Mirka,

The thermal relief issue has bugged me too. The rule of thumb that I
hear often is that you want to have a maximum sum of .160in of 1oz
copper trace width going into a pth (.080in for 2oz).  This means that
if you have 4x 1oz power planes that you want to connect to one solder
pin with a 4 spoke thermal relief design, then the trace should be:

.160 in / (4 planes * 4 spokes)= .010in wide.

Unfortunately, there not as much agreement on thermal relief "spoke"
length.  Most people I have checked with use .010 to .020in lengths.

Another guideline is to use 4 spokes (per layer) whenever possible to
give a uniform temperature distribution around the PTH.  3, 2 and 1
spoke can also be used when necessary.

*********************************************************************

I have developed the following equations based on simple physics, (no
testing).  There are three issues of concern:

1) The temperature rise when soldering
2) The temperature rise under electrical load
3) The voltage drop under electrical load

For the first case, the equations are based on simple 1D heat
transfer:

Q= (K a dT)/L = K dT (a/L)

where Q is the input heat power, K is the thermal conductivity of the
material, dT is the temperature difference and a, L are the cross
section area and length.

The actual values to use in this equation are an area of debate.  K is
easy.  I use K=10 (W/inC) in mixed SI and english units for Copper.
dT is the temperature difference between the ambient temperature of
the board and the solder joint at reflow.  Lets ASSUME 100C ambient
(mild preheat) and 250C soldering temperature for a dT of 150C for
example.  Lets also ASSUME that we are using a 25W hand soldering
iron. These assumptions can be changed around to suit your conditions.

We can now rearrange the above equation to solve for geometry:

(a/L)= K dT/Q   or  L= a K dT/Q

If we use the .160" of 1oz copper trace width as a rule for the area
we can calculate the length needed as:

L= .160in * .00135in * 10 W/inC * 150C / 25W = .013in

This falls in the .010 to .020in range that I mentioned above.  Note:
larger dT and smaller Q assumptions increase length.

The second concern, can be evaluated by changing the equation.  The
electrical resistance in a copper trace is:

R= pL/a

where R is ohms, and p= 6.7E-7 ohm*in.  The power dissipated is:

Q= i^2R

Combining these equations and solving for dT:

dT=(p/K)*(iL/a)^2

(p/K)=6.7E-8 in^2C/A^2 or to allow for minimum plating thickness and
trace width, I use (p/K)=1E-7 in^2C/A^2

For example, the temperature rise in a 15A .160in 1oz copper thermal
relief spoke .013in long is:

dT= 1E-7 in^2C/A^2 *(15A *.013in /(.160in *.00135in))^2= 0.08C

This is a very small temperature rise and it shows how good copper is
for this use.

The final concern is voltage drop:

V= iR= ipL/a

For our example above:

V= 15A * 6.7E-7 ohm*in *.013in / (.160in *.00135in)= .0006 volts


***********************************************************************
Based on all this analysis, my department tends to use a 4 spoke
design with .015in wide by .010in long spokes on outer layers and
.020in long spokes on inner layers.  We have set up a common aperature
table and will use this when we have 4 or fewer layers.  The increased
temperature rise and voltage drop when there are fewer layers are
still small.

Any comments on this analysis???  I hope it helps :)

--
Karl Sweitzer                     voice: 716.47.77546
Eastman Kodak Company             pager: 716.25.33681
800 Lee Road                        fax: 716.47.77293
Rochester, NY 14650-3118         mailto:[log in to unmask]

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