Assume that we tested FD2510B05W5 at 80oC and 50oC.
- 200 test samples at 80oC
- 200 test samples at the same period
- 1008 hours tested
- 0.09A current
- 3 fans fail at 80oC test. 1 at 960th hour; 1 at 984th hour; and 1 at 1008th hour. And there was 1 fan failed during 50oC test at 1008th hour.
Formula: A.F. = e (DH/K) x (1/273 + TL - 1/273 +TH)
A.F. is Acceleration Factor that is number of fan failed at TL test divided by number of fan failed at TH.
273 is absolute temperature
e is natural log
DH is Activation Energy
K is Bottzmann's Constant = 8.623x10-5
TL is tested temperature low = 50oC
TH is tested temperature high = 80oC
1.Find out the DH:
Plug in K= 8.623x10-5, TL =50oC, TH =80oC. And, A.F. =3/1.
Thus, 3/1 = e (DH / 8.623x10-5) x (1/273+50 - 1/273+80)
So, DH = 0.36
Note: DH is different from a fan manufacturer to another and from time to time, since it is related to the fan technology and quality technology of the fan manufacturer. Somehow, it is "know-how" of a fan manufacturer.
2.Find out the total test hour at 80oC (T):
The total test hours at 80oC (T)
= 197 x 1008 + 960 +684 + 1008 = 201528 hours
3.Find out the MTTF hours at 80oC at 90% confidence level:
MTTF hours at 90% confidence level = T/r,
Where r is found from the GEM Table as "Attachment B" r = 6.6808
So, MTTF at 80oC, at 90% confidence level = 201528/6.6808 = 30165 hours
4.Find out l (failure rate) at 80oC:
l = 1/MTTF
l = 1/30165 = 3.32 x 10-5
5.Then, we are able to find out MTTF hours at other temperature level at 90% confidence level.
For example, now we want to find the MTTF hours at 60oC at 90% confidence level:
60oC to 80oC A.F. =e(0.36/8.623 x 10-5) x ( 1/273+60 - 1/273 + 80)
So, now A.F. = 2.033
Thus, MTTF at 60oC = 30162 x 2.033 = 61325 hours
Furthermore, l = 1/61325 = 1.63 x 10-5