看楼主这个问题比较急,花了些时间,帮楼主验证和计算了一下.请参考(数据表中红色的是,我重新计算的,不过有些数据不明,可能有误,但与楼主数据较接近.):
组数 | ss | hours | devicehourscase | temp | AFforEa@0.55ev | acceleratedhours | 1 | 22 | 5200 | 114400 | 85 | 12.95 | 13.04391189 | 1481323 | 1492224 | 2 | 10 | 2100 | 21000 | 100 | 26.51 | 26.75488468 | 556647 | 561852.6 | 3 | 10 | 800 | 8000 | 100 | 26.51 | 26.75488468 | 212056 | 214039.1 | 4 | 22 | 10000 | 220000 | 95 | 21.01 | 21.1948681 | 4622614 | 4662871 | 5 | 18 | 5000 | 90000 | 95 | 21.01 | 21.1948681 | 1891069 | 1907538 | 6 | 25 | 5000 | 125000 | 110 | 41.42 | 41.8623 | 5178004 | 5232788 | 7 | 1920 | 200 | 384000 | 100 | 26.51 | 26.75488468 | 10178683 | 10273876 | 8 | 768 | 800 | 614400 | 95 | 21.01 | 21.1948681 | 12909699 | 13022127 | | | | 1576800 | 40 | | | 37030095 | 37367314 | | | | | | | | | | 0failures | | 60%cl | FITrate= | 25 | MTTF= | | 404126325 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | AF=exp{(Ea/k)*[(1/Tu)-(1/Ts)]} | | | | | 第一个AF | 13.04391189 | =EXP(0.55*10^6/8.6*(1/(25+273)-1/(85+273))) | | | | | | | | | | | | | | | | A= | 1.832581484 | =CHIINV(1-0.6,2*(0+1)) | 定时单侧 | | | | | | | | | | | | | | | MTTF= | 40413040.64 | =2*37030095/A | | 你的MTTF多了一位了 | | | FIT= | 24.74448802 | | | | | |
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