1. What It Models

Quantity: The Weibull distribution gives the fraction of a population that has failed by time $t$ , $F (t) = 1 - exp [- (t / η)^{β}]$ , with a scale $η$ , the characteristic life at which 63.2 per cent have failed, and a shape $β$ (NIST; the symbols differ between authors and software, see the Q/A).
Shape and the bathtub: The hazard falls with time for $β < 1$ , is constant for $β = 1$ , where the Weibull becomes the exponential model of the Failure Rate page, and rises for $β > 1$ , so one family covers early failures, the useful life and wear-out (NIST; Failure Rate page).
Weakest link: The Weibull is an extreme value distribution, the time to the first of many competing failure processes, which explains its use for capacitors, dielectrics and material strength and leads to the area scaling of oxide breakdown (NIST; background knowledge for the scaling).
Waiting time: A three-parameter form shifts the start of the time scale by a location $μ$ before which no failure occurs (NIST).
Limits: Real breakdown data can bend away from a straight Weibull plot through thickness fluctuations or clustering of defects, so the low tail is not always Weibull (Costa et al.; Salvador et al.).
Scope: This page covers the distribution, its estimation from failure data and its scaling. The lognormal alternative has its own page, temperature and voltage acceleration have their own pages, and the failure rate and FIT are on the Failure Rate page.

2. Usage

Plotting: Order the failure times, assign median ranks, exactly or with Benard’s $(j - 0.3) / (N + 0.4)$ , and plot $ln [- ln (1 - F)]$ against $ln t$ ; the slope is $β$ and $η$ is read at 63.2 per cent (ReliaWiki).
Estimating: For a handful of failures, median rank regression is less biased in $β$ than maximum likelihood, which overestimated $β$ by about 50 per cent with three failures; with many failures and with censoring, maximum likelihood is preferred (Olteanu and Freeman; ReliaWiki for suspensions).
Reading the shape: $β < 1$ points to early failures and a need for screening, $β \approx 1$ to random failures, and $β > 1$ to wear-out; a large $β$ means a narrow distribution (Failure Rate page; NIST).
Extrapolating: Low percentiles follow $t_{p} \approx η p^{1/ β}$ , so an error in $β$ moves the ppm life by orders of magnitude; check the tail with enough failures before relying on it (Costa et al.; computed in the table).
Scaling: Convert test-structure results to a product by the area or element count, $η_{2} / η_{1} = (A_{1} / A_{2})^{1/ β}$ , keeping $β$ (background knowledge; CMOS Logic Transistors page for oxide breakdown).
Zero failures: With a shape known from earlier data, a test without failures still bounds $η$ (Weibayes, background knowledge).

Weibull Functions

Name	Formula	Related Formulae	Variables/Constants
Weibull distribution and hazard	$F (t) = 1 - e^{- (t / η)^{β}}, R (t) = e^{- (t / η)^{β}}$	$h (t) = \frac{β}{η} (\frac{t}{η})^{β - 1}, H (t) = (\frac{t}{η})^{β}, f (t) = h (t) R (t)$	$F (t) : c u m u l a t i v e f r a c t i o n f ai l e d b y t im e t, R (t) : r e l iabi l i t y$ $η : sc a l e (c ha r a c t er i s t i c l i f e, 63.2 p er ce n t f ai l e d)$ $β : s ha p e, h (t) : ha z a r d (f ai l u r e r a t e), H (t) : c u m u l a t i v e ha z a r d, f (t) : d e n s i t y; N I S T w r i t es t h e sc a l e a s α an d t h e s ha p e a s γ, n o t es t ha t so m e so f tw a r e u ses β f or t h e c ha r a c t er i s t i c l i f e an d α f or t h e s ha p e an d t ha t so m e a u t h or s u se θ = α^{γ}, an d s t a t es t ha t γ = 1 g i v es t h e e x p o n e n t ia l m o d e l w i t h α = 1/ λ = M T T F an d γ = 2 t h e R a y l e i g h d i s t r ib u t i o n; t h e W e ib u l l i s an e x t r e m e v a l u e d i s t r ib u t i o n f or t h e w e ak es t l ink o f man y co m p e t in g f ai l u r e p r ocesses, w hi c h ma y e x pl ain i t s s u ccess f or c a p a c i t or s, ba l l b e a r in g s, r e l a y s an d ma t er ia l s t r e n g t h (N I S T)$ $(co m p u t e d : F (η) = 1 - e^{- 1} = 0.632 f or an y β; h f a l l s w i t h t im e f or β < 1, i s co n s t an t f or β = 1 an d r i ses f or β > 1, t h e t h r ee r e g i o n s o f t h e ba t h t u b c u r v e)$
Mean, median and percentiles	$\overset{ˉ}{t} = η Γ (1 + \frac{1}{β}), t_{50} = η (ln 2)^{1/ β}$	$t_{p} = η [- ln (1 - p)]^{1/ β} \approx η p^{1/ β} (p ≪ 1), σ^{2} = η^{2} [Γ (1 + \frac{2}{β}) - Γ^{2} (1 + \frac{1}{β})]$	$\overset{ˉ}{t} : m e an l i f e$ $Γ : g amma f u n c t i o n (Γ (N) = (N - 1)! f or in t e g er N)$ $t_{50} : m e d ian$ $t_{p} : t im e t o a f ai l e d f r a c t i o n p$ $σ^{2} : v a r ian ce (N I S T)$ $(co m p u t e d : \overset{ˉ}{t} / η = 2.00, 1.00, 0.89, 0.89 an d 0.90 an d t_{50} / η = 0.48, 0.69, 0.81, 0.83 an d 0.90 f or β = 0.5, 1, 1.7, 2 an d 3.5; t_{1 pp m} / η = 1 \times 1 0^{- 12}, 1 \times 1 0^{- 6}, 3.0 \times 1 0^{- 4}, 1 \times 1 0^{- 3} an d 0.019, so t h e l o w t ai l d e p e n d s f a r m or e o n β t han t h e m e an d oes)$
Three-parameter Weibull	$F (t) = 1 - exp [- (\frac{t - μ}{η})^{β}] (t \geq μ)$	$F (t) = 0 (t < μ)$	$μ : w ai t in g t im e (l oc a t i o n) b e f or e w hi c h n o f ai l u r e c an occ u r$ $η : sc a l e$ $β : s ha p e; N I S T : t h e t im e sc a l e s t a r t s a t μ, so μ i s s u b t r a c t e d f r o m a l l f ai l u r e an d r e a d o u t t im es an d t h e tw o - p a r am e t er f or m u l a s t h e n a ppl y$ $(co m p u t e d : n o n e)$

Estimation from Failure Data

Name	Formula	Related Formulae	Variables/Constants
Median ranks and the Weibull probability plot	$0.50 = \sum_{k = j}^{N} (k N) Z^{k} (1 - Z)^{N - k}, M R_{j} \approx \frac{j - 0.3}{N + 0.4}$	$ln [- ln (1 - F)] = β ln t - β ln η, η = exp (- \frac{a}{β})$	$Z : e x a c t m e d ian r ank o f t h e j - t h o f N or d er e d f ai l u r es$ $M R_{j} : m e d ian r ank$ $j : f ai l u r e or d er$ $N : s am pl e s i z e$ $F : c u m u l a t i v e f r a c t i o n f ai l e d$ $β : s l o p e o f t h e pl o t$ $a : in t er ce pt o f y = a + β x w i t h x = ln t (r ank r e g r ess i o n o n Y); R e l iaW ik i : t h e e x a c t r ank n ee d s a n u m er i c a l so l u t i o n or t h e F - d i s t r ib u t i o n f or m, B e na r d^{'} s f or m u l a i s q u i c k an d l ess a cc u r a t e, r e g r ess i o n o n X minimi z es h or i z o n t a l d e v ia t i o n s, η c an a l so b e r e a d a t F = 63.2 p er ce n t, t h e s i x e x a c t r ank s o f t h e i r e x am pl e a r e 10.91, 26.44, 42.14, 57.86, 73.56 an d 89.10 p er ce n t, an d w i t h s u s p e n s i o n s t h e r ank a d j u s t m e n t g a v e β = 0.81 an d η = 11400 h w hi l e M L E g a v e 1.33 an d 6920 h or 0.93 an d 21300 h f or tw o s u s p e n s i o n p a tt er n s$ $(co m p u t e d : B e na r d g i v es 10.94 an d 89.06 p er ce n t f or j = 1 an d 6 o f N = 6, w i t hin 0.04 p er ce n t o f t h e e x a c t 10.91 an d 89.10)$
Bias of the shape estimate for small samples	$bias (\hat{β}) = \frac{E [ β ^ ] - β}{β}$	$\hat{β}_{M L E} \to β (n \to \infty)$	$\hat{β} : es t ima t e d s ha p e, E [\hat{β}] : i t s m e an o v er man y s im u l a t e d s am pl es$ $β : t r u e s ha p e$ $n : n u mb er o f f ai l u r es; O l t e an u an d F r ee man (tw o - p a r am e t er W e ib u l l, β = 0.5, η = 333.3) : u n ce n sor e d M L E g i v es 0.755, 0.621, 0.552 an d 0.520 an d m e d ian r ank r e g r ess i o n (M R R) 0.521, 0.501, 0.493 an d 0.492 f or 3, 5, 10 an d 25 f ai l u r es, f or β = 1 M L E i s 51.7 p er ce n t hi g h a t 3 f ai l u r es an d 3.9 p er ce n t a t 25 a g ain s t 5.0 an d - 1.7 p er ce n t f or M R R, w i t h h e a v y r i g h t ce n sor in g (100 u ni t s) M L E an d M R R a r e 11.7 an d 11.1 p er ce n t hi g h a t 3 f ai l u r es, an d M L E i s e x p ec t e d t o d o mina t e a s t h e s am pl e g r o w s$ $(co m p u t e d : u n ce n sor e d bia s a t β = 0.5 o f + 50.9, + 24.3, + 10.5 an d + 3.9 p er ce n t f or M L E a g ain s t + 4.1, + 0.3, - 1.3 an d - 1.6 p er ce n t f or M R R)$
Zero-failure demonstration with an assumed shape	$η_{L} = [\frac{\sum _{i} t _{i}^{β}}{- l n ( 1 - C )}]^{1/ β}$	$- ln (0.40) = 0.916, - ln (0.10) = 2.303$	$η_{L} : l o w er co n f i d e n ce l imi t o f t h e c ha r a c t er i s t i c l i f e$ $t_{i} : t es t t im e o f u ni t i (n o f ai l u r es)$ $β : s ha p e a ss u m e d f r o m e a r l i er d a t a$ $C : co n f i d e n ce l e v e l (0.916 a t 60 p er ce n t an d 2.303 a t 90 p er ce n t) (W e iba y es m e t h o d, ba c k g r o u n d k n o w l e d g e); w i t h β = 1 i t r e d u ces t o t h e c hi - s q u a r e b o u n d o f t h e F ai l u r e R a t e p a g e$ $(co m p u t e d : 20 u ni t s f or 1000 h w i t h o u t f ai l u r e an d β = 2 (a ss u m e d) g i v e η_{L} = 2950 h a t 90 p er ce n t an d 4670 h a t 60 p er ce n t co n f i d e n ce)$

Scaling and Departures

Name	Formula	Related Formulae	Variables/Constants
Area and weakest-link scaling	$R_{A_{2}} (t) = [R_{A_{1}} (t)]^{A_{2} / A_{1}}, \frac{η _{2}}{η _{1}} = (\frac{A _{1}}{A _{2}})^{1/ β}$	$F_{c hi p} = 1 - (1 - F_{t es t})^{A_{c hi p} / A_{t es t}} \approx \frac{A _{c hi p}}{A _{t es t}} F_{t es t}$	$R_{A} : r e l iabi l i t y o f an a r e a (or n u mb er o f e l e m e n t s) A$ $A_{1} an d A_{2} : t es t an d t a r g e t a r e a s$ $η_{1} an d η_{2} : t h e i r c ha r a c t er i s t i c l i v es$ $β : s ha p e (u n c han g e d b y t h e sc a l in g)$ $F_{c hi p} an d F_{t es t} : f ai l e d f r a c t i o n s o f a c hi p an d o f a t es t s t r u c t u r e (w e ak es t - l ink co n se q u e n ce o f t h e W e ib u l l, ba c k g r o u n d k n o w l e d g e; u se d f or o x i d e b r e ak d o w n o n t h e C M O S L o g i c T r an s i s t or s p a g e); N I S T : t h e W e ib u l l a r i ses a s t h e w e ak es t l ink o f man y co m p e t in g p r ocesses; S t a t hi s : o x i d e b r e ak d o w n f o l l o w s f r o m a p er co l a t i o n p a t h o f d e f ec t s, an d f e w er t han 1 p er ce n t o f t h e p a t h s l e a d t o d es t r u c t i v e b r e ak d o w n$ $(co m p u t e d : f or a 1 \times 1 0^{4} t im es l a r g er a r e a η f a l l s t o 1 \times 1 0^{- 4}, 2.2 \times 1 0^{- 3} an d 0.046 o f t h e t es t v a l u e f or β = 1, 1.5 an d 3, an d a t es t - s t r u c t u r e f r a c t i o n o f 1 \times 1 0^{- 6} b eco m es 1.0 p er ce n t o f c hi p s)$
Departures from the Weibull at the tails	$F_{C} (t) = 1 - [1 + \frac{1}{α} (\frac{t}{τ})^{β}]^{- α} \to 1 - e^{- (t / τ)^{β}} (α \to \infty)$	$F_{q} (t) = 1 - exp_{q} [- (\frac{t}{α _{q}})^{β}]$	$F_{C} : t im e - d e p e n d e n t c l u s t er in g (B u r r X I I) d i s t r ib u t i o n$ $α : c l u s t er in g f a c t or$ $τ : sc a l e (63 r d p er ce n t i l e o f t h e W e ib u l l l imi t)$ $β : s ha p e$ $F_{q} : q - W e ib u l l d i s t r ib u t i o n$ $exp_{q} : q - e x p o n e n t ia l$ $α_{q} : i t s sc a l e; S a l v a d or e t a l . : t h e c l u s t er in g m o d e l co in c i d es w i t h t h e W e ib u l l a t t h e l o w es t p er ce n t i l es b u t d i f f er s s t r o n g l y a t t h e hi g h es t o n es, w i t h t h e sc a l e f o l l o w in g τ = τ_{0} e^{- γ (V - I R_{S})}; C os t a e t a l . : t hi c k n ess f l u c t u a t i o n s b e n d t h e W e ib u l l pl o t, t h e q - W e ib u l l f i t s 6.9 nm S i O 2 d a t a a t 12.7 MV/cm f or 1 \times 1 0^{- 4} an d 1 \times 1 0^{- 5} m m^{2} a r e a s an d sc a l es b e tt er w i t h a r e a, an d e x t r a p o l a t in g a s l o p e f i tt e d a t hi g h p er ce n t i l es t o l o w p er ce n t i l es c an l e a d t o ser i o u s er r or s$ $(co m p u t e d : a t t = τ w i t h β = 1 an d α = 2, F_{C} = 0.556 a g ain s t 0.632 f or t h e W e ib u l l)$

Important Questions

Ques: Is beta the shape or the scale of a Weibull distribution?

It depends on the source, which is a common cause of errors. NIST writes the scale, or characteristic life, as $α$ and the shape as $γ$ , and warns that some software uses $β$ for the characteristic life and $α$ for the shape, while some authors use $θ = α^{γ}$ as the scale. Most reliability texts and the semiconductor literature, and this page, use $β$ for the shape and $η$ for the scale, while the Failure Rate page follows NIST’s $γ$ , and dielectric breakdown papers use $β$ as the Weibull slope with the scale written as $x_{0}$ , $τ$ or $t_{63}$ (Costa et al.; Salvador et al.). The safe check is the definition: the scale is the time at which $F = 0.632$ , and the shape is the slope of $ln [- ln (1 - F)]$ against $ln t$ . A value quoted without its definition should not be reused. Reference: NIST/SEMATECH e-Handbook of Statistical Methods, Section 8.1.6.2 ‘Weibull’ (fetch-tool summary); U. M. S. Costa, V. N. Freire, L. C. Malacarne, R. S. Mendes, S. Picoli Jr., E. A. de Vasconcelos and E. F. da Silva Jr., ‘An Improved Description of the Dielectric Breakdown in Oxides Based on a Generalized Weibull distribution’, arXiv:cond-mat/0602515 (2008 version) (fetch-tool summary); E. Salvador, R. Rodriguez and E. Miranda, ‘SPICE simulation of the time-dependent clustering model for dielectric breakdown’, Solid-State Electronics 215, 108895 (2024) (fetch-tool summary).

Ques: Why is extrapolating a Weibull fit to ppm levels risky?

Because the low tail depends very strongly on the shape. From $t_{p} \approx η p^{1/ β}$ , the time to 1 ppm is $1 \times 1 0^{- 6}$ of the characteristic life for $β = 1$ , $1 \times 1 0^{- 3}$ for $β = 2$ and 0.019 for $β = 3.5$ (computed), so a small error in $β$ changes a ppm life by orders of magnitude. Costa et al. add that thickness fluctuations bend the Weibull plot of oxide breakdown, and that extrapolating a slope fitted at high percentiles to low ones can lead to serious errors. Salvador et al. note that the clustering model agrees with the Weibull at the lowest percentiles but not at the highest, so a fit dominated by the upper part of the data can also mislead in the opposite direction. The safe practice is to fit the tail with enough failures. Reference: U. M. S. Costa, V. N. Freire, L. C. Malacarne, R. S. Mendes, S. Picoli Jr., E. A. de Vasconcelos and E. F. da Silva Jr., ‘An Improved Description of the Dielectric Breakdown in Oxides Based on a Generalized Weibull distribution’, arXiv:cond-mat/0602515 (2008 version) (fetch-tool summary); E. Salvador, R. Rodriguez and E. Miranda, ‘SPICE simulation of the time-dependent clustering model for dielectric breakdown’, Solid-State Electronics 215, 108895 (2024) (fetch-tool summary); NIST/SEMATECH e-Handbook of Statistical Methods, Section 8.1.6.2 ‘Weibull’ (fetch-tool summary).

Ques: Should a Weibull fit use rank regression or maximum likelihood?

For few failures, rank regression; for many, maximum likelihood. Olteanu and Freeman simulated two-parameter Weibull samples with $β = 0.5$ and found that maximum likelihood overestimated the shape by 50.9 per cent with 3 failures, 24.3 per cent with 5 and 3.9 per cent with 25, while median rank regression stayed within about 4 per cent (computed from their means). For $β = 1$ the pattern was the same, 51.7 against 5.0 per cent at 3 failures. With heavy right censoring both methods were about 11 per cent high at 3 failures, and the authors expect maximum likelihood to dominate as the sample grows, since it is asymptotically superior. ReliaWiki shows a further difference: rank regression ignores where suspensions occur, while maximum likelihood uses it. Reference: D. Olteanu and L. Freeman, ‘Technical Report on the Evaluation of Median Rank Regression and Maximum Likelihood Estimation Techniques for a Two-Parameter Weibull Distribution’, Virginia Tech, Department of Statistics Technical Report 08-4 (fetch-tool summary); HBK ReliaSoft, ReliaWiki, ‘Parameter Estimation: Median Ranks’ and rank regression sections (fetch-tool summary).

Ques: Is Benard’s approximation good enough for median ranks?

For plotting, yes. ReliaWiki gives the exact median rank as the solution of a cumulative binomial equation at 50 per cent, which needs a numerical method or the F-distribution, and calls Benard’s $(j - 0.3) / (N + 0.4)$ a quick and less accurate approximation. For their six-failure example the exact ranks are 10.91 and 89.10 per cent for the first and last failures, while Benard gives 10.94 and 89.06 per cent (computed), a difference of 0.04 per cent that is invisible on a probability plot and small against the scatter of six failures. The approximation matters more for very small samples and when suspensions require adjusted ranks, which ReliaWiki handles with mean order numbers before applying the rank formula. Reference: HBK ReliaSoft, ReliaWiki, ‘Parameter Estimation: Median Ranks’ and rank regression sections (fetch-tool summary).

Ques: How does the Weibull scale from a test structure to a whole chip?

Through the weakest link. If a chip has $A_{2} / A_{1}$ times the area, or the number of elements, of a test structure and each part fails independently, its reliability is the test reliability to that power, so the characteristic life scales as $(A_{1} / A_{2})^{1/ β}$ while the shape stays the same (background knowledge, following from the weakest-link origin that NIST gives). For an area ratio of $1 \times 1 0^{4}$ , the characteristic life falls to $1 \times 1 0^{- 4}$ of the test value for $β = 1$ , $2.2 \times 1 0^{- 3}$ for $β = 1.5$ and 0.046 for $β = 3$ , and a test-structure fraction of 1 ppm becomes 1.0 per cent of chips (computed). This is why oxide breakdown with a low shape, typical of thin oxides, is so sensitive to the total gate area. Reference: NIST/SEMATECH e-Handbook of Statistical Methods, Section 8.1.6.2 ‘Weibull’ (fetch-tool summary); CMOS Logic Transistors page of this vault; J. H. Stathis, IBM Research, ‘Physical and Predictive Models of Ultra Thin Oxide Reliability in CMOS Devices and Circuits’ (abstract only; venue and year not given).

Ques: How can a test with no failures still bound a Weibull life?

If the shape is known from earlier data, a zero-failure test bounds the characteristic life through the Weibayes relation $η_{L} = [\sum t_{i}^{β} / (- ln (1 - C))]^{1/ β}$ (background knowledge). With 20 units run for 1000 $h$ each and an assumed $β = 2$ , the lower limit is 2950 $h$ at 90 per cent and 4670 $h$ at 60 per cent confidence (computed). For $β = 1$ the relation reduces to the constant-rate bound with $- ln (1 - C)$ equal to half the chi-square value for two degrees of freedom, the zero-failure case of the Failure Rate page. The result depends directly on the assumed shape: a wrong $β$ gives a wrong bound, so the shape must come from the same failure mechanism, ideally from earlier tests of the same product. Reference: Failure Rate and the Bathtub Curve page of this vault; NIST/SEMATECH e-Handbook of Statistical Methods, Section 8.1.6.2 ‘Weibull’ (fetch-tool summary).

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Weibull Distribution