Hi John,
The underlying assumption in Eq. 16 is an adequate sample size (32).
What you are doing is not partitioning a component, however, but daisy-chaining components together.
In this case, what you did is correct, because you record the first failed component of 10, and so your eq. should read like this:
Nf(50%,component) = Nf(50%,m daisy-chained components) * [m]^(1/BETA)
= 800 * (10)^1/4
= 1422
Plotting things like this on a Weibull graph typically is very helpful.
Werner
-----Original Message-----
From: Nieznanski, John A - SSD <[log in to unmask]>
To: [log in to unmask]
Sent: Thu, 18 Dec 2008 12:43 pm
Subject: [TN] to apply partition correction on individual daisy chains or not?
Hi TechNet List Gurus,
If I have several different serial daisy chains undergoing accelerated
reliability testing per IPC-SM-785, can I apply partition correction (IPC-SM-785
Eq. 16) to calculate an "effective" Nf @test (X%) for each daisy chain? Each
chain only has a single component type, but there are different numbers of
components in each chain.
For example, if the first chain has 10 two-terminal leadless SMT components
(assume BETA=4), and the first IPC-SM-785 failure occurs at 800 cycles, then per
Eq. 16:
Nf (component) = Nf (m partitions) * [1/m] ^ (1/BETA)
= 800 * (10) ^ 1/4
= 1422
So with 10 components in the first partition or daisy chain, then Nf @test
(component) = Nf @test (10%) = 1422.
A more conservative analysis neglecting the partition correction would be to say
Nf (10%) = 800.
Which way makes the most sense?
From Nf @test (10%) and BETA, I can calculate Nf @test (50%), then apply the
acceleration factors to get Nf @use (50%), then extrapolate to get Nf @use (1%)
and Nf @use (0.1%).
Best regards,
John Nieznanski
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