Applied Reliability Page 1 APPLIED RELIABILITY. Techniques for Reliability Analysis

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1 Applied Reliability Page 1 APPLIED RELIABILITY Techniques for Reliability Analysis with Applied Reliability Tools (ART) (an EXCEL Add-In) and JMP Software AM216 Class 4 Notes Santa Clara University Copyright David C. Trindade, Ph.D. STAT-TECH Spring 2010

2 Applied Reliability Page 2 Review of AR-1 Continued Reliability Data Plotting Properties of Straight Lines Rectification Probability Plotting of Various Distributions Median Ranks Lifetime Distribution Probability Plots in JMP Multicensored Data Kaplan-Meier Product Limit Estimation Example Accelerated Testing True Acceleration Linear Acceleration Models Acceleration Factor Transformations for various Distributions Running Accelerated Tests Temperature Acceleration and Arrhenius Model Activation Energy Eyring Models Determining Model Parameters Matrix Reliability Studies and Example Planning Guidelines Accelerated Analysis in JMP Degradation Modeling Sample Sizes for Accelerated Testing Degradation Analysis in JMP

3 Applied Reliability Page 3 Reliability Data Plotting Graphical Procedures for Checking Models By suitable transformations or by the use of special purpose charting paper, called probability paper, failure data may be plotted such that a linear appearance is a check on the appropriateness of the model distribution. Also graphical procedures often permit alternative estimates of population parameters, especially when censoring occurs.

4 Applied Reliability Page 4 Reliability Data Plotting Properties of Straight Lines Slope m : m rise run y x 2 2 y x 1 1 y x Equation for straight line : y = mx + b where b is the intercept since y = b at x = 0 P 2 (x 2,y 2 ) y (rise) P 1 (x 1,y 1 ) x (run) Straight Line Plot

5 Applied Reliability Page 5 Reliability Data Plotting Rectification Ohm s Law V = IR can be plotted as a straight line y = mx + b by the following associations : V = y I = x R = m b = 0 Plotting voltage V vs. current I should produce a straight line through the origin with slope equal to the constant resistance R. V Slope = R I

6 Applied Reliability Page 6 Reliability Data Plotting Rectification The Gas Law is pv = RT. A plot of pressure vs. volume produces the following graph : p T=Constant Gas Law Plot v If we let y = p, x = 1/v, then the Gas Law becomes A plot of p vs. 1/v should be linear with slope RT. Knowing T, we can estimate the gas constant R from the slope of the line. p 1 p RT v Slope = RT T=Constant 1/v

7 Applied Reliability Page 7 Class Project Rectification The CDF for the exponential distribution is : F t 1 e t Suggest a transformation to an (x,y) plot such that if y is plotted versus x on linear-linear graph paper, then a straight line should result if the failure times follow an exponential distribution. What information does the slope of such a line provide?

8 Applied Reliability Page 8 Class Project Rectification of Exponential Distribution Begin with the CDF formula : Rewrite as F t Take logs of both sides If the exponential distribution holds, a plot of y ln1 Fˆ 1 F straight line with slope. 1 e t t e t Ft t ln1 t versus x=t should appear as a ln 1 F t Slope = t

9 k Applied Reliability Page 9 Table for Probability Plot Exponential Distribution Failure Count Failure Times CDF Estimate Ft ˆ () Transformation ln[1 Ft ˆ( )] X Y 1 t 1 1/n -ln(1-1/n) 2 t 2 2/n -ln(1-2/n) 3 t 3 3/n -ln(1-3/n) k t k k/n -ln(1-k/n) Total of n units on stress

10 Applied Reliability Page 10 Probability Plotting in ART Exponential Distribution Exact Data Example Enter data into column(s). Under Reliability Plotting, select Exponential (Exact Times). Provide needed information. Click OK.

11 Applied Reliability Page 11 Reliability Data Plotting Rectification of Weibull Distribution Rewrite the CDF formula as F Take logs of both sides t / c t 1 e m t / c t e m Take logs of both sides again ln 1 F ln 1 F t t / c m ln1 Ft mlnt mlnc If we plot y ln ln1 F t versus x lnt and the Weibull distribution holds, we should get a straight line with slope m and intercept mlnc. ln ln 1 F t Slope = m intercept = mln c c e intercept/m ln 1 ln t

12 Applied Reliability Page 12 Reliability Data Plotting Rectification Weibull Plots: Rectification can be done in several ways : 1. Plot on linear-linear paper ln ln1 Fˆ vs. ln t 2. Plot on log-log paper ln1 Fˆ vs. t 3. Plot on Weibull probability paper Fˆ vs. t

13 Applied Reliability Page 13 Graphical Techniques Weibull Rectification Examples Transformation Method on Log-Log Paper Specially Designed Weibull Probability Paper

14 Applied Reliability Page 14 Probability Plotting in ART Weibull Distribution Readout Example Enter data into column(s). Under Reliability Plotting, select Weibull (Readout Data). Provide needed information. Click OK.

15 Applied Reliability Page 15 Reliability Data Plotting Rectification for Lognormal Distribution Start with the transformation formula to get the standard normal variate z lnt lnt z s Rewrite the equation as lnt 1 Since z Fˆ set x 1 Fˆ z and y = ln t to get straight line y = mx + b with slope s and intercept lnt 50. Tˆ 50 sz lnt To estimate 50, use the time at z = 0 (or CDF = 50%). To estimate slope, use ˆ s lntˆ / Tˆ ln t Slope = sigma intercept = ln T 50 z = 0 z

16 Applied Reliability Page 16 Reliability Data Plotting Rectification Lognormal Plots: Rectification can be done in several ways : 1) Plot on linear-linear paper z 1 F vs. ln t 2) Plot on linear-log (semi-log) paper z vs. t 3) Plot on normal probability paper F vs. ln t 4) Plot on lognormal probability paper F vs. t

17 Applied Reliability Page 17 Probability Plotting in ART Lognormal Distribution Readout Example Enter data into column(s). Under Reliability Plotting, select Lognormal (Readout Data). Provide needed information. Click OK.

18 Applied Reliability Page 18 Reliability Data Plotting Plotting Positions For interval data, the CDF F(t) estimate is the cumulative number of failures divided by the sample. For exact failure times, this estimate is not optimal. Consider one unit on stress. A plotting position of 1/1=100%, for the failure time of a single unit represents the 100th percentile of the failure distribution. That makes no sense. A single failure time should be closer to the mean or median. For two units on stress, the plotting positions for two ordered failure times would be the 50th and 100th percentiles, which would be unlikely. There are several alternative estimates for plotting positions. The general format is i n a b where i is the failure number, n is the sample size, and a and b are constants. Two common formulas are: i 1 2 n i n 1 We recommend another form called median ranks.

19 Applied Reliability Page 19 Reliability Data Plotting Estimating the CDF F(t): Median Ranks Estimating the CDF F(t) at the ith failure time is best done by using median ranks. An exact formula is available. An approximation is to use: i. Fti 0 3 n 0. 4 where n is the starting number of units under test. Note if one unit is stressed to failure, the estimate is Ft Similarly, if two units are stressed to failure, the CDF plotting position estimates are Ft Ft

20 Applied Reliability Page 20 Lognormal Probability Paper

21 Applied Reliability Page 21 Lognormal Probability Paper

22 Applied Reliability Page 22 Multicensored Data Kaplan-Meier Product Limit Estimation Multicensored data occurs in many situations: in medicine where individuals are added to a study of the effectiveness of a drug after favorable initial results in warranty analysis where systems are installed throughout the year in stress studies where other failure modes show up in reliability where different experiments are combined Kaplan-Meier PLE Let t 2 >t 1. The probability of surviving to time t 2 is equal to the probability of surviving to time t 1 multiplied by the probability of surviving to t 2 given survival to time t 1. In equation form : P T t P T t P T t T t We can calculate conditional probability of survival in each interval and multiplication of such terms gives probability of surviving entire time period of included intervals.

23 Applied Reliability Page 23 Kaplan-Meier PLE Example : There are two groups of identical disk drives. Group one consists of 50 units that were installed in the field on January 1, Group two has 150 units installed on January 1, 2002, one year later. During 2001, there was one reported failure (Group 1). In 2002, there were 7 reported failures among the total units: 3 in group one and 4 in group two. Thus, during the second year, group one had 3 failures Group 1: Group 2: What is the probability of surviving two years? The probability of surviving two years in the field is thus: PT Units removed from observation through censoring are called losses.

24 Applied Reliability Page 24 Class Project Multi-Censored Data : Kaplan Meier Estimator A computer manufacturer had three major shipments to customers during the past year. At customer A, on the 31st of January, 100 new computers were installed. At customer B, 83 days later, 200 hundred new computers started operating. Finally, 170 days following (253 days after customer A), another 150 new computers began running at customer C. All computers run 24 hours per day, seven days per week. By the year end, Customer A had reported failures on three computers, occurring at 512, 2417 and 7,012 hours. Customer B had failures at 3,250 and 5,997 hours. Customer C reported one 105 hour failure. Failing computers were not replaced. Estimate the CDF at each failure time for all installed computers using the KM PLE. A X X X 8,040 hours ( = 335 days) B 200 X X 6,048 hours ( = 252 days) 105 C 150 X 1,248 hours ( = 52 days)

25 Applied Reliability Page 25 Class Project Worksheet Multi-Censored Data : Kaplan Meier Estimator Time to Failure (hrs) Survival Probability KM Product Limit Estimator CDF

26 CDF (%) Applied Reliability Page 26 Class Project KM PL Estimator CDF Plot Time (Hrs)

27 Applied Reliability Page 27 Kaplan-Meier Estimation in ART In ART, select Kaplan-Meier Estimation. Directions are provided. Click OK.

28 Applied Reliability Page 28 Probability Plots in JMP Exact Times Set up data table with three columns: failure times, type (failure 0 or censored 1 observations), and frequency. Under Analyze, select Reliability and Survival > Life Distribution. Enter required information for Time to Event, Censor, and Freq as shown. Click Run Model.

29 Applied Reliability Page 29 Weibull and Lognormal Probability Plots EDF is shown as Nonparametric Distribution. Click LogNormal and Weibull boxes.

30 Applied Reliability Page 30 Weibull and Lognormal Probability Plots For Probability Plots, click radial scale buttons.

31 Applied Reliability Page 31 The Need for Accelerated Testing The Tyranny of Numbers If we wanted to demonstrate a failure rate of no more than 100 FITS at a 90% confidence level, and if we could run the units under stress for 2,500 hours or nearly 3.5 months, and if we allowed no failures, the reliability study would require over 9,200 units!

32 Applied Reliability Page 32 Accelerated Testing What is Accelerated Testing? Stressing at higher than normal conditions to make failures occur earlier. What is true acceleration? True acceleration occurs when levels of increasing stress cause things to happen faster. The failure mechanisms are exactly the same as seen under normal stress, only the time scale has been changed. Imagine a video tape in fast play mode. True acceleration is, therefore, just a transformation of the time scale.

33 Applied Reliability Page 33 Physical Acceleration Models Linear Acceleration If we know the life distribution of units operating at a high stress, and we apply an appropriate time scale transformation to a lower stress condition, we can derive the life distribution at that lower stress. The transformations are almost always restricted to simple constant multipliers of the time scale. When every time of failure is multiplied by the same constant value to obtain the projected results at another operating stress, we have linear acceleration. In other words, failure time low stress = AF x failure time high stress where AF is called an acceleration factor.

34 Applied Reliability Page 34 Acceleration Factor The acceleration factor can be defined as the ratio of the time to a given CDF value under low stress conditions to the time to that same CDF value under high stress conditions, that is: AF t t S S low high F S low F S high For true, linear acceleration, the acceleration factor, once determined, is the same for any CDF percentile value. Thus, AF t50 t S low 10S low t t 50 Shigh 10 Shigh t t 1 1 S low Shigh For example, it takes 1,000 hours to reach 10% failures under high stress. If the AF is 25 to field use, it should take 25,000 hours to achieve 10% failures in the field.

35 Applied Reliability Page 35 Time Transformations for Lifetime Distributions Exponential u s / AF Exponential distribution is the only case where the failure rate varies inversely with the acceleration factor. Not true in general. Weibull c AF c U m U m Lognormal For both the Weibull and the lognormal distributions, equality of shape parameters is a requirement for linear acceleration. This stipulation translates into parallel slopes for different stress levels on probability plots. S S T AF T s 50U 50S U s S

36 Applied Reliability Page 36 Checklist of Events for Accelerated Modeling 1) Run at least two stress cells per accelerating factor under different accelerated conditions. 2) Confirm the failure distribution, for example, lognormal or Weibull, for each set of conditions. Probability plots are useful here. 3) For each cell, estimate the failure distribution parameters, and verify that shape parameters across cells are statistically equal. 4) Choose the acceleration model for the mechanism involved, check graphically, and estimate the parameters of the model. 5) For the specified field conditions, apply the appropriate acceleration multiplier to transform the scale parameter (e.g., T 50 or characteristic life) under stress to the scale parameter in the field. 6) Use the failure distribution model and the common shape parameter along with the field scale parameter to estimate the field failure distribution.

37 Applied Reliability Page 37 Temperature Acceleration The Arrhenius Model The Arrhenius model states that the log of the time to a given CDF value is proportional to an activation energy (E A or H) and inversely proportional to the temperature in degrees Kelvin (which is degrees Celsius). Express the relationship for T 50 as T A e H / kt where A 50 is a constant, and the Boltzmann constant k = eV/ o K. The acceleration factor between temperatures T 1 and T 2 becomes : AF We can simplify the above form to where TF is tabulated (in units of ev -1 ). T50 at T 1 1 e H k T at T 50 AF 2 2 e H TF / 1/ T 1/ T 1 2

38 Ac c ele ra tio n F a c to r Applied Reliability Page 38 Tables for Arrhenius Model Lower Temperature TF Values for Calculating Acceleration Factor. Higher Temperature T 2 ( o C) T 1 ( o C) A c c e le ra t io n F a c t or v s. T e mp e r a tu re 1 0,0 0 0, ,0 0 0,0 0 0 R e lativ e t o 55 o C fo r a R an g e o f Ac tiv a tio n E n erg ie s E A , , , T e m p e r atu re ( o C) AF for Various Temperatures and E A for Field at 55 o C

39 Applied Reliability Page 39 Class Project Acceleration Factors Devices are stressed at a junction temperature T j of 135 o C. Field usage is specified at T j = 55 o C. 1) For a failure mechanism having an activation energy of 0.8 ev, determine the AF. 2) If the failure distribution is lognormal, and the stress T 50 is 500 hours, what is the field T 50? 3) If the stress sigma is 1.0, what is the field sigma?

40 Applied Reliability Page 40 Class Project Acceleration Factors : Activation Energy Using the TF table in the notes, and the equation, AF e HxTF determine the effect of doubling an activation energy from 0.65 ev to 1.3eV, for stress and use temperatures of 125 o C and 45 o C, respectively.

41 Applied Reliability Page 41 Arrhenius Model in ART Select Accelerated Testing > Arrhenius Acceleration Factors. Enter information. Click Calculate.

42 Applied Reliability Page 42 Activation Energy The activation energy is specific to a given failure mechanism and must be determined empirically or obtained from reliability literature. The higher the number, the greater the acceleration by temperature. Some typical values are shown below. Mechanism H Gate Oxide Electromigration Corrosion Mobile Ion Movement ev ev ev ev

43 Applied Reliability Page 43 The Eyring Model for Acceleration The Eyring model is a general solution to the problem of additional stresses. The Eyring model for temperature and a second stress S 1 is: T AT e e 50 H/ kt B C/ T S Additional stress terms may be added. For example, a temperature and voltage acceleration model can be written from the Eyring model as where H kt T Ae V / 50 = 0, C = 0, S 1 = lnv, and B = - 1 Note e B C / T S ln e V ln e V V 1

44 Applied Reliability Page 44 Class Project Accelerated Modeling Radiation Dosage Example Components made by a certain manufacturer have a failure mechanism described by an exponential distribution. Typical use temperature is 45 o C. An accelerated life test was run on a sample of 100 units. The radiation dosage was a factor of 10 over normal use exposure. A model for the time to failure as a function of dosage is : T = AD -alpha, where A is a constant, D is the radiation dosage rate, and alpha is an empirically determined constant equal to % failures occurred on life test, and the stress MTTF estimate was 1,800 hrs. What is the estimate of the MTTF under normal use conditions?

45 Applied Reliability Page 45 Acceleration Models Other Examples Electromigration T AJ n e H / k T 50 where temperature (T) and current density (J) are accelerating conditions. Mechanical Stress Cycle Failures (Coffin Manson Equation) N f Af T T max Note the acceleration factor is the ratio of the number of cycles under field conditions to cycles under stress. For example, consider temperature cycle testing. If one stress is at -65 to 150 o C and a second is at 0 to 125 o C, with equal cycle frequency, f, the acceleration factor from high to low stress is: AF

46 Applied Reliability Page 46 Model Parameters Empirical Determination by Rectification We ll use the electromigration formula T AJ n e H / k T 50 Take logs of both sides to get 1 ln T50 ln A n ln J H k T For fixed current density J, plot of lnt 50 vs. 1 kt should produce a straight line plot with slope H. For fixed temperature T, plot of lnt 50 vs. ln J should give a straight line plot with slope -n.

47 Applied Reliability Page 47 Matrix Reliability Studies To provide data for estimating parameters, we ll run a matrix series of reliability experiments. Temperature T 1 T 2 T 3 Current Density J 1 J 2 J 3 T 5011 T 5012 T 5013 T 5021 T 5022 T 5023 T 5031 T 5032 T 5033 Plot, for each row, lnt 50 vs. 1/kT to estimate H from the slope of each line. Obtain three estimates. Plot, for each column, lnt 50 vs. ln J to estimate -n from the slope of each line. Obtain three estimates For the three estimates of H, if all are nearly equal, use average or median. Similarly, average the three estimate of n. A better approach, to get one H and one n, is to perform multiple regression analysis using y = lnt 50, x 1 = ln J, and x 2 = 1/kT

48 Applied Reliability Page 48 Matrix Example ln T 50 = ln A - n ln J + H/kT T 50 Cell Estimates from Probability Plots Current Density (amps/cm2) Temperature(oC) J ,329, , , ,800 59,900 14, ,200 14,760 4,915 I/kT H Estimates ln J n estimates Average H Average n Y = X X 2 REGRESSION ANALYSIS SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 9 Table for Regression Using EXCEL Data Analysis Tools X1=1/kT X2=ln J Y=ln T ANOVA df SS MS F Significance F Regression E-09 Residual Total Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept X1=1/kT X2=ln J

49 Applied Reliability Page 49 Planning Matrix Reliability Studies A few rules of thumb to keep in mind : 1) Allocate the units to the individual cells so that you obtain approximately an equal number of failures per cell. Dividing the units available for stress equally among cells is not advised. The higher stress cells will produce higher cumulative percent failures than lower stress cells. If more units are not used in the lower stress cells, there may not be any failures. 2) Plot each cell data on probability paper to check fit to the assumed failure distribution. The shape parameters for each cell should be roughly the same. Verify using MLE methods. 3) Best estimates of acceleration model parameters are obtained by using multiple regression with T 50 s weighted by the inverse variances of the T 50 estimates of each cell.

50 Applied Reliability Page 50 Class Project Accelerated Modeling : Lognormal Case Components made by a certain manufacturer have a failure mechanism described by a lognormal distribution. Typical use temperature is 35 o C. An accelerated life test was run employing three different temperature cells: 175 o C, 150 o C, and 125 o C. Thus (1/kT Kelvin )=25.89, 27.42, and respectively. Analysis of the data in each cell confirmed a lognormal distribution with common sigma of 3.9. The T 50 s estimated for the three cells were, respectively, 75, 150, and 500 hours. Plot ln T 50 vs. 1/kT. Determine the activation energy from the slope. 7 ln T 50 6 T( o C) 1/kT T 50 ln T Project a field usage T /kT

51 Applied Reliability Page 51 What AF to Use? To project field T 50, we can use one of the three acceleration factors between different stress temperatures and field 35 o C (activation energy = 0.58eV) Which is correct? Is there a better way to do this projection? Answer: Use the regression model with calculated slope and intercept to predict T 50 at field temperature.

52 Applied Reliability Page 52 Determining Slope and Intercept in Spreadsheet Results =intercept(y-range; x-range) =slope(y-range; x-range) Note with slope and intercept, we can project T 50 at any temperature using the model T 50 = e [intercept +(slope)/kt] At use temperature, kt = ( )x( ) = So field T 50 is predicted as T 50 = e [ (0.5822)/( )] = e (11.11 ) = 66,732 hrs Compare to previous T 50 estimate using only one cell.

53 Applied Reliability Page 53 Accelerated Test Example JMP Analysis Company ABC has developed a new integrated circuit and wishes to estimate the reliability of the product after one year (8,766 hours) under use conditions at 55 C. An accelerated stress is run for 2000 hours on a total of 400 units. Four temperature cells are run, with temperatures 85, 100, 125, and 150 C. Allocated sample sizes are 160, 120, 80, and 40, respectively. Readouts are conducted at 0, 24, 48, 96, 168, 336, 500, 750, and 1000 hours. The results are shown in the next table.

54 Applied Reliability Page 54 Accelerated Test Results

55 Applied Reliability Page 55 JMP Data Table

56 Applied Reliability Page 56 JMP Fit Life by X Model Selecting Analyze > Reliability and Survival > Fit Life by X, we then fill in the dialog box.

57 Applied Reliability Page 57 Fit Life by X Results: Scatterplot Not too informative: shows each temperature cell had the same readouts starting at 24, 48, 96, 1000 hours.

58 Applied Reliability Page 58 Fit Life by X Results: Nonparametric Overlay Adjusting scale limits, display shows nonparametric fraction failures at the end of each readout interval on a lognormal probability plot scale. Lognormal looks promising for the failure distribution.

59 Applied Reliability Page 59 Fit Life by X Diagnostics Lognormal fit looks good. Constrained equal slopes looks good also. Regression acceleration model appears good also.

60 Applied Reliability Page 60 Fit Life by X Distribution Profiler At field use temperature of 55 C, the expected fraction failing after one year (8766 hours) is about 18%.

61 Applied Reliability Page 61 Fit Life by X: Density Curves and Quantile Lines Quantile lines at 0.1,0.2, and 0.5.

62 Applied Reliability Page 62 Fit Life by X: Quantile Profiler Time to reach 5% fraction failure at use conditions.

63 Applied Reliability Page 63 Fit Life by X: Acceleration Factor Profiler Acceleration factor between 125 and 85 C. E A estimate. Common s estimate. Equation to estimate lognormal mean at any temperature.

64 Measurement Applied Reliability Page 64 Degradation Modeling Analysis Without Failures Using Parametric Data Assume there exists some function of a measurable parameter that is changing in time. Then, for each test unit at each readout, it is possible to plot the degradation of the actual value of a parameter in time. The time to failure for each unit can be estimated by extrapolation using the time to cross the failure point D or, equivalently, by dividing the distance to failure D by the rate of degradation for each unit. The derived failure times form the estimated life distribution. Repeat this procedure for every unit in every stress cell to fit an acceleration model. x x x x x x x x xx x xx x Failure Limit T r1 T r2 t 1 t 2 t 3 t 4 t 5 Time

65 Applied Reliability Page 65 Degradation Example A resistor used in a power supply is known to drift over time, degrading eventually to failure. Higher temperatures produce faster degradation. At a 30% change, the power supply can no longer operate. Preliminary results indicate an Arrhenius model applies. Two stress cells, each with ten resistors, are run at 105 o C and 125 o C. Percent change in resistance from time zero is measured for each unit at 24, 96, and 168 hours. Estimate H and project an average failure rate over 100,000 hours of field life at use temperature of 30 o C.

66 Percent from Zero Applied Reliability Page 66 Data for Degradation Example Degradation Plots Hr 96 Hr 168 Hr

67 Applied Reliability Page 67 Least Squares Estimation Through Origin Method 1 To estimate the rate of degradation for each unit, we use the formula for the least squares slope through the origin: xy 2 x where x is the readout time and y is the percent change. Then we divide 30% (the distance D to failure) by each rate to get the estimated time to failure t f. Assuming lognormal times to failure, we take the natural logs to calculate the parameters of the lognormal distribution. Since we have complete data on all units, the formulas are lnt / n T e i 50 s stdev( lnt i )

68 Applied Reliability Page 68 Degradation Calculations 105 o Data 125 o Data Component 24 Hr 96 Hr 168 Hr Tf lntf 24 Hr 96 Hr 168 Hr Tf lntf Average of ln(tf) 7.0 Average of ln(tf) 5.7 T T Sigma 0.85 Sigma 0.76 Pooled Sigma 0.81 Example calculation: Component 1, 105ºC Data xy x Tf

69 Applied Reliability Page 69 Activation Energy for Degradation Example To estimate H, use H T kln T T1 1 T So, AF between 125ºC and 30ºC is 2,009. T 50 estimate at 30ºC is 2009x309 = 621,000 hrs. Sigma estimate is Lognormal CDF at 100K hours is 1.2%. AFR(100K) is estimated at 120 FITs. These methods are very powerful and may provide valuable information with small numbers of units. If the drift mechanism is understood, degradation analysis can be a very useful tool.

70 Applied Reliability Page 70 Degradation Analysis in JMP Data Table

71 Applied Reliability Page 71 JMP Graph Builder Plot Failure Level

72 Applied Reliability Page 72 Least Squares Estimation of Failure Times To estimate the time to failure for each unit, we must first estimate the rate of degradation R i for each unit, measured by the slope of each line. Since the change for each unit is zero at time zero, we use the formula for the least squares slope through the origin: R xy x i i i 2 i where x i is the readout time and y i is the percent change. Alternatively, an approximate estimate of the rate is (168 hour parameter readout value)/168. Then, the time to 30% change is found by dividing 30 by the degradation rate for each unit. In JMP, this estimate is found by using JMP s formula capabilities in new columns.

73 Applied Reliability Page 73 Subset of Degradation Data Table with Derived Columns

74 Applied Reliability Page 74 Degradation Analysis Using Fit Life by X Platform on Subset Table

75 Applied Reliability Page 75 Degradation Analysis Using Fit Life by X Platform: Plots

76 Applied Reliability Page 76 Degradation Analysis Using Fit Life by X Platform: Statistics Activation Energy E A Common Sigma Arrhenius Temperature Acceleration Formula

77 Applied Reliability Page 77 Distribution Profiler Degradation Analysis Using Fit Life by X Platform: Profilers Hazard Profiler CDF Estimate at 100,000 hours Hazard Rate Estimate at 100,000 hours

78 Applied Reliability Page 78 Sample Sizes for Accelerated Testing To demonstrate that a product satisfies a stated average failure rate (AFR) under a Weibull or lognormal distribution, we refer to The ASQC Basic References in Quality Control, Volume 15: How to Determine Sample Size and Estimate Failure Rate in Life Testing, by E.C. Moura. Procedure: Weibull (assumes shape parameter m known): Calculate the probability of failure associated with the AFR requirement for the time period T related to the AFR: F(T ) 1 e AFRT Calculate the population characteristic life that will satisfy the AFR requirement: T c 1 1 F(T ) m

79 Applied Reliability Page 79 Sample Sizes for Accelerated Testing (Weibull Distribution) For the specified stress time, T s, Estimate the CDF for the equivalent field time T e under accelerated conditions T e F T T e s AF 1 e Obtain the chi-square distribution percentile 2 2r 2, 1 that corresponds to the desired confidence probability with 2r+2 degrees of freedom, where r is the allowed number of failures. Te c m Calculate the minimum sample size using the formula n 2 2r 2, 1 2F T e

80 Applied Reliability Page 80 Example of Sample Size Calculation for Weibull Distribution The failure times for a specific capacitor follow a Weibull distribution with shape parameter m = 1.5. The failure specification is that an AFR of 0.1%/Khr cannot be exceeded during the first 10,000 hours. For a temperature stress of 125 C for 1000 hours, an acceleration factor of AF = 30 is applicable. Allowing up to one failure, how many units are required to satisfy the AFR with 90% confidence? Solution: F c F n 10, ln , 000 / 215, , e 2 F 1 e 10, ; , x ,

81 Applied Reliability Page 81 Sample Sizes for Accelerated Testing (Lognormal Distribution) Procedure: Lognormal (assumes shape parameter s known): Calculate the probability of failure associated with the AFR requirement for the time period T related to the AFR: AFRT F(T ) 1 e For the normal distribution, obtain the standard normal Z variate that corresponds to this CDF Calculate the lognormal T50 that will satisfy the AFR requirement T 50 Te sz For the specified stress time, T s, Estimate the CDF for the equivalent field time T e under accelerated conditions F T FT AF e ln T AF / T s s 50 s

82 Applied Reliability Page 82 Sample Sizes for Accelerated Testing (Lognormal Distribution) Obtain the chi-square distribution percentile 2 2r 2, 1 that corresponds to the desired confidence probability with 2r+2 degrees of freedom, where r is the allowed number of failures. Calculate the minimum sample size using the formula n 2 2r 2, 1 2F T e

83 Applied Reliability Page 83 Example of Sample Size Calculation for Lognormal Distribution The failure times for a specific inductor follow a lognormal distribution with shape parameter s = 1.5. The failure specification is that an AFR of 0.1%/Khr cannot be exceeded during the first 10,000 hours. For a temperature stress of 125 C for 1000 hours, an acceleration factor of AF = 30 is applicable. Allowing up to one failure, how many units are required to satisfy the AFR with 90% confidence? F Solution: 10, 000 T 50 10, 000e F n 2 F 30, e ln 2 4; , , 000 x , , 000/ 328,

84 Applied Reliability Page 84 Appendix

85 Applied Reliability Page 85 Class Project Worksheet Multi-Censored Data : Kaplan Meier Estimator Time to Failure (hrs) Survival Probabilit y KM Product Limit Estimator CDF (%) Censoring Time Censoring Time (449/450) = = (448/449) = = (298/299) = = (297/298) = = (296/297) = = (97/98) = = Censoring Time

86 Applied Reliability Page 86 Class Project Acceleration Factors Devices are stressed at a junction temperature T j 135 O C. of Field usage is specified at T j = 55 O C. 1) For a failure mechanism having an activation energy of 0.8 ev, determine the AF. H TF AF e e 0. 8 ( 6. 9 ) 250 2) If the failure distribution is lognormal, and the stress T 50 is 500 hours, what is the field T 50? field T 50 = AF(stress T 50 ) = 250(500) = 125,000 hrs 3) If the stress sigma is 1.0, what is the field sigma? field sigma = stress sigma = 1.0

87 Applied Reliability Page 87 Class Project Acceleration Factors : Activation Energy Using the TF table in the notes, and the equation, AF e HxTF determine the effect of doubling an activation energy from 0.65 ev to 1.3eV, for stress and use temperatures of 125 o C and 45 o C, respectively. HxTF 0. 65( 7. 3) AF e e 115 HxTF 1. 3( 7. 3) AF e e 13, 227 Note that = 13,225. Thus, doubling H squares AF. x AF e e ( 0. 65) x e e e AF

88 Applied Reliability Page 88 Class Project Accelerated Modeling Radiation Dosage Example Components made by a certain manufacturer have a failure mechanism described by an exponential distribution. Typical use temperature is 45 o C. An accelerated life test was run on a sample of 100 units. The radiation dosage was a factor of 10 over normal use exposure. A model for the time to failure as a function of dosage is : T = AD -alpha, where A is a constant, D is the radiation dosage rate, and alpha is an empirically determined constant equal to % failures occurred on life test, and the stress MTTF estimate was 1,800 hrs. What is the estimate of the MTTF under normal use conditions? AF TU DU T 1. 5 D S S a lp h a MTTF U = AF(MTTF S ) = 31.6(1800) = 56,921 hrs

89 Applied Reliability Page 89 Class Project Accelerated Modeling : Lognormal Case Components made by a certain manufacturer have a failure mechanism described by a lognormal distribution. Typical use temperature is 35 o C. An accelerated life test was run employing three different temperature cells: 175 o C, 150 o C, and 125 o C. Thus (1/kT Kelvin )=25.89, 27.42, and respectively. Analysis of the data in each cell confirmed a lognormal distribution with common sigma of 3.9. The T 50 s estimated for the three cells were, respectively, 75, 150, and 500 hours. Plot ln T 50 vs. 1/kT. Determine the activation energy from the slope. 7 ln T 50 T( o C) 1/kT T 50 ln T 50 6 x x x slope = /kT