MIL-HDBK-217电路实例计算MTBF

可靠度工程簡介姚煥樟-41–-harry.yauMIL–HDBK–217實例計算簡介(一)參考電路：可靠度工程簡介姚煥樟-42–-harry.yau(二)零件計數法可靠度預估(PARTSCOUNT)1.用於設計初期估計裝備總失效率，適於招標或欲簽訂單時的可靠度預估。2.必須資料a.零件屬性與型式b.該零件使用數目c.零件品質水準d.裝置所在的環境3.零件失效率計算：I=nλepuip=Σni(λgπθ)I=1λg=零件基本的失效率(f/106hrs)πθ=零件的品質因數Ni=第i類零件的數量4.例題計算：Gf=40℃(Ta)a.λr=5個×(0.018f/106hrs×3)=0.27f/106hrsb.λce=4個×(0.290f/106hrs×3)=3.48f/106hrsc.λtr=2個×(0.016f/106hrs×10)=0.32f/106hrsd.λdiode=2個×(0.0031f/106hrs×6)=0.031f/106hrse.λled=1個×(0.033f/106hrs×10)=0.33f/106hrsλequip=λr+λce+λtr+λdiodeλled=4.431f/106hrsMTBF==×106=225682H故零件計數法得到MTBF=225682H(三)零件應力分析法可靠度預估(PARTSTRESSANALYSIS)1.TRANSISTOR類a.求Q1之Imax=V/R3=10V/39K=0.256mAPw=V×I=0.5V×0.256=0.128mW1λ14.431可靠度工程簡介姚煥樟-43–-harry.yauI取RMS值時(0.256/1.414=0.18mW)Pw=0.18×(10-0.18×39)=0.54mWST=0.54mW/400mW=0.0014(C.F.)C.F=(Tmax–25)/150=100–25/150=0.5T=400C+(175–100)=1150C查T=1150C,ST=0.1得λb=0.0019f/106hrs查πe=5.8(Gf)πa=1.5(Linear)πq=12(PLASTIC)πr=1(1watt)πs2=AppliedVce/Vceo=10V/60V=20%得0.3πc=1.0(Singletransistor)λp=λb(πe×πa×πq×πr×πs2×πc)λp=0.0019(5.8×1.5×12×1×0.3×1.0)=0.05951f/106hrsb.求Q2之Imax=(Vdc–Vf)/1K=8mAPw=V×I=8×0.5=4mWI取RMS值時(8/1.414=5.65mA)Pw=5.65×(10-2–5.65×1)=13.3mWST=13.3mW/400mW=0.033(C.F.)C.F.同Q1所以T=1150C查T=1150C，ST=0.1得λb=0.0019f/106hrs查πe=5.8(Gf)πa=1.5(Linear)πq=12(PLASTIC)πr=1(1watt)πs2=AppliedVce/Vceo=10V/60V=20%得0.3πc=1.0(Singletransistor)λpq2=λb(πe×πa×πq×πr×πs2×πc)λpq2=0.0019(5.8×1.5×12×1×0.3×1.0)=0.05951f/106hrsC.求晶體之總失效率可靠度工程簡介姚煥樟-44–-harry.yauλp(Q)=λpq1+λpq2=0.05951+0.05951=0.11902f/106hrs2.LEDλp(led)=λb×πe(環境)×πt(溫度)×πq(品質)查表得λb=0.00065f/106hrs(SingleLED)πt=[LED=Ta+300C=Tj=700C]查表得430πe=2.4(Gf)πq=0.5(LOWER)，1.0(PLASTIC)λp(led)=0.00065×430×1.0×2.4=0.6708f/106hrs3.Diodeλp(diode)=λb×πe×πr×πq×πa×πc×πs2ST=Imax/110mA=0.256/110=0.002C.F.=1所以查表得λb=0.00025πe=3.9(Gf)πr=1.0(1Amp)πq=7.5(Lower)πa=1.0(Analogcircuit500mA)πc=1.0(metallurgicallybonded)πs2=appliedVr/ratedVr×100%=10/35=30%=0.7λp(diode)=0.00025×3.9×1.0×7.5×1.0×1.0×0.7×2個=0.01024f/106hrs4.Resistera.R1Prl=I2×R1=[10V/(R1+R2)]2×680K=0.14mWST=0.14mW/250mW=0.0056(ST=0.1T=400C)查表得λb=0.00088f/106hrsπe=2.4(Gf)πr=1.1(0.1-1Mohm)πq=15.0(Lower)可靠度工程簡介姚煥樟-45–-harry.yauλp(r1)=λb×πe×πr×πq=0.00088×2.4×15×1.1=0.03485f/106hrsb.R2Pr2=[10/(680+43)]2×43=0.008mWST=0.008/250mW=0.00033查表得λb=0.00088f/106hrsπe=2.4(Gf)πr=1.0(0–100Kohm)πq=15.0(Lower)λp(r2)=λb×πe×πr×πq=0.00088×2.4×15×1.0=0.03168f/106hrsc.R3同上方法求得λp(r3)=λb×πe×πr×πq=0.00088×2.4×15×1.0=0.03168f/106hrsd.R4同上方法求得λp(r4)=λb×πe×πr×πq=0.00088×2.4×15×1.0=0.03168f/106hrse.R5Pr5=[(10-2)/1K]2×1K=64mWST=64mW/250mW=0.256取ST=0.3T=40oC查表得λb=0.0011f/106hrsπe=2.4(Gf)πr=1.0(0–100Kohm)πq=15.0(Lower)λp(r5)=λb×πe×πr×πq=0.0011×2.4×15×1.0=0.0396f/106hrs可靠度工程簡介姚煥樟-46–-harry.yauf.求電阻之總失效率λp(r)=λp(rl)+λp(r2)+λp(r3)+λp(r4)+λp(r5)=0.03485+0.03168+0.03168+0.03168+0.0396=0.1695f/106hrsCE(電解電容器)以25V電解計算時---(括號內則以16V電解計算時)a.C1ST=Vapplied/Vrated=5/25=0.2---(5/16=0.312)λb=0.019-------------(0.025)πe=2.4(Gf)πq=10(lower)πcv=0.4(3.2uF)λp(c1)=λb×πe×πq×πcv=0.019×2.4×10×0.4=0.1824f/106hrs-----------------(0.24)b.C2ST=0.02V/25V=0.001--------------(0.001)λb=0.018-------------(0.018)πe=2.4(Gf)πq=10(lower)πcv=0.4(3.2uF)λp(c1)=λb×πe×πq×πcv=0.018×2.4×10×0.4=0.1728f/106hrs------------(0.1728)c.C3ST=10/25=0.4--------------(0.7)λb=0.025-------------(0.055)πe=2.4(Gf)πq=10(lower)πcv=0.4(3.2uF)λp(c1)=λb×πe×πq×πcv可靠度工程簡介姚煥樟-47–-harry.yau=0.025×2.4×10×0.4=0.24f/106hrs------------(0.528)d.C4ST=5/25=0.2------------(0.4)λb=0.019-------------(0.025)πe=2.4(Gf)πq=10(lower)πcv=0.4(3.2uF)λp(c1)=λb×πe×πq×πcv=0.019×2.4×10×0.4=0.1824f/106hrs--------------(0.24)e.求電解電容器總失效率λp(c)=λp(cl)+λp(c2)+λp(c2)+λp(c3)+λp(c4)=0.1824+0.1728+0.24+0.1824=0.7776f/106hrs------------------(1.1808)5.電路總失效率λp(total)=λp(c)+λp(r)+λp(diode)+λp(led)+λp(Q)=0.7776+0.1695+0.01024+0.6708+0.11902=1.74716f/106hrs---------------(2.15036)MTBF=1/λp(total)=1/1.74716×106=572357Hrs---------(以25V-電解電容計算求得)=(465038Hrs)(以16V-電解電容計算求得)故以應力分析法計算得:MTBT:572357H---------(用25v電解電容)MTBT:465038H---------(用16v電解電容)