反激式变压器设计

反激式变压器设计 例 : 60watts ADAPTER POWER MAIN X'FMR INPUT : 90 ~ 264 Vac 47 ~ 63 HZ ; OUTPUT : DC 19V 0 ~ 3.16A ; Vcc = 12 VDC 0.1A η? 0.83 ; f s = 70KHZ ; Duty cylce over 50% ?t ?40o ( 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 面) @ 60W ; X'FMR限高 21mm. CASE Surface Temperature ? 78? . Note : Constant Voltage & Current Design (UC3843AD) Step1. 选择CORE材质,确定?B 本例为ADAPTER DESIGN,由于该类型机散热效果差,故选择CORE材质应考量高Bs,低损耗 PC40 or PC44为优选, 对比TDK 及高μi材质,结合成本考量,在此选用Ferrite Core, 以TDK 之 DATA BOOK, 可知 PC44材质单位密度 相关参数如下: μi = 2400 ? 25% Pvc = 300KW / m2 @100KHZ ,100? Bs = 390mT Br = 60mT @ 100? Tc = 215? 为防止X'FMR出现瞬态饱和效应, 此例以低?B设计. 选 ?B = 60%Bm, 即?B = 0.6 * (390 - 60) = 198mT ?0.2 Step2 确定Core Size和 Type. 1> 求core AP以确定 size AP= AW*Ae=(Pt*104)/(2ΔB*fs*J*Ku) = [(60/0.83+60)*104]/(2*0.2*70*103*400*0.2) = 0.59cm4 式中 Pt = Po /η +Po 传递功率; J : 电流密度 A / cm2 (300~500) ; Ku: 绕组系数 0.2 ~ 0.5 . 2> 形状及规格确定. 形状由外部尺寸,可配合BOBBIN, EMI要求等决定,规格可参考AP值及形状要求而决定, 结合上 述原则, 查阅TDK之DATA BOOK,可知RM10, LP32/13, EPC30均可满足上述要求,但RM10和 EPC30可用绕线容积均小于LP32/13,在此选用LP32/13 PC44,其参数如下: Ae = 70.3 mm2 Aw = 125.3mm2 AL = 2630?25% le = 64.0mm AP = 0.88 cm4 Ve = 4498mm3 Pt = 164W ( forward ) Step3 估算临界电流 IOB ( DCM / CCM BOUNDARY ) 本例以IL达80% Iomax时为临界点设计变压器. 即 : IOB = 80%*Io(max) = 0.8*3.16 = 2.528 A Step4 求匝数比 n n = [VIN(min) / (Vo + Vf)] * [Dmax / (1-Dmax)] VIN(min) = 90*?2 - 20 = 107V = [107 / (19 + 0.6)] *[0.5 / (1- 0.5)] = 5.5 ? 6 匝比 n 可取 5 或 6,在此取 6 以降低铁损,但铜损将有所增加. CHECK Dmax: Dmax = n (Vo +Vf) / [VINmin + n (Vo + Vf)]= 6*(19 + 0.6) /[107 + 6*(19 + 0.6)] = 0.52 Step5 求CCM / DCM临 ΔISB = 2IOB / (1-Dmax) = 2*2.528 / (1-0.52) = 10.533 Step6 计算次级电感 Ls 及原边电感 Lp Ls = (Vo + Vf)(1-Dmax) * Ts / ΔISB = (19+0.6) * (1-0.52) * (1/70000) / 10=12.76uH Lp = n2 Ls = 62 * 12.76 = 459.4 uH ? 460 此电感值为临界电感,若需电路工作于CCM,则可增大此值,若需工作于DCM则可适当调小 此值. Step7 求CCM时副边峰值电流Δisp Io(max) = (2ΔIs + ΔISB) * (1- Dmax) / 2 ΔIs = Io(max) / (1-Dmax) - (ΔISB / 2 ) ΔIsp = ΔISB +ΔIs = Io(max) / (1-Dmax) + (ΔISB/2) = 3.16 / (1-0.52) + 10.533 / 2=11.85A Step8 求CCM时原边峰值电流ΔIpp ΔIpp = ΔIsp / n = 11.85 / 6 = 1.975 A Step9 确定Np、Ns 1> Np Np = Lp * ΔIpp / (ΔB* Ae) = 460*1.975 / (0.2*70.3) = 64.6 Ts 因计算结果为分数匝,考虑兼顾原、副边绕组匝数取整,使变压器一、二次绕组有相同的安匝值,故调整 Np = 60Ts OR Np = 66Ts 考量在设定匝数比n时,已有铜损增加,为尽量平衡Pfe与Pcu,在此先选 Np = 60 Ts. 2> Ns Ns = Np / n = 60 / 6 = 10 Ts 3> Nvcc 求每匝伏特数Va Va = (Vo + Vf) / Ns = (19+0.6) / 10 = 1.96 V/Ts ? Nvcc = (Vcc + Vf) / Va =(12+1)/1.96=6.6 Step10 计算AIR GAP lg = Np2*μo*Ae / Lp = 602*4*3.14*10-7*70.3 / 0.46 = 0.69 mm Step11 计算线径dw 1> dwp Awp = Iprms / J Iprms = Po / η / VIN(min) = 60/0.83/107 = 0.676A Awp = 0.676 / 4 J取4A / mm2 or 5A / mm2 Φ0.35mm*2) = 0.169 (取 2> dws Aws = Io / J = 3.16 / 4 (Φ1.0 mm) 量可绕性及趋肤效应,采用多线并绕,单线不应大于Φ0.4, Φ0.4之Aw= 0.126mm2, 則 0.79 / 0.126 = 6.27 6 (即Ns采用Φ0.4 * 6) 3> dwvcc Awvcc = Iv / J = 0.1 /4 上述绕组线径均以4A / mm2之计算,以降低铜损,若结构设计时线包过胖,可适当调整J之取值. 4> 估算铜窗占有率. 0.4Aw ?Np*rp*π(1/2dwp)2 + Ns*rs*π(1/2dws)2 + Nvcc*rv*π(1/2dwv)2 0.4Aw ?60*2*3.14*(0.35/2)2+10*6*3.14+(0.4/2)2+7*3.14*(0.18/2)2 ? 11.54 + 7.54 + 0.178 = 19.26 0.4 * 125.3 = 50.12 50.12 > 19.26 OK Step12 估算损耗、温升 1> 求出各绕组之线长. 2> 求出各绕组之RDC和Rac @100? 3> 求各绕组之损耗功率 4> 加总各绕组之功率损耗(求出Total值) 如 : Np = 60Ts , LP32/13BOBBIN绕线平均匝长 4.33cm 则 INP = 60*4.33 = 259.8 cm Ns = 10Ts 则 INS = 10*4.33 = 43.3 cm Nvcc = 7Ts 則 INvc = 7 * 4.33 = 30.31cm 查线阻表可知 : Φ0.35mm WIRE RDC = 0.00268Ω/cm @ 100? Φ0.40mm WIRE RDC = 0.00203 Ω/cm @ 100? Φ0.18mm WIRE RDC = 0.0106 Ω/cm @ 100? R@100? = 1.4*R@20? 求副边各电流值. 已知Io = 3.16A. 副边平均峰值电流 : Ispa = Io / (1-Dmax ) = 3.16 / (1- 0.52) = 6.583A (1-Dmax)*I2spa〕 = ?(1- 0.52)*6.5832 = 4.56A 副边直流有效电流 : Isrms = ?〔 副边交流有效电流 : Isac = ?(I2srms - Io2) = ?(4.562-3.162) = 3.29A 求原边各电流值 : ? Np*Ip = Ns*Is 原边平均峰值电流 : Ippa = Ispa / n = 6.58 / 6 = 1.097A 原边直流有效电流 : Iprms = Dmax * Ippa = 1.097 * 0.52 = 0.57A 原边交流有效电流 : Ipac = ?D*I2ppa = 1.097*?0.52 = 0.79A 求各绕组交、直流电阻. 原边 : RPDC = ( lNp * 0.00268 ) / 2 = 0.348Ω Rpac = 1.6RPDC = 0.557Ω 副边 : RSDC = ( lNS*0.00203 ) /6 = 0.0146Ω Rsac = 1.6RSDC = 0.0243Ω Vcc绕组 : RDC =30.31*0.0106 = 0.321Ω 计算各绕组交直流损耗: 副边直流损 : PSDC = Io2RSDC = 3.162 * 0.0146 = 0.146W 交流损 : Psac = I2sac*Rsac = 3.292*0.0234 = 0.253W Total : Ps = 0.146 + 0.253 = 0.399 W 原边直流损 : PPDC = Irms2RPDC = 0.572 * 0.348 = 0.113W 交流損 : Ppac = I2pac*Rpac = 0.792*0.557 = 0.348W 忽略Vcc绕组损耗(因其电流甚小) Total Pp = 0.461W 总的线圈损耗 : Pcu = Pc + Pp = 0.399 + 0.461 = 0.86 W 2> 计算铁损 PFe 查TDK DATA BOOK可知PC44材之?B = 0.2T 时,Pv = 0.025W / cm2 LP32 / 13之Ve = 4.498cm3 PFe = Pv * Ve = 0.025 * 4.498 = 0.112W 3> Ptotal = Pcu + PFe = 0.6 + 0.112 = 0.972 W 4> 估算温升 ?t 依经验公式 ?t = 23.5PΣ/?Ap = 23.5 * 0.972 / ?0.88 = 24.3 ? 估算之温升?t小于SPEC,设计OK. Step13 结构设计 查LP32 / 13 BOBBIN之绕线幅宽为 21.8mm. 考量安规距离之沿面距离不小于6.4mm. 为减小LK提高效率,采用三明治结构,其结构如下 : X'FMR结构 : Np #1 3.2 / 3.2 2 -- A Φ0.35 * 2 30 1L SHI #2 3.2 / 3.2 SHI- 4 2mils * 12 1 3L Ns #3 3.2 / 3.2 8.9 - 6.7 Φ0.4 * 6 10 3L SHI #4 3.2 / 3.2 SHI- 4 2mils * 12 1 1L Np #5 3.2 / 3.2 A -- 1 Φ0.35 * 2 30 1L Nvcc #6 3.2 / 3.2 3 -- 4 Φ0.18 7 2L #7 连 结 两 A 点 2L