Reading a Datasheet

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.

Let’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.  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:

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

Stress Temperature = 85°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 Boltzmann’s constant = 8.617×10^-5eV/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 = °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 IC’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 ≃ 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 isn’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 (10⁹) 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.

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Related:

Sample Size – success testing (article)

life testing question (article)

Accelerated life testing first steps (article)