[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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