Fan reliability can be evaluated in several ways. The data for a life test can be plotted as a cumulative distribution that shows the total fraction of fans failing up to any operating time. Fig. lisa sample of
cumulative distribution,which was stopped at 8,400 hours after l8outof 48 fans had failed.

Figure 1: Sample cumulative distribution function, Weibull vs. Empirical with 95% confidence bands
(Reference : IB5 Corp., May 1996, Vol.2, No.2, Electronics Cooling)

Some vendors provide life expectancy data to customers based on the exponential assumption.

However, life test data does not support the use of the exponential distribution. Nevertheless the past experimentation fitting has shown that the Weibull distribution provides a good fit to real fan life data.

Cumulative Distribution Function, F(t) of Weibull distribution is a below:

Normally, L10 was introduced a life expectancy parameters by fan vendors. That means the tenth percentiles under an assumed life distribution such as the Weibull. Sometimes vendors will also quote the Mean Time To Failure (MTTF) then we need to figure out the correlation between L_ and MTTF by following equations:

 

After we have verified the correlation between L,, and MTTF, we also need to know how long should a sample size be tested to determine with 90% confidence level that L,. greater than or equal to expectancy value at a test temperature without failure (x=O) Here we introduce the Poisson Distribution to estimate.

P(x, I) = ((A Vt
P(O,t) _((At)eA)/O!_
R(t)
MTTF= aT (1+1/fl)
...t=aI(Brc)/nlle
t1 MTTF/T(l+I/fl)I X I(B,)/n1
where Br;c, is Poisson Distribution Factor

Normally on the condition of 90% confidence level and 0 failure then Br;c = 2.303.
Then we introduce Takes Martin Marietta Model to estimate Life at different environment stress.

AF -IVu/IuI x2(Ta-Tu)/10
where
AF: Acceleration Factor
Va: Actual Testing Voltage
Vu Rating Voltage
Ta: Actual Testing Temperature
Tu: Rating Temperature
if Va = Vu
then AF (t) =2(Ta-Tu)/10

Then we can define the Required Test Time (t) with zero failure is as below:
t = [MTTF/T (1+1/fl) ] x[ (Br;c)/n ] 1/d/2(Ta-Tu)/10

Where MTTF is an expectancy value.