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 > > This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. Visit our web site at: http://www.ewh.ieee.org/soc/emcs/pstc/ To cancel your subscription, send mail to: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: emc-p...@hypercom.com Dave Heald: davehe...@attbi.com For policy questions, send mail to: Richard Nute: ri...@ieee.org Jim Bacher: j.bac...@ieee.org Archive is being moved, we will announce when it is back on-line. All emc-pstc postings are archived and searchable on the web at: http://www.ieeecommunities.org/emc-pstc