary Reduce winding losses by reducing magnetic field (or equivently the mmf) seen by conductors in winding. Not possible in an inductor. MMF N pri I pri = N sec I sec Simple two-section transformer winding situation. 0 b w P S P P 4 S P S P 4 Division into multiple sections reduces MMF and hence eddy current losses. MMF MMF N pri I pri 0 0 N pri I pri N pri I pri 4 N pri I pri 4 Magnetics - 36

Optimization of Solid Conductor Windings Normalized Power Dissipation 500 100 10 m = 0 m = 18 m = 16 m = 14 m = 1 m = 10 m = 9 m = 8 m = 7 m = 6 m = 5 m = 4 m = 3 Nomalized power dissipation = P w F R R dc R dc,h=δ (I rms ) = R dc,h=δ h Conductor height/diameter F l = copper layer factor h w F l h F l = b/b o for rectangular conductors F l = d/d o for round conductors h = effective conductor height π h = d for round conductors 4 m = number of layers δ 1 dc loss Locus of minimum total loss = 1.5 dc loss m = m = 1 0.6 0.1 1 10 d b d o b o N turns per layer = l h w N turns per layer = l h w d o h w b o h/δ Magnetics - 37

Transformer Leakage Inductance Transformer leakage inductance causes overvoltages across power switches at turn-off. Leakage inductance caused by magnetic flu which does not completely link primary and secondary windings. Linear variation of mmf in winding window indicates spatial variation of magnetic flu in the window and thus incomplete flu linkage of primary and secondary windings. N pri I pri H window = H leak = ; 0 < < b h w b w / w Direction and relative magnitude of leakage magnetic field. h w MMF Primary = current into page Secondary = current out of page N pri I pri = N sec I sec N pri I pri H leak = (1 - /b h w ) ; b w / < < b w w L leak (I pri ) = 1 µ o (H leak ) dv V w Volume element dv = h w l w ()d ; l w () equals the length of the conductor turn located at position. Assume a mean turn length l w 8a for double-e core independent of. b w / L leak (I pri ) = () 1 N pri I pri µ o [ ] h h w b w l w d w 0 0 b w µ o (N pri ) l w b w L leak = 3 p h w If winding is split into p+1 sections, with p > 1, leakage inductance is greatly reduced. Magnetics - 38

Volt-Amp (Power) Rating - Basis of Transformer Design Input design specifications Rated rms primary voltage V pri Design proceedure starting point - transformer V-A rating S S = V pri I pri + V sec I sec = V pri I pri Rated rms primary current I pri Turns ratio N pri /N sec Operating frequency f Maimum temperatures T s and T a V pri = N pri dφ dt = N pri A core ω B ac N pri A core ω B ac S = V pri I pri = A cu,pri = k cu A w N pri ; I pri = J rms A cu,pri J rms A cu,pri Design consists of the following: Selection of core geometric shape and size N pri A core ω B ac S = V pri I pri = J rms k cu A w N pri Core material S = V pri I pri = 4.4 k cu f A core A w J rms B ac Winding conductor geometric shape and size Number of turns in primary and secondary windings. Equation relates input specifications (left-hand side) to core and winding parameters (right-hand side). Desired design procedure will consist of a systematic, single-pass method of selecting k cu, A core, A w, J rms, and B ac. Magnetics - 39

Core Database - Basic Transformer Design Tool Interactive core database (spreadsheet-based) key to a single pass tramsformer design procedure. User enters input specifications from converter design requirements. Type of conductor for windings (round wire, Leitz wire, or rectangular wire or foil) must be made so that copper fill factor k cu is known. Spreadsheet calculates capability of all cores in database and displays smallest size core of each type that meets V- I specification. Also can be designed to calculate (and display as desired) design output parameters including J rms, B, A cu,pri, A cu,sec, N pri, N sec, and leakage inductance.. Multiple iterations of core material and winding conductor choices can be quickly done to aid in selection of most appropriate tranformer design. Information on all core types, sizes, and materials must be stored on spreadsheet. Info includes dimensions, A w, A core, surface area of assembled transformer, and loss data for all materials of interest. Pre-stored information combined with user inputs to produce performance data for each core in spreadsheet. Sample of partial output shown below. Core No. Material AP = Aw A c Rθ T=60 C Psp @ T s =100 C Jrms @ T s =100 C & P sp B^rated @ T s =100 C & 100 khz. k cu f J rms B^ AP (f = 100kHz) 8 3F3.1 cm 4 9.8 C/W 37 mw/cm 3 (3.3/ k cu ) R dc R ac A/mm 170 mt.610 3 k cu R dc R ac [V-A] Magnetics - 40

Details of Interactive Transformer Core Database Calculations User inputs: V pri, I pri, turns ratio N dc / N sec, f, T s, T a, and k cu Stored information (static, independent of converter requirements) Core dimensions, A w, A core, V c, V w, surface area, mean turn length, mean magnetic path length, etc. Quantitative core loss formulas for all materials of interest including approimate temperature dependence. Calculation of core capabilities 1. Compute converter-required stored energy value: S = V pri I pri. Compute allowable specific power dissipation P sp = [T s - T a ] /{ R θsa [V c + V w ]}. R θsa = h/a s or calculated interactively using input temperatures and formulas for convective and radiative heat transfer from Heat Sink chapter. 3. Compute allowable flu density P sp = k f b [B ac ] d and current density P sp = k cu ρ cu {J rms }. 4. Compute core capabilities 4.4 f k cu A w A core B ac J rms Calculation transformer parameters. 1. Calculate number of primary turns N pri = V pri /{π f A cpre B ac } and secondary turns N sec = V sec /{π f A cpre B ac }. Calculate winding conductor areas assuming low frequencies or use of Leitz wire A cu,pri = [k cu A w ]/[ N pri ] and A cu,sec = [k cu A w ]/[ N sec ] Magnetics - 41

Details of Interactive Transformer Core Database Calculations (cont.) 3. Calculate winding areas assuming eddy current/proimity effect is important Only solid conductors, round wires or rectangular wires (foils), used.j rms = [{P sp R dc }/{R ac k cu r cu }] 1/ Conductor dimensions must simultaneously satisfy area requirements and requirements of normalized power dissipation versus normalized conductor dimensions. May require change in choice of conductor shape. Most likely will require choice of foils (rectangular shapes). Several iterations may be needed to find proper combinations of dimensions, number of turns per layer, and number of layers and sections. Best illustrated by a specific design eample. 4. Estimate leakage inductance L leak = {µ o {N pri } l w b w }/ {3 p h w } 5. Estimate S ma = 4.4 k cu f A core A w J rms B ac 6. If S ma > S = V pri I pri reduce S ma and save on copper cost, weight, and volume. If N pri w A c B ac > V pri, reduce S ma by reducing N pri and N sec. If J rms A cu, pri > I rms, reduce A cu,pri and A cu, sec. If S > S ma by only a moderate amount (10-0%) and smaller than S ma of net core size, increase S ma of present core size. Increase I rms (and thus winding power dissipation) as needed.temperature T s will increase a modest amount above design limit, but may be preferable to going to larger core size. Magnetics - 4

Single Pass Transformer Design Procedure Start Enter design inputs into core database Eamine database outputs & select core Yes Neglect skin, proimity effects? No Select wires Iterative selection of conductor type/size. No Estimate S ma. Too large? Yes Remove turns Finish Magnetics - 43

Transformer Design Eample Design inputs V pri = 300 V rms ; I rms = 4 A rms Turns ratio n = 4 Operating frequency f = 100 khz T s = 100 C and T a = 40 C V - I rating S = (300 V rms)(4 A rms) = 100 watts Core material, shape, and size. Use 3F3 ferrite because it has largest performance factor at 100 khz. Use double-e core. Relatively easy to fabricate winding. Core volt-amp rating =,600 k cu R dc R ac Use solid rectangular conductor for windings because of high frequency. Thus k cu = 0.6 and R ac /R dc = 1.5. 0.6 Core volt-amp capability =,600 1.5 = 1644 watts. > 100 watt transformer rating. Size is adequate. Using core database, R θ = 9.8 C/W and P sp = 40 mw/cm 3. Flu density and number of primary and secondary turns. From core database, B ac = 170 mt. 300 N pri = (1.510-4 m )(π)(10 5 Hz)(0.17 T) = 6.5 4. Rounded down to 4 to increase fleibility in designing sectionalized transformer winding. N sec = 4 6 = 6. From core database J rms = 3.3 (0.6)(1.5) = 3.5 A/mm. 4 A rms A cu,pri = 3.5 A rms/mm = 1.15 mm A cu,sec = (4)(1.15 mm ) = 4.6 mm Magnetics - 44

Transformer Design Eample (cont.) Primary and secondary conductor areas - proimity effect/eddy currents included. Assume rectangular (foil) conductors with k cu = 0.6 and layer factor F l = 0.9. Iterate to find compatible foil thicknesses and number of winding sections. 1st iteration - assume a single primary section and a single secondary section and each section having single turn per layer. Primary has 4 layers and secondary has 6 layers. Primary layer height h pri = = 1.15 mm (0.9)(0 mm) = 0.064 mm A cu,pri F l h w Normalized primary conductor height F l h pri 0.9 (0.064 mm) φ = d = (0.4 mm) = 0.5 ; δ = 0.4 mm in copper at100 khz and 100 C. Optimum normalized primary conductor height φ = 0.3 so primary winding design is satisfactory. Secondary layer height h sec = = 4.6 mm (0.9)(0 mm) 0.6 mm. A cu,sec F l h w Normalized secondary conductor height φ = F l h sec d = 0.9 (0.6 mm) (0.4 mm) = 1 However a si layer section has an optimum φ = 0.6. A two layer section has an optimum φ = 1. nd iteration needed. nd iteration - sectionalize the windings. Use a secondary of 3 sections, each having two layers, of height h sec = 0.6 mm. Secondary must have single turn per layer. Two turns per layer would require h sec = 0.5 mm and thus φ =. Eamination of normalized power dissipation curves shows no optimum φ =. Magnetics - 45

Transformer Design Eample (cont.) P 6 S 3 P 3 S 3 P 3 S 3 P 6 Three secondary sections requires four primary sections. Turns/ layer Two outer primary sections would have 4/6 = 4 turns each and the inner two sections would have 4/3 = 8 turns each. Need to determine number of turns per layer and hence number of layers per section. h pri No. of Layers φ Optimum φ 1 0.064 mm 8 0.5 0.45 0.18 mm 4 0.5 0.6 4 0.6 mm 1 1 Use four turns per layer. Two interior primary sections have two layers and optimum value of φ. Two outer sections have one layer each and φ not optimum, but only results in slight increase in loss above the minimum. Leakage inductance L leak = (4π10-9 )(4) (8)(0.7)(1) (3)(6) () = 0. µh Sectionalizing increases capacitance between windings and thus lowers the transformer self-resonant frequency. S ma = 1644 watts Rated value of S = 100 watts only marginally smaller than S ma. Little to be gained in reducing S ma to S unless a large number of transformer of this design are to be fabricated. Magnetics - 46

Iterative Transformer Design Procedure Start Assemble design inputs Compute V I pri pri Choose core size using initial values of J and B Find allowable power dissipation density P sp Find corrected core flu density B ac Find corrected current density J rms Design windings A( cu,pri, N pri, A cu,sec, N sec ) Corrected V-I rating greater than V pri I pri? No Select larger core size Find maimum V - I rating Set S ma to desired S Yes Estimate leakage inductance End Iterative design procedure essentially consists of constructing the core database until a suitable core is found. Choose core material and shape and conductor type as usual. Use V - I rating to find an initial area product A w A c and thus an initial core size. Use initial values of J rms = -4 A/mm and B ac = 50-100 mt. Use initial core size estimate (value of a in double-e core eample) to find corrected values of J rms and B ac and thus corrected value of 4.4 f k cu J rms B^ A w A core. Compare 4.4 f k cu J rms B^ A w A core with V pri I pri and iterate as needed into proper size is found. Magnetics - 47

Simple, Non-optimal Transformer Design Method Start Assemble design inputs and compute required V pri I pri Assemble design inputs Choose core geometry and core material based on considerations discussed previously. Compute V pri I pri Determine core size using assumed values of J and B Assume J rms = -4 A/mm and B ac = 50-100 mt and use V pri I pri = 4.4 f k cu J rms B ac A w A core to find the required area product A w A core and thus the core size. Assumed values of J rms and B ac based on eperience. Design winding ( A, cu,pri N, pri A cu,sec, N sec ) Complete design of transformer as indicated. Set S ma to desired S Select larger core size Check power dissipation and surface temperature using assumed values of J rms and B ac. If dissipation or temperature are ecessive, select a larger core size and repeat design steps until dissipation/temperature are acceptable. No Check power dissipation and surface temperature. Ecessive?. Done Yes Procedure is so-called area product method. Useful in situations where only one ore two transformers are to be built and size/weight considerations are secondary to rapid construction and testing.. Magnetics - 48