Calculating string amperages [RE-wrenches]

Sat Feb 9 08:58:15 PST 2008

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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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