Chapter 15 Transformer Design
Chapter 15 Transformer DesignSome more advanced design issues, not considered in previouschapter:n1 : n2 Inclusion of core loss i(t)1 Selection of operating fluxv2(t)v1(t)density to optimize total loss –Multiple winding design: as inthe coupled-inductor case,allocate the available windowarea among several windings–R1R2 A transformer designprocedureik (t)vk(t)How switching frequencyaffects transformer sizeFundamentals of Power Electronicsi2(t)–: nk1RkChapter 15: Transformer design
Chapter 15 Transformer Design15.1Transformer design: Basic constraints15.2A step-by-step transformer design procedure15.3Examples15.4AC inductor design15.5SummaryFundamentals of Power Electronics2Chapter 15: Transformer design
15.1 Transformer Design:Basic ConstraintsCore lossP fe K fe( B) β A c l mTypical 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 constraintFundamentals of Power Electronics3Chapter 15: Transformer design
Flux densityConstraint #2v1(t)Flux density B(t) is related to theapplied winding voltage accordingto Faraday’s Law. Denote the voltseconds applied to the primarywinding during the positive portionof v1(t) as 1:area λ1t1t2t2λ1 v1(t)dtt1This 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 is B λ12n 1A cFundamentals of Power ElectronicsTo attain a given flux density,the primary turns should bechosen according ton1 4λ12 BA cChapter 15: Transformer designt
Copper lossConstraint #3 Allocate window area between windings in optimum manner, asdescribed in previous section Total copper loss is then equal toPcu with2 21 totρ(MLT)n IWAK ukI tot Σj 1njn1 I jEliminate n1, using result of previous slide:ρ λ 21 I 2totPcu 4K u(MLT )W A A 2c1 B2Note that copper loss decreases rapidly as B is increasedFundamentals of Power Electronics5Chapter 15: Transformer design
Total power loss4. Ptot Pcu PfePowerlossCoPtot Pfe Pcufess PCoss P cr loppePtotre loThere is a value of Bthat minimizes the totalpower lossuP fe K fe( B) β A c l mOptimum Bρ λ 21 I 2totPcu 4K uFundamentals of Power Electronics(MLT )W A A 2c1 B6 B2Chapter 15: Transformer design
5. Find optimum flux density BGiven thatPtot Pfe PcuThen, at the B that minimizes Ptot, we can writedP fedPcudPtot 0d( B) d( B) d( B)Note: optimum does not necessarily occur where Pfe Pcu. Rather, itoccurs wheredP fedPcu –d( B)d( B)Fundamentals of Power Electronics7Chapter 15: Transformer design
Take derivatives of core and copper lossP fe K fe( B) A c l mβdP fe βK fe ( B)d( B)β–1Now, substitute into B ρλ 21 I 2tot2K uFundamentals of Power ElectronicsAclmρ λ 21 I 2totPcu 4K u(MLT )W A A 2cρλ 21 I 2totdPcu –2d( B)4K udP fedPcu –d( B)d( B)1β 2(MLT )13W A A c l m βK fe81 B2(MLT)–3( B)W A A 2cand solve for B:Optimum B for agiven core andapplicationChapter 15: Transformer design
Total lossSubstitute optimum B into expressions for Pcu and Pfe. The total loss is:Ptot A c l m K feρλ 21 I 2tot2β 24K u(MLT )W A A 2cββ 2β2–ββ 2 β22β 2Rearrange as follows:WA Ac2(β – 1)/β(MLT )l m2/ββ2–ββ 2– β2Left side: terms depend on coregeometryFundamentals of Power Electronics2β 2β 2β2/β ρλ 21 I 2tot K fe4K u Ptotβ 2 /βRight side: terms depend onspecifications of the application9Chapter 15: Transformer design
The core geometrical constant KgfeDefineK gfe WA Ac2(β – 1)/β(MLT)l m2/ββ2–ββ 2– β22β 2β 2βDesign procedure: select a core that satisfies2/βK gfe ρλ 21 I 2tot K fe4K u Ptotβ 2 /βAppendix 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 lossFundamentals of Power Electronics10Chapter 15: Transformer design
15.2 Step-by-steptransformer design procedureThe following quantities are specified, using the units noted:Wire effective resistivity ( -cm)Total rms winding current, ref to priItot(A)Desired turns ratiosn2/n1, n3/n1, etc.Applied pri volt-sec 1(V-sec)(W)Allowed total power dissipationPtotWinding fill factorKuCore loss exponent Core loss coefficientKfe(W/cm3T )Other quantities and their dimensions:Core cross-sectional areaAcCore window areaWAMean length per turnMLTMagnetic path lengthleWire areasAw1, Peak ac flux density BFundamentals of Power Electronics11(cm2)(cm2)(cm)(cm)(cm2)(T)Chapter 15: Transformer design
Procedure1.Determine core size2/βK gfe ρλ 21 I 2tot K fe4K u Ptotβ 2 /β10 8Select 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.Fundamentals of Power Electronics12Chapter 15: Transformer design
2.Evaluate peak ac flux densityρλ 21 I 2tot (MLT )1 B 102K u W A A 3c l m βK fe1β 28At 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 2Fundamentals of Power Electronics13Chapter 15: Transformer design
3. and 4.Evaluate turnsPrimary turns:n1 λ110 42 BA cChoose secondary turns according todesired turns ratios:Fundamentals of Power Electronicsn2 n1n2n1n3 n1n3n114Chapter 15: Transformer design
5. and 6.Choose wire sizesFraction of window areaassigned to each winding:Choose wire sizes accordingto:n 1I 1α1 n 1I totn 2I 2α2 n 1I totα 1K uWAn1α 2K uWAA w2 n2A w1 n kI kαk n 1I totFundamentals of Power Electronics15Chapter 15: Transformer design
Check: computed transformer modelPredicted magnetizinginductance, referred to primary:n1 : n2i1(t)µn 21 A cLM lmiM (t)i2(t)LMPeak magnetizing current:λ1i M, pk 2L MR1R2Predicted winding resistances:ik(t)ρn 1(MLT)A w1ρn (MLT)R2 2A w2R1 Fundamentals of Power Electronics: nk16RkChapter 15: Transformer design
15.4.1Example 1: Single-output isolatedCuk converterIg4A vC1(t) –– vC2(t) 25 V –20 A –VgI v1(t)v2(t) V5V––i1(t)n:1i2(t)100 Wfs 200 kHzD 0.5n 5Ku 0.5Allow Ptot 0.25 WUse a ferrite pot core, with Magnetics Inc. P material. Lossparameters at 200 kHz areKfe 24.7Fundamentals of Power Electronics 2.617Chapter 15: Transformer design
Waveformsv1(t)VC1Applied primary voltseconds:Area λ1D'TsDTsi1(t)λ 1 DTsVc1 (0.5) (5 µsec ) (25 V) 62.5 V–µsec– nVC2Applied primary rmscurrent:I/nI1 – Igi2(t)2 D' I g2 4AApplied secondary rmscurrent:I 2 nI 1 20 AITotal rms windingcurrent:I tot I 1 1n I 2 8 A– nIgFundamentals of Power ElectronicsD nI18Chapter 15: Transformer design
Choose core size(1.724 10 – 6)(62.5 10 – 6) 2(8) 2(24.7) 2/2.68K gfe 104 (0.5) (0.25) 4.6/2.6 0.00295Pot core data of Appendix D lists 2213 pot core withKgfe 0.0049Next smaller pot core is not large enough.Fundamentals of Power Electronics19Chapter 15: Transformer design
Evaluate peak ac flux density1/4.6(1.724 10 – 6 )(62.5 10 – 6 ) 2(8) 2(4.42)1 B 1032 (0.5)(0.297)(0.635) (3.15) (2.6)(24.7)8 0.0858 TeslaThis is much less than the saturation flux density of approximately0.35 T. Values of B in the vicinity of 0.1 T are typical for ferritedesigns that operate at frequencies in the vicinity of 100 kHz.Fundamentals of Power Electronics20Chapter 15: Transformer design
Evaluate turns–6(62.5 10)n 1 10 42(0.0858)(0.635) 5.74 turnsn1n 2 n 1.15 turnsIn practice, we might selectn1 5andn2 1This would lead to a slightly higher flux density and slightly higherloss.Fundamentals of Power Electronics21Chapter 15: Transformer design
Determine wire sizesFraction of window area allocated to each winding:α1 4A8Aα2 15(Since, in this example, the ratio ofwinding rms currents is equal to theturns ratio, equal areas areallocated to each winding) 0.520 A8A 0.5From wire table,Appendix D:Wire areas:(0.5)(0.5)(0.297) 14.8 10 – 3 cm 2(5)(0.5)(0.5)(0.297)A w2 74.2 10 – 3 cm 2(1)A w1 Fundamentals of Power Electronics22AWG #16AWG #9Chapter 15: Transformer design
Wire sizes: discussionPrimary5 turns #16 AWGSecondary1 turn #9 AWG Very large conductors! One turn of #9 AWG is not a practical solutionSome alternatives Use foil windings Use Litz wire or parallel strands of wireFundamentals of Power Electronics23Chapter 15: Transformer design
Effect of switching frequency on transformer sizefor this P-material Cuk converter Bmax , TeslaPot core size36220.02025 kHz50 kHz100 kHz200 kHz250 kHz400 kHz500 kHz1000 kHzSwitching frequency As switching frequency isincreased from 25 kHz to250 kHz, core size isdramatically reducedFundamentals of Power Electronics As switching frequency isincreased from 400 kHz to1 MHz, core sizeincreases24Chapter 15: Transformer design
15.3.2Example 2Multiple-Output Full-Bridge Buck ConverterQ1D1Q3T1D3n1 :I5V: n2160 V –100 A D5 Vgi2a(t)i1(t) v1(t)5V–Q2D2Q4D6i2b(t): n2D4I15V: n3i3a(t)150 kHzTransformer frequency75 kHzTurns ratio110:5:15Optimize transformer atD 0.75Fundamentals of Power Electronics15 A D7Switching frequency–15 VD8i2b(t)–: n325Chapter 15: Transformer design
Other transformer design detailsUse Magnetics, Inc. ferrite P material. Loss parameters at 75 kHz:Kfe 7.6 W/T cm3 2.6Use E-E core shapeAssume fill factor ofKu 0.25(reduced fill factor accounts for added insulation requiredin multiple-output off-line application)Allow transformer total power loss ofPtot 4 W(approximately 0.5% of total output power)Use copper wire, with 1.724·10–6 -cmFundamentals of Power Electronics26Chapter 15: Transformer design
Applied transformer waveformsv1(t)T1D3n1 :: n2Area λ1 Vg DTsVgi2a(t)0D5 – Vgi1(t) v (t)1nn2I 5V 3 I 15Vn1n1i1(t)–D40D6i2b(t)0: n2: n3i3a(t)D7–i2a(t)nn2I 5V 3 I 15Vn1n1I5V0.5I5V0D8i2b(t)i3a(t)I15V0.5I15V0: n30Fundamentals of Power ElectronicsDTs27TsTs DTs 2TstChapter 15: Transformer design
Applied primary volt-secondsv1(t)VgArea λ1 Vg DTs00– Vgλ 1 DTsVg (0.75) (6.67 µsec ) (160 V) 800 V–µsecFundamentals of Power Electronics28Chapter 15: Transformer design
Applied primary rms currenti1(t)nn2I 5V 3 I 15Vn1n10–nn2I 5V 3 I 15Vn1n1n2n3I 1 n I 5V n I 15V11Fundamentals of Power Electronics29D 5.7 AChapter 15: Transformer design
Applied rms current, secondary windingsi2a(t)I5V0.5I5V0i3a(t)I15V0.5I15V00DTsTsTs DTs 2TstI 2 12 I 5V 1 D 66.1 AI 3 12 I 15V 1 D 9.9 AFundamentals of Power Electronics30Chapter 15: Transformer design
ItotRMS currents, summed over all windings and referred to primaryI tot Σall 5windingsnjn2n3n1 I j I 1 2 n1 I 2 2 n1 I 3 5.7 A 5 66.1 A 15 9.9 A110110 14.4 AFundamentals of Power Electronics31Chapter 15: Transformer design
Select core size(1.724 10 – 6)(800 10 – 6) 2(14.4) 2(7.6) 2/2.68K gfe 104 (0.25) (4) 4.6/2.6 0.00937From Appendix DAFundamentals of Power Electronics32Chapter 15: Transformer design
Evaluate ac flux density BEq. (15.20):2 2ρλ81I tot (MLT)1Bmax 102K u WAA 3c l m βK fe1β 2Plug in values:1/4.6(1.724 10 – 6 )(800 10 – 6 ) 2(14.4) 2(8.5)1 B 1032(0.25)(1.1)(1.27) (7.7) (2.6)(7.6)8 0.23 TeslaThis is less than the saturation flux density of approximately 0.35 TFundamentals of Power Electronics33Chapter 15: Transformer design
Evaluate turnsChoose n1 according to Eq. (15.21):n1 λ110 42 BA cTo obtain desired turns ratioof(800 10 – 6)n 1 102(0.23)(1.27) 13.7 turns4110:5:15we might round the actualturns toChoose secondary turnsaccording to desired turns ratios:22:1:3Increased n1 would lead to5n2 n 0.62 turns110 1 Less core loss More copper loss15n 1.87 turnsn3 110 1Fundamentals of Power ElectronicsRounding the number of turns Increased total loss34Chapter 15: Transformer design
Loss calculationwith rounded turnsWith n1 22, the flux density will be reduced to(800 10 – 6 ) B 10 4 0.143 Tesla2(22)(1.27)The resulting losses will bePfe (7.6)(0.143) 2.6(1.27)(7.7) 0.47 W(8.5)(1.724 10 – 6)(800 10 – 6) 2(14.4) 281Pcu 104 (0.25)(1.1)(1.27) 2 (0.143) 2 5.4 WPtot Pfe Pcu 5.9 WWhich exceeds design goal of 4 W by 50%. So use next larger coresize: EE50.Fundamentals of Power Electronics35Chapter 15: Transformer design
Calculations with EE50Repeat previous calculations for EE50 core size. Results: B 0.14 T, n1 12, Ptot 2.3 WAgain round n1 to 22. Then B 0.08 T, Pcu 3.89 W, Pfe 0.23 W, Ptot 4.12 WWhich is close enough to 4 W.Fundamentals of Power Electronics36Chapter 15: Transformer design
Wire sizes for EE50 designWindow allocationsWire gaugesα1 I1 5.7 0.396I tot 14.4α2 n 2I 2 5 66.1 0.209n 1I tot 110 14.4α 1K uWA (0.396)(0.25)(1.78) 8.0 10 – 3 cm 2(22)n1 AWG #19αKW(0.209)(0.25)(1.78) 93.0 10 – 3 cm 2A w2 2 u A (1)n2n 3I 3 15 9.9 0.094n 1I tot 110 14.4 AWG #8αKW(0.094)(0.25)(1.78) 13.9 10 – 3 cm 2A w3 3 u A (3)n3α3 A w1 AWG #16Might actually use foil or Litz wire for secondary windingsFundamentals of Power Electronics37Chapter 15: Transformer design
Discussion: Transformer design Process is iterative because of round-off of physical number ofturns and, to a lesser extent, other quantitiesEffect of proximity loss– Not included in design process yet– Requires additional iterations Can modify procedure as follows:– After a design has been calculated, determine number of layers ineach winding and then compute proximity loss– Alter effective resistivity of wire to compensate: define eff Pcu/Pdc where Pcu is the total copper loss (including proximityeffects) and Pdc is the copper loss predicted by the dc resistance.– Apply transformer design procedure using this effective wireresistivity, and compute proximity loss in the resulting design.Further iterations may be necessary if the specifications are notmet.Fundamentals of Power Electronics38Chapter 15: Transformer design
15.4 AC Inductor Designi(t)Window area WA Core mean lengthper turn (MLT )v(t)LCoreCore areaAcnturnsAir gaplg–Wire resistivity ρFill factor Kuv(t)Area λt1t2i(t)tDesign a single-winding inductor, havingan air gap, accounting for core loss(note that the previous design procedure ofthis chapter did not employ an air gap, andinductance was not a specification)Fundamentals of Power Electronics39Chapter 15: Transformer design
Outline of key equationsObtain specified inductance:Total loss is minimized whenµ0 Acn2L lgRelationship betweenapplied volt-seconds andpeak ac flux density:2 2 B ρλ I2K u(MLT )13W A A c l m βK fe1β 2Must select core that satisfies2/β B λ2nA cK gfe ρλ 2I 2K fe2K u Ptotβ 2 /βCopper loss (using dcresistance):See Section 15.4.2 for step-by-stepdesign equationsρn 2(MLT ) 2Pcu IK uW AFundamentals of Power Electronics40Chapter 15: Transformer design
Fundamentals of Power Electronics Chapter 15: Transformer design3 15.1 Transformer Design: Basic Constraints Core loss Typical 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 flux density So increasing B causes core loss to increase rapidly
3. Instrument transformer: Used in relay and protection purpose in different instruments in industries . . Current transformer (CT) . Potential transformer (PT) . Open circuit and Short circuit Test on transformer . These two transformer tests are performed to find the parameters of equivalent circuit of transformer and losses of the transformer.
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
Transformer Design & Design Parameters - Ronnie Minhaz, P.Eng. Transformer Consulting Services Inc. Power Transmission Distribution Transformer Consulting Services Inc. Generator Step-Up Auto-transformer Step-down pads transformer transformer 115/10 or 20 kV 500/230 230/13.8 132 345/161 161 161 230/115 132 230 230/132 115 345 69 500 34 GENERATION TRANSMISSION SUB-TRANSMISSION DISTRIBUTION .
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
Transformer Lab 1. Objectives: 1.1 Comparison of the ideal transformer versus the physical transformer 1.2 Measure some of the circuit parameters of a physical transformer to determine how they affect transformer performance. 1.3 Investigate the ideal transformer and ca
transformer there are hysteresis and eddy current losses in transformer core. Theory of transformer on no-load, and having no winding resistance and no leakage reactance of transformer Let us consider one electrical transformer with only core losses. That means it has only core losses but no copper lose and no leakage reactance of transformer.
Step 13: Now click on the 2-Winding Transformer icon . Place the 2-winding transformer in the same way that you placed the previous two components. Join the primary of the transformer to the Main Bus. Double click the transformer icon and set the following properties: On the Info Tab o Change the transformer ID to "Main Transformer".
the distribution magnetic flux in the transformer. It has been used ANSYS Package Version 11 to model the distribution transformer.Table (1) shows the data of distribution transformer. 3.1 Transformer Geometry (Building and meshing): The transformer study is 250 kVA, three phase distribution core type "stacked core" transformer.
