Question 9.4: Accelerated temperature cycling of solder joints on a 256 pi...
Accelerated temperature cycling of solder joints on a 256 pin quad flat package (QFP) was carried out at a frequency of 1 cycle/45 min over the range -40 to 125 °C. The Weibull cumulative failure distribution was found to be of the form F_s(t)=1-exp[-(t/42,900)^{1.27}] , where t is the time in minutes.This chip will be used in a computer, where it undergoes 1 thermal cycle per day and is exposed to ΔT(u) = 85 °C. If it is assumed that MTTF(u)(85 °C)/MTTF(s)(125 °C) = 1.7.
1. What is the acceleration factor?
2. Predict the failure rate under use conditions.
3. Derive an expression for F_u (t).
1. Substituting in Eqn (9.38),
AF=\frac{N(u)}{N(s)}=\left[\frac{f(u)}{f(s)}\right]^{1/3}\left(\frac{\Delta T(s)}{\Delta T(u)} \right) ^2\left(\frac{MTTF(u)}{MTTF(s)}\right) noting that 1 cycle/45 min = 32 cycles/day, AF = (1/32) ^{1/3} (165/85)^{ 2} (1.7) = 2.02.
2. From Eqn 4.42
and the definition of Weibull failure rate (Eqn 4.15),
\lambda(t)=\frac{\beta t^{\beta-1}}{\alpha^\beta}
\lambda_u(t)=(1/2.02)^{1.27}\times(1.27/42,900)(t/42,900)^{0.27} \ or \ 0.0,000,121(t/42,900)^{0.27} .
3. From Eqn 4.39,
F_u(t)=(1/AF)\times\left\{AF\times F_s(t/AF)\right\} =F_s(t/AF)
F_u(t)=F_s(t/AF).Therefore,F_u(t)=1-exp[-(t/2.02 \times 42,900)^{1.27}]=1-exp[-(t/86,700)^{1.27}].