Hi John,
No, N50 is the mean cyclic life for a large number of identical components. For practical reasons, the 'large' number is reduced to a minimum of 32 to maintain some semblance of statistical robustness. Anything less than that at the confidence band gets much too wide.
Sometimes you see N63.2, which simply is an artifact of the Weibull distribution being based on 'e'. It does not have any physical meaning—so N50 is preferred.
Of course, when you do a DfR-procedure (see IPC-D-279), you do not design for N50, but something like N1 (consumer products), N0.1 (industrial equ.), N0.01 (aircraft) depending on the severity of the possible consequences of failure.
N50 has nothing to do with sample size.
However, in what you are doing there may be the danger that your accelerated testing produces tighter failure distributions with larger values of beta—thus you would be too optimistic.
On the other hand, taking the first failure as your N50 might be quite conservative.
Werner
-----Original Message-----
From: Nieznanski, John A - SSD <[log in to unmask]>
To: 'Werner engelmaier' <[log in to unmask]>; [log in to unmask] <[log in to unmask]>
Sent: Thu, 18 Dec 2008 2:33 pm
Subject: RE: [TN] to apply partition correction on individual daisy chains or not?
Hi Werner,
Thanks for confirming this calculation.
Is it correct to conclude that the reason
you are calling this N50 (and not som
e other percentage) is because I have
limited samples of each component type and thus the first failure on the single
chain of “m” components essentially defines the characteristic
lifetime (or N50) for all “m” components on the chain?
I have some chains with as few as 8
components and other chains which as many as 40 components. These also need to
be analyzed appropriately, so I want to confirm the analysis approach for all the
chains in my test.
If this isn’t the reason, please let
me know how we justify the claim of calling this N50.
Thanks again.
John
From: Werner
engelmaier [mailto:[log in to unmask]]
Sent: Thursday, December 18, 2008
1:38 PM
To: [log in to unmask]; Nieznanski,
John A - SSD
Subject: Re: [TN] to apply
partition correction on individual daisy chains or not?
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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