DOUG,
        COULD YOU RUN US OFF COPIES OF THIS INFO?  WE'VE BEEN USING 
     IPC-D-275 FOR FIGURING CURRENT CAPACITY.
                                ROB


______________________________ Reply Separator _________________________________
Subject: Re: des re: hot trace width
Author:  [log in to unmask] at internet-mail
Date:    4/9/96 9:49 AM


     
     There are TWO issues here.  One - current carrying ability over a long 
     period of time, two - current carrying ability over a short period of 
     time.
     
     For extended use, basic good engineering practices say, the trace has 
     to be able to carry the current for which it's fused.  As if it were 
     solid wire.  Look up AMAPACITY in the Electrical Engineer's Handbook.  
     Find the current you need at the max. temp you'll experience in the 
     Underwriter's Laboratories column.  Cross over to the gauge wire for 
     it.  Look up the SQUARE MILS of the wire.  From your known trace 
     thickness calculate your width.
     
     
     Short time stressing for wires should be under a second.  Fuses 
     typically react well under a second.
     
     Be careful if you are doing trace calculations for Primary voltage 
     carrying traces.
     
     Doug McKean
     ADC Video Systems
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______________________________ Reply Separator _________________________________
Subject: des re: hot trace width
Author:  [log in to unmask] at internet-mail 
Date:    4/8/96 4:10 PM
     
     
Andy Pedersen asked about trace with to current capacity requirements.
     
In addition to Kevin L. Seaman's variables (conductor width, thickness 
and current) the board layout has an impact also, in a flex design, 
The test folks caused a short (get to blame them _this_ time ;) )
that put 4 amps on a .013 trace.  Where the trace was near other 
copper, the board survived, but de-laminated in more open areas. 
(and we still don't realy know how _long_ it was shorted) 
(The trace is still connected & not shorted to any near by traces)
     
In "Electromechanical Design" by Ronald A Walsh, pg 255, 6.6.10 
"Fusing Time-current for copper connections" he quotes an equation 
by I.M. Onderdonk:
(let's see if I get this right..)
     
33(I/A)**2 * S = log10(((Tm-Ta)/234+Ta)+1)
     
I=A * ((log(((Tm-Ta)/234+Ta)+1))/(33 * S)) **.5
     
I=current, Ampers, A=conductor area, cir mils, S=time current applied, 
seconds, Tm=melting point of copper in dec C, Ta= ambient Temp in deg C.
     
Has any one used this?  I haven't tested it out ( we try to keep the 
smoke _in_ the parts don't you know ;)  ).
     
Bill Gaines
AeroJet (for $)         IHPVA   (for speed & fun) 
IEPS    (for info)      SCCA    (for trophys)    
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