applied reliabilty class notes in powerpointapplied reliability page 15 spotting a defective...

Applied Reliability

APPLIED

RELIABILITY

Techniques for Reliability

Analysis

with

Applied Reliability Tools (ART)

(an EXCEL Add-In)

and

JMP® Software

AM216 Class 5 Notes

Santa Clara University

Copyright David C. Trindade, Ph.D.

STAT-TECH ®

Spring 2010

Applied Reliability

AM216 Class 5 Notes

• Accelerated Testing(continued from Class 4 Notes)

– Accelerated Test Example (Analysis in JMP)

– Degradation Modeling

– Sample Sizes for Accelerated Testing

• System Models

– Series System

– Parallel System

– Analysis of Complex Systems

– Standby Redundancy

• Defective Subpopulations

– Graphical Analysis

– Mortals and Immortals

– Models

– Case Study

– Class Project Example

• Modeling the Field Reliability

– Evolution of Methods

– General Reliability Model

– AMD Example

Applied Reliability

System ModelsSeries System

Consider a system made up with n components in

series. If the i th component has reliability Ri (t),

the system reliability is the product of the individual

reliabilities, that is,

R t R t R t R ts n( ) ... 1 2

which we denote with the capital “pi” symbol for

multiplication

R t R ts i

i

n

1

The system CDF, in terms of the individual CDF’s, is

The system failure rate is the sum of the individual

component failure rates. The system failure rate

is higher than the highest individual failure rate.

F t F ts

i

n

i

1 11

Applied Reliability

System ModelsParallel System

Consider a system made up with n components in

parallel. The system CDF is the product of the

individual CDF’s, that is,

The system reliability is

System failure rates are no longer additive (in

fact, the system failure rate is smaller than the

smallest individual failure rate), but must be

calculated using basic definitions.

F t F ts

i

n

i

1

R t R ts

i

n

i

1 11

Applied Reliability

System Failure RateTwo Parallel Components

A component has CDF F(t) and a failure rate h(t).

Two components are used in parallel in a system.

Determine the failure rate of the system.

SOLUTION

The CDF for the two components in parallel is F2(t)

and the PDF, by differentiation, is 2F(t)f(t). The

failure rate of the system is

f t

F t

F t f t

F t

F t

F t

f t

F t

F t

F th t

s

s1

2

1

2

1 1

2

1

2

h ts

The result shows that the system failure rate is a

factor 2F/(1+F) times the component failure rate. The

smaller the component CDF, the bigger the

improvement. Redundancy makes a larger difference

in early life, and much less difference later on.

Applied Reliability

Class ProjectSystem Models

A) A component has reliability R(t) = 0.99.

Twenty-five components in series form a

system. Calculate the system reliability.

B) A component has reliability R(t) = 0.95

Three components in parallel form a system.

Calculate the system reliability.

Applied Reliability

Reliability Block Diagrams

For components in series:

For components in parallel:

A B

A

B

Applied Reliability

Example of Series-Parallel

System: Big Rig

GH

JI

CD

FE

A

B

Trailer Cab

G

H

I

J

E

F

C

D

B A

Reliability Block Diagram (RBD)

Applied Reliability

Class ProjectComplex Systems

A system consists of seven units: A, B, C, D, E, G, H. For the system to function unit A and either unit B or C and either D and E together or G and H together must be working. Draw the reliability block diagram for this setup.

Write the equation for the CDF of the system in terms of the individual component reliabilities, that is, the Ri, where i = A, B, C, ..., G, H. Hint: Consider the three subsystems:A alone; B with C; and D,E,G,H.

Applied Reliability

Standby Versus Active

Redundancy

In contrast to active parallel redundancy, there is

standby redundancy in which the second

component is idle until needed. Assuming perfect

switching and no degradation of the idle

component, standby redundancy results in higher

reliability and less maintenance costs than active

parallel redundancy. An illustration, assuming

exponentially distributed failure times, is shown

below.

System Failure Rates (2 Components)

0

0.002

0.004

0.006

0.008

0.01

0.012

0 50 100 150 200 250 300 350 400

Single

Parallel

Standby

Applied Reliability

Series, Parallel Reliability in

ARTIn ART, select System Reliability... Enter necessary

information. Click OK.

Applied Reliability

Reliability ExperimentConsider . . .

We test 100 units for 1,000 hours. There are 30

failures by 500 hours, but no more by the end of

test.

Question : Are we dealing with two

populations or just censored data ?

Question : If we continue the test, will we see

only a few more failures, or will the other 70 fail

with the same life distribution ?

Applied Reliability

Defect ModelsMortals versus Immortals

The usual assumption in reliability analysis is that

all units can fail for a specific mechanism. If a

defective subpopulation exists, only a fraction of

the units containing the defect may be susceptible

to failure. These are called mortals.

Units without the fatal flaw do not fail. These are

called immortals.

The model for the total population of mortals and

immortals becomes :

CDF = (fraction mortals) x CDF(mortals)

Reliability analysis focuses on the life distribution of

the defective subpopulation and the mortal fraction.

Applied Reliability

Example of a Defective

Subpopulation

A Processing Problem

Suppose we have 25 wafers in a lot, but only two

wafers are contaminated with mobile ions due to a

processing error.

If components are assembled from the 25 wafers,

assuming equal yield per wafer, only 2/25= 8% of

the components can have the fatal “defect” that

makes failure possible.

The components from the non-contaminated

wafers will not fail for this mechanism since they

are defect free; that is, we have a defective

subpopulation.

Applied Reliability

Spotting a Defective

SubpopulationGraphical Analysis

Assume that a specified failure mode follows a

lognormal distribution.

Plot the data on lognormal graph paper. If instead of

following a straight line, the points seem to curve

away from the cumulative percent axis, it’s a signal

that a defective subpopulation may be present.

If test is run long enough, expect plot to bend over

asymptotic to cumulative percent line that represents

proportion of defectives in the sample.

Applied Reliability

Defective SubpopulationsGraphical Analysis

Plot based on total sample (mortals and immortals).

Plot based only on mortal subpopulation.

Applied Reliability

Defect ModelMortals and Immortals

The observed CDF Fobs(t) is

Fobs(t) = p Fm(t)

where Fm(t) is the CDF of the mortals and p is the fraction of mortals (units with the fatal defect) in the total sample size.

For example, if there are 25 % mortals in the population, and the mortal CDF at time t is 40%, then we would expect to observe about

0.25x0.40 = 0.10

or 10% failures in the total random sample at time t.

Applied Reliability

Major Computer

Manufacturer Reliability Data

Gate Oxide Fails

Time (hours) 24 48 168 500 1000

Rejects 201 23 1 1 1

Sample Size 58,000 57,392 10,000 2,000 1,999

Censored 407 47,369 7,999 0 1,998

Analysis by Company Using Lognormal Distribution

T50: 1.149E32 hours Sigma: 26.175

Applied Reliability

What Do These

Numbers Mean?

Plus and minus 3 sigma range of time to failure distribution extends from 33 seconds to 1.66E62 years !

It takes seconds to get to 0.1% cumulative failures, but over 412,000 hours (that is, 47 years) to get to 1.00% !

Assuming everything can fail is misleading and unnecessary.

Analysis by Company Using Lognormal Distribution

T50 : 1.149E32 hours Sigma : 26.175

Applied Reliability

Modeling with

Defective Subpopulations

The same data, assuming 99% of the failures have

occurred by 48 hours, can be modeled by a fraction

defective subpopulation of 227/58,000 = 0.39% and

a lognormal distribution of failure times for the

mortals T50 =10.6 hours and sigma = 0.68.

Practically 100% of failures occur by 168 hours. Any

failures thereafter are probably not related to the

defective subpopulation. For example, handling

induced failures are a possibility.

Applied Reliability

Defective Subpopulation

Models

If we don’t consider mortals vs. immortals, we will

incorrectly assume that all units can fail.

Projections of field reliability will be biased

unless we identify the limited defective units.

Applied Reliability

Statistical Reliability

Analysis and Modeling:

A Case Study

Analysis of Reliability Data

with Failures from a

Defective Subpopulation

Applied Reliability

Reliability StudyBackground

One lot of a device type with initial burn-in results at 168 hours, 125

oC :

Over 50% fallout due to bake recoverable failures

Since other lots, with similar manufacturing, might have escaped to a few customers, we needed to assess the field impact.

We were able to impound this lot, containing about 300 devices not burned-in.

Applied Reliability

Reliability StudyDesign

Two static stresses:

179 Units : 125oC ambient

90 Units : 150oC ambient

30 Units: Control

Frequent readouts at 2, 4, 8, 16, 32, 48, 68, 92,

116 hours

Applied Reliability

Purpose of StudyReliability Modeling

• Determine if fraction defective (mortals) model applies

• Determine failure distribution (lognormal, parameters)

• Determine if true acceleration is present

• Determine activation energy for acceleration factors

• Determine recovery kinetics with and without bake

- Is 24 hours at 150oC necessary?

- Do devices recover at room temperature?

Applied Reliability

Modeling Procedure

Statistical Analysis Plan

• Analyze cumulative percent failures plot versus

time, both linear and probability plots.

• Estimate fraction mortals for stress cells. Test

for significant difference.

• Plot fallout of mortals (reduced sample size) on

lognormal probability graph. Check for linearity

and equality of slopes.

• Run maximum likelihood analysis. Test for

equality of shape factors (sigmas). Estimate

single sigma. Estimate median life T50 for both

cells.

• Check model fit against original data.

Applied Reliability

Reliability StudyBake Recoverable Failures

L in e a r P lo t o f C u m u la tiv e F a i lur e s Ve rsu s T im e

0 %

1 0 %

2 0 %

3 0 %

4 0 %

5 0 %

6 0 %

7 0 %

8 0 %

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

Stre s s Tim e (Pow e r on H our s )

Cu

mu

lati

ve

Pe

rc

en

t

1 50oC 1 25oC

Sam ple S iz e s : 1 5 0 oC =9 0 ; 1 2 5o C = 1 7 9

Applied Reliability


P r o b a bi li ty Pl o ts (N o A d ju s tm e n t fo r M o r ta ls)

-2 .5

- 2

-1 .5

- 1

-0 .5

0

0 .5

1

0 1 2 3 4 5

Ln (Tim e t o Fa ilure )

Sta

nd

ar

d N

orm

al

Va

ria

te:

Z

1 5 0 o C 1 2 5 o C

Sa m pl e S iz e s : 1 5 0 o C = 9 0 ; 1 2 5 o C = 1 7 9

Applied Reliability


Probability Plot (Adjusted for Mortals)

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Ln(T im e to Fa ilure )

Sta

nd

ard

No

rma

l V

ari

ate

: Z

150oC 125oC

M orta l S a m ple S iz e s: 150oC = 64; 125oC = 113

Applied Reliability

GENLNEST

ENTER NUMBER OF CELLS: 2

CHOOSE CONF. LIMIT FOR BOUND IN PERCENT: 90

ENTER ANY EXACT TIMES OF FAILURE FOR CELL 1

ENTER START AND ENDPOINT OF ALL READOUT INTERVALS (INCLUDE ZERO’S)

SPREAD 2 4 8 16 32 48 68 92 116

ENTER CORRESPONDING NUMBERS OF FAILS PER INTERVAL (INCLUDE ZERO’S)

34 6 21 2 0 0 0 1 0

ENTER TIMES ALL FAILED UNITS WERE REMOVED FROM TEST (INCLUDING END OF TEST)

116

ENTER CORRESPONDING NUMBERS REMOVED

0

ENTER ANY EXACT TIMES OF FAILURE FOR CELL 2

ENTER START AND ENDPOINT OF ALL READOUT INTERVALS (INCLUDE ZERO’S)

SPREAD 2 4 8 16 32 48 68 92 116

ENTER CORRESPONDING NUMBERS OF FAILS PER INTERVAL (INCLUDE ZERO’S)

5 0 36 8 42 7 3 4 3

ENTER TIMES ALL FAILED UNITS WERE REMOVED FROM TEST (INCLUDING END OF TEST)

16 116

ENTER CORRESPONDING NUMBERS REMOVED

2 3

MAXIMUM LIKELIHOOD ESTIMATES

VARIANCE VARIANCE COVARIANCE

CELL T50 SIGMA MU SIGMA MU MU SIGMA

1 1.90 1.208 .444 .0322 .0373e-1 .643e-2

2 15.08 1.060 2.714 .0059 .0104e-3 .266e-5

ESTIMATE BOUNDS (90 PERCENT CONFIDENCE)

NUM. NUM.

CELL ON TEST FAIL T50 LOW T50 UP SIGMA LOW SIGMA UP

1 64 64 1.38 2.63 .909 1.508

2 113 108 12.74 17.86 .933 1.187

WANT EQUAL T50’S OR SIGMAS OR BOTH IN SOME CELLS (Y/N)?

Y

CELLS: 1 2

TYPE 1 FOR EQUAL SIGMA’S, 2 FOR EQUAL MU’S, 3 FOR BOTH THE SAME: 1

THE ASSUMPTION OF QUAL SIGMA’S CAN NOT BE REJECTED AT THE 95 PERCENT LEVEL.

UNDER THIS ASSUMPTION, RESULTS LIKE OBSERVED OCCUR ABOUT 41.9 PERCENT OF THE TIME.

(THE SMALLER THIS PERCENT, THE LESS LIKELY THE ASSUMPTION.)

MAXIMUM LIKELIHOOD ESTIMATES

VARIANCE VARIANCE COVARIANCE

CELL T50 SIGMA MU SIGMA MU MU SIGMA

1 2.02 1.090 .704 .0051 .0247e-2 .538e-3

2 15.08 1.090 1.713 .0051 .0110e-2 .250e-5

ESTIMATE BOUNDS (90 PERCENT CONFIDENCE)

NUM. NUM.

CELL ON TEST FAIL T50 LOW T50 UP SIGMA LOW SIGMA UP

1 64 64 1.56 2.63 .972 1.207

2 113 108 12.68 17.54 .972 1.207

WANT EQUAL T50’S OR SIGMAS OR BOTH IN SOME CELLS (Y/N)?

N

APL PROGRAM FOR MLE

Applied Reliability


Model Fit to Actual

0%

10%

20%

30%

40%

50%

60%

70%

80%

0 20 40 60 80 1 00 1 20 1 40

Tim e (P ow er o n Ho ur s)

Cu

mu

ma

tiv

e P

er

ce

nt

Fa

ilu

re

s

1 50oC

1 25oC

M LE F it: 1 50oC

M LE F it: 1 25oC

Applied Reliability

Projection to Field ConditionsAcceleration Statistics

• Estimate acceleration factor between two stress cells : AF = 15.08 / 2.02 = 7.465

• Estimate activation energy, based on Tj’s, 35

oC above ambient: EA = 1.375 eV

• Estimate field T50 based on Tj at 55oC ambient

: field T50 = 18,288 hours

• Using field T50, sigma = 1.090, lognormaldistribution:

-project fallout and failure rates for various mortal fractions

-use customer field data to determine which mortal fraction applies

Applied Reliability

Projection to Field UseBake Recoverable Fails

Pr o je cted F ie ld F a llo u t w ith Va rio u s M o rta l

P er cen tag es

0%

2%

4%

6%

8%

1 0%

1 2%

1 4%

1 6%

1 8%

2 0%

0 2 4 6 8 1 0

T ime i n F ie ld ( K H o ur s )

Cu

mu

lativ

e P

erc

en

t

5%

1 0%

2 0%

3 0%

4 0%

5 0%

6 6%

Applied Reliability

A Note of Caution

Analysis When Mortals Are Present

Since the analysis which took into account the presence of a defective subpopulation, parameterestimates were accurate. The two customers, notified of the affected lots, used analysis for decisions on how to treat remaining product in field.

If assessment is not done correctly and there is a low incidence of mortals, the T50’s and sigma’s for a lognormal distribution may become very large and inaccurate.

Applied Reliability

A Side BenefitScreening a Wearout Mechanism

Note that it may be possible to screen a wearout failure mechanism if only a subpopulation of the units are mortal for that mechanism and sufficient acceleration is obtainable.

See Trindade paper “Can Burn-in Screen Wearout Mechanism? Reliability Models of Defective Subpopulations - A Case Study” in 29th Annual Proceedings of Reliability Physics Symposium (1991)

Applied Reliability

Class ProjectDefect Models

50 components are put on stress. Readouts are at

10, 25, 50, 100, 200, 500, and 1,000 hours. The

failure counts at the respective readouts are 2, 2,

4, 5, 4, 3, and 0.

1. Estimate the CDF for all units using the table

below with n = 50.

2. Plot the data on Weibull probability paper on

the next page.

Does the data appear distributed according to a

Weibull distribution or does a defect model seem

possible?

Time

Cum #

Fails

CDF Est

All Units

(%)

10 2

25 4

50 8

100 13

200 17

500 20

1000 20

Applied Reliability

Weibull Probability Paper

Applied Reliability

Note: “Percent Failure” scale on Weibull Probability paper is faint. Values are 99.9, 98.0, 90.0,

70.0, 50.0, 30.0, 20.0, 10.0, 5.0, 2.0, 1.0, 0.5, 0.2, 0.1, etc.

Applied Reliability

Class ProjectDefect Model Estimates

Weibull Parameter Estimates for Mortal Population:

Characteristic Life (c) __________

Shape Parameter (m) __________

F t e t c

m

( ) / 1

How could we confirm that the Weibull model for

the mortal population fits the data? We estimate

the CDF at three times and compare to

observations.

Time

Mortal

CDF

(Weibull

Model)

Mortal

Fraction

Model

CDF for

All Units

Empirical

CDF All

Units

25 0.221 0.4

100 0.632 0.4

1000 1.000 0.4

Applied Reliability

Defective Subpopulations in

ARTEnter failure information (readout times, cumulative

failures) into columns. Under ART, select Defective

Subpopulations… Enter required information. Click OK.

Applied Reliability

System Models

A General Model for the

Field Reliability of

Integrated Circuits

An Evolution in the Projection

of Field Failure Rates

Applied Reliability

Failure Rate CalculationsPrimitive Method

Assumptions

• Constant failure rate

• Single overall activation

energy

• Ambient temperatures

• No separation of failure modes

Applied Reliability

Primitive MethodProblems with Calculations

Example

100 units are stressed for 1,000 hours at 125oC.

Assume no self heating. One unit fails at 10 hours for

mechanism with EA of 1.0 eV. Second unit fails at 500

hours for failure mechanism with EA of 0.5 eV.

Primitive Method Calculation

Overall average activation energy : 0.75 eV

Acceleration Factor (125oC to 55

oC): AF = 106

IFR (constant) at 55oC :

[1E9x2/(10+500+98x1000)]/AF = 192 FITS

Applied Reliability

Primitive MethodComparative Calculation

Individual Analysis by Failure Mechanism

Mechanism 1: EA = 1.0 eV, AF = 501

IFR (constant) at 55oC:

[1E9/(10+500+98x1000)]/AF = 20 FITS

Mechanism 2: EA = 0.5 eV, AF = 22,

IFR (constant) at 55oC:

[1E9/(10+500+98x1000)]/AF = 461 FITS

Total IFR = 481 FITS

Applied Reliability

Failure Rate CalculationsLater Improved Method

• Early failures (infant mortality) reported separately

• Long-term life modeled with activation energy

specific to failure mechanisms

• Constant failure rate for long term life

• Temperature acceleration calculated with junction

temperatures

Applied Reliability

Later MethodProblems

• Defective subpopulations not adequately

modeled

• Competing failure modes not adequately

modeled with constant failure rate

• Zero rejects and unidentified mechanisms

often not treated

• Bathtub curve approximated in flat region only

because of constant failure rate

Applied Reliability

An Alternative Model

Three categories of possible failures:

Test Escapes


Competing Failure Mechanisms

The three D’s:

Dead

Defective

Deficient

Applied Reliability

Non-Functional Test Escapes

Dead on arrival (DOA)

Quality issue

Inadequate testing at manufacturer

or damaged after testing prior to

customer receipt

Rejects “discovered” at customer;

called mistakenly reliability failures

Assume zero in model

Applied Reliability


There are proportions of the total population at risk

of failure. Defective units are called mortals. The

ones without the defect are called immortals.

Defective subpopulations are generally associated

with processing problems.

There are physical reasons why defective

subpopulations should exist.

Always question the assumption (common in the

traditional approach) that any observed failure type

will eventually affect all other devices.

Applied Reliability

Competing Risks

There are failure mechanisms that can affect all

units.

We call these mechanisms competing risks

because several different types may exist and any

one can cause the unit to fail.

These mechanisms are typically associated with

design, processing, or material problems.

We model the failures using Weibull or Lognormal

distributions

Applied Reliability

General Reliability Model

• Activation energies are specific to failure

mechanisms.

• Zero rejects and unidentified

mechanisms are included.

• Generates complete bathtub curve!

F F F FT e d N 1

where

FN = 1 - R1R2. . . RN

Applied Reliability

General Reliability Model In

Use at AMD

AMD Reliability Brochure 1994 Data

Applied Reliability

AMD Reliability Brochure 1994 Data

Applied Reliability

Appendix

Applied Reliability

Class ProjectSystem Models

A) A component has reliability R(t) = 0.99.

Twenty-five components in series form a

system. Calculate the system reliability.

Rs(t) = (0.99)25 = 0.778 or 77.8%

B) A component has reliability R(t) = 0.95

Three components in parallel form a system.

Calculate the system reliability.

Rs(t) = 1- (1- 0.95)3 = 0.9999 or 99.99%

Applied Reliability

Class ProjectComplex Systems

A system consists of seven units: A, B, C, D, E, G, H.

For the system to function unit A and either unit B or C

and either D and E together or G and H together must

be working. Draw the reliability block diagram for this

setup.

Write the equation for the CDF of the system in

terms of the individual component reliabilities, that is,

the Ri, where i = A, B, C, ..., G, H. Hint: Consider the

three subsystems:A alone; B with C; and D,E,G,H.

1) RA

2) RBC=1- (1- RB )(1- RC )

3) RDEGH = 1- (1- RDE )(1- RGH )

= 1- (1- RDRE )(1- RGRH )

The system CDF is

FS = 1 - RS = 1 - RA RBC RDEGH

A

B

C

D E

G H

Applied Reliability

Class ProjectDefect Models

1. Estimate the proportion defective p and the

number of mortals in the sample. Fill in the mortal

CDF column in the table below.

2. Plot the data for the mortal subpopulation on

the same sheet of paper. Does the fit look

reasonable?

4. Estimate the characteristic life c = T63, the 63rd

percentile.

5. Estimate the shape parameter m by drawing a

line perpendicular to the “best fit by eye line”

through the estimation point on the Weibull paper

and reading the beta estimation scale.

Time

Cum #

Fails

CDF Est All

Units (%)

CDF Est

Mortals (%)

10 2 2/50 = 4%

25 4 4/50 = 8%

50 8 8/50 = 16%

100 13 13/50 = 26%

200 17 17/50 = 34%

500 20 20/50 = 40%

1000 20 20/50 = 40%

Applied Reliability

Class Project

Defect Model Example

Time

Cum #

Fails

CDF Est All

Units (%)

CDF Est

Mortals (%)

10 2 2/50 = 4% 2/20 = 10%

25 4 4/50 = 8% 4/20 = 20%

50 8 8/50 = 16% 8/20 = 40%

100 13 13/50 = 26% 13/20 = 65%

200 17 17/50 = 34% 17/20 = 85%

500 20 20/50 = 40% 20/20 = 100%

1000 20 20/50 = 40% 20/20 = 100%

n = 50

Estimated mortal fraction, p : 0.40 or 40%

CDF estimate for mortals is based on

sample size of defective subpopulation.

Applied Reliability

Weibull Probability Plot

Applied Reliability

Class ProjectDefect Model Example

Model Check

Weibull Parameter Estimates for Mortal Population :

Characteristic Life (c) ___ 100 ______

Shape Parameter (m) ___ 1.0 ______

F t e t c

m

( ) / 1

Time

Mortal

CDF

(Weibull

Model)

Mortal

Fraction

Model

CDF for

All Units

Empirical

CDF All

Units

25 0.221 0.4 0.088 0.08

100 0.632 0.4 0.253 0.26

1000 1.000 0.4 0.400 0.40

Applied Reliability

Defect Model Example

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 100 200 300 400 500 600 700 800 900 1000

Times (Hrs)

CD

F

Class Project

Defect Model

p x Weibull CDF Plot

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