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April 1998

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Subject:
From:
Scott Decker <[log in to unmask]>
Reply To:
DesignerCouncil E-Mail Forum.
Date:
Fri, 24 Apr 1998 06:10:09 -0700
Content-Type:
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JoAnn Amerson wrote:

> > From:          [log in to unmask]
> >
> >      Example:  Assume finished plated through hole size of 0.034+/-0.003";
> >                minimum annular ring requirement of 0.002"; and, FA = 0.014"
> >
> >      Therefore:  P = (0.034 + 0.005) + 0.014 + 2(0.002) = 0.057"
> >
> >
> >      Using the information from the above example, I would answer your
> >      questions as follows:
> >
> >      #1  Maximum finished hole size = 0.034 + 0.003 = 0.037
> >
> >      #2  Value to add to nominal hole size = 0.005 + 0.014 + 0.004 = 0.023
>
> This thread is of interest to me.  We don't use a standard formula
> here but I would like to try to have one.  One of my
> responsibilities is to develop a "guideline" (the word "standard"
> isn't allowed) on creating component footprints.  Right now I can't
> justify the pad sizes we are using.  If I had a formula I could start
> using that and feel beter about the numbers I come up with.
>
> I was doing fine with Mr. Wolf's explanation until he popped in real
> numbers.  He started with .034±.003".  Then he states P = (0.034 +
> 0.005) Where did that .005 come from?  Or was that a typo?
>
> Anyone care to help me?  I know it's early, and it's Friday, but
> surely someone has a few brains cells they can show off this morning.
>
> TIA,
> Jo
>
> JoAnn L. Amerson
> Design Librarian
> Red Lion Controls, Inc.
> E-mail: [log in to unmask]
> Voice: (717) 767-6961 ext 6308
>
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 JoAnn,
  The drill hole size number that was mentioned, (.005") is what most board manufactures use as
an
oversize amount. You have to drill the hole larger than the finished size to account for
plating. This
.005" is .0025 on each side and when they plate .001" or so in the hole, presto a finished hole
size of
what you specified... One thing you do have to take into consideration is that this .005" should
be from
the plus side of your tolerance. If the hole is .035 +/- .003 the drilled hole at the board shop
will be
.038 + .005 = .0043. Your pad size needs to be able to handle this oversize for the annular
ring.
Dang.... I was trying to post this to the group but I hate Netscape 4.XXX..... I hope this helps
in you
quest for a stand..... I mean GUIDELINE at your place..

--
         Scott Decker
      AKA: PadMasterson
Praegitzer Design On Location at
 Enterprise Server Group CO3
      Intel Corporation
     Ph: (503)-677-6582

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