From: [log in to unmask] (Karl Bates)
To: [log in to unmask]
Organization: Westell, Inc.
Date: Fri, 11 Oct 1996 15:13:03 -0500
Subject: Sheet Resistivity of 1 OZ. Copper
>Technet people, I have an engineer who would like verification on his
>calculations to this question.
>If any of you would have any knowledge of this, please post to the forum.
>Thank you,
>Karl Bates
>[log in to unmask]
>The issue I need resolved is the Sheet Resistivity Characteristic of 1 oz. copper.
>I am astounded that none of the PC design manuals in our possession has any
>numbers about this.
Karl/Mike, IMO (in my opinion) it's only been in the last several years that the printed board
industry (in the main) has gotten into "serious" printed circuit design (electrical characteristics, not the
process). Up until we (as an industry) got above 1-15 MHZ (depending on application) most printed
boards (with the exception of power/ground distribution) were "printed wiring" (see IPC's-T-50 for
definitions). "True" printed circuit designs were a very specialized nitch part of our industry
>Let me describe a hypothetical problem.
>Suppose I need a copper trace of exactly 0.1 ohms, plus or minus a reasonable
>percentage. How would I do that?
>It seems that there should be standard curves available giving resistance per unit
>length for various trace widths.
I'll give you a sample later - it's easier/more general than that, you don't have to use any "curves",
all you'll need is a four-function calculator.
>Mathematically, though, it can be expressed as
>R =3D Rs X (L/W)
>where:
> - R is the Resistance of the trace.
> - Rs is the sheet resistance of the copper layer in ohms/square
> - L is the trace length
> - W is the trace width.
>Rs is simply the Bulk resistivity of copper divided by the layer's thickness.
>For 1 oz. copper, the published thickness is 0.035 mm (ANSI/IPC-L-108B,
>REVB, Table 2), and the bulk resistivity is 1.724e-6 ohm*cm (ITT Reference
>Data for Radio Engineers, 1975, page 421, Table 12).
Technically, it's not ohm*cm (Ohms times cm), it's ohm-cm
>The ratio of these two numbers gives a sheet resistivity of .4926 ohms/square.
>But as confident I am in my math, I would feel a lot better if someone else has
>published that result also. Besides, there may be second order effects that I have
>not anticipated.
Your sheet resistance is off by three-orders-of-magnitude (1000), it should be 492.6 microhms/sq.
The calculation follows.
>Do any of your contacts have an answer or a reference?
With best regards,
Michael A. Banak
**************************************************************
Karl/Mike, you're off to a good start, there are easier ways, the following will help and be a
suggestion/method to consider on how to get there.
Over 20 years ago (in my former life at LLNL) we were having similar problems in the printed circuit
design of conductive patterns to specific electrical characteristics. So, in the case of conductor-
resistance we translated the method of "sheet resistance" from the integrated circuit and hybrid circuit
(pre-MCM-C's) world to printed board design. Then about 10 years ago, true "printed circuit
design" requirements was becoming a need by industry. At that time, the IPC convinced me to stop
complaining and do something, like to conduct a series of workshops and "afternoon tutorials"
entitled "Electrical Characteristics of Printed Board Materials", which we did for about 6-years. In
these workshops/tutorials, one of the subjects was the use of sheet resistance for determining the
resistance of printed board conductive patterns. The following is a brief summary of that
presentation's material.
As you mentioned, the resistance of a conductor is a function of it's resistivity, length and cross-
sectional area. The general form is:
R = k p L/A = k p L / W*T
Where R is the conductor's resistance
k is a catch-all constant for the units
p is represents the Greek letter rho which is the material property "resistivity" and is generally
expressed in Ohm-cm, and we'll use your number 1.724 microhm-cm.
L is the length of the conductor, A is the cross-sectional area, W is the width, and T is the thickness.
The resistivity is generally expressed in Ohm-cm, and is the resistance between opposite faces of a
cube of material that is 1-cm cube.
Let's deviate for a moment to review series and parallel resistor networks/calculations. Using the 1-
cm cube of material, if we series two cubes of material we have twice the resistance, if we series "n"
number of 1-cm cubes, we have n*cube resistance. The general formula is Re = R1 + R2 +R3
+....+Rn. When we parallel two resistors, we now have two parallel paths for the current to flow (in
fluid flow, this would be equivalent to "equal" pipes in parallel) therefore the resistance would be 0.5
the resistance of the 1-cm cube. The general formula for parallel resistance is 1/Re = 1/R1 + 1/R2 +
..... + 1/Rn.
Now to develop the concept of "Ohms/sq." using copper as a simple example. Looking "down" on
the 1-cm cube, we have a "square" that is 1-cm on a side and the thickness is 1 cm, (opposite) face-
to-face we have 1.7+ microhms of resistance. If we put two squares in series we now have 2-
squares of series resistance which results in 3.4+ microhms of resistance; if we put "n" in series, we
have n*1.7+ microhms of resistance (face-to-face, "n" squares apart). Now if we take 2 ea of the
"resistors" with 2-squares in series and parallel the length, we have two equal resistors in parallel and
the resistance is halved. (In theory) as long as the "thickness" of the material remains constant, as long
as L = W, it is a square and the "sheet resistance" (for a specified thickness of material) is a constant
and is expressed in "Ohms/sq." This holds true if the L=W= to submicrometers or cyberspace kms,
as long as it is "square" the face-to-face resistance of "opposite sides" of the square is always the
same.
R = k p L / W*T which simplifies to k p / T, when L=W
for convenience, the conversion to "micrometer" units is provided:
Remember, p (resistivity) is given in cm's) so we'll need to adjust k convert from cm to micrometers.
1 cm = 10e4 micrometers
Doing the math, R = k p / T * 10e-4, for T in micrometers.
Which reduces to R (Ohms/sq.) = 1.724e-2 / T, for copper and T in micrometers.
As you mentioned, 35 micrometers thick (1 oz/sq-ft) Cu the sheet resistance is 492.6 microhms/sq.
For is 70 micrometers,(2 oz) Cu the sheet resistance is 246.3 microhms/sq. (Comment, as a simple
"rule-of-thumb" for quick determinations I use 600 microhms/sq. for 35 micrometer Cu foil and 300
micrometers/sq for 70 micrometer Cu foil. Easy to remember and accounts for typical manufacturing
tolerances as mentioned below.)
Now to example the use of the "Ohms/sq." concept. In your posting, you mentioned a 0.1 Ohm
resistor, assuming 35 micrometer Cu foil. Divide the resistance you want by the sheet resistance in
Ohms/sq. (0.1 mohms / 0.4926 mohms per sq. = 0.2030 squares long). So the L/W aspect ratio for
you resistor is ~0.2 of about 1/5 (that's 5-times as wide as it is long). Another example, assume you
want your conductor to have an electrical resistance of 1 Ohm and the average "effective" conductor
thickness is 70 micrometer; then the calculation is 1 Ohm / 246 microhms per sq. = (about) 4000
squares long, for a 50 micrometer (0.002 inch) wide conductor the length would be ~200 mm (8
inches) long.
Working backwards for "quality" personnel, all they've got to do is to measure the length and width
of the conductor, determine the L/W aspect ration (number of squares), determine the Cu conductor
thickness and multiply the number of squares times the sheet resistance --- then they've got a
"reasonably" close approximation.
Two of the major "delta's" that need to be controlled are (total) conductor thickness and conductor
width. In general, "narrow" conductors are more effected by conductor width tolerances and "etch
factors", wide conductors are more effected by conductor thickness. It's a simple "percentage"
calculation. If you are "serious" about printed circuit design and manufacturing, you will not be able
to use the default "unless otherwise specified" conductor width/thickness conductor resistance due to
the allowable "unless otherwise specified" material / process / manufacturing tolerances contained in
IPC specifications. For example, in IPC's-RB-276, the "default unless otherwise specified"
conductor width tolerance for Class 1 products is +/- 30 % and for Class 2 & 3 products is +/- 20
% of the conductor width specified in/on the master drawing. You're going to have a major concern
with conductor thickness, especially for outer layers in controlling the foil and plating thickness to give
you the "total" conductor thickness you'll need. Needless to say, you'll need to have good
communication of the requirements well documented in the master drawing and well qualified
manufacturer's of printed boards.
Other delta's -- in general, most conductive material "overplatings" and coatings do not significantly
effect the resistance of the conductor. Au & Ag are about the same resistance as Cu and are
"normally" very thin coatings, therefore the effect is a small percentage. Fused solder or Sn/Pb has a
electrical resistivity of about 10 times that of Cu; their coatings are generally a max. of about 30% of
Cu conductor thickness and a 30% of 10X is less than a 3% error - so you can generally (for all
practical purposes) forget/neglect them.
Last "delta" thought - all of the above assumes the operating frequency is <~1MHz and "skin-depth"
is not a design concern because the "sheet resistance" is a function of "skin-depth", and the material
properties for "Metal Resistivities" is a "0" (zero) frequency (dc - direct current) value.
I've avoided thermal rise, current carrying capacity and power dissipation because (IMO) this
response is too long now.
Hope this helps, if not you can contact me below.
Ralph Hersey
Ralph Hersey & Associates
PHN/FAX 510.454.9605
e-mail: [log in to unmask]
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