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February 2004

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(Leadfree Electronics Assembly Forum)
Date:
Fri, 20 Feb 2004 12:36:51 EST
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Hello Pascal,
From a generic point of view, my preference would be for your second approach 
using "t110/t60", i.e. based on field reliability goals.  However, it was not 
clear to me what mechanism(s) you want to accelerate and my question is where 
does the 0.7eV activation energy come from? If this is for creep, 0.7eV looks 
kind of high but I'd have to double-check on that.
Interestingly, near-eutectic SnBi did quite well (in accelerated thermal 
cycling) in the NCMS lead-free project of the early 90's.

With best regards,
Jean-Paul


In a message dated 2/20/2004 9:29:41 AM Eastern Standard Time, 
[log in to unmask] writes:
As alternative to SnPb solder, we sometimes see application of low-melting
solders, such as the ones based on the SnBi system. Of course, these can
only be used in applications in a low tmperature range. The question is now
how would we test  reliability for these joints, standard test @ 150 °C for
1000 hours will obviously not work. Is there a standard available?
If not, does anyone has an idea?
We have identified two approaches :
Standard testing would take place at 30 degrees below the melting point,
this would translate in the case of SnBi (Tm=138) to ~ 110°C. (Using the
homologuous temperature would result in the same temp.)

1) Assuming an activation energy of 0.7 eV (= 1.12-19 J), the ratio between
the test times at the two temperatures is: t150/t110 = exp{E/kB(1/T1-1/T2
)}, kB is the Boltzman constant, temperatures T1,2 in Kelvin. The result
is: t150/t110 = 0.135, meaning that at 110°C you have to extend the 1000
hour test to 7400 hours.
2) Another approach is to start from the field specifications: 5 years
continuous use at 60°C (5*365*24 = 43800 hours) and translate this to "use"
at 110°C. We then get in the same way: t110/t60 = 0.041, meaning to test at
110°C for 1800 hours.
pproach it from field specifications, e.g. 5 years continuous service at 60
°C (5*365*24 = 43800 hours) and translate this to "use" at 110°C. We then
get in the same way: t110/t60 = 0.041, meaning to test at 110°C for 1800
hours.


My preference is for the second test since this is based on field
applications. Any ideas?

_____________________________________
Jean-Paul Clech
EPSI Inc.
Home page URL: http://jpclech.com
P.O. Box 1522
Montclair, NJ 07042
USA
tel: 973-746-3796
fax: 973-655-0815

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