Rick,

You can start your fuse design using the equations I gave a about a
month ago (5/16/98) for calculating thermal relief temperature rise
and voltage drop.  This time you need to find the temperature rise up
to the melting point of copper for a conductor that you want to fuse
at 5-6 amps.  The main problem is that you also have heat transfer to
the printed board, solder (below and above its melting point), and air
around the board.  To really do a good job your should probably have a
detailed thermal model of this whole system.

For a first order analysis I would use the following equations from
the previous posting:

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

where dT the temperature difference, (p/K)=6.7E-8 in^2C/A^2, i is
current and L/a is the ratio of conductor length to area (in inches).

The length L here is the length to the center of a reduced conductor
(fuse) width, ( i.e., half the total conductor length) and dT is the
temperature difference between the ambient temperature of the board
and the melting point of copper (1083 C). Solving this conductor half
length (L/2):

L/2= a*sqrt(dT K/p)/i

For your problem at 6 amps,  T ambient = 65C, 2 oz conductor thickness
= 0.0027 inch and an assumed width of 0.003 inch, the half-lenght is:

L/2= (.0027*.003)inch^2 *sqrt((1083-65)C * 6.7E-8(units))/6 amp=
L/2= 0.166 inch, L=.332 inch

The electrical resistance in a copper conductor is:

R= pL/a

where R is ohms, and p= 6.7E-7 ohm*inch.  For this conductor

R= 6.7E-7 ohm*inch * .332 inch / (.0027*.003)inch^2= 27E-3 ohm

about 5 times what you wanted.  You cant have your fuse and get the
resistance you want as well!

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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