[SI-LIST] Re: Effects of steam?

Dear John Barrett,

I don't have exact numbers, but I think the problem will be much worse that
you imagine.   Here is my reasoning. 

(1) Water can (apparently) induce some types of lamination failure. To
address that problem, board manufacturers long ago agreed to a standard
delamination test for water absorbtion.  In this test, if I remember it
correctly, the finished pcb is subjected to a hot, high humidity
environment.  Then you try to pull the board apart (de-laminating it) with
suction cups.  Anyway, after processing, the board can also be weighed to
determine the percentage water absorption.  That's what you want to know. I
remember figures like 1% being rated as "OK" for pcb laminates (please check
the number with your board supplier).

(2) I do not know the density of your board material, but if I did, I would
take the percentage water absorption by weight and from it compute the
percentage water BY VOLUME, which is what counts.  Call that percentage
"alpha".

(3) The complex dielectric permittivity of your wet board may be
approximated by the following equation. This equation takes into account
both the real and imaginary parts of permittivity.  
    E(wet) = (1-alpha)E[polyimide] + (alpha)E[water]

(4) Now we get to the interesting part. The complex dielectric permittivity
of water has a magnitude of about 80 (eighty), and a terrible loss tangent
(don't recall how bad -- but this is part of the reason microwave ovens work
so well).  Since the magnitude of E[water] is so huge (80), it doesn't take
much of an "alpha" to significantly degrade the properties of your laminate.


For example, (please don’t quote these numbers this is just an example):
Park Nelco N7000 polyimide resin/e-glass:  E[polyimide] = 3.8*(1 - 0.016j)
Water:  E[water] = 80*(1 - 0.2j)  (just a guess for this example)
Alpha=0.01 (one-percent absorbtion)
Result: E[wet] = 0.99*3.8*(1 - 0.016j) + 0.01*80*(1 - 0.2j)  
               = 3.762 - 0.0608j       + 0.80 - 0.16j
               = 4.562 - .2208  
The loss tan of the result is .2208/4.562 = 0.048, triple the original
figure.

The water component, even if it doesn't change the overall dielectric
constant very much, can change the loss tangent a lot. 

RELATED ISSUE: People who make low-loss capacitors keep their materials dry.
That seems closely related to your issue. The following designers guide for
capacitors says, "Water in printed circuit laminates is responsible for a
form of dielectric absorption called “hook” that causes many problems for
users of high impedance and high frequency attenuators."
        http://www.designers-guide.org/Modeling/da.pdf 


I hope these brief comments are helpful to you.

If anyone has the dielectric loss numbers for water, I'd like to see them,
please. 

Best regards,
Dr. Howard Johnson, Signal Consulting Inc.,
tel +1 509-997-0505,  howie03@xxxxxxxxxx
www.sigcon.com -- High-Speed Digital Design seminars, publications and films
 




-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of John Barrett
Sent: Monday, November 24, 2008 4:09 AM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Effects of steam?

This one is a bit off the beaten track but I want to operate an organic
circuit board (no devices, just gold conductors on polyimide), in an
autoclave with high pressure (3 bar), high temperature (135C) non-saturated
steam. There's no condensation and so no real water about. What would be the
effect of steam absorption on the dielectric constant and the loss tangent
up to 5GHz? 
 

Non-saturated steam, i.e. dry, invisible steam, has dielectric properties
very close to air so, if it were to penetrate a porous material e.g. porous
alumina, then it would just displace the air in the pores and cause little
variation in electrical parameters. With something like polyimide, which I
presume absorbs moisture inter-molecularly, there are measureable shifts in
dielectric properties with variations in moisture content caused by humidity
variations at normal temperatures, and I think that this is caused by the
release of extra charges, but what are the effects in 135C, 3bar steam? I
could measure the effects with resonator test structures and a VNA but I
would need to find and invest in high frequency connectors and cables able
to withstand the autoclave atmosphere - not cheap, I imagine, even if they
exist. Perhaps there are dielectric experts on the list who might be able to
speculate from a materials science viewpoint?

 

Regards and thanks,

 

John

 

Séan de Baróid                     

Ionad um Raidichórais Inoiriúnaithe
Roinn na hInnealtóireachta Leictreonaí
Institiúd Teicneolaíochta Chorcaí    

                          
________________
John Barrett

Centre for Adaptive Wireless Systems
Department of Electronic Engineering
Cork Institute of Technology

"Beidh fáilte roimh freagra as Gaeilge"

 


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