Calculating string amperages [RE-wrenches]

Bill Brooks bill at brooksolar.com
Sat Feb 9 21:46:52 PST 2008


Ray,

The only use of the paralleling connector is for an on-grid PV system
needing only two series strings in parallel. Beyond that they have no other
application that I know of.

The issue is not whether a PV array acts as a blocking diode at night. Newer
modules do better with this. The issue is a fault in the wiring driving
current from the battery back through the modules. Some charge controllers
may do better than others preventing backfeed, but I have never seen this
presented as a feature. The popular feature is shutdown at night to prevent
slow loss of power through the resistor of the array at night (the array
operating in the 4th quadrant of the IV curve). This has little to do with
the fuse sizing issue.

Bill.

-----Original Message-----
From: Ray Walters [mailto:walters at taosnet.com] 
Sent: Saturday, February 09, 2008 8:58 AM
To: RE-wrenches at topica.com
Cc: Bill Brooks
Subject: Re: Calculating string amperages [RE-wrenches]



I'm passing Bill's  post to the list for everyone. Thanks BIll.
My understanding is that one of the basic functions of a charge  
controller is to prevent back feed of current to the modules. A  
normal functioning solar module becomes a resistor without sunlight  
on it.
Would one of the Manus care to respond?
Also, this comes back to the original start of this thread: where and  
what are the parallels MC connectors for, if we can't use them?

Ray

Please also see Bill's post below:

On Feb 9, 2008, at 2:23 AM, Bill Brooks wrote:

> Ray,
>
> I don't think my postings are working anymore, so I haven't been  
> responding
> on this and other issues. John does a good job of explaining the  
> issues.
> I've never heard that charge controllers are only one-way devices  
> so that
> would need to be confirmed by each manufacturer. Be careful about  
> making
> broad assumptions about products just because other products (i.e.  
> certain
> inverters) have specific features (backfeed protection).
>
> Your calculation of overcurrent size left out a very important  
> point that
> John tried to make. The total current to the module is the sum of  
> the 15-amp
> breaker plus the 6.375 amps from the paralleled string. This means the
> module is exposed to 21.375 amps (decimal places are silly) which  
> exceeds
> the 15-amp rating of the module.
>
> Also, the first 1.25 multiplier from 690.8(A) is not for cloud edge
> enhancement. Instead it is a general code fudge factor to adjust  
> the rating
> of the module for certain locations (high altitude) and times of  
> the year
> (October) when the irradiance can sustain at 1200 W/m2 or more for  
> hours.
> I've seen cloud edge enhancement last for 15 minutes in very rare  
> instances,
> but not for hours--and edge enhancement can exceed 140%-plus you  
> can have
> snow reflection on top of that.
>
> As it turns out, nearly all fuses and circuit breakers we use in PV  
> systems
> are rated 100% continuous duty, so the second 1.25 multiplier from  
> 690.8(B)
> is not absolutely required. However, the fuse and breaker  
> manufacturers
> still recommend only running their overcurrent protection at 75% of
> rating--read the design guides from Littlefuse and Bussmann for
> confirmation. In the field we have historically had lots of  
> problems with
> fuses blowing when only sized at 1.25xIsc. These fatigue failures  
> are due to
> high current and high ambient temperature (and fuse holder  
> temperature).
>
> You are obviously a very conscientious and intelligent installer,  
> and I
> don't want to come across too strongly on this issue, but I don't  
> want you
> to mislead yourself into a design concept that does not comply with  
> the NEC
> or with good practice. I agree that is sucks to have to put a  
> breaker on
> every pair of 72-cell modules, that is life and it ain't fair.
>
> Bill.
>
> -----Original Message-----
> From: Ray Walters [mailto:walters at taosnet.com]
> Sent: Friday, February 08, 2008 11:11 AM
> To: RE-wrenches at topica.com
> Subject: Re: Calculating string amperages [RE-wrenches]
>
>
>
> Well we do have some discrepancies here. William was saying we could
> go to 3 parallel Solar World strings, John Berdner here says only
> one, and I'm still saying 2 is good.
> John I have 2 issues with your excellent analysis.
> 1) 690.8 B) (1) exception: "Circuits containing an assembly, together
> with its overcurrent device(s), that is listed for continuous
> operation at 100% of its rating shall be permitted to be utilized at
> 100% of its rating."
> This means that with the right breaker, we don't have to use the 1.56
> multiplier for array current, only 1.25 which is for edge of cloud
> effect, etc. I have seen Solar world modules exceed their nameplate
> rating, so this multiplier is appropriate.
> 2 x 5.1 A x 1.25 = 12.75 Amps on a 15A breaker that is rated for 100%
> duty is acceptable. (most Outback and Midnite solar offerings are
> 100% rated BTW)
>
> 2) All this argument is based on the possibility of back feed from
> the grid or batteries, but my understanding is that both grid tied
> inverters, and charge controllers prevent back feed. So I don't think
> this is applicable to most work that we are doing. (I am aware of
> relay type controllers APT... that could allow backfeed possibly?) I
> suppose with large enough arrays, the backfeed would be from adjacent
> subarrays?
>
> So unless I'm missing something (please help me out here) I think I'm
> going to keep paralleling the SW165s in lower voltage off grid
> systems. I like to set up 4 modules per breaker, with the modules
> wired 2 in series, 2 parallels. If I'm running series strings over 48
> v nom., I'll take another look at not paralleling.
>
> Ray Walters
>
> On Feb 7, 2008, at 3:10 PM, John Berdner wrote:
>
>>
>> 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 >>>
>>
>
>

R. Walters
Solarray.com
NABCEP # 04170442	




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