Applied Reliability Page 1 APPLIED RELIABILITY. Techniques for Reliability Analysis

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



S

T

•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

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,

Properties of Straight Lines

Equation for straight line :

y = mx + b

where b is the intercept since y = b at x = 0

x

y

x

x

y

y

run

rise

m















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.

Slope = R V

Reliability Data Plotting

Rectification

The Gas Law is pv = RT. A plot of pressure vs. volume produces the following graph :

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.

T=Constant p

v

Gas Law Plot

Class Project

Rectification

The CDF for the exponential distribution is :

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 ?

 

e

t

Class Project

Rectification of Exponential Distribution

Rewrite as

Take logs of both sides

If the exponential distribution holds, a plot of

versus x=t should appear as a straight line with slope .

Table for Probability Plot

Exponential Distribution

ˆ ( )

F t

ˆ

ln[1

F t

( )]





.

.

.

.

.

.

.

.

.

.

.

.

Exponential Distribution

Exact Data Example

Rectification of Weibull Distribution

Rewrite the CDF formula

as

Take logs of both sides

Take logs of both sides again

If we plot versus

and the Weibull distribution holds, we should get a straight line with slope m and intercept .

Reliability Data Plotting

Rectification

Rectification can be done in several ways : 1. Plot on linear-linear paper

vs. ln t 2. Plot on log-log paper

vs. t

Graphical Techniques

Weibull Rectification Examples

Weibull Distribution

Readout Example

Enter data into column(s). Under Reliability Plotting, select Weibull (Readout Data). Provide needed

Reliability Data Plotting

Rectification for Lognormal Distribution

Start with the transformation formula to get the standard normal variate z

Rewrite the equation as

Since set and

y = ln t to get straight line y = mx + b with slope s

and intercept lnT_{50 .}

s

 

ln

ln

t



s

z



T

 

F

z

x





ˆ



To estimate , use the time at z = 0 (or CDF = 50%). To estimate slope, use

Rectification

Rectification can be done in several ways : 1) Plot on linear-linear paper

2) Plot on linear-log (semi-log) paper

3) Plot on normal probability paper

4) Plot on lognormal probability paper

Lognormal Distribution

Readout Example

Plotting Positions

We recommend another form called median ranks.

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

where i is the failure number, n is the sample size, and a and b are constants. Two common formulas are:

Reliability Data Plotting

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:

where n is the starting number of units under test. Note if one unit is stressed to failure, the estimate is

Similarly, if two units are stressed to failure, the CDF plotting position estimates are

 



.

.

F t

i

n







0 3

0 4

 

 

 

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 t₂>t₁. The probability of surviving to time t₂ is equal to the probability of surviving to time t₁ multiplied by the probability of surviving to t₂ given survival to time t₁. In equation form :

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.







 



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, 2001. 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.

2001 2002

Group 1: 50 |---1---|---3---|

Group 2: 150 |---4---|

What is the probability of surviving two years?

The probability of surviving two years in the field is thus:





0

.

915

49

46

200

195

2









T

P

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.

Class Project Worksheet

Multi-Censored Data : Kaplan Meier Estimator

Kaplan-Meier Estimation in

ART

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

Probability Plots

Probability Plots

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

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

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

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



















For true, linear acceleration, the acceleration

factor, once determined, is the same for any CDF percentile value. Thus,

AF

t

t

t

t

t

t







Time Transformations for

Lifetime Distributions

Exponential

Exponential distribution is the only case where the

failure rate varies inversely with the acceleration factor. Not true in general.

Weibull

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

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., T₅₀ 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

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_Aor H) and inversely

proportional to the temperature in degrees Kelvin

(which is 273.16 + degrees Celsius). Express the relationship for T₅₀ as

where A₅₀ is a constant, and the Boltzmann

constant k = 0.00008617eV/o_{K. The acceleration}

factor between temperatures T₁ and T₂ becomes :

We can simplify the above form to

where TF is tabulated (in units of eV-1_).

Lower Temperature T₁ (oC) 65 75 85 95 105 115 125 135 145 155 25 4.6 5.6 6.5 7.4 8.2 9.0 9.8 10.5 11.2 11.8 35 3.3 4.3 5.3 6.1 7.0 7.8 8.5 9.2 9.9 10.6 45 2.2 3.1 4.1 5.0 5.8 6.6 7.3 8.0 8.7 9.4 55 1.0 2.0 3.0 3.8 4.7 5.5 6.2 6.9 7.6 8.3 65 0.99 1.9 2.8 3.6 4.4 5.2 5.9 6.6 7.2 75 0.93 1.8 2.6 3.4 4.2 4.9 5.6 6.2 85 0.88 1.7 2.5 3.3 4.0 4.6 5.3 95 0.83 1.6 2.4 3.1 3.8 4.4 105 0.79 1.5 2.3 2.9 3.6 115 0.75 1.5 2.1 2.8 125 0.71 1.4 2.0 135 0.68 1.3 145 0.65 Higher Temperature T₂ (oC)

TF Values for Calculating Acceleration Factor.

A c c e le ra t io n F a c t or v s . T e m p e r a tu re 1 1 0 1 0 0 1 ,0 0 0 1 0 ,0 0 0 1 0 0 ,0 0 0 1 ,0 0 0 ,0 0 0 1 0 ,0 0 0 ,0 0 0 5 0 6 0 7 0 8 0 9 0 1 0 0 1 1 0 1 2 0 1 3 0 1 4 0 1 5 0 1 6 0 1 7 0 1 8 0 1 9 0 2 0 0 T e m p e r atu r e (oC) A c c e le r a ti o n F a c to r 1 . 4 EA

R e la tiv e t o 55 oC fo r a R a n g e o f Ac tiv a tio n E n e r g ie s

1 . 2 1 . 0 0 . 8 0 . 6 0 . 4 0 . 2 0 . 0

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 T₅₀ is 500 hours, what is the field T₅₀? 3) If the stress sigma is 1.0, what is the field

Class Project

Acceleration Factors : Activation Energy

AF



e

determine the effect of doubling an activation

energy from 0.65 eV to 1.3eV, for stress and use

Arrhenius Model in ART

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



Gate Oxide 0.2 - 0.3 eV

Electromigration 0.5 - 1.2 eV

Corrosion 0.3 - 0.6 eV

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 S₁ is:

 





T



AT e

e

Additional stress terms may be added. For example, a temperature and voltage

acceleration model can be written from the Eyring

Accelerated Modeling

Radiation Dosage Example

Components made by a certain manufacturer have a failure mechanism described by an exponential

distribution. Typical use temperature is 45oC.

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

where A is a constant, D is the radiation dosage rate, and alpha is an empirically determined constant equal to 1.5.

10% failures occurred on life test, and the stress

MTTF estimate was 1,800 hrs.

Acceleration Models

Other Examples

Electromigration

where temperature (T) and current density (J) are accelerating conditions.

Mechanical Stress Cycle Failures (Coffin Manson Equation)

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 150o _{C and a} second is at 0 to 125o _{C, with equal cycle frequency, f,} the acceleration factor from high to low stress is:

Model Parameters

We‟ll use the electromigration formula

T



AJ

e

Take logs of both sides to get

ln

T

ln

A

n

ln

J

H

k

1

















T

For fixed current density J, plot of lnT₅₀ vs. should produce a straight line plot with slope H.

For fixed temperature T, plot of lnT₅₀ vs. ln J should give a straight line plot with slope -n.

1

kT

Matrix Reliability Studies

To provide data for estimating parameters, we‟ll run a matrix series of reliability experiments.

T₅₀₁₁ T₅₀₁₂ T₅₀₁₃ T₅₀₂₁ T₅₀₂₂ T₅₀₂₃ T₅₀₃₁ T₅₀₃₂ T₅₀₃₃ Temperature Current Density J₁ J₂ J₃ T₁ T₂ T₃

Plot, for each row, lnT₅₀ vs. 1/kT to estimate H

from the slope of each line. Obtain three estimates. Plot, for each column, lnT₅₀ 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

Table for Regression Using EXCEL Data Analysis Tools X1=1/kT X2=ln J Y=ln T50 31.1 4.61 14.1 29.2 4.61 12.8 27.4 4.61 11.6 31.1 5.52 12.4 29.2 5.52 11.0 27.4 5.52 9.6 31.1 6.21 11.1 29.2 6.21 9.6 27.4 6.21 8.5

REGRESSION ANALYSIS SUMMARY OUTPUT

Regression Statistics Multiple R 0.9990 R Square 0.9979 Adjusted R Square 0.9972 Standard Error 0.0932 Observations 9 ANOVA df SS MS F Significance F Regression 2 24.977 12.488 1437.289 9.037E-09 Residual 6 0.052 0.009 Total 8 25.029

Coefficients Standard Error t Stat P-value Lower 95% Upper 95%

Intercept 0.951 0.655 1.452 0.197 -0.652 2.554 X1=1/kT 0.712 0.021 34.615 0.000 0.662 0.762 X2=ln J -1.941 0.047 -40.944 0.000 -2.058 -1.825 Y = ₀+ ₁X₁ + ₂X₂ ln T₅₀ = ln A - n ln J + H/kT I/kT H Estimates ln J 31.1 29.2 27.4 4.61 14.1 12.8 11.6 0.679 5.52 12.4 11.0 9.6 0.760 6.21 11.1 9.6 8.5 0.708 n estimates -1.9 -2.0 -1.9 Average H Average n -1.9 0.716 Current Density (amps/cm2) Temperature(oC) J 100 125 150 100 1,329,000 362,200 109,100 250 242,800 59,900 14,760 500 66,200 14,760 4,915

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

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 35oC.

An accelerated life test was run employing three

different temperature cells: 175oC, 150oC, and 125oC. Thus (1/kT_Kelvin)=25.89, 27.42, and 29.15 respectively. Analysis of the data in each cell confirmed a lognormal distribution with common

sigma of 3.9. The T₅₀‟s estimated for the three cells

were, respectively, 75, 150, and 500 hours. Plot

ln T₅₀ vs. 1/kT. Determine the activation energy

from the slope.

T(o_{C) 1/kT T}

50 ln T50 175 25.89 75 4.32 150 27.42 150 5.01 125 29.15 500 6.21

Project a field usage T₅₀.

What AF to Use?

To project field T

, we can use one of the

three acceleration factors between different

stress temperatures and field 35

_C

(activation energy = 0.58eV)

Which is correct? Is there a better way to

do this projection?

Intercept in Spreadsheet

=intercept(y-range; x-range) =slope(y-range; x-range)

Results

Note with slope and intercept, we can project T₅₀ at

any temperature using the model

T₅₀ = e[intercept +(slope)/kT]

At use temperature, kT = (0.00008617)x(273.16+35) =0.02655. So field T₅₀ is predicted as

T₅₀ = e[-10.82+(0.5822)/(0.02655)] = e(11.11 )

=

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.

JMP Fit Life by X Model

Selecting Analyze > Reliability and Survival

> Fit Life by X, we then fill in the dialog

Fit Life by X Results:

Scatterplot

Nonparametric Overlay

Adjusting scale limits, display shows

nonparametric fraction failures at the end of

each readout interval on a lognormal

Fit Life by X Diagnostics

Lognormal fit looks good. Constrained equal slopes looks good also. Regression

Fit Life by X Distribution

Profiler

At field use temperature of 55° C, the

Fit Life by X: Density Curves

and Quantile Lines

Fit Life by X: Quantile

Profiler

Fit Life by X: Acceleration

Factor Profiler

Acceleration factor between 125° and 85° C.

E

estimate.

Common

s

estimate.

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.

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 105o_{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

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:

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

Activation Energy for

Degradation Example

To estimate H, use

So, AF between 125ºC and 30ºC is 2,009.

T₅₀ estimate at 30ºC is 2009x309 = 621,000 hrs.

Sigma estimate is 0.81.

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.

JMP Graph Builder Plot

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:

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.

Degradation Analysis

Using Fit Life by X

Platform: Profilers

CDF Estimate at

100,000 hours

Distribution

Profiler

Hazard Profiler

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:

Calculate the population characteristic life that will satisfy the AFR requirement:

Accelerated Testing

(Weibull Distribution)

For the specified stress time, T_s, Estimate the CDF for the equivalent field time T_e under accelerated conditions

Obtain the chi-square distribution percentile

that corresponds to the desired confidence

probability with 2r+2 degrees of freedom, where r is the allowed number of failures.

Calculation for Weibull

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:

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

Accelerated Testing

(Lognormal Distribution)

Obtain the chi-square distribution percentile

that corresponds to the desired confidence

probability with 2r+2 degrees of freedom, where r is the allowed number of failures.

Calculation for Lognormal

Distribution

The failure times for a specific inductor follow a

lognormal distribution with shape parameter s =

Class Project Worksheet

Multi-Censored Data : Kaplan Meier Estimator

Class Project

Acceleration Factors

Devices are stressed at a junction temperature T_j of

135O_C.

Field usage is specified at T_j = 55O_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 T₅₀ is 500 hours, what is the field T₅₀?

field T₅₀ = AF(stress T₅₀) = 250(500) = 125,000 hrs

3) If the stress sigma is 1.0, what is the field sigma?

field sigma = stress sigma = 1.0

Class Project

Acceleration Factors : Activation Energy

Using the TF table in the notes, and the equation,

AF



e

determine the effect of doubling an activation

energy from 0.65 eV to 1.3eV, for stress and use

temperatures of 125oC and 45oC, respectively.

Note that 1152 _{= 13,225. Thus, doubling} _H

Accelerated Modeling

Radiation Dosage Example

Components made by a certain manufacturer have a failure mechanism described by an exponential

distribution. Typical use temperature is 45oC.

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

where A is a constant, D is the radiation dosage rate, and alpha is an empirically determined constant equal to 1.5.

10% 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 ?

25 26 27 28 29 30 ln T 50 1/kT 4 5 6 7

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 35oC.

An accelerated life test was run employing three

different temperature cells: 175oC, 150oC, and 125oC. Thus (1/kT_Kelvin)=25.89, 27.42, and 29.15 respectively. Analysis of the data in each cell confirmed a lognormal distribution with common

sigma of 3.9. The T₅₀‟s estimated for the three cells

were, respectively, 75, 150, and 500 hours. Plot

ln T₅₀ vs. 1/kT. Determine the activation energy

from the slope.