Mail*Link(r) SMTP FWD>DES/FAB: hole wall thickness
Susan inquired about plated-through hole (PTH), hole wall thickness via the
following inquiry:
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Date: 4/9/96 2:46 PM
From: [log in to unmask]
What document has instructions on how to
determine minimum thickness of plating in
plated through holes?
I have a customer who has a design with
4 oz copper on surface conductors and
2 mils of copper in holes and they want
to know if the 2 mils is really necessary and
don't know how to calculate needs.
Thanks for your help
Susan Mansilla
Robisan Lab
317-353-6249
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Response
The IPC's D-275 & RB-276 both establish the avg. and minimum Cu PTH wall
thickness, and is "illustrated" in the IPC-A-600 Acceptability of Printed
Boards document. In the case of Class 1 printed boards, the "avg" PTH Cu
thickness is 20 micrometers with local/isolated area of 13 micrometers
minimum, with anything less being considered a void. For Class 2 & 3 products
the "avg" PTH Cu thickness is 25 micrometers with local/isolated ares of 20
micrometers, with anything less being considered a void. These numbers and
acceptance criteria have been in use for about 20 years and have been
acceptable for most products and applications.
There are two main reasons for thicker PTH Cu plating requirements in the PTH
barrel: One is called reliability, the second is electrical characteristics:
#1 Reliability
Depending on your (or your customers) reliability requirements based on
operating or storage thermal ranges, and/or long-term life requirements, the
traditional MIL-SPEC and IPC Class 3 PTH Cu plating thicknesses are not
adequate. A some of us in the industry with "serious" high-rel Class 3
product reliability specify the Cu plating thickness shall average in the
range of 35-40 micrometers and no local areas shall measure less than 25
micrometers. The additional Cu plating makes for a "stronger" PTH barrel and
virtually eliminates barrel cracking / failures for most environments -- and
it doesn't cost that much more.
#2 Electrical
We, like your customer, sometimes need to have "thick" non-traditional
conductor thickness due to electrical design requirements. We have some
designs that require 245-254 micrometer [7 oz/sqft-10 millinch] thick
laminated Cu thicknesses prior to plating. I assume you customer is worried
about voltage drops and thermal rise in the conductive patterns on/in the
printed board. In order for you customer to achieve the same electrical
performance, they would have to increase the conductor widths by 3-4 times in
order to use 35 micrometer thick Cu foil, and to use 70 micrometer thick Cu
foil it would require a 2-3X increase in conductor widths. This could
significantly increase the overall size of a printed board.
Now to comment on your PTH concern:
An easy way to analyze the PTH is to look at it from an electrical resistance
point of view and the easiest is using the concept of "sheet" electrical
resistance in units of "Ohms per square". Sheet resistance is a very easy way
to analyze printed board conductive patterns because the beauty of sheet
resistance is that as long as the conductor is the same thickness, the
resistance from one side of a square to the opposite side of a square is a
constant and is the "Ohms per square" resistance. Sheet resistance is derived
from the resistivity of the material, which for Cu is 1.7 microhm-cm, and is
the electrical resistance between the opposite faces of a 1 cm cube. Keeping
the length and width of the cube a constant, as we reduce the thickness
(height) of the cube, the resistance increases. For example if the height of
the cube is reduced from 1 cm to 0.5 cm the resistance increases to 3.4
microhms, if we reduce it to 0.1 cm (1 mm) the resistance is increased to 17
microhms. In a more generalized form, the resistance in microhms between
opposite sides of a square shaped copper conductor is 1.7e4 divided by the
thickness of the Cu in micrometers. The following is a sampling of some of
the sheet resistances for the indicated Cu thicknesses:
25 micrometer [1 millinch] ~680 microhms/sq
35 micrometer [1 oz/sqft Cu] ~485 microhms/sq
50 micrometer [2 millinch] ~340 microhms/sq
70 micrometer [2 oz/sqft Cu] ~240 microhms/sq
140 micrometer [4 oz/sqft Cu] ~120 microhms/sq
190 micrometer [4 oz/sqft + 5 mils Cu] ~90 microhms/sq
Your customers "outer" layers of conductive patterns will consist of the 140
micrometer laminated foil plus at least 50 micrometers of electroplated Cu due
to the PTH process for an effective surface conductor thickness of about 190
micrometers which would have a sheet resistance of ~90 microhms/sq. A
conductor 2 squares long (twice as long as wide) would have an electrical
resistance of ~180 microhms, and a conductor 100 squares long would have a
resistance of 9 milliohms. On the other hand, a conductor 0.5 squares long
would be 45 microhms and one 0.1 squares long would be 9 microhms.
Now to examine the PTH. Assume the diameter of the hole is 1/3 the thickness
of the printed board. The circumference of the hole is Pi times the diameter
of the hole, or for all practical purposes, ~3 times the diameter of the hole.
Therefore, from one surface land to the opposite surface land there is one
square of resistance in the PTH hole barrel due to the 3 squares in parallel
around the circumference of the hole and the length of the hole being 3 times
the diameter. This results in the one square. Now you indicated the Cu PTH
thickness was 50 micrometers therefore, the electrical resistance of the hole
would be about 340 microhms. I would suspect though that you do not have a
3:1 aspect ration hole (length/diameter ratio), I would guess the diameter is
1-2 times the board thickness, and the l/d is more probably in the range of
1/3 or 1/6. In this case the electrical parallel resistance formula is used,
and for equal valued resistors (which we have in ohms/sq), the equivalent
resistance is the resistor value divided by the number of resistors in
parallel. So for the 50 micrometer thick Cu, the 340 microhms/sq would be be
divided by 3 or 6 (in this example) so the resistance of the PTH would
respectively be ~110 or ~58 microhms land-to-land, and is in the range of the
resistance of one square of outer conductor resistance (in ohms/sq). So you
custome has design the printed board so that if the PTH aspect rations are 1:1
or less the PTH has about the same electrical resistance as the surface
condutors.
The other sheet resistance values were supplied for additional information so
that if you want you can do some other modeling of the electrical network of
interconnections and conductive patterns.
CAUTIONARY NOTE: The above is good for current limited applications, if the
PTH should have an electrical resistance greater than the conductors and is
subjected to high pulse currents, the high resistance will function as a
fuseable link and blow out in a cloud of smoke.
Hope this helped to explain some of you customer's design requirements
Ralph Hersey
Lawrence Livermore National Laboratory
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