Re: [SI-LIST] : Trace impedance

Dennis Tomlinson (det@tellabs.com)
Fri, 13 Nov 1998 14:16:11 -0600

allen_wang@asus.com.tw wrote:
>
> Dear Dennis:
>
> How about the heat dissipation for those traces?
>
> Allen

Hi Allen & list,

I've been looking in my "Reference Data for Radio Engineers", 5th Ed., 1974.
On page 5-33, shown as Fig. 24, is a nomograph re-printed from MIL-STD-275B.
This nomograph shows (for surface conductors) the relationship between current,
temperature rise, conductor weight, and conductor width.

Elsewhere in the same book, or in the "CRC Handbook of Chemistry and Physics,"
the temperature coefficient for the resistivity of copper is found to be
0.39%/deg-C. If ambient temperature and conductor rise are known, this can be
used to modify the milli-Ohms per square value, or equivalently, the trace
resistance. Then, knowing current and trace resistance, voltage drop can be
re-calculated.

This information does not provide a single, closed form solution to the problem,
but it does provide enough information to support an iterative closed loop
calculation scenario (provided sufficient requirements from the design are
at hand).

Regards,

Dennis

ps. Someone once told me that nomograph is a Greek word whose meaning is:
impossible to understand;-)

>
> ______________________________ Reply Separator _________________________________
> Subject: Re: [SI-LIST] : Trace impedance
> Author: Dennis Tomlinson <det@tellabs.com> at internet
> Date: 11/9/98 1:40 PM
>
> C.C. Chiu wrote:
> >
> > hi! anyone who knows how to calculate the PCB trace width which flow with
> 1A,2A.......
> > thxs!!
> >
>
> 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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