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August 2005

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Subject:
From:
"Valerie St.Cyr" <[log in to unmask]>
Reply To:
TechNet E-Mail Forum <[log in to unmask]>, [log in to unmask]
Date:
Tue, 2 Aug 2005 15:29:35 -0400
Content-Type:
text/plain
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text/plain (125 lines)
6th paragraph typo: "is sequentially laminated" should be "isn't
sequentially laminated". Sorry.






"Valerie St.Cyr" <[log in to unmask]>
Sent by: TechNet <[log in to unmask]>
08/02/2005 03:23 PM
Please respond to TechNet E-Mail Forum; Please respond to valerie.st.cyr


        To:     [log in to unmask]
        cc:
        Subject:        Re: [TN] Required capture pad size for drilled holes?


Hal,

This actually is not a simple question with a simple answer ... here is
why. The answer depends on several factors: the thickness of the board;
the number of layers; the type of layers; the amount of copper ...

The drill tends to wander off center as it goes through the board, with
the centerpoint of the exit opening different from the center point of the
entry opening. So, for each layer it intersects on the way down, the
accuracy gets progressively worse; therefore, the shorter the distance it
travels in the z-axis, the more accurate the location at any point in the
z-axis.

Signal layers, especially half-ounce copper foil signals, which are
lightly populated will distort more than heavy copper planes, which means
the pads may end up further away from their artwork nominal locations.

A big factor that is usually not understood by designers is that a hole
described as .010 +0/-.010" is not a .010" hole. If your board is thick,
the fabricator will want to drill it with the largest drill size your pad
selection will allow - and you want them to do that too, because 1) it
saves you money and 2) it is more reliable. So if you have an .062" board
they can drill it with a .010" bit and it will finish at about .007"
nominally (and you will pay for a .007" hole even though the quote comes
back with .010" holes on it!). But if you have a .125" thick board they
would prefer not to drill it with a .010" bit. They would like to drill it
with a .014" (give or take a half mil) bit; and that hole would finish at
.010" nominally.

So what is really the deciding factor here is not the finished hole size
you put on the drawing, but the pad size you put in your padstack. And the
thicker the board the larger should be the pad. If we assume a +/-.005"
allowance for layer registration (a common value for the state of volume
production) and you want +.002" of annular ring guarantee, then you need a
minimum of .007" annular over *drill size*, or drill + .014". But you
don't know what the drill size is. Typically the drill size is +.004" over
the finished hole size. So the .010" hole drilled with a .014" bit into a
pad with a  .002" annular ring requirement, needs a .028" pad, minimum.

You can get that pad size down, but only if your board isn't too thick ...
or is sequentially laminated ... or has a lot of sparse signal layers ...

Unless this is a plain vanilla board, if you really need Class III, you
should pull in an applications engineer from your fab house to recommend
the optimal pad sizes for your type of board.

Regards,

Valerie





Hal Winslow <[log in to unmask]>
Sent by: TechNet <[log in to unmask]>
08/02/2005 03:02 PM
Please respond to TechNet E-Mail Forum; Please respond to Hal Winslow


        To:     [log in to unmask]
        cc:
        Subject:        [TN] Required capture pad size for drilled holes?
I'm looking for what fab houses would consider to be the minimum sized
capture pad for a drilled hole in a PCB if acceptability is specified
per IPC 600 Class 3.

For instance, when specifying a .010 via (+0, -.010 finished hole size)
what size pad should I design in so that you can hit the minimum .002
annular ring required for Class 3?

I believe that I have heard drill size plus .014 for the pad size.

Bonus points will be awarded if there is a difference between inner and
outer layer pad size and you detail that for me...

Thanks!

Hal Winslow


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