Question 4.6: Suppose that ICs are used in a critical application at 40°C....
Suppose that ICs are used in a critical application at 40°C. The manufacturer selects a sample population of 1000 chips, ages them at 150°C for 2 years (t_{a}=1.75 \times 10^{4} h) and no failures are observed. Typical data for other similar products suggest that the activation energy for failure is 0.4 eV and β = 0.25. What upper bound value for the failure rate, \lambda_{1}(40^{\circ}C) , can be predicted with a confidence of 90% (C = 0.9)?
First the AF is evaluated from Eqn (4.36):
AF=\frac{Rate(T_{2})}{Rate(T_{1})}=\frac{MTTF(T_{1})}{MTTF(T_{2})} =exp\left[\frac{E}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}} \right) \right] 4.36
AF=exp\left[\frac{0.4}{8.62 \times 10^{-5}}\left(\frac{1}{313}-\frac{1}{423} \right) \right] =47.2 .
After substituting in Eqn (4.54),
\lambda_{1}(T_{u})=\frac{\beta\ln [1/(1-c(0))]}{N(AFt_{a})^{\beta}} 4.54
\lambda_{1}(T_{u})=\frac{0.25\ln \frac{1}{(1-0.9)}}{10^{3}(47.2 \times 1.75 \times 10^{4})^{0.25}}=1.91 \times 10^{-5}/h or 19,094 FITs Compared to the decreasing hazard rate of the Weibull model, the use of the exponential distribution is easily derived. Under conditions where no failures are observed during testing, \lambda(t)=\lambda_{1}(T_{u})=const , and it is easily shown that.
\lambda=(N.AF.t_{a})^{-1}\times\ln [1/(1-C(0))] 4.55
Using the same data above in Example 4.6, the assumption of an exponential distribution would result in \lambda=2.7FITs, quite a difference.