Now that is an answer, full of information. And rather than asking for more
details before giving any answer at all you indicate what the missing
details are and fill them in with appropriate assumptions, allowing the
questioner to replace them with details fitting his situation.
Such responses make reading this forum more fun.
Thanks,
Ahne.
-----Original Message-----
From: TechNet [mailto:[log in to unmask]] On Behalf Of Rob Legg
Sent: Monday, April 22, 2002 15:34
To: [log in to unmask]
Subject: Re: [TN] Trace width to carry high current
I'm assuming this is a safety ground trace test, and that failure is
represented by an open circuit / visible discoloration/damage or detectable
outgassing.
The test connection interface will contribute to the temperature rise if the
junction is poor.This effect is best reduced by using a soldered connection
at the point of stimulator contact to the circuit under test.
You have also neglected to mention copper weight.
***I will assume 3oz, as this is a test being performed on a power circuit.
***I will also assume that the track is on only one layer, and that the
layer is not buried, although it makes little difference in this case.
In one second, the only effect on temperature rise is the thermal capacity
of the heating element, as determined by it's material's specific heat, the
material's mass and the power being absorbed. If the substrate material is
FR4, the thermal capacity is so low as to have no beneficial effect over
that of the copper alone. Ceramic or filled substrates would contribute to
the effective mass.
***I will assume a permitted temperature rise that exceeds the UL normal
rise limit in non-accessible locations, by a factor of 50%, as this is a
temporary excursion limited by the vaporization temperature of the
***assumed FR4 substrate.
The specific heat of copper is .092cal/gdegC.
The permitted temperature rise from room temperature is
95degC x1.5 = 137 degrees for a hotspot temperature of 160degC.
Maximum permitted energy input will be:
.092 x 137 = 12.6 cal/g
or 4.19 x 12.6 = 52.8 watt.sec/g
or 52watts/gram in the one second time interval.
The power dissipation of the trace is I^2.R.
I=100A
R=K x L x W
K=1.61E-4 for 3oz copper and
L/W is the length/width ratio in 'squares'.
The mass of the trace is equal to .105mm x L x W x 8.89E-3g/mm^3
.105mm is the thickness of 3oz copper
8.89E-3g/mm^3 is the density of copper
Combine the two equations
52.8 = 1.61E-4 x L x (100)^2 / 0.105 x L x W^2 x8.89E-3
Minimum trace width W= 5.76 mm, with the length having no effect.
Note: If the conduction interval exceeds 1sec, the temperature of the track
will continue to rise at the rate of 130degC/sec.
If on the other hand, you have a trace width already laid out, you can use
the same method to figure out what the transient rise will actually be in
one or two seconds.
RL
PS: useful copper track resistance numbers -
Copper Resistance: Rc= r . L/HW r = 1.7241WcmE-6 or
6.786WinchE-7
Tempco Tc= +.039%/°C
R = K L/W
K @ 1oz (H=.0014 or .035mm) = 4.84E-4
@ 2oz (H=.0028 or .070mm) = 2.43E-4
@ 3oz (H=.0042 or .105mm) = 1.61E-4
@ 4oz (H=.0056 or .140mm) = 1.21E-4
@ foil .01 = 6.79E-5
@ foil .02 = 3.39E-5
@ foil .05 = 1.37E-5
@ bar .10 = 6.8E-6
Note: K is traditionally expressed as ohms/square, as L/W of the 'printed'
square is 1.
The resistivity of 1oz copper can therefore be considered as 0.48
milliohms/sq.
1oz PTH .062L .0017H
dia. 0.02 W=0.0613 R=0.48mW
dia. 0.04 W=0.1257 R=0.24mW
dia. 0.06 W=0.1885 R=0.159mW
L/W 2oz 3oz 4oz
1 .243 .16 .121 mW
10 2.43 1.61 1.21 mW
100 24.3 16.1 12.1 mW
----- Original Message -----
From: <[log in to unmask]>
To: "TechNet E-Mail Forum." <[log in to unmask]>; "Rob Legg" <[log in to unmask]>
Sent: Monday, April 22, 2002 4:01 PM
Subject: Re: [TN] Trace width to carry high current
>
>
> Rob,
>
> Thanks for the response.
> It's an AC 50-60HZ, at room temperature, one time only, no repetition.
>
> Thanks,
> Patrick
>
>
>
>
>
> Rob Legg <[log in to unmask]> on 04/22/2002 12:33:03 PM
>
> Please respond to "TechNet E-Mail Forum." <[log in to unmask]>; Please
respond to
> Rob Legg <[log in to unmask]>
>
> To: [log in to unmask]
> cc: (bcc: Patrick Lam/SEL)
> Subject: Re: [TN] Trace width to carry high current
>
>
>
> Depends on start temperature repetition rate and failure criteria.
>
> RL
>
> ----- Original Message -----
> From: <[log in to unmask]>
> To: <[log in to unmask]>
> Sent: Monday, April 22, 2002 1:13 PM
> Subject: [TN] Trace width to carry high current
>
>
> > Dear Technetters,
> >
> > Can anybody tell me the total trace width required to carry 100 amp for
1
> second
> > between two connections that is onch inch apart. Should the trace(s)
> better to
> > be on the outer layers than inner layers to avoid cold solder joints?
> >
> > Any information is appreciated.
> >
> > Thanks,
> > Patrick
> >
> > This e-mail may contain SEL confidential information. The opinions
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> >
>
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