RE: [SI-LIST] : Current flow limit for wire bonding

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From: Aubrey_Sparkman@Dell.com
Date: Thu Oct 12 2000 - 11:18:22 PDT


Maybe I'm not stating the obvious...
Wire bonds CAN carry more current per cross-sectional area than that allowed
by suppliers like Belden and others who supply 12 AWG wire. The difference
is in the enviroment. Large wires are specified to work in metal conduit in
buildings; the conduit traps heat. Wirebond wires are connected to Silicon
on one end (very good heat conductor) and a metal bar on the other end which
is an even better conductor of heat. Therefore I would not expect
MIL-M-38510 to agree with the NEMA standard.

Aubrey Sparkman
Signal Integrity
Aubrey_Sparkman@Dell.com
(512) 723-3592

> -----Original Message-----
> From: Bob Lewandowski [mailto:Bob.Lewandowski@Vixel.com]
> Sent: Tuesday, October 10, 2000 2:41 PM
> To: Brian Seol
> Cc: si-list@silab.eng.sun.com
> Subject: Re: [SI-LIST] : Current flow limit for wire bonding
>
>
> I found the MIL spec based on a web search. What I quoted
> was directly from an
> on-line copy of MIL-M-38510. I can't vouch for it's
> accuracy. The document I
> saw looked like a scanned-in copy.
>
> The constant applied is dependent on the length of the wire.
> The 30,000 value
> was for a 40 mil length between attachment points, and the
> 20,500 value was for
> length greater than 40 mils. I don't have any other
> information than that.
>
> I found the same 10,244 value for copper in my old "Reference
> Data for Radio
> Engineers" under 'Fusing currents of wires'. The last
> sentence says: "Owing to
> the wide variety of factors that influence the rate of heat
> loss, these figures
> must be considered as only approximations."
>
> ---Bob Lewandowski
>
> Brian Seol wrote:
>
> > Bob, Doug, John, and Won,
> >
> > Thank you for your reply. These are very helpful for my work.
> > Bob, I have a question regarding the constant term you gave.
> > You mentioned k = 30,000 or 20,500 for copper. This is
> quite different
> > than that from Doug Brooks' papers and Standard Handbook for
> > Electrical Engineers
> (http://home.earthlink.net/~jimlux/hv/fuses.htm).
> > They suggested k = 10,244 (d in inches) or 80 (d in mm) for copper.
> > Could you please confirm this?
> >
> > Brian Seol
> > Tessera Inc.
> >
> > -----Original Message-----
> > From: Bob Lewandowski [mailto:Bob.Lewandowski@Vixel.com]
> > Sent: Wednesday, October 04, 2000 6:35 PM
> > To: Won Chang
> > Cc: si-list@silab.eng.sun.com
> > Subject: Re: [SI-LIST] : Current flow limit for wire bonding
> >
> > Minor problem, the equation from MIL-M-38510 is I =
> k*d^(3/2), {three
> > halves
> > power, not two thirds}
> >
> > k = 30000 for gold or copper with a bond-to-bond length <=
> 0.040 in. (0.1
> > cm).
> > k = 20500 for gold or copper with a bond-to-bond length
> >0.040 in. (0.1cm).
> > (For Al wire k = 22000, and 15200, respectively)
> > d = wire diameter in inches.
> > I = DC or rms current.
> >
> > For a 1 mil gold wire < 0.04" long, I < 0.95A; > 0.04"
> long I < .65A.
> >
> > ---Bob Lewandowski
> > Vixel Corp.
> >
> > Won Chang wrote:
> >
> > > Another reference is MIL-M-38510, Military Specification,
> General Spec for
> > > Microcircuits. Check section 3.5.5.3 where you can find
> an equation:
> > >
> > > I=kd2/3 (d to the two thirds power)
> > >
> > > where: I: maximum allowed current in amperes
> > > d: bonding wire diameter in inches
> > > k: a constant (dependening upon wire
> material and length)
> > >
> > > I hope this helps.
> > >
> > > Won
> > >
> > > -----Original Message-----
> > > From: jrbarnes@lexmark.com [mailto:jrbarnes@lexmark.com]
> > > Sent: Wednesday, October 04, 2000 12:12 PM
> > > To: si-list@silab.eng.sun.com
> > > Subject: Re: [SI-LIST] : Current flow limit for wire bonding
> > >
> > > Brian,
> > > In researching the ampacity (current-carrying capacity)
> of wires and
> > printed
> > > circuit board (PCB) traces last year, I ran across the
> Onderdonk equation
> > > for
> > > the fusing (opening) current of wires as a function of
> time. I found
> > three
> > > sites on the Internet with the identical information on
> designing fuses
> > > using
> > > the Preece Equation and the Onderdonk Equation:
> > > * http://home.earthlink.net/~jimlux/hv/fuses.htm
> > > * http://www.2.ozland.net.au/users/egel/fuses.htm
> > > * http://www2.murray.net.au/users/egel/fuses.htm
> > >
> > > Gold isn't listed in the table, but its melting point of
> 1064C is very
> > > close to
> > > that of copper (1083C). The Onderdonk equation suggests
> that both metals
> > > thus
> > > should behave about the same, letting us use the Preece Equation
> > >
> > > I = 80.0 * (0.025^1.5) = 316mA
> > >
> > > I would not feel comfortable running more than 200mA or
> so continuous
> > > through
> > > each wire, myself.
> > >
> > > A. J. Rainal's paper "Current-Carrying Capacity of
> Fine-Line Printed
> > > Conductors"
> > > (Bell System Technical Journal,
> > > Vol. 60 no. 7, September 1981, p. 1375-1388) pointed out
> that for printed
> > > circuit board (PCB) traces there is a
> > > "runaway" current:
> > > * Below which the trace's temperature rise will stabilize.
> > > * But above which, if the current is held constant, the trace's
> > temperature
> > > will continue to rise until it melts open
> > > or sets the board on fire.
> > >
> > > This is due to the increase in copper's electrical resistance with
> > > temperature,
> > > versus conduction/convection/
> > > radiation carrying the dissipated power away. Since the
> electrical
> > > resistance
> > > of almost all metals increases with
> > > temperature, I would expect to see a similar behavior for
> your gold wires.
> > >
> > > John Barnes
> Advisory
> > > Engineer
> > > Lexmark
> International
> > >
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