About the FIT calculation

[Steven Chen]

For the laser diodes reliability analysis, most people refer to "GR-468-CORE" documents to get the analysis methods. Among the reliability parameters, the FIT calculation make me confuse for long time.
In the section "5.19.2 Wear-Out Failure Rate" suggest us use the Goldthwaite Curves to get the FIT number and give us one case for FIT calculation.
I write the case in follows from the GR-468-CORE" document in section 5.19.2:
"For example, a laser diode life test might give ML=1hours (calculated for a 40C normal operation temperature), with =1.0. After 20 years (1.75hours) of operation, the expected failure rate will be ~430FITs"
I do not understand how to get the 430FIT from the Goldthwaite Curves.
For another issues in the same case, from the trend of curve with =1.0, after 10 years, the FIT number is greater than that of 20 years. It seems unreasonable.
I looking forward to hearing the way to calculate the FIT clearly.
Thanks a lot,

Steven Chen

[rwherrick]

Steven,
You bring up a number of issues. Let me deal with them in turn.

1) What the heck does this confusing example on page 5-19 of Telcordia GR-468, release 1 mean? This is so sloppy almost no one could understand it without the arrows on the graph, and even then you need to guess the sigma. I believe what they meant is an example with a median life of 1 million years, and a sigma of ~0.9, and the time of interest is 20 years (175,200h). So the normalized time is 0.175. Please see the discussion below in #4 if you really want to use the curves. Goldthwaite curves have some use in understanding failure distributions, but otherwise are nearly as obsolete as slide rules. Nobody uses them for doing calculations nowadays. By the way, they never could have got two significant digits of accuracy (“430 FITS”) in their answer using Goldthwaite curves. Not that most reliability data allows you that kind of accuracy as anything more than a computational exercise.
2) How do I calculate the FIT rate? (the easy, but more expensive, way) This can be rather easy if you assume a constant failure rate applies. The Chi-squared distribution is the most popular method. Since this is “the laser diode test forum”, a constant failure rate is a very poor assumption for most lasers. Most people use lognormal distributions, and depending on the failure mechanism, sigma varies from 0.2 to as high as 1.5, with values of 0.5 to 0.9 being the most common. This tells you the “variation in strength” of your parts, with small sigmas occurring where the parts all fail at nearly the same time, and large sigmas occurring where they are broadly distributed in time. The other variable is “mu”, which is the natural log of the median time to failure (i.e., ln[TT50%F]). This will vary widely depending on the stress conditions. A typical value for a VCSEL under normal operating conditions might be 15, which would correspond to a median time to failure of 3.3 million hours. But for a more interesting example, let’s say it’s only 13, due to poor laser quality or harsh service. The easy way to calculate FITs is to open Reliasoft’s Weibull++, open a new data sheet, click on the “lognormal tab” at the upper right, and click the calculate button. If you have data, it will analyze it. If you don’t it will ask you for mu and sigma. You enter 13 and 0.6, and click “okay”, and then you can plot out the failure rate versus time. Or if you just want the results for a few different times, you can use the “Quick Calculation Pad” (QCP). So for example, if you want the failure rate after 10 years, you enter “87600h”, and get “1.994E-7” as an answer. Multiply by 1E9 to get 199.4 FITs.
3) What if I don’t have access to Weibull++ (a $600 program)? Excel has a number of useful built-in utilities, including “Weibull”, “LogNormDist”, and “LogInv”. But they are for solving difficult integrals, not straightforward formulae. To calculate f(t), use the complicated equation: =((F15*C15*SQRT(6.283))^-1)*EXP((-0.5*F15^-2)*(LN(C15)-E15)^2), where C15 is time in hours, F15 is sigma, and E15 is mu (or adjust you cell names accordingly). This looks complicated; look in a reference book or on the web for the equation, and you’ll see where this came from. The “FIT rate” is really the hazard rate, and when the cumulative failures are less than a few percent, f(t) ~ h(t). When it is larger, you’ll need to divide f(t) by 1/(S(t)) to get the correct result, where S(t) is the fraction surviving, and can be calculated using the LogNormDist function with the same three inputs mentioned above. Go through the math, and you’ll find that f(t) = 1.987E-7, F(t) = 0.35%, S(t) = R(t) = 0.9965 and h(t) = 1.994E-7, just as with the Reliasoft program.
4) Do you really want to calculate the FIT rate with the Goldthwaite curves? The normalized time is 87,600h / 442,413h = 0.20. Read off the graph, and you’ll get something around 9E7 if you have a detailed graph. Divide by the median time to failure (4E5), and you’ll get something around 200 FITs.

You may find some details on this from my class at SJ State University: http://www.engr.sjsu.edu/mae/faculty%20and%20staff/webpages/herrick/herrick.php.
I believe this was covered in HW 2.

[Steven Chen]

Thank you so much.
Steven