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January 1997

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Subject:
From:
Don Vischulis <[log in to unmask]>
Date:
Tue, 28 Jan 1997 22:11:35 -0600
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Ralph Hersey wrote:
> 
> [log in to unmask] wrote:
> >
> > A related question to current carrying capacity in circuits:
> >
> > Need a method of calculating current carrying capacity in screened silver
> > polymer traces on 0.007" thick polyester membrane circuits.  Resistance of
> > the polymer per supplier data sheet is 0.015 ohms/square at 0.001" applied
> > thickness.  Typical width of screened trace is 0.025", and the actual
> > thickness applied in most real designs is 0.0005".  Length of the trace
> > varies with application.
> > Is there a formula for determining how much current can be carried by the
> > circuit?  Voltage is typically 5 VDC.
> >
> > ***************************************************************************
> 
> Dear cbyxbee--
> 
> I'm glad someone is manufacturing "true" printed circuits, using a printed conductive
> pattern  process as well as having electrical requirements.  The terms and acronyms for
> PCB's  (polychlorinated biphenols???) and PWB's are consistently mis-used by so many in
> our industry, perhaps some day we'll read, understand, and use the preferred terms in
> the IPC's-T-50 -- so off the soap box of my hang-ups and to address your inquiry.
> 
> You mentioned the sheet resistance of you silver (Ag) loaded polymer is 15 mohm/sq for a
> conductor thickness of 25 micrometers (per the manufacturers spec. sheet).  You also
> indicated you experience a conductor thickness of 13 micrometers, the design width of
> the conductor is 0.64 mm, and is applied to a polymeric film base material.
> 
> IMO, you've got one main unknown -- the acceptable "thermal rise" above ambient to
> identify.
> 
> A very good approximation can be derived by using (preferredly) the "flex", or as
> an alternative the rigid printed board current carrying capacity and thermal rise
> above ambient charts with appropriate compensations and interpretations.
> 
> The Key -- we have to assume you've got good quality "in-process" control on the
> screened contuctor width, thickness, and composition of the Ag paste (in other
> words you've got good control of the conductor's cross-sectional area, AND
> CONDUCTANCE / RESISTANCE OF THE DEPOSITED CONDUCTIVE MATERIAL), if not all bets are off.
> 
> As a rule of thumb, the conductivity of silver is slightly better than copper.
> Therefore, the available tables/charts for "copper" conductors may be used with minimum
> risk.
> 
> For convenience, we'll use Figure 3-4, on page 10, in the IPC's-D-275, and unfortunately
> the chart is in customary, USA dumb units, oz/sq. ft, etc.  So to make it easier to use
> the chart, the cross-sectional area of your conductor is about 12 sq. mils (inch) (25 X
> 0.5 mils).
> 
> Using Figure 3-4a, (we don't need Figure 3-4b because we know the cross-sectional area)
> we approximate the 12 sq mil area on the ordinate axis and go up to the desired "thermal
> rise" above ambient curve in the chart.  The allowable current for a thermal rise of 10
> deg. "C" above ambient would be about 0.75 A.  I suspect that because you're using a
> "polyester" base material you'll want to limit thermal rise to 10-20 deg. C above
> ambient.
> 
> If you do not have a method of transferring heat from the conductor to something,
> then you may want to "derate" the current by approx. 50% (like a MLB) and use
> Figure 3-4c instead.  Then again, if your printed board is mounted in good thermal
> contact with a cooling sink then you could increase the current density.
> 
> I assume you are aware of the possibility of "electromigration" of silver (I believe
> this has been mentioned a few times on "technet").
> 
> If you need some more info. or want some more comments/thoughs, feel free to
> contact me as indicated below.
> 
> --
> Ralph Hersey,
> 
> Ralph Hersey & Associates
> PHN/FAX 510.454.9805
> e-mail:  [log in to unmask]
> 
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Dear Ralph & cbyxbee

The concern for the temperature rise above ambient is always a concern
for reliable operation of a circuit.  My experience indicates that the
voltage (I*R) drop through the circuit is quite often the limiting
factor.  Either the voltage drop will prevent proper operation of an
electronic device or will produce noise in the circuit.

>From the circuit parameters given, a calculation of the resistance per
inch of the circuit is possible.  I don't recall how to calculate it,
but an electrical engineer or someone involved in hybrid circuits should
be able to produce the calculation.

Don Vischulis
[log in to unmask]

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