Fundamentals of Power Electronics Chapter 14: Inductor design1Chapter 14 Inductor Design14.1Filter inductor design constraints14.2A step-by-step design procedure14.3Multiple-winding magnetics design using theKg method14.4 Examples14.5Summary of key pointsFundamentals of Power Electronics Chapter 14: Inductor design214.1 Filter inductor design constraintsPcu = Irms2 RObjective:Design inductor having a given inductance L,which carries worst-case current Imax without saturating,and which has a given winding resistance R, or, equivalently,exhibits a worst-case copper loss ofLRi(t)+–Li(t)i(t)t0DTsTsI∆iLExample: filter inductor in CCM buck converterFundamentals of Power Electronics Chapter 14: Inductor design3Assumed filter inductor geometrySolve magnetic circuit:Air gapreluctanceRgnturnsi(t)ΦCore reluctance Rc+v(t)–+–ni(t)Φ(t)RcRgFc+ –Rc=lcµcAcRg=lgµ0Acni = Φ Rc+ Rgni ≈ΦRgUsually Rc < Rg and henceFundamentals of Power Electronics Chapter 14: Inductor design414.1.1 Constraint: maximum flux densityGiven a peak winding current Imax, it is desired to operate the core fluxdensity at a peak value Bmax. The value of Bmax is chosen to be lessthan the worst-case saturation flux density Bsat of the core material.From solution of magnetic circuit:Let I = Imax and B = Bmax :This is constraint #1. The turns ratio n and air gap length lg areunknown.ni = BAcRgnImax= BmaxAcRg= Bmaxlgµ0Fundamentals of Power Electronics Chapter 14: Inductor design514.1.2 Constraint: inductanceMust obtain specified inductance L. We know that the inductance isThis is constraint #2. The turns ratio n, core area Ac, and air gap lengthlg are unknown.L=n2Rg=µ0Acn2lgFundamentals of Power Electronics Chapter 14: Inductor design614.1.3 Constraint: winding areacore windowarea WAwire bare areaAWcoreWire must fit through core window (i.e., hole in center of core)nAWTotal area ofcopper in window:KuWAArea available for windingconductors:Third design constraint:KuWA≥ nAWFundamentals of Power Electronics Chapter 14: Inductor design7The window utilization factor Kualso called the “fill factor”Ku is the fraction of the core window area that is filled by copperMechanisms that cause Ku to be less than 1:• Round wire does not pack perfectly, which reduces Ku by afactor of 0.7 to 0.55 depending on winding technique•Insulation reduces Ku by a factor of 0.95 to 0.65, depending onwire size and type of insulation•Bobbin uses some window area•Additional insulation may be required between windingsTypical values of Ku :0.5 for simple low-voltage inductor0.25 to 0.3 for off-line transformer0.05 to 0.2 for high-voltage transformer (multiple kV)0.65 for low-voltage foil-winding inductorFundamentals of Power Electronics Chapter 14: Inductor design814.1.4 Winding resistanceThe resistance of the winding iswhere  is the resistivity of the conductor material, lb is the length ofthe wire, and AW is the wire bare area. The resistivity of copper atroom temperature is 1.72410–6 -cm. The length of the wire comprisingan n-turn winding can be expressed aswhere (MLT) is the mean-length-per-turn of the winding. The mean-length-per-turn is a function of the core geometry. The aboveequations can be combined to obtain the fourth constraint:R = ρn (MLT)AWR= ρlbAWlb= n(MLT)Fundamentals of Power Electronics Chapter 14: Inductor design914.1.5 The core geometrical constant KgThe four constraints:R = ρn (MLT)AWKuWA≥ nAWThese equations involve the quantitiesAc, WA, and MLT, which are functions of the core geometry,Imax, Bmax , µ0, L, Ku, R, and , which are given specifications orother known quantities, andn, lg, and AW, which are unknowns.Eliminate the three unknowns, leading to a single equation involvingthe remaining quantities.nImax= BmaxAcRg= Bmaxlgµ0L=n2Rg=µ0Acn2lgFundamentals of Power Electronics Chapter 14: Inductor design10Core geometrical constant KgAc2WA(MLT)≥ρL2Imax2Bmax2RKuElimination of n, lg, and AW leads to• Right-hand side: specifications or other known quantities• Left-hand side: function of only core geometrySo we must choose a core whose geometry satisfies the aboveequation.The core geometrical constant Kg is defined asKg=Ac2WA(MLT)Fundamentals of Power Electronics Chapter 14: Inductor design11DiscussionKg=Ac2WA(MLT)≥ρL2Imax2Bmax2RKuKg is a figure-of-merit that describes the effective electrical size of magneticcores, in applications where the following quantities are specified:• Copper loss•Maximum flux densityHow specifications affect the core size:A smaller core can be used by increasingBmax  use core material having higher BsatR  allow more copper lossHow the core geometry affects electrical capabilities: A larger Kg can be obtained by increase ofAc  more iron core material, orWA  larger window and more copperFundamentals of Power Electronics Chapter 14: Inductor design1214.2 A step-by-step procedureTh e following quantities are specified, using the units noted:Wire resistivity  (-cm)Peak winding current Imax(A )Inductance L (H)Winding resistance R ()Winding fill factor KuCore maximum flux density Bmax(T)The core dimensi ons are expressed in cm:Core cross-sectional area Ac(cm2)Core window area WA(cm2)Mean length per turn MLT (cm)Th e use of centimeters rather than meters requires that appropriatefactors be added to the design equat ion s.Fundamentals of Power Electronics Chapter 14: Inductor design13Determine core sizeKg≥ρL2Imax2Bmax2RKu108(cm5)Choose a core which is large enough to satisfy this inequality(see Appendix D for magnetics design tables).Note the values of Ac, WA, and MLT for this core.Fundamentals of Power Electronics Chapter 14: Inductor design14Determine air gap lengthwith Ac expressed in cm2. µ0 = 410–7 H/m.The air gap length is given in meters.The value expressed above is approximate, and neglects fringing fluxand other nonidealities.lg=µ0LImax2Bmax2Ac104(m)Fundamentals of Power Electronics Chapter 14: Inductor design15ALCore manufacturers sell gapped cores. Rather than specifying the airgap length, the equivalent quantity AL is used.AL is equal to the inductance, in mH, obtained with a winding of 1000turns.When AL is specified, it is the core manufacturer’s responsibility toobtain the correct gap length.The required AL is given by:AL=10Bmax2Ac2LImax2(mH/1000 turns)L = ALn210–9(Henries)Units:Ac cm2,L Henries,Bmax Tesla.Fundamentals of Power Electronics Chapter 14: