Fundamentals of Power Electronics Chapter 15: Transformer design1Chapter 15 Transformer DesignSome more advanced design issues, not considered in previouschapter:• Inclusion of core loss• Selection of operating fluxdensity to optimize total loss• Multiple winding design: as inthe coupled-inductor case,allocate the available windowarea among several windings•A transformer designprocedure•How switching frequencyaffects transformer sizen1 : n2: nkR1R2Rk+v1(t)–+v2(t)–+vk(t)–i1(t)i2(t)ik(t)Fundamentals of Power Electronics Chapter 15: Transformer design2Chapter 15 Transformer Design15.1 Transformer design: Basic constraints15.2 A step-by-step transformer design procedure15.3 Examples15.4 AC inductor design15.5 SummaryFundamentals of Power Electronics Chapter 15: Transformer design315.1 Transformer Design:Basic ConstraintsCore lossTypical value of  for ferrite materials: 2.6 or 2.7B is the peak value of the ac component of B(t), i.e., the peak ac fluxdensitySo increasing B causes core loss to increase rapidlyThis is the first constraintPfe= Kfe(∆B)βAclmFundamentals of Power Electronics Chapter 15: Transformer design4Flux densityConstraint #2Flux density B(t) is related to theapplied winding voltage accordingto Faraday’s Law. Denote the volt-seconds applied to the primarywinding during the positive portionof v1(t) as 1:λ1= v1(t)dtt1t2This causes the flux to change fromits negative peak to its positive peak.From Faraday’s law, the peak valueof the ac component of flux density isTo attain a given flux density,the primary turns should bechosen according toarea λ1v1(t)t1t2t∆B =λ12n1Acn1=λ12∆BAcFundamentals of Power Electronics Chapter 15: Transformer design5Copper lossConstraint #3• Allocate window area between windings in optimum manner, asdescribed in previous section• Total copper loss is then equal toPcu=ρ(MLT)n12Itot2WAKuItot=njn1IjΣj =1kwithEliminate n1, using result of previous slide:Note that copper loss decreases rapidly as B is increasedPcu=ρλ12Itot24Ku(MLT)WAAc21∆B2Fundamentals of Power Electronics Chapter 15: Transformer design6Total power loss4. Ptot = Pcu + PfePtot= Pfe+ PcuThere is a value of Bthat minimizes the totalpower lossPcu=ρλ12Itot24Ku(MLT)WAAc21∆B2Pfe= Kfe(∆B)βAclm∆BPowerlossPtotCopper loss PcuCore loss PfeOptimum ∆BFundamentals of Power Electronics Chapter 15: Transformer design75. Find optimum flux density BPtot= Pfe+ PcuGiven thatThen, at the B that minimizes Ptot, we can writeNote: optimum does not necessarily occur where Pfe = Pcu. Rather, itoccurs wheredPtotd(∆B)=dPfed(∆B)+dPcud(∆B)=0dPfed(∆B)=–dPcud(∆B)Fundamentals of Power Electronics Chapter 15: Transformer design8Take derivatives of core and copper lossNow, substitute intoand solve for B:Optimum B for agiven core andapplicationPcu=ρλ12Itot24Ku(MLT)WAAc21∆B2Pfe= Kfe(∆B)βAclmdPfed(∆B)= βKfe(∆B)β –1AclmdPcud(∆B)=–2ρλ12Itot24Ku(MLT)WAAc2(∆B)–3dPfed(∆B)=–dPcud(∆B)∆B =ρλ12Itot22Ku(MLT)WAAc3lm1βKfe1β +2Fundamentals of Power Electronics Chapter 15: Transformer design9Total lossSubstitute optimum B into expressions for Pcu and Pfe. The total loss is:Rearrange as follows:Left side: terms depend on coregeometryRight side: terms depend onspecifications of the applicationPtot= AclmKfe2β +2ρλ12Itot24Ku(MLT)WAAc2ββ +2β2–ββ +2+β22β +2WAAc2(β –1)/β(MLT)lm2/ββ2–ββ +2+β22β +2–β +2β=ρλ12Itot2Kfe2/β4KuPtotβ +2/βFundamentals of Power Electronics Chapter 15: Transformer design10The core geometrical constant KgfeDefineDesign procedure: select a core that satisfiesAppendix D lists the values of Kgfe for common ferrite coresKgfe is similar to the Kg geometrical constant used in Chapter 14:• Kg is used when Bmax is specified• Kgfe is used when B is to be chosen to minimize total lossKgfe=WAAc2(β –1)/β(MLT)lm2/ββ2–ββ +2+β22β +2–β +2βKgfe≥ρλ12Itot2Kfe2/β4KuPtotβ +2/βFundamentals of Power Electronics Chapter 15: Transformer design1115.2 Step-by-steptransformer design procedureTh e following quantiti es are specifi ed, u sing t he u nits n oted:Wire effective resistivity  (-cm)Total rms winding current, ref to pri Itot(A )Desired turns r atios n2/n1, n3/n1, etc.Applied pri volt-sec 1(V - sec)Allowed total power dissipation Ptot(W)Winding fill factor KuCore loss exponent Core loss coefficient Kfe(W/cm3T)Other qu antities and their dimensions:Core cross-sectional ar ea Ac(cm2)Core window area WA(cm2)Mean length per turn MLT (cm)Magnetic path length le (cm)Wire areas Aw1, … (cm2)Peak ac f lux density B (T)Fundamentals of Power Electronics Chapter 15: Transformer design12Procedure1. Determine core sizeSelect a core from Appendix D that satisfies this inequality.It may be possible to reduce the core size by choosing a core materialthat has lower loss, i.e., lower Kfe.Kgfe≥ρλ12Itot2Kfe2/β4KuPtotβ +2/β108Fundamentals of Power Electronics Chapter 15: Transformer design132. Evaluate peak ac flux densityAt this point, one should check whether the saturation flux density isexceeded. If the core operates with a flux dc bias Bdc, then B + Bdcshould be less than the saturation flux density Bsat.If the core will saturate, then there are two choices:• Specify B using the Kg method of Chapter 14, or• Choose a core material having greater core loss, then repeatsteps 1 and 2∆B =108ρλ12Itot22Ku(MLT)WAAc3lm1βKfe1β +2Fundamentals of Power Electronics Chapter 15: Transformer design143. and 4. Evaluate turnsPrimary turns:Choose secondary turns according todesired turns ratios:n2= n1n2n1n3= n1n3n1n1=λ12∆BAc104Fundamentals of Power Electronics Chapter 15: Transformer design155. and 6. Choose wire sizesα1=n1I1n1Itotα2=n2I2n1Itotαk=nkIkn1ItotFraction of window areaassigned to each winding:Choose wire sizes accordingto:Aw1≤α1KuWAn1Aw2≤α2KuWAn2Fundamentals of Power Electronics Chapter 15: Transformer design16Check: computed transformer modeliM, pk=λ12LMR1=ρn1(MLT)Aw1R2=ρn2(MLT)Aw2Predicted magnetizinginductance, referred to primary:Peak magnetizing current:Predicted winding resistances:n1 : n2: nkR1R2Rki1(t)i2(t)ik(t)LMiM(t)LM=µn12AclmFundamentals of Power Electronics Chapter 15: Transformer design1715.4.1 Example 1: Single-output isolatedCuk converter100 W fs = 200 kHzD = 0.5 n = 5Ku = 0.5 Allow Ptot = 0.25 WUse a ferrite pot core, with Magnetics Inc. P material. Lossparameters at 200 kHz areKfe =