Transformer and Inductor Design Handbook Fourth Edition © 2011 by Taylor and Francis Group, LLC Transformer and Inductor Design Handbook Fourth Edition Colonel Wm T McLyman Boca Raton London New York CRC Press is an imprint of the Taylor & Francis Group, an informa business © 2011 by Taylor and Francis Group, LLC CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2011 by Taylor and Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S Government works Printed in the United States of America on acid-free paper 10 International Standard Book Number-13: 978-1-4398-3688-0 (Ebook-PDF) This book contains information obtained from authentic and highly regarded sources Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint Except as permitted under U.S Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400 CCC is a not-for-profit organization that provides licenses and registration for a variety of users For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com © 2011 by Taylor and Francis Group, LLC To My Wife, Bonnie © 2011 by Taylor and Francis Group, LLC Contents Foreword  ix Preface  xi Acknowledgements  xiii About the Author  xv Symbols  xvii Chapter Fundamentals of Magnetics 1-1 Chapter Magnetic Materials and Their Characteristics .2-1 Chapter Magnetic Cores .3-1 Chapter Window Utilization, Magnet Wire, and Insulation 4-1 Chapter Transformer Design Trade-Offs .5-1 Chapter Transformer-Inductor Efficiency, Regulation, and Temperature Rise 6-1 Chapter Power Transformer Design .7-1 Chapter DC Inductor Design, Using Gapped Cores .8-1 Chapter DC Inductor Design, Using Powder Cores .9-1 Chapter 10 AC Inductor Design 10-1 Chapter 11 Constant Voltage Transformer (CVT) 11-1 Chapter 12 Three-Phase Transformer Design 12-1 vii © 2011 by Taylor and Francis Group, LLC Contents viii Chapter 13 Flyback Converters, Transformer Design .13-1 Chapter 14 Forward Converter, Transformer Design, and Output Inductor Design .14-1 Chapter 15 Input Filter Design 15-1 Chapter 16 Current Transformer Design 16-1 Chapter 17 Winding Capacitance and Leakage Inductance 17-1 Chapter 18 Quiet Converter Design 18-1 Chapter 19 Rotary Transformer Design 19-1 Chapter 20 Planar Transformers and Inductors 20-1 Chapter 21 Derivations for the Design Equations 21-1 Chapter 22 Autotransformer Design .22-1 Chapter 23 Common-Mode Inductor Design 23-1 Chapter 24 Series Saturable Reactor Design 24-1 Chapter 25 Self-Saturating, Magnetic Amplifiers 25-1 Chapter 26 Designing Inductors for a Given Resistance 26-1 Index  I-1 © 2011 by Taylor and Francis Group, LLC Foreword Colonel McLyman is a well-known author, lecturer and magnetic circuit designer His previous books on transformer and inductor design, magnetic core characteristics and design methods for converter circuits have been widely used by magnetics circuit designers In his 4th edition, Colonel McLyman has combined and updated the information found in his previous books He has also added five new subjects such as autotransformer design, common-mode inductor design, series saturable reactor design, self-saturating magnetic amplifier and designing inductors for a given resistance The author covers magnetic design theory with all of the relevant formulas He has complete information on all of the magnetic materials and core characteristics along with the real world, step-by-step design examples This book is a must for engineers doing magnetic design Whether you are working on high “rel” state of the art design or high volume, or low cost production, this book will help you Thanks Colonel for a well-done, useful book Robert G Noah Application Engineering Manager (Retired) Magnetics, Division of Spang and Company Pittsburgh, Pennsylvania ix © 2011 by Taylor and Francis Group, LLC Preface I have had many requests to update my book Transformer and Inductor Design Handbook, because of the way power electronics has changed in the past few years I have been requested to add and expand on the present Chapters There are now twenty-six Chapters The new Chapters are autotransformer design, common-mode inductor design, series saturable reactor design, self-saturating magnetic amplifier and designing inductors for a given resistance, all with step-by-step design examples This book offers a practical approach with design examples for design engineers and system engineers in the electronics industry, as well as the aerospace industry While there are other books available on electronic transformers, none of them seem to have been written with the user’s viewpoint in mind The material in this book is organized so that the design engineer, student engineer or technician, starting at the beginning of the book and continuing through the end, will gain a comprehensive knowledge of the state of the art in transformer and inductor design The more experienced engineers and system engineers will find this book a useful tool when designing or evaluating transformers and inductors Transformers are to be found in virtually all electronic circuits This book can easily be used to design lightweight, high-frequency aerospace transformers or low-frequency commercial transformers It is, therefore, a design manual The conversion process in power electronics requires the use of transformers, components that frequently are the heaviest and bulkiest item in the conversion circuit Transformer components also have a significant effect on the overall performance and efficiency of the system Accordingly, the design of such transformers has an important influence on overall system weight, power conversion efficiency, and cost Because of the interdependence and interaction of these parameters, judicious trade-offs are necessary to achieve design optimization Manufacturers have for years assigned numeric codes to their cores to indicate their power-handling ability This method assigns to each core a number called the area product, Ap, that is the product of its window area, Wa, and core cross-section area, Ac These numbers are used by core suppliers to summarize dimensional and electrical properties in their catalogs The product of the window area, Wa, and the core area, Ac, gives the area product, Ap, a dimension to the fourth power I have developed a new equation for the power-handling ability of the core, the core geometry, Kg The core geometry, Kg, has a dimension to the fifth power This new equation gives engineers faster and tighter control of their design The core geometry coefficient, Kg, is a relatively new concept, and magnetic core manufacturers are now beginning to put it in their catalogs Because of their significance, the area product, Ap, and the core geometry, Kg, are treated extensively in this handbook A great deal of other information is also presented for the convenience of the designer Much of the material is in tabular form to assist the designer in making the trade-offs best suited for the particular application in a minimum amount of time xi © 2011 by Taylor and Francis Group, LLC xii Preface Designers have used various approaches in arriving at suitable transformer and inductor designs For example, in many cases a rule of thumb used for dealing with current density is that a good working level is 1000 circular mils per ampere This is satisfactory in many instances; however, the wire size used to meet this requirement may produce a heavier and bulkier inductor than desired or required The information presented here will make it possible to avoid the use of this and other rules of thumb, and to develop a more economical and better design The author or the publisher assumes no responsibility for any infringement of patent or other rights of third parties that may result from the use of circuits, systems, or processes described or referred to in this handbook I wish to thank the manufacturers represented in this book for their assistance in supplying technical data Colonel Wm T McLyman © 2011 by Taylor and Francis Group, LLC Acknowledgements In gathering the material for this book, I have been fortunate in having the assistance and cooperation of ­several companies and many colleagues As the author, I wish to express my gratitude to all of them The list is too long to mention them all However, there are some individuals and companies whose contributions have been significant Colleagues that have retired from Magnetics include Robert Noah and Harry Savisky who helped so greatly with the editing of the final draft Other contributions were given by my ­colleagues at Magnetics, Lowell Bosley and his staff with the sending of up-to-date catalogs and sample cores I would like to thank colleagues at Micrometals Corp., Jim Cox and Dale Nicol, and George Orenchak of TSC International I would like to give a special thanks to Richard (Oz) Ozenbaugh of Linear Magnetics Corp for his assistance in the detailed derivations of many of the equations and his efforts in checking all the design examples I would also like to give special thanks to Steve Freeman of Rodon Products, Inc and Charles Barnett of Leightner Electronics, Inc for building and testing all of the magnetic components used in the design examples There are individuals I would like to thank: Dr Vatche Vorperian of Jet Propulsion Laboratory (JPL) for his help in generating and clarifying equations for the Quiet Converter; Jerry Fridenberg of Fridenberg Research, Inc for modeling circuits on his SPICE program; Dr Gene Wester of (JPL) for his inputs and Kit Sum for his assistance in the energy storage equations I also want to thank the late Robert Yahiro for his help and encouragement over the years xiii © 2011 by Taylor and Francis Group, LLC DC Inductor Design, Using Gapped Cores 8-12 As the air gap increases, the flux across the gap fringes more and more Some of the fringing flux strikes the core, perpendicular to the strip or tape, and sets up eddy currents, which cause additional losses in the core If the gap dimension gets too large, the fringing flux will strike the copper winding and produce eddy currents, generating heat, just like an induction heater The fringing flux will jump the gap and produce eddy currents, in both the core and winding, as shown in Figure 8-12 The inductance, L, computed in Equation [8-10], does not include the effect of the fringing flux The value of inductance, L´, corrected for fringing flux is: L´ = 0.4 π N F Ac (10 −8 lg ), [henrys] [8-12] Eddy Currents Core Winding + + + + + + + + + + lg/2 Fringing Flux Magnetic Path Figure 8-12.  Fringing Flux Around the Gap of an Inductor The effect permeability may be calculated from the following Equation: µe = µm  lg  1+  µ  MPL  m [8-13] Where, μm, is the material permeability Inductors Inductors that carry direct current are used frequently in a wide variety of ground, air, and space applications Selection of the best magnetic core for an inductor frequently involves a trial-and-error type of calculation © 2011 by Taylor and Francis Group, LLC Relationship of, Ap, to Inductor’s Energy-Handling Capability 8-13 The author has developed a simplified method of designing optimum, dc carrying inductors with gapped cores This method allows the engineer to select the proper core that will provide correct copper loss, and make allowances for fringing flux, without relying on trial-and-error and the use of the cumbersome Hanna’s curves Rather than discuss the various methods used by transformer designers, the author believes it is more useful to consider typical design problems, and to work out solutions using the approach based upon newly formulated relationships Two gapped core designs will be compared To compare their merits, the first design example will use the core geometry, Kg, and the second design will use the area product, Ap Inductors, designed in this handbook, are banded together with phosphor bronze banding material, or held together with aluminum brackets The use of steel banding material, or brackets that bridge the gap are not recommended, because the use of steel across the gap is called shorting the gap When the gap is shorted, the inductance will increase from the calculated value Relationship of, Ap, to Inductor’s Energy-Handling Capability The energy-handling capability of a core is related to its area product, Ap, by the Equation: Ap = ( ), ( Energy ) 10 Bm J K u [cm ] [8-14] Where: Energy is in watt-seconds Bm is the flux density, teslas J is the current density, amps-per-cm2 Ku is the window utilization factor (See Chapter 4) From the above, it can be seen that factors, such as flux density, Bm, window utilization factor, Ku, (which defines the maximum space that may be used by the copper in the window), and the current density, J, which controls the copper loss, all impact the area product, Ap The energy-handling capability of a core is derived from: © 2011 by Taylor and Francis Group, LLC Energy = LI2 , [watt-seconds] [8-15] DC Inductor Design, Using Gapped Cores 8-14 Relationship of, Kg, to Inductor’s Energy-Handling Capability Inductors, like transformers, are designed for a given temperature rise They can also be designed for a given regulation The regulation and energy handling ability of a core is related to two constants: Energy ) ( α= , [%] K g Ke [8-16] Where, α, is the regulation, %: The constant, Kg, is determined by the core geometry: Kg = Wa Ac2 K u , [cm ] MLT [8-17] The constant, Ke, is determined by the magnetic and electrical operating conditions: ( ) K e = 0.145 Po Bpk 10 −4 [8-18] The peak operating flux density, Bpk, is: Bpk = Bdc + Bac , [teslas] [8-19] From the above, it can be seen that the flux density, Bpk, is the predominant factor governing size The output power, Po, is defined in Figure 8-13 Po( L1) = V( 01) I( 01) L1 Iin Po( L 2) = V( 02) I( 02) L2 Q1 I(01) [8-20] I(02) + Vin CR1 V(01) C1 + CR2 C2 + + Vo − − Figure 8-13.  Defining the Inductor Output Power © 2011 by Taylor and Francis Group, LLC V(02) Io Gapped Inductor Design Example Using the Core Geometry, Kg, Approach Gapped Inductor Design Example Using the Core Geometry, Kg, Approach Step No 1: Design a linear dc inductor with the following specifications: Inductance, L = 0.0025 henrys dc current, Io = 1.5 amps ac current, ΔI = 0.2 amps Output power, Po = 100 watts Regulation, α = 1.0 % Ripple Frequency = 200kHz Operating flux density, Bm = 0.22 tesla Core Material = ferrite Window utilization, Ku = 0.4 10 Temperature rise goal, Tr = 25°C Step No 2: Calculate the peak current, Ipk I pk = I o + ∆I , [amps] I pk = (1.5) + ( 0.2 ) , [ampps] I pk = 1.6, [amps] Step No 3: Calculate the energy-handling capability Energy = Energy = L I pk , [watt-seconds] ( 0.0025)(1.6 )2 , [watt-seconds] Energy = 0.0032, [watt-seconds] Step No 4: Calculate the electrical conditions coefficient, Ke ( K e = 0.145 Po Bm2 10 −4 ) ( K e = 0.145(100 )( 0.22 ) 10 −4 © 2011 by Taylor and Francis Group, LLC K e = 0.0000702 ) 8-15 DC Inductor Design, Using Gapped Cores 8-16 Step No 5: Calculate the core geometry coefficient, Kg Kg = Kg = ( Energy )2 , Ke α [cm ] ( 0.0032 )2 , [cm ] ( 0.0000702 )(1.0 ) K g = 0.146, [cm ] Step No 6: Select an ETD ferrite core from Chapter The data listed is the closest core to the calculated core geometry, Kg Core Number = ETD-39 Magnetic Path Length, MPL = 9.22 cm Core Weight, Wtfe = 60 grams Mean Length Turn, MLT = 8.3 cm Iron Area, Ac = 1.252 cm2 Window Area, Wa = 2.34 cm2 Area Product, Ap = 2.93 cm4 Core Geometry, Kg = 0.177 cm5 Surface Area, At = 69.9 cm2 10 Material, P = 2500μ 11 Millihenrys-per-1000 Turns, AL = 3295 mh 12 Winding Length, G = 2.84 cm Step No 7: Calculate the current density, J, using the area product Equation, Ap J= J= ( ), [amps-per-cm ] ( ) , [amps-perr-cm ] ( Energy ) 10 Bm Ap K u ( 0.0032 ) 10 ( 0.22 )( 2.93)( 0.4 ) J = 248, [amps-per-cm ] Step No 8: Calculate the rms current, Irms I rms = I o2 + ∆I , [amps] I rms = © 2011 by Taylor and Francis Group, LLC (1.5)2 + ( 0.2 )2 , [amps] I rms = 1.51, [amps] Gapped Inductor Design Example Using the Core Geometry, Kg, Approach 8-17 Step No 9: Calculate the required bare wire area, Aw(B) AW ( B ) = I rms , [cm ] J AW ( B ) = (1.51) , [cm ] ( 248 ) AW ( B ) = 0.00609, [cm ] Step No 10: Select a wire from the Wire Table in Chapter If the area is not within 10%, take the next smallest size Also, record the micro-ohms per centimeter AWG = # 19 Bare, AW ( B ) = 0.00653, [cm ] Insulated, AW = 0.00754, [cm ]  µΩ    = 264, [micro-ohm/cm] cm  Step No 11: Calculate the effective window area, Wa(eff), using the window area found in Step A typical value for, S3, is 0.75, as shown in Chapter Wa(eff ) = Wa S3 , [cm ] Wa(eff ) = ( 2.34 )( 0.75) , [cm ] Wa(eff ) = 1.76, [cm ] Step No 12: Calculate the number turns possible, N, using the insulated wire area, Aw, found in Step 10 A typical value for, S2, is 0.6, as shown in Chapter N= Wa (eff ) S2 , [turns] AW N= (1.76 )( 0.60 ) , [turns] ( 0.00754 ) N = 140, [turns] © 2011 by Taylor and Francis Group, LLC DC Inductor Design, Using Gapped Cores 8-18 Step No 13: Calculate the required gap, lg lg = lg = ( 0.4 π N Ac 10 −8 L ) −  MPL  , [cm]  µ  m (1.26 )(140 )2 (1.25)(10 −8 )  9.22  , [cm] −  2500  ( 0.0025) lg = 0.120, [cm] Step No 14: Calculate the equivalent gap in mils mils = cm ( 393.7 ) mils = ( 0.120 )( 393.7 ) mils = 47.2 use 50 Step No 15: Calculate the fringing flux factor, F  2G  lg ln  Ac  lg  F = 1+ F = 1+ ( 0.120 )  ( 2.84 )  ln  0.120  1.25 F = 1.41 Step No 16: Calculate the new number of turns, Nn, by inserting the fringing flux, F Nn = Nn = lg L ( 0.4 π Ac F 10 −8 ) , [turns] ( 0.120 )( 0.0025) , [turns] (1.26 )(1.25)(1.41)(10 −8 ) N n = 116, [turns] Step No 17: Calculate the winding resistance, R L Use the MLT, from Step 6, and the micro-ohm per centimeter from Step 10  µΩ  10 −6 , [ohms] RL = ( MLT ) ( N n )   cm  ( ( ) ) RL = (8.3)(116 )( 264 ) 10 −6 , [ohms] © 2011 by Taylor and Francis Group, LLC RL = 0.254, [ohms] Gapped Inductor Design Example Using the Core Geometry, Kg, Approach 8-19 Step No 18: Calculate the copper loss, Pcu Pcu = I rms RL , [watts] Pcu = (1.51) ( 0.254 ) , [watts] Pcu = 0.579, [watts] Step No 19: Calculate the regulation, α α= Pcu (100 ) , [%] Po α= ( 0.579 ) %] (100 ) , [% (100 ) α = 0.579, [%] Step No 20: Calculate the ac flux density, Bac  ∆I  0.4 π N n F   10 −4  2 , [teslas] Bac =  MPL  lg +   µ m  ( Bac = ) 0.2  −4  10  , [teslas]  9.22  ( 0.120 ) +   2500  ( (1.26 )(116 )(1.41) ) Bac = 0.0167, [teslas] Step No 21: Calculate the watts per kilogram for ferrite, P, material in Chapter Watts per kilogram can be written in milliwatts per gram (n) mW/g = k f ( m ) Bac (1.663) mW/g = ( 0.00004855)( 200000 ) ( 0.0167 )(2.62) mW/g = 0.468 Step No 22: Calculate the core loss, Pfe ( ) Pfe = ( mW/g ) (Wtfe ) 10 −3 , [watts] ( ) Pfe = ( 0.468 ) ( 60 ) 10 −3 , [watts] © 2011 by Taylor and Francis Group, LLC Pfe = 0.0281, [watts] DC Inductor Design, Using Gapped Cores 8-20 Step No 23: Calculate the total loss, copper plus iron, PΣ PΣ = Pfe + Pcu , [watts] PΣ = ( 0.0281) + ( 0.579 ) , [watts] PΣ = 0.607, [watts] Step No 24: Calculate the watt density, ψ The surface area, At, can be found in Step ψ= PΣ , [watts/cm ] At ψ= ( 0.607 ) , [watts/cm ] ( 69.9) ψ = 0.00868, [watts/cm ] Step No 25: Calculate the temperature rise, Tr ( 0.826) Tr = 450 ( ψ ) , [°C] ( 0.826) Tr = 450 ( 0.00868 ) , [°C] Tr = 8.92, [°C] Step No 26: Calculate the peak flux density, Bpk Bpk ∆I   0.4 π N n F  I dc +  10 −4  2 = , [teslas]  MPL  lg +   µ m  Bpk = ( ) (1.26 )(116 )(1.41)(1.6 )(10 −4 ) 9.22  ( 0.127 ) +   2500  , [teslas] Bpk = 0.252, [teslas] Note:  The big advantage in using the core geometry design procedure is that the wire current density is calculated When using the area product design procedure, the current density is an estimate, at best In this next design the same current density will be used as in the core geometry design © 2011 by Taylor and Francis Group, LLC Gapped Inductor Design Example Using the Area Product, Ap, Approach Gapped Inductor Design Example Using the Area Product, Ap, Approach Step No 1: Design a linear dc inductor with the following specifications: Inductance, L = 0.0025 henrys dc current, Io = 1.5 amps ac current, ΔI = 0.2 amps Output power, Po = 100 watts Current Density, J = 250 amps-per-cm2 Ripple Frequency = 200kHz Operating flux density, Bm = 0.22 tesla Core Material = ferrite Window utilization, Ku = 0.4 10 Temperature rise goal, Tr = 25°C Step No 2: Calculate the peak current, Ipk I pk = I o + ∆I , [amps] I pk = (1.5) + ( 0.2 ) , [ampps] I pk = 1.6, [amps] Step No 3: Calculate the energy-handling capability Energy = Energy = L I pk , [watt-seconds] ( 0.0025)(1.6 )2 , [watt-seconds] Energy = 0.0032, [watt-seconds] Step No 4: Calculate the area product, Ap Ap = Ap = © 2011 by Taylor and Francis Group, LLC ( ), ( Energy ) 10 Bm J K u ( ), ( 0.0032 ) 10 ( 0.22 )( 248 )( 0.4 ) Ap = 2.93, [cm ] [cm ] [cm ] 8-21 DC Inductor Design, Using Gapped Cores 8-22 Step No 5: Select an ETD ferrite core from Chapter The data listed is the closest core to the calculated area product, Ap Core Number = ETD-39 Magnetic Path Length, MPL = 9.22 cm Core Weight, Wtfe = 60 grams Mean Length Turn, MLT = 8.3 cm Iron Area, Ac = 1.252 cm2 Window Area, Wa = 2.34 cm2 Area Product, Ap = 2.93 cm4 Core Geometry, Kg = 0.177 cm5 Surface Area, At = 69.9 cm2 10 Material, P = 2500μ 11 Millihenrys-per-1k, AL = 3295 mh 12 Winding Length, G = 2.84 cm Step No 6: Calculate the rms current, Irms I rms = I o2 + ∆I , [amps] I rms = (1.5)2 + ( 0.2 )2 , [amps] I rms = 1.51, [amps] Step No 7: Calculate the required bare wire area, Aw(B) AW ( B ) = I rms , [cm ] J AW ( B ) = (1.51) , [cm ] ( 248 ) AW ( B ) = 0.00609, [cm ] Step No 8: Select a wire from the Wire Table in Chapter If the area is not within 10%, take the next smallest size Also, record micro-ohms per centimeter AWG = # 19 Bare, AW ( B ) = 0.00653, [cm ] Insulated, AW = 0.00754, [cm ] © 2011 by Taylor and Francis Group, LLC  µΩ    = 264, [micro-ohm/cm] cm  Gapped Inductor Design Example Using the Area Product, Ap, Approach 8-23 Step No 9: Calculate the effective window area, Wa(eff) Use the window area found in Step A typical value for, S3, is 0.75, as shown in Chapter Wa(eff ) = Wa S3 , [cm ] Wa(eff ) = ( 2.34 )( 0.75) , [cm ] Wa(eff ) = 1.76, [cm ] Step No 10: Calculate the number turns possible, N, using the insulated wire area, Aw, found in Step A typical value for, S2, is 0.6, as shown in Chapter N= Wa (eff ) S2 , [turns] AW N= (1.76 )( 0.60 ) , [turns] ( 0.00754 ) N = 140, [turns] Step No 11: Calculate the required gap, lg lg = lg = ( 0.4 π N Ac 10 −8 L ) −  MPL  ,  µ  m [cm] (1.26 )(140 )2 (1.25)(10 −8 )  9.22  , [cm] −  2500  ( 0.0025) lg = 0.120, [cm] Step No 12: Calculate the equivalent gap in mils mils = cm ( 393.7 ) mils = ( 0.120 )( 393.7 ) mils = 47.2 use 50 Step No 13: Calculate the fringing flux factor, F  2G  lg ln Ac  lg  F = 1+ F = 1+ © 2011 by Taylor and Francis Group, LLC ( 0.120 )  ( 2.84 )  ln F = 1.41 1.25  0.120  DC Inductor Design, Using Gapped Cores 8-24 Step No 14: Calculate the new number of turns, Nn, by inserting the fringing flux, F Nn = Nn = lg L ( 0.4 π Ac F 10 −8 ) , [turns] ( 0.120 )( 0.0025) , [turns] (1.26 )(1.25)(1.41)(10 −8 ) N n = 116, [turns] Step No 15: Calculate the winding resistance, R L Use the, MLT, from Step 5, and the micro-ohm per centimeter, from Step 10  µΩ  10 −6 , [ohms] RL = ( MLT ) ( N n )   cm  ( ( ) ) RL = (8.3)(116 )( 264 ) 10 −6 , [ohms] RL = 0.254, [ohms] Step No 16: Calculate the copper loss, Pcu Pcu = I rms RL , [watts] Pcu = (1.51) ( 0.254 ) , [watts] Pcu = 0.579, [watts] Step No 17: Calculate the regulation, α α= Pcu (100 ) , [%] Po α= ( 0.579 ) %] (100 ) , [% (100 ) α = 0.579, [%] Step No 18: Calculate the ac flux density, Bac  ∆I  0.4 π N n F   10 −4  2 , [teslas] Bac =  MPL  lg +   µ m  ( Bac = © 2011 by Taylor and Francis Group, LLC ) 0.2  −4  10  , [teslas] 9.22  ( 0.120 ) +   2500  (1.26 )(116 )(1.41) Bac = 0.0167, [teslas] ( ) Gapped Inductor Design Example Using the Area Product, Ap, Approach 8-25 Step No 19: Calculate the watts per kilogram for ferrite, P, material in Chapter Watts per ­k ilogram can be written in milliwatts per gram (n) mW/g = k f ( m ) Bac (1.663) mW/g = ( 0.00004855)( 200000 ) ( 0.0167 )(2.62) mW/g = 0.468 Step No 20: Calculate the core loss, Pfe ( ) Pfe = ( mW/g ) (Wtfe ) 10 −3 , [watts] ( ) Pfe = ( 0.468 ) ( 60 ) 10 −3 , [watts] Pfe = 0.0281, [watts] Step No 21: Calculate the total loss copper plus iron, PΣ PΣ = Pfe + Pcu , [watts] PΣ = ( 0.0281) + ( 0.579 ) , [watts] PΣ = 0.607, [watts] Step No 22: Calculate the watt density, ψ The surface area, At, can be found in Step ψ= PΣ , [watts/cm ] At ψ= ( 0.607 ) , [watts/cm ] ( 69.9) ψ = 0.00868, [watts/cm ] Step No 23: Calculate the temperature rise, Tr ( 0.826) Tr = 450 ( ψ ) ( 0.826) Tr = 450 ( 0.00868 ) Tr = 8.92, [°C] © 2011 by Taylor and Francis Group, LLC , [°C] , [°C] DC Inductor Design, Using Gapped Cores 8-26 Step No 24: Calculate the peak flux density, Bpk Bpk ∆I   0.4 π N n F  I dc +  10 −4  2 = , [teslas]  MPL  lg +   µ m  Bpk = ( ) (1.26 )(116 )(1.41)(1.6 )(10 −4 ) ( 0.127 ) +  9.22   2500  , [teslas] Bpk = 0.252, [teslas] Step No 25: Calculate the effective permeability, μe Knowing the effective permeability, the ETD-39 ferrite core can be ordered with a built-in gap µe = µe = µm  lg  µ 1+   MPL  m ( 2500 )  ( 0.120 )  1+  ( 2500 )  9.22  µ e = 74.5 use 75 Step No 26: Calculate the window utilization, Ku Ku = N n Aw( B) Wa Ku = (116 )( 0.00653) ( 2.34 ) K u = 0.324 © 2011 by Taylor and Francis Group, LLC ... gram (n) mW/g = k f ( m ) Bac (1 .4 1) mW/g = ( 0.000788 )( 20000 ) ( 0.0215 )( 2 .2 4) mW/g = 0.168 Step No 20: Calculate the core loss, Pfe ( ) = ( 0.168 ) ( 34.9 ) (1 0 ) , Pfe = ( mW/g ) (Wtfe ) 10... [oersteds] MPL H= (1 .26 )( 256 )( 1 .6 ) , [oersteds] (8 .9 5) H = 57.7, [oersteds] Step No 25: Calculate the window utilization, Ku Ku = Ku = N L( new ) Aw( B) # 20 Wa (( 256 )( 0.0051 9)) ( 3.94 ) K u = 0.337... Wa (eff ) S2 , [turns] AW N= ( 2.96 )( 0.60 ) , [turns] ( 0.00606 ) N = 293, [turns] Step No 13: Calculate the required core permeability, μ µ∆ = µ∆ = ( ) Bm ( MPL ) 10 0.4 πWa J K u ( 0.30 )( 8 .95)