|
楼主 |
发表于 2012-1-16 18:03:56
|
显示全部楼层
放了三天都没有顶,是不是都放假啦,我自己没办法做了一个,不知道哪里不妥,还望知道的指点一二呀!
我假设产品平均寿命5年,而且MTBF也约等于MTTF,Ea=0.6,常温25C/50%H,加速试验75C/85%RH
这样算下来AF=46
再根据单信置下限和采用截尾试验方法,C=95%的可信度可以计算得到大约10个样品在12天之内75C/85%RH的条件下没有失效,MTBF=MTTF=平均寿命5年
推算过程如下:
Derivationprocessasbelow:
1. Weselect“FixedTruncationtime”testmethodandused“OneSided(LowerLimit)”measures.
2. MTTF=MTBF=5years*365days*24hours=43800hours
3. Failednumberr=0(nosamplesfailedduringTHBconditiontest)
4. α=5%(acceptableriskoferror)
5. C=(θL≤θ≤θu)=1-α=95%(confidencelevel)
6. θ=Redcircledformula
whenr=0afterTHBfinished,wesetupTHBtime=Ty
Ty=T/10moudles
AccordingtoDistributionofX2(CHI-SQUARE)(MIL-HDBK-338BTABLE8.3-11)
Whileα=0.05r=0X2=5.991
T=43800Hours*5.991/2=131203Hours
SoTy=13120Hours
Upisnormaltemperatureandhumidity(25C/50%RH),weneedgettheAF(accelerationfactor)valueforhightemperatureandhighhumiditytest.
AccordingtoArrheniusModelmeasuredformula
AF=exp{(Ea/k)*[(1/Tu)-(1/Ts)]+(RHu^n-RHs^n)}
SetupEa=0.6Ev(Asthepowerofmovementandshort-termorshort-circuit),Tu=25C,Ts=75C,RHu=50%RHs=85%
AF=EXP(0.6*((1/298)-(1/348))*10^5/8.6+(0.85^2-0.50^2)=46
SoneedtesttimeTs=Ty/AF=13120/46=285Hours≈12days(sowecansaywehad95%trustthesemodulescanworkonwell5years)
Becommensuratewithupmeasured,ifr≤1(atfinishedtimepointfailed1pcssamples),C=95%,MTTF=5years,
SoTs=446hour≈19days |
|