Solaredge optimizer output current

HoosierSparky

Senior Member
Location
Scottsdale AZ
BIG "discussion" going on about sizing conductors from a junction box to an inverter. There are multiple stings of 12+ modules with Solaredge optimizers with a max Imax of 15A. This is a CONTINUOUS current. PV wire in free air is used for the strings to J-box. From the J-box conductors are run in a single conduit after changing to THHN-2. Here is where the "discussion" gets interesting. Do you apply 210.19(A)(1)(a) (125%) to the conductors for continuous current? This is in addition to applying derates for conduit fill and temperature above the roof.

For example:
1617634206203.png
15A x 125% = 18.75A if applying 210.19.
Next, #10awg @ 90C = 40A
Temp at .5" above roof (Phoenix) = 122F + 40F = 162F =0.50 derate
40A x 0.50 = 20A.
6 conductors in a conduit = 0.80 derate
20A x 0.80 = 16A

16A < 18.75A, so #10 is too small. #8awg would work. HOWEVER, if the application of 210.19 isn't necessary, then #10 would work. Does 690.8(A)(3) apply or even 690.9(B)?
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
My understanding is that in general the 125% continuous factor applies only at the terminations, where you use the termination temperature ampacity and don't apply the correction and adjustment factors. And in the conduit you use the correction and adjustment factors but don't need to use the 125% continuous factor.

When combining more than 2 solaredge strings, is fusing required? There used to be a white paper from Solaredge on that, but it doesn't seem to be available anymore.

Cheers, Wayne
 

pv_n00b

Senior Member
Location
CA, USA
The continuous current adjustment and the conditions of use adjustments are not applied together. If they are the result is usually an oversized conductor. This is described in chapter 2 and 3 but it is reproduced in 690.8(B) (1) and (2) for the PV folk.
 

pv_n00b

Senior Member
Location
CA, USA
15A x 125% = 18.75A if applying 210.19.
Next, #10awg @ 90C = 40A
Temp at .5" above roof (Phoenix) = 122F + 40F = 162F =0.50 derate
40A x 0.50 = 20A.
6 conductors in a conduit = 0.80 derate
20A x 0.80 = 16A
#10 continuous ampacity = 40/1.25 = 32A

#10 COU = 40 x 0.50 x 0.80 = 16A

#10 is good because 15A< 32A & 15A < 16A. The other check to make is the terminal temperature, which is the #10 ampacity @ 75deg divided by 1.25.
 

HoosierSparky

Senior Member
Location
Scottsdale AZ
My understanding is that in general the 125% continuous factor applies only at the terminations, where you use the termination temperature ampacity and don't apply the correction and adjustment factors. And in the conduit you use the correction and adjustment factors but don't need to use the 125% continuous factor.

When combining more than 2 solaredge strings, is fusing required? There used to be a white paper from Solaredge on that, but it doesn't seem to be available anymore.

Cheers, Wayne
I think you are incorrect in the application of the 125%. No where does it state that it applies to the terminations. It applies to the conductor. Terminations are limited by their size and temp ratings. So...............If the current is continuous, which it is due to the current limiting by the optimizer, it would follow that 125% should be applied to the current supplied (15A x 125% = 18.75A). Or continuous current should not be considered due to the current limiting?
 

HoosierSparky

Senior Member
Location
Scottsdale AZ
The continuous current adjustment and the conditions of use adjustments are not applied together. If they are the result is usually an oversized conductor. This is described in chapter 2 and 3 but it is reproduced in 690.8(B) (1) and (2) for the PV folk.
The way I read 690.8(B) I use the larger of value from (1) or (2). (1) says to apply 125% to the sum of the modules (15A x 125% = 18.75A) and (2) says to make all corrections and adjustments to determine max current (40A x .0.5 x 0.80 =16A). So now I have a choice of 18.75A < 30A (Max current allowed for #10awg w/ no additional derates) or 15A from the optimizers < 16A after applying corrections and adjustments makes #10 OK.

Where did the KISS principal go? :)
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Not a solar guy, so I may be missing the actual basis of the 125% requirement in solar installations, however:

In general, the requirement that conductors be rated at 125% of continuous loading derives from breaker limitations. Breakers are not permitted to be loaded to more than 80% of their trip rating by continuous loading, but they must protect the conductors. So if you have an 80A continuous load, you are required to use a 100A breaker, which then forces you to use 100A conductors. But _if_ you use 100% rated breakers then you are permitted to load the conductors up to 100% of their ampacity.

So the question here is: are there circuit breakers here that have this 80% restriction?
Is there another aspect of the code that introduces an 80% restriction?

If not, then conductors can be used at the full 100% of their ampacity, which is of course adjusted by ambient temperature and other conductors sharing the same raceway.

-Jon
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
I think you are incorrect in the application of the 125%.
Your next post with the description of 690.8(B) says the same thing I did--one check is with the 125% continuous factor, and no adjustment or correction (this is implicitly at the termination and limited to the termination temperature, as per your comparison 18.75A < 30A), the other check is without the 125% continuous factor, with adjustment and correction factors, and without the termination temperature limitation (which represents the conductors away from the terminations.)

The point is that you never have to apply the 125% and the adjustment and correction factors simultaneously, i.e. you never do the computation 40A * 0.5 * 0.8 / 125% = 12.8A

Cheers, Wayne
 

Designer101

Member
Location
California
Occupation
Solar Designer
BIG "discussion" going on about sizing conductors from a junction box to an inverter. There are multiple stings of 12+ modules with Solaredge optimizers with a max Imax of 15A. This is a CONTINUOUS current. PV wire in free air is used for the strings to J-box. From the J-box conductors are run in a single conduit after changing to THHN-2. Here is where the "discussion" gets interesting. Do you apply 210.19(A)(1)(a) (125%) to the conductors for continuous current? This is in addition to applying derates for conduit fill and temperature above the roof.

For example:
View attachment 2556079
15A x 125% = 18.75A if applying 210.19.
Next, #10awg @ 90C = 40A
Temp at .5" above roof (Phoenix) = 122F + 40F = 162F =0.50 derate
40A x 0.50 = 20A.
6 conductors in a conduit = 0.80 derate
20A x 0.80 = 16A

16A < 18.75A, so #10 is too small. #8awg would work. HOWEVER, if the application of 210.19 isn't necessary, then #10 would work. Does 690.8(A)(3) apply or even 690.9(B)?
I understand your problems here, even I have come across with same issue before, solarabcs.org gives you an interactive map , you don't have to use 0.5" though, use 3.5" above roof and the temperature will be little less, so you can use different temperature derate factor, I usually use 0.65 and I am in palm springs CA, same whether as Phoenix. i never get any questions from AHJ.
 

HoosierSparky

Senior Member
Location
Scottsdale AZ
I understand your problems here, even I have come across with same issue before, solarabcs.org gives you an interactive map , you don't have to use 0.5" though, use 3.5" above roof and the temperature will be little less, so you can use different temperature derate factor, I usually use 0.65 and I am in palm springs CA, same whether as Phoenix. i never get any questions from AHJ.
The 122F is from an ASHRAE table. As the AHJ reviewer, I go with what is on the plans. I can't engineer it, just "grade" it. Pass/fail!
1617821201407.png
 

Designer101

Member
Location
California
Occupation
Solar Designer
The 122F is from an ASHRAE table. As the AHJ reviewer, I go with what is on the plans. I can't engineer it, just "grade" it. Pass/fail!
View attachment 2556099
ya 122 F IS the max temperature you have to take that value but 40 degrees adder is not needed when we install conductors 7/8 inch above roof top.( its actually 33 degrees c adder per 2014 NEC)
Also many industry professional agree that if we apply conditions of use (here temperature factor and conduit fill factor), then we don't need to multiply 1.25 percent to the optimizers current.
I understand you might be plan reviewer and i appreciate the post you made to make things clarify.
 

HoosierSparky

Senior Member
Location
Scottsdale AZ
ya 122 F IS the max temperature you have to take that value but 40 degrees adder is not needed when we install conductors 7/8 inch above roof top.( its actually 33 degrees c adder per 2014 NEC)
Also many industry professional agree that if we apply conditions of use (here temperature factor and conduit fill factor), then we don't need to multiply 1.25 percent to the optimizers current.
I understand you might be plan reviewer and i appreciate the post you made to make things clarify.
I could not find where you get the 7/8" for the adder. Per 2014 NEC Table 310.12(B)(3)(c) ABOVE the roof (1/2" - 3 1/2") the adder IS 40F. If you read some of the previous responses, 690.8 (B) 1 & 2 allows the application of 125% to the current from the optimizers for continuous current.

The point I have been attempting to get across is when there is a hard and fast rule (125% adder for continuous current) it all changes when you get to 690. Clarity is a wonderful thing! :)
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
Per 2014 NEC Table 310.12(B)(3)(c) ABOVE the roof (1/2" - 3 1/2") the adder IS 40F.
Are you still under the 2014 NEC? This changed in the 2017 NEC to what Designer101 quoted. It was determined that the 2014 provision was excessively conservative.

The point I have been attempting to get across is when there is a hard and fast rule (125% adder for continuous current) it all changes when you get to 690.
Not applying both the 125% current factor and the correction and adjustment factors simultaneously is not a 690-only provision, it is the general rule.

Cheers, Wayne
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
Not applying both the 125% current factor and the correction and adjustment factors simultaneously is not a 690-only provision, it is the general rule.
P.S. I've been a little lazy and haven't previously provided the code citations. See, for example, 215.2(A)(1) on sizing feeders. Note that the 125% factor occurs only in (a), and that the adjustment and correction factors occur only in (b). I expect there's a similar section in 210.

Cheers, Wayne
 
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