Hi Mr. Chiu,
Since the currents you refer to are 1A, 2A, .. I assume you are asking a
question about DC trace resistance and not characteristic impedance. That
is the question I will answer. My apologies if I am wrong.
First, let me define a square trace. A trace which is of width w and length
l = w is said to be a one square trace. A trace of width w and length l = 10*w
is then a 10 square trace. As an illustration, consider the following:
PCB trace segment rated as a 5 square trace
_________________________________________________ ___
| | | | | | |
| | | | | | |
| | | | | | w
| | | | | | |
| | | | | | |
|_________|_________|_________|_________|_________| __|_
|____w____|
In tabular form, some numerical examples would look like:
Trace width (mils) Number of squares per routed inch
------------------ ---------------------------------
10 100
20 50
30 33.333
50 20
100 10
w 1000/w
With this concept, the DC resistance of routed trace can then
be calculated from the following table based on copper weight:
Cu wt. in ounces /in^2 thickness (in.) milli-Ohms/square
---------------------- ---------------- -----------------
1/2 0.0007 1
1 0.0014 0.5
2 0.0028 0.25
These values are based on the conductivity of copper, and are nearly
exact at room temperature. The process employed in the manufacture of
PCB's often cause thickness to be slightly less than above. A simple
proportion can be used to re-scale the mill-Ohms/square appropriately.
As an example of how to use the above tables, assume I have 6 inches
of 20 mil wide trace routed with 1 oz/in^2 Cu. The number of squares in
this trace is:
(6 in)*(50 squares/in) = 300 squares.
The total trace resistance is given by:
(300 squares)*(0.5 milli-Ohms/square) = 150 milli-Ohms
With this information, augmented by requirements from the design
such as:
1. How much DC voltage drop you can afford,
2. How much routed trace length is expected, and
3. What Cu weights can or will be used,
you can choose trace width appropriately.
On the other hand, if I have misunderstood your question, I should
now be very embarrassed!
Cheers,
Dennis
ps. Thanks to Tom Stamm who taught this to me when I was a fledgling
engineer, more concerned with which end of the soldering
iron to hang onto.
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