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Failure Rates 2
- Thread starter RobV
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Try these people:
Stuart
Stuart
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2
- #4
rollingrock
Electrical
- Nov 8, 2003
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The original poster might want to contact the Reliability Analysis Center ( and see about getting a copy of their publication NPRD-95 "Nonelectronic Parts Data".
For the last poster (mooimooi), you may find MIL-HDBK-338, "Electronic Reliability Design Handbook" helpful. In chapter five of the handbook, it derives the exponential form of the reliability function assuming the hazard rate equals the failure rate and that the failure rate is constant (these assumptions are not always valid but are often used for simplistic, first-cut analysis). The hazard rate is defined in terms of
h(t) = (-1/R(t))*(dR(t)/dt)
and so from linear ordinary differential equations theory one can solve this to get R(t) = exp(-h(t)*t) = exp(-lambda*t) if h(t) = lambda = constant. Note that for very small values of lambda the Taylor series expansion of the exponential function exp(x) = 1+x+(x^2)/2!+(x^3)/3!+... can be used to approximate the probability of failure, which is F(t) = 1-R(t) as simply lambda, given the above assumptions. You get that by 1-exp(-x)=x-(x^2)/2!+(x^3)/3-... and all the second-order and higher terms vanish for very small values of x (for our application, "x" is lambda).
Anyway, you can get MIL-HDBK-338 in .pdf format from
and though it is a huge document (4.7Mb, over 1000 pages) it is definitely worth skimming through to identify directly applicable parts.
Good luck,
-MC
ps. You might want to check my math. I'm submitting this at 1am while riding out an episode of insomnia.
For the last poster (mooimooi), you may find MIL-HDBK-338, "Electronic Reliability Design Handbook" helpful. In chapter five of the handbook, it derives the exponential form of the reliability function assuming the hazard rate equals the failure rate and that the failure rate is constant (these assumptions are not always valid but are often used for simplistic, first-cut analysis). The hazard rate is defined in terms of
h(t) = (-1/R(t))*(dR(t)/dt)
and so from linear ordinary differential equations theory one can solve this to get R(t) = exp(-h(t)*t) = exp(-lambda*t) if h(t) = lambda = constant. Note that for very small values of lambda the Taylor series expansion of the exponential function exp(x) = 1+x+(x^2)/2!+(x^3)/3!+... can be used to approximate the probability of failure, which is F(t) = 1-R(t) as simply lambda, given the above assumptions. You get that by 1-exp(-x)=x-(x^2)/2!+(x^3)/3-... and all the second-order and higher terms vanish for very small values of x (for our application, "x" is lambda).
Anyway, you can get MIL-HDBK-338 in .pdf format from
and though it is a huge document (4.7Mb, over 1000 pages) it is definitely worth skimming through to identify directly applicable parts.
Good luck,
-MC
ps. You might want to check my math. I'm submitting this at 1am while riding out an episode of insomnia.
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