Calculating string amperages [RE-wrenches]

[John Berdner]

William/Wrenches:
 
The "other" they are talking about in 690.8 is the same as the "n-1"
parallel strings in the fuses white paper I did.
 
Consider a system with n strings in parallel feeding a single over
current device.
If you were to create a fault in one of the strings the faulted string
does not contribute back flow current to itself.
The fault current comes from the other strings that are connected in
parallel, i.e. n-1, and from current flowing backwards through the fuse.
  
For the n-1, or other parallel connected, strings we have to add a
safety factor of 1.25 for high irradiance. We do not have to consider
the second safety factor of 1.25 for continuous duty.  (Why the fault is
not considered as possibly continuous still alludes me.)
The resulting fault current is as described in 690.8: Ifault =
((n-1)*Isc*1.25 )+the rating of the over current fuse.
Since the Code requires the over current fuse to be sized for a minimum
of n*Isc*1.56 this leads to a minimum wire ampacity / UL series fuse
rating of >= ((n-1)*1sc*1.25)+(n*Isc*1.56).  
 
Using the example from another recent post with two parallel SolarWorld
modules with an Isc of 5.1 Amps and a UL series fuse rating of 15 Amps:
 
1) The fault current from the "other parallel connected strings" is
1*5.1*1.25 = 6.375 Amps.
2) The minimum over current fuse size / wire ampacity required by Code
for 2 strings would be 2*5.1*1.56 =  15.91 Amps. Unfortunately the
standard fuse sizes are 15 Amps (too small) and 20 Amps (next size
larger) so the Code would require a 20 Amp fuse.
3) The total available fault current would then be 6.375 Amps for the
modules plus another 20 Amps for the fuse or 26.375 Amps. 
 
Since the UL series fuse size for the module is 15 Amps there is no
Code / UL compliant way to connect 2 parallel strings of this module to
a single fuse in an application where it is possible for current to back
feed the fault through the fuse.  
 
The minimum fault current of 26.375 Amps also applies to the wire
ampacity of the module interconnecting wires per 690.8.  Since most
modules have AWG 12 leads this would not pass 690.8 regardless of the UL
series fuse rating.  With a possible fault current of 26.375 Amps AWG 10
is marginal and likely would not fly after ambient temperature derating
even with 90C wire.  Modules with AWG 8 leads ? 
 
I hope this helped to add clarity rather than muddy the waters.
 
Best Regards,
 
John Berdner  


>>> wrmiller at charter.net 2/5/2008 01:23:54 PM >>>

Jay:

Begin citation:

690.8 (D) Sizing of Module Interconnection Conductors. Where a single 
overcurrent device is used to protect a set of two or more 
parallel-connected module circuits, the ampacity of each of the module

interconnection conductors shall not be less than the sum of the rating
of 
the single fuse plus 125 percent of the short-circuit current from the

other parallel-connected modules.

End citation

Note above where it says "other."

What is not totally clear to me is that in an installation without a 
combiner, i.e. connecting leads through a Square DU HU361 directly to
the 
inverter, are we meeting the "single overcurrent device" test of the
above 
section? It seems to me that the internal current limitation 
characteristic of the modules makes the overcurrent protection
inapplicable.

This code interpretation is explained pretty thoroughly and
convincingly by 
John Berdner in the SMA white paper. If you have lost the link, I have
it 
on our web site:

http://mpandc.com/resources/resources.html 

This paper has been universally accepted in the industry and has passed

muster by the local building inspector code wonk guys. I used to submit
a 
copy with permit applications to forestall questions.

William Miller


At 08:05 AM 2/5/2008, you wrote:

>Hi William,
>
>Where is this subtraction of 1 string located in the code?
>
>thanks,
>
>jay
>peltz power
>On Feb 4, 2008, at 10:18 PM, William Miller wrote:


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