Reading a datasheet to determine a reliability value may take some investigative work. Whenever I see a fit rate based on failure-free testing, I am curious about how they did the testing and the calculations.

Letâ€™s explore just the calculations for a CMOS component that has undergone the JEDEC standard JESD47 High-Temperature Operating Life Test. The testing preconditioned 231 samples and operated them at both high temperature and high voltages to accelerated the testing.Â The test ran for 1008 hours. Then the calculations used a 60% confidence to determine the expected maximum failure rate. They calculated a FIT rate of 123.

Exploring the acceleration first, we have two accelerating stresses, temperature, and voltage. We are given the testing and assumed use conditions as:

OperatingÂ Temperature = 50Â°C and Operating Voltage = 2.8V

Stress Temperature = 85Â°C and Stress Voltage = 3.1V

The overall acceleration factor, AF_{overall} is the product of the temperature and voltage acceleration factors, assuming there is no interaction of the stresses and failure mechanisms.

$$ \large\displaystyle A{{F}_{overall}}=A{{F}_{temperature}}\times A{{F}_{voltage}}$$

The overall acceleration factor, AFoverall is the product of the temperature and voltage acceleration factors, assuming there is no interaction of the stresses and failure mechanisms.

$$ \large\displaystyle A{{F}_{T}}={{e}^{\frac{{{E}_{a}}}{k}\left( \frac{1}{{{T}_{o}}}-\frac{1}{{{T}_{s}}} \right)}}$$

where

E_{a} is the activation energy

k is Boltzmannâ€™s constant = 8.617×10^{-5}eV/K

T_{o} is the operating (use) temperature in Kelvin

T_{s} is the stress (testing) temperature in Kelvin

Kelvin is absolute temperature and is converted from the Centigrade scale by K = Â°C + 273.15

A major consideration when using the Arrhenius equation is the activation energy, E_{a}. In this case, the vendor chose 0.6 eV/K which corresponds to a failure mechanism of charge loss, time dependent dielectric breakdown and electrolytic corrosion. The range of common activation energies for CMOS includes values from 0.1 to 1.1, and depending on the dominant or expected failure mechanism(s) may significantly change the acceleration factor calculation. Without further information about the expected failure mechanisms, we are assuming the activation energy is appropriate.

The calculation then is

$$ \large\displaystyle A{{F}_{T}}={{e}^{\frac{0.6}{8.617\times {{10}^{-5}}}\left( \frac{1}{323}-\frac{1}{358} \right)}}=8.2$$

The AF_{voltage} uses empirically derived relationship for the life to thin oxide voltage stress

$$ \large\displaystyle A{{F}_{T}}={{e}^{\beta \left( {{V}_{s}}-{{V}_{o}} \right)}}$$

where

? = empirically derived parameter, often within 4 to 6 volt^{-1}

V_{s} is the stress (testing) voltage

V_{o} is the operating (use) voltage

This model is for the specific failure mechanisms related to the pinholes and contamination of the gate oxide in CMOS ICâ€™s . The purity and cleanliness is important and applying a test voltage assists in identifying these early failure modes.

The calculation then is

$$ \large\displaystyle A{{F}_{T}}={{e}^{4.5\left( 3.1-2.8 \right)}}=3.9$$

Therefore the overall AF is 8.2 x 3.9 = 31.98 â‰ƒ 32.

The next step is to calculate the failure rate. The testing has 231 units survive without failure for the 1008 hours of testing. The common estimate for failure rate is the total number of failures divided by the total hours of testing (number of samples times the duration in hours of the test). In this case, there are no failures, thus resulting in a failure rate of zero, which isnâ€™t likely.

Using the upper confidence level for an estimate of the failure rate, we can set a bound of the expected failure rate. The Poisson distribution is useful in this case, since the calculation of a binomial distribution is tedious, and the Poisson approximation works in this case. The probability of failure, p, is small for a sample during each hour, plus the samples size, n, is large, yet the product np < 5 (making the normal approximation inappropriate.)

The datasheet uses a confidence level of 60%, which is low (making the estimated failure rate lower than if using a higher confidence level). Given ? = 0.6 in this case, we can use the Poisson distribution to calculate the upper bound of the demonstrated failure rate. The Poisson distribution is stated as

$$ \large\displaystyle P\left( x \right)=\frac{{{\mu }^{x}}{{e}^{-\mu }}}{x!}$$

where

x is the number of observed failures

? is the mean and equal to n?

P(x) is the Poisson statistic at a specific confidence level, ?.

Given,

n is the number of sample-hours tested (total time)

? is the failure rate, ? = ?/n

Setting P(0) equal to (1-?) as the probability of observing zero failures with 60% confidence, we can solve for the corresponding failure rate.

$$ \large\displaystyle \begin{array}{l}P\left( 0 \right)=\frac{{{\mu }^{0}}{{e}^{-\mu }}}{0!}={{e}^{-\mu }}=1-0.6\\\mu =-\ln \left( 0.4 \right)-0.916\end{array}$$

To calculate the 60% upper confidence of the failure rate, then we use

$$ \large\displaystyle \lambda =\frac{\mu }{n}=\frac{0.916}{231\times 1008\times 32}=1.23\times {{10}^{-7}}$$

Converting to units of FITS (failure per billion (10^{9}) hours) we get a 123 FIT rate. BTW, using 90% confidence, ?=2.3 and ? = 310 FIT rate, or almost three times higher.

And, this is only the failure rate. To determine Reliability at seven years of use, for example, we use the exponential distribution reliability function.

For 7 years of continuous operation, we have time, t, as 8760 hours per year x 7 years for 61,320 hours. And, the failure rate, ?, is 1.23×10^{-7} failures per hour. Therefore,

Therefore, the reliability at 7 years is

$$ \large\displaystyle R\left( t \right)={{e}^{-\mu }}={{e}^{-\left( 1.23\times {{10}^{-7}} \right)\left( 61320 \right)}}=0.992$$

Thus expecting less than 1% of units to fail over the 7 years of operation.

Related:

Sample Size – success testingÂ (article)

life testing questionÂ (article)

Accelerated life testing first stepsÂ (article)

Syed Hussain says

Hi Fred,

What’s the significance of last equation that you used for 7 years operation and what it’s called?

For acceleration factor calculation, do we really need to consider Voltage Acceleration? Can we just use temp or thermal acceleration factor for FIT rate, In your case if we just use temp acceleration factor for FIT rate then the difference of FIT rate would be significant. Whatâ€™s your thought on this? Typically, customer only interested to know eV and CL factors for FIT rates

Thanks

Syed

Fred Schenkelberg says

Hi Syed,

Good questions. the last equation calculates the probability of successfully operating over 7 years, it is the reliability function. Without it we have information on the failure rate, not the probability of successfully operating over the time period of interest.

The AF calculations are always related to the failure mechanisms involved. voltage accelerates some things, and temperature other failure mechanisms. Just using temperature misses many important failure mechanisms and therefore underestimates the expected life performance.

FIT like MTBF is a failure rate, it is rarely constant and rarely useful by itself. Convert to reliability over the period of time of interest, say first month, warranty period and useful life period.

Activation energy is important for thermally accelerated chemical reaction based failure mechanisms. If that is not the case (say solder joint fatigue and thermal cycling stress) then educate the customers by providing sufficient information for the correct understanding of your product. Confidence levels are great, not by themselves though. You should include sample size, assumed distributions and CL.

cheers,

Fred

Syed Hussain says

Thanks Fred for your details response, very helpful.

I have couple of more questions though. Please respond if you find time

Besides determining the 7 years of use through exponential distribution. I am interested to know how you would calculate the useful life based on 123 FIT that you calculated. According to my calculation, it should be like that (please correct me if I am wrong):

AF x Stress hrs.

32 x 1008/365×24 = 3.68 years (probability of useful life)

Also, for Voltage Acceleration: How did you come up with 4.5V = ?

What technology are you referring to?

Syed

Fred Schenkelberg says

Hi Syed,

A FIT number is a failure rate value, it is not directly used with acceleration factors. FIT like MTBF is the probability of failure per unit time, often hours. It means nothing related to how long the product is going to last directly.

You need to use the appropriate distribution’s reliability function to estimate probability of success over a time period.

Acceleration a factors are to convert between test conditions and use conditions – often used to move the characteristic life of the distribution and assuming the same failure mechanism(s) apply that the slope will remain the same.

The 4.5 eV is from the reference which I seem to have not listed, Lloyd W. Condra’s book Reliability Improvement with Design of Experiments and relates specifically to specific failure mechanisms related to the pinholes and contamination of the gate oxide in CMOS ICâ€™s. There are many papers related to estimating activation energy and you should either determine the value directly with your experiments or use one that is as close as possible to the actual failure mechanism you are modeling.

And, looking at the original post today, it appears some of the formulas are not showing Greek letters. Hope your browser is treating you better. I’ll update the graphics and hope to fix that issue soon.