可靠性寿命试验

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1、可靠性寿命试验可靠性试验定义和目的为测定，验证，评价，分析或者提高产品可靠性而进行的为测定，验证，评价，分析或者提高产品可靠性而进行的试验成为可靠性试验。试验成为可靠性试验。目的：目的：1.1.确定电子产品的可靠特征量。确定电子产品的可靠特征量。2.2.产品研制定型时，进行鉴定试验，考核是否达到可预订的产品研制定型时，进行鉴定试验，考核是否达到可预订的可靠性指标。可靠性指标。3.3.研究产品失效机理研究产品失效机理。按试验目的分类表可靠性鉴定试验确定产品可靠性特征量而进行的试验寿命试验评价分析产品的寿命特征量而进行的试验耐久性试验考察产品性能与所加应力的关系而进行一定时间的试验筛选试验剔除早期

2、失效产品或选择一定特性的产品而进行的试验可靠性增长试验采取纠正措施，系统地并永久地消除某些失效机理，使电子产品可靠性达到或者超过预定可靠性要求的试验。环境试验环境试验机械试验机械试验振动,冲击,离心加速度,跌落,弯曲,抗拉强度等温度试验温度试验高温,低温，温度交变湿热试验湿热试验恒定湿热，交变湿热试验密封性试验密封性试验粗检(氟碳油),细检(氦质谱,放射性示踪）特殊试验特殊试验盐雾,霉菌,低气压,超高真空,红外谱检测,X射线检测,辐射等综合试验综合试验低温/低压,低温/振动,高温/振动,振动/温循/潮湿试验等组合试验组合试验温度-湿度-气压试验寿命试验寿命试验长期寿命试验长期寿命试验长期贮存寿

3、命试验长期工作寿命试验-连续工作寿命试验(动态,静态) - 间隙工作寿命试验加速寿命试验加速寿命试验恒定应力加速试验,步进应力加速试验等现场使用试验现场使用试验实际工作试验，现场贮存试验，现场环境试验按试验项目的分类表寿命试验1.1.全寿命试验全寿命试验2.2.截尾试验截尾试验a)a)定时截尾：实验到规定的时间而中止定时截尾：实验到规定的时间而中止b)b)定数截尾：实验到出现规定的故障数或失效定数截尾：实验到出现规定的故障数或失效数时而中止数时而中止如受测试的样品数量为如受测试的样品数量为n n，则存在下列截尾试验方法：，则存在下列截尾试验方法：(n,(n,无无, ,时时)-)-取取n n个产

4、品进行无替换定时截尾寿命试验个产品进行无替换定时截尾寿命试验(n,(n,无无, ,数数)-)-取取n n个产品进行无替换定数截尾寿命试验个产品进行无替换定数截尾寿命试验(n,(n,有有, ,时时)-)-取取n n个产品进行有替换定时截尾寿命试验个产品进行有替换定时截尾寿命试验(n,(n,有有, ,数数)-)-取取n n个产品进行有替换定数截尾寿命试验个产品进行有替换定数截尾寿命试验寿命试验参数的数值分析估算法1.参数点估计参数点估计当母体的分布类型已知，常用极大似然法来寻求未知参数的当母体的分布类型已知，常用极大似然法来寻求未知参数的估计。估计。极大似然估计极大似然估计: Maximum li

5、kelihood estimation (MLE)设母体的待估参数为设母体的待估参数为, ,它可以取很多值它可以取很多值. .我们在我们在的一切可能值之中选的一切可能值之中选出一个使子样观察结果出现的概率的最大值出一个使子样观察结果出现的概率的最大值的估算的估算, ,记为记为 , ,并称并称 为为的极大似然估计的极大似然估计(MLE).(MLE).我们把已经发生的事件，看成最可能出现的事件，即认为它我们把已经发生的事件，看成最可能出现的事件，即认为它具有最大的概率。具有最大的概率。3p2(1-p)3p2(1-p)=3(0.1)2x0.9=0.0273p2(1-p)=3(0.2)2x0.8=0.

6、096设有一批量很大的产品，预计它们的不合格率设有一批量很大的产品，预计它们的不合格率p p只可能是只可能是0.10.1或或0.2.0.2.为了确定这批产品的不合格率为了确定这批产品的不合格率p p到底是多少，从中任取到底是多少，从中任取3 3件产品件产品, ,发现有两件不合格发现有两件不合格, ,一件合格一件合格, ,如何根据这个字样观察值对如何根据这个字样观察值对p p作出估计作出估计? ?四种截尾寿命试验四种截尾寿命试验,在指数分布场合在指数分布场合,失效率失效率和平均寿和平均寿命的命的的的MLE可以统一成一个公式可以统一成一个公式: =r/T, =T/r例例1: 从寿命服从指数分布的产

7、品里随机抽取从寿命服从指数分布的产品里随机抽取7个样品进行个样品进行(n,(n,无无, ,时) )试验, ,截尾截尾时间为700hrs.700hrs.试验结束束时, ,失效失效时间按按分分别为650,450,120,530,600,450(650,450,120,530,600,450(小小时）, ,求求和平均寿命和平均寿命的的的的MLE.解：解：n=7, r=6 T=650+450+120+530+600+450+(7-6)*700=3500hrs 的的MLE为为: r/T=6/3500=0.00171/hr的的MLE为为: T/r=3500/6=583.3 hr 例例2: 从寿命服从指数分

8、布的产品里随机抽取从寿命服从指数分布的产品里随机抽取7个样品进行个样品进行(n,(n,有有, ,时) )试验, ,截尾截尾时间为700hrs.700hrs.试验结果如下果如下, ,求求和平均和平均寿命的寿命的的的MLE.解解: n=7, r=8 T=4900hrs 的的MLE为为: r/T=8/4900=0.001633/hr的的MLE为为: T/r=4900/8=612.5hr01204506006005305304506501003005007001357t.U L P(L U )=1- =0.92. 参数区间估计参数区间估计= 1-CC: 置信区间/置信度: 风险度/显著性水平T: 总实

9、验时间n: 样本数量2r+2: 自由度2 : 二项式分布/卡方分布定时结尾双侧区间估计定时结尾双侧区间估计定时结尾单侧区间估计定时结尾单侧区间估计,例例3: 从寿命服从指数分布的产品里随机抽取从寿命服从指数分布的产品里随机抽取7个样品进行个样品进行(n,(n,无无, ,时) )试验, ,截尾截尾时间为700hrs.700hrs.试验结束束时, ,失效失效时间按按分分别为650,450,120,530,600,450(650,450,120,530,600,450(小小时）, ,求求1.1.置信水平置信水平为90%90%的平均寿命的平均寿命的单侧置信下限置信下限. .2.2.置信度置信度为80%

10、80%双双侧置信区置信区间. .3.3.若若试验中到截尾中到截尾时间700hrs700hrs时都无失效都无失效发生生, ,如何求平均寿如何求平均寿命命.解：解：n=7, r=6 T=650+450+120+530+600+450+(7-6)*700=3500hrs 置信水平置信水平为90%的的单侧置信下限置信下限为L=332.32hrs置信度置信度为80%双双侧置信区置信区间L, U为332.32,1110.44例例4: 设产品寿命服从指数分布设产品寿命服从指数分布.抽其抽其n个样品进行无替换的定个样品进行无替换的定时截尾寿命试验时截尾寿命试验。如果在。如果在测试中无失效中无失效发生，那么在置

11、信生，那么在置信水平水平0.950.95下，下，为了了满足平均寿命的足平均寿命的单侧置信下限置信下限为1000hrs,1000hrs,总的的试验时间应为多少？多少？解：解： 置信水平为95%的单侧置信下限为1000hrs总的试验时间为2995.5hrs.平均寿命的最大似然估计平均寿命的最大似然估计(MLE)和置信区间和置信区间例例5: 某产品寿命符合指数分布某产品寿命符合指数分布,抽其抽其20个样品进行个样品进行500小时的无替换可靠小时的无替换可靠性试验性试验.在实验期间有在实验期间有5个产品失效个产品失效.它们的失效时间分别为它们的失效时间分别为110,180,300,410,480小时小

12、时,现要求工作时间为现要求工作时间为2小时小时,10小时小时,50小时时置信小时时置信水平为水平为0.9的可靠度置信区间的可靠度置信区间.1. 可靠度可靠度R(t)=1-F(t)= e-t(t0)=e-t/MTBF我们用来表示MTBF,那么无论是哪种截尾寿命,平均寿命的置信水平为90%的置信区区间(L, U)是可以求得的,且满足P(L U)=0.9P(e-t/L e-t/e-t/u)=0.9RL(t)=e-t/L RU(t)=e-t/U2.根据无替换的定时截尾寿命试验下根据无替换的定时截尾寿命试验下MTBF计算为计算为:=(110+180+300+410+480+(20-5)*500)/5=1

13、796 hours3. L =CL* =0.476*1796=854.9 hrs U =CU*=2.538*1796=4558.2 hrs4. RL(2)=e-t/L= e-2/854.9= 0.9977 RU(2)=e-t/U= e-2/4558.2= 0.9996 (RL(10), RU(10) =(0.9884,0.9978), (RL(50), RU(50) =(0.9432,0.9891) 加速寿命试验加速寿命试验目的:1.较短时间内用较少的样品估计高可靠性器件的可靠性水平;2.短时间内提供测试结果,检验器件设计,工艺等改进的效果,同时可以比较不同工艺和器件结构的利弊;3.短时间内暴

14、露器件或电路的失效形式,结合失效分析,了解失效的机理,反馈到设计制造部门加以纠正.同时可以合理制定筛选条件和测试规范.机理：加大应力(诸如热应力,电应力,机械应力等)的办法,加快产品失效,缩短测试时间,运用加速寿命模型,估计出产品在正常工作应力下的可靠性特征.S0, S1,S2Sn在遇多种失效机理的情况下，就应到选择那种对产品失效机理起促进最大的应力作为加速应力。加速失效进程，但是不能改变失效机理加速失效进程，但是不能改变失效机理. .微电子器件参数的退化是由器件内部物理和化学变化引起的，当这种变化积累到一定程度时即发生失效，退化经历的时间即产品的寿命.Arrhenius Model（阿伦尼斯

15、模型）（阿伦尼斯模型）The max use environment is 45C, the max burn-in environment is 80C The targeted failure has an activation energy of 0.60eV. The nominal customer end-use ambient temperature is 25C the thermal rise of the UUT above ambient in the customer environment is 15C The fluctuation of ambient tempe

16、rature in the customer environment is 5C. The UUT is continuously powered at full load during the duration of the burn-in test.Tu = 45C = 318.15 KTt = 80C = 353.15 KAT = exp 0.60/0.00008623 x (1/318.15)-(1/ 353.15 ) = 8.74Coffin-Manson ModelACMD = (TA/Tu)2 x (fA/fu) = Du/DADu= DA*(TA/Tu)2 x (fA/fu)A

17、CMD = Acceleration factor for number of cycles (dimensionless)TA = Thermal cycle temperature change in accelerated environment (K)Tu = Thermal cycle temperature change in use environment (K)fu = Frequency of thermal cycles in use environment (cycles/day)fA = Frequency of thermal cycles in accelerate

18、d environment (cycles/day)Du = Time duration in use environment (days)DA = Time duration in accelerated environment (days)T max=80C, T mix=-10C, 3 hours with 3 complete thermal cycles in the customer end-use environment.T max in the accelerated environment= 80CT max in the use environment = 45CThe n

19、ominal customer end-use ambient temperature= 25CThe thermal rise of the UUT above ambient in the customer environment is 15C The fluctuation of ambient temperature in the customer environment is 5C. The customer end-use temperature change is 20C and the customer end-use environment classification is

20、 “computers”.TA = 80C-(-10C)+273.15= 90KTu = 45C- 25C=20Kfu = 4 cycles/dayfA = 24 cycles/dayACMD= (90K/20K)2*(24/4)=121.5ACMD = (TA/Tu)2 x (fA/fu) = Du/DADu= DA*(TA/Tu)2 x (fA/fu)Norris-Landzberg ModelANL = (TA / Tu)2 (fu / fA)1/3 e1414 *(1/Tu)-(1/TA) )Where:ANL = Acceleration factor for number of c

21、ycles (dimensionless)TA = Thermal cycle temperature change in accelerated environment (K)Tu = Thermal cycle temperature change in use environment (K)fu = Frequency of thermal cycles in use environment (cycles/day)fA = Frequency of thermal cycles in accelerated environment (cycles/day)Tu = Maximum UU

22、T temperature in use environment (K)TA = Maximum UUT temperature in accelerated environment (K)ANLD = (TA / Tu)2 (fu / fA)1/3 e1414 *(1/Tu)-(1/TA) ) *(fA/fu)Where:ANLD = Acceleration factor for time duration (dimensionless)TA = Thermal cycle temperature change in accelerated environment (K)Tu = Ther

23、mal cycle temperature change in use environment (K)fu = Frequency of thermal cycles in use environment (cycles/day)fA = Frequency of thermal cycles in accelerated environment (cycles/day)Tu = Maximum UUT temperature in use environment (K)TA = Maximum UUT temperature in accelerated environment (K)MCA

24、TT #: 20101201Issue Title: Mayzie Intermittent System Freeze Issue in FieldDate: Dec-13-2010Description:Dell end user report that many Mayzie MB (Total 14pcs) failed with “system freeze” issue under idle state in AMF& EMEA site from Aug to Nov. Details show at the attached file in backup. After FA,

25、3pcs filed return MBs can be duplicated the failure symptom and the failure focus onOld Mayzie MB without 6 caps & with Renesas MOSFETFailure can only be duplicated with Clarkdale CPU, but cant with Lynnfield CPU.PPID for 3pcs MBs list as below: CN0X744K7082105O00FTA02 / CN0X744K7082105S00QIA02 / N0

26、X744K7082105S00SOA02Champion: Loewi.loTeam member:Jin.Jiang, Cristine.Lee, James.Wu, Dragon, Tiger, Catherine.Yang, Hevil.Du, Zilong.Lin, Helen.Wong, Phoebe.Liu, Robert, LilyContainment Action:1.Mitac have stopped using Renesas low-side MOSFET & transferred to Fairchild for Mayzie fresh build since

27、Jun 4th. Done on Jun 42.Mitac have phased in Clarkdale CPU at Mayzie EBT station since Jun 4th but dont found “system freeze” issue on production line till Dec 13th. Done from Jun 43.Double confirm un-pulled quantity for old Mayzie MB (Without 6 caps) still in WW site to predict the impact on Dec 7.

28、Total 238pcs, return all of them back to MSL for rework to new MB (To add 6 caps). Done on Dec 74.To double confirm if “system freeze” impact Mayzie new MB (with 6 caps), Mitac take 1pcs Mayzie new MB (With 6 caps) to run idle test under normal temperature (For 142 hours) and low temperature (5 for

29、28 hours), no “system freeze” issue found. Done on Dec 5Root Cause: Design Issue. See the FA summary report for details in back up 1.VTT_CPU_PG have noise from CPU sink current, which cause “system freeze/hang” when running under idle state2.Mayie CAD test cad didnt test C-state idle status in CAD t

30、est (Both NPI qualification and sustaining stage).CharacterizeContainCauseCorrective Action:1.Phase in ECR to add 6 MLCCs 0.1uF cap x3 (C524, C526, C528) and 22nF cap x3 (C523, C525, C527) on Mayzie old MB to improve the performance of VTT_CPU_PG signal on Jul 12010. Done on Jul 12.For Mayzie, use R

31、AD test (4pcs boards with 6 caps) to test idle status to cover Mayzie CAD test cad not test idle status issue. For following projects, Mitac will remind Dell to add “idle test” into CAD test cad (Such as D&C CAD test in the attached file). Done on Dec 173.For Mayzie old MB (Without 6 caps) in WW hub

32、s or Service hubs, Mitac will be responsible for “system freeze issue” units. The failure units can go through RMA process (For WW hub section) or swap for improved MB (For Service hub section) before improved MB being pulled in Dell factory/consumed in end customer. Done from Jun 2CorrectiveClosure

33、Preventive Action: 1.Lesson Learned for CPU qualification: Need to do more in-depth NUDD(DFMEA) analysis for new CPU qualification. On goingIdentify the CTF ckt/part in the concerned building block.Identify the CTF design variable/parameter of the concerned parts.Identify the worst case with the hel

34、p of the simulation to minimize the test case/sample with no sacrifice of test validity . Owner: R&D Due day: Development stage for each NPI projects2. Add idle test in CAD test to cover this issue for future projects and for new CPU qualification. On goingOverall check idle test status for current

35、NPI (Diamas & Coaster) & Sustaining projects (Horton & Mayzie) to add idle test if not. (Diamas & Coaster Yes; Horton & Mayzie No)For following new NPI projects, add idle test in CAD test For new CPU qualification, add idle test in CAD test Owner: Project leader/Sustaining PM Due day: Development st

36、age for NPI project/CPU qualification stageVerification: 1.Take 30pcs Mayzie new MB (With 6 caps) to run idle test for 50 hours at normal temperature under Lynnfield & Clarkdale CPU respectively to identify “system freeze” not impact on new MB (Have add 6 caps), Till Jan 4, finish 30pcs with (18pcs

37、with Clarkdale CPU and 12pcs with Lynnfield CPU), no failure found. Done on 2011/1/42.Take 3pcs Mayzie failure boards (old MB but to add 6 caps) and 1pcs Mayzie good board (with 6 caps) to run RAD test (High (40 ) /low (5 ) temperature test) for 72 hours to indentify “To add 6 caps” can fix “system

38、freeze” issue. It finished and no failure found. Done on Dec 173.Take Mayzie L10 system (10 sets running 66 days) to run burn-in/aging test to identify the solution (add 6 caps) efficiency on “system freeze” issue in following 2 years (3 years after Mayzie RTSed on 2009/9). Pending Owner: Sustaining

39、 PM Cristine.Lee Due day: 2011/5/1 (Will start at about week 2/14 after PSU qual and CNY)4.Track the performance (If any “system freeze” issue) for Mayzie new MB (With 6 caps) in WW site and field in following 3 months. On going Owner: QE Phoebe.Liu Due day: 2011/3/31Reviewer: Dell Customer (John. Yu)Review Results: TBCABCACBABC