Chris,
I have estimated this type of thing in the past assuming adiabatic
conditions. That is, the energy put into the material heats it and no energy
is lost during the heating. This gives the worse-case temperature rise. So,
what is needed is the electrical resistance of the material and the specific
heat of the material. Let's look at the case of a trace sized to handle 25
amps continuously with a 40 degree C rise.
http://www.geocities.com/CapeCanaveral/Lab/9643/TraceWidth.htm
We will pulse it with 200 amps for 20 ms and see what the heat rise is
assuming adiabatic conditions:
The trace is 500 mils (1.27 cm) wide, 1 inch (2.54 cm) long, and is 1 oz
copper (1.4 mils or 3.55 x10^-3 cm). The resistance is 1 milliohm. The
energy absorbed is (I^2)(R)(t) = 0.8 joules. The density of copper is 8.96
gr/cm^3. The mass of this trace is 0.103 gr. The specific heat of copper is
0.386 J/gr*C. The specific heat of this trace is therefore 0.0398J/C which
gives a heat rise of 20 degrees C for a 200 amp, 20ms pulse. I have
neglected the change in resistance and specific heat with temp.
I have actually viewed the voltage drop across a metal line as it was heated
by a pulse. From this one can plot the temperature versus time. The real
issue, I think are the vias and vias with heat reliefs. How many do we use?
We can calculate the vias the same way and come up with a recommendation. Of
course, it would be great to check this with experiments. I would be
interested in doing this if someone here wants to partner on the project. I
can have a test board designed and built and do the pulsing. What I need are
standards, suggestions, circuit breaker data, and any other help (such as
researching to see if we are reinventing the wheel. The results would then
be published in Compliance, Conformity, or Printed Circuit design magazine.
Dave Cuthbert
Micron Technology
From: Chris Maxwell [mailto:chris.maxw...@nettest.com]
Sent: Tuesday, February 04, 2003 6:32 AM
To: drcuthbert; John Woodgate; emc-p...@majordomo.ieee.org
Subject: RE: EN60950 protective conductor test (was Re: Circuit Breaker
Tripping Dring Fault Tests)
Exactly! There is lots of data and tables available on the web for steady
state current; but I haven't found any sources that would give the (I^2)(t)
values for wires or PCB traces. Such tables would take a great deal of
mystery out of this subject. Right now, the best guess is to go by steady
state current rating; but there must be faults in this. A PCB trace that
can handle 10 Amps of steady state current has a totally different geometry
than a wire that can handle 10 Amps of steady state current. This would
make heat dissipation different; and I would assume that it would make the
fusing characteristics (I^2)(t) slightly different as well.
Chris Maxwell | Design Engineer - Optical Division
email chris.maxw...@nettest.com | dir +1 315 266 5128 | fax +1 315 797 8024
NetTest | 6 Rhoads Drive, Utica, NY 13502 | USA
web www.nettest.com | tel +1 315 797 4449 |
> -----Original Message-----
> From: drcuthbert [SMTP:drcuthb...@micron.com]
> Sent: Monday, February 03, 2003 7:50 PM
> To: 'John Woodgate'; emc-p...@majordomo.ieee.org
> Subject: RE: EN60950 protective conductor test (was Re: Circuit
Breaker Tripping Dring Fault Tests)
>
>
> What is needed is the I squared t rating of the breaker. Then the (I^2)(t)
> rating of the PCB. Then you know if the PCB can take it.
>
> Dave Cuthbert
>
>
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