Hi All,
A good reference on the above subject can be found in Giovani Leonida's book,
"The Handbook of Printed Circuit Design, Manufacture, Components and Assembly
etc. etc. etc. per attached
Subj: Trace width vs Current Carrying Ability
To: [log in to unmask]
(Amira Mohammed [log in to unmask])
From: [log in to unmask] (Mary Sugden)
Hi All,
A good reference on the above subject can be found in Giovani
Leonida's book, "The Handbook of Printed Circuit Design,
Manufacture, Components and Assembly. The following is a
small excerpt. I hope it proves helpful.
P.S. I typed this with two untrained fingers. If something
does not make sense, call me for a fax copy.
BEGIN EXCERPT...
The resistance of a printed circuit conductor is dependent on
the cross-sectional area of the conductor, the length of the
conductor, the specific resistivity of the metal, temperature
and frequency of the signal. For the following explanations,
electroplated copper is used in all of the examples and assumes
a rectangular cross section, rather than a round wire.
The factors are;
p = Resistivity of Copper
L = Length of conductor
H = Thickness of conductor
W = Width of conductor
T = Temperature
f = Frequency
The resistivity of electroplated copper " p " is assumed to be
.693 x 10-6. This is 2% greater than the resistivity of pure
annealed electrolytic copper. Any contribution from metallic
coatings such as tin-lead, nickel, gold or other finishes can
be ignored, except for very high frequency applications.
At high frequencies, the current does not use the full cross-
sectional area of the conductor and tends to travel only on
the outer surface. This phenomenon is referred to as
‘skin effect. Since the high frequency currents use less of
the cross-sectional area, the net result is a higher resistance
per unit length.
At 25 degrees C, the resistance of a conductor R is calculated as;
L
R = p ----
W H
R = Resistance in Ohms
p = Resistivity of Copper (.693 x 10-6 )
L = Length of conductor in inches
W = Width of conductor in inches
H = Thickness of conductor in mils (x10-3)
For standard sheet copper foil, the following expression can be used.
1 Oz/Ft Sq = 1.4 mils R = 0.50 x 10-3 (L/W)
2 Oz/Ft Sq = 2.8 mils R = 0.25 x 10-3 (L/W)
For example, a conductor .005" Wide (W) by 12" long (L), using
1 Oz/Ft2 copper, the resistance (R) is calculated as...
R = 0.50 x 10-3 (12"/.005") R = 1.2 Ohms
When calculating the resistance of a printed circuit conductor,
care must be exercised to include the effects of;
Conductor width reduction due to horizontal etch
Uneven copper thickness on outer layers due to
variations in plating thickness
Conductor width reduction due to nicks and pits
in the copper foil
For necked down traces or other conductors with a non-uniform
width, it is possible to calculate the resistance of the various
portions of a trace and then add these together as resistance
in series.
For calculating the resistance of a conductor with a trapezoidal
shape such as a necked down trace, a first approximation can
be made by determining arithmetic average width using the
following...
WA = (WMIN + WMAX) /2
where, WMIN = Minimum conductor width
WMAX = Maximum conductor width
WA = Averaged conductor width
If the ratio of WMAX to WMIN is less than 1.5, the error is
negligible. The greater the ratio becomes, the larger the
error becomes. If needed, a more accurate value can be
determined by calculating the logarithmic average.
1
1 / WA = ----------- ln (Wmax / WMIN)
Wmax - WMIN
where, WMIN = Minimum conductor width
WMAX = Maximum conductor width
WA = Averaged conductor width
ln = natural logarithm
For example, for a trapezoidal conductor with a maximum width
of .050" and a minimum width of .020" the logarithmic average
is calculated as...
1
------------ ln (.050/.020) = 1.25 ln .10 = .468"
.050 - .020
This value is 7% greater than the arithmetic average.
If WMIN = 1mm and WMAX = 5mm, the ratio is 1:5 and the
difference is much greater.
Arithmetic ave = 1 / WA = 2 / (5 + 1) = 3.3mm
Logarithmic ave = 1 / WA = (1/4) ln 5 = 0.25 x 1.61 = 4.0mm
The two values differ by 21%. The arithmetic average produces
larger values for WA than the logarithmic approach, and of
course lower values of resistance.
The foregoing is valid only at a temperature of 25 degrees C.
When current flows in a conductor, the Joule effect causes an
increase in the conductor temperature. The resistance of the
copper increases as temperature increases. When warranted,
the calculations should include considerations for temperature
rise in the conductor due to Joule effect, ambient heating and
the local heating effects of hot components.
Joule Effect:
If at a temperature of T1, a conductor has a resistance of R1,
at temperature T2 its resistance R2 will be...
R2 = [ 1+ a (T2 - T1) ] R1
Where, a is the coefficient of thermal resistivity of copper.
This also changes with temperature. For the temperature range
usual in electronics an average value can be used.
a = 0.0040 degrees C x 10-1
a = 0.0022 degrees F x 10-1
To be more precise, the coefficient can be expressed as a
function of temperature by the empirical equation:
0.0045
a = ------------- where, T is in degress C ...or,
1 + (T/180)
0.0045
a = -------------- where, T is in degrees F
1 + (T-32) /324
Knowing T1 and T2, the two relevant values, a1 and a2, can be
calculated and an arithmetic average can be used. A precise
calculation requires detailed data on the copper resistivity.
In most cases the required precision is limited and sufficient
accuracy can be achieved with the following empirical equation.
235 + T2
R2 = R1 ----------- where T is in degrees C
235 + T1
END OF EXCERPT...
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