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1. Senior Member
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Originally Posted by ggunn
You don't. You find it in the 90 degree column and look to the left. The temperature rating of a conductor is actually the rating of its insulation (copper is copper), but you don't want to overheat your 75 degree rated terminals, either.

The number in the 75 degree column tells you the maximum current in a conductor to keep the temperature below 75 degrees, but it has nothing to do with the rating of the insulation. A #10 copper wire with 35A in it gets it to 75 degrees irrespective of the rating of the insulation.

The COU derating of the conductor @ 90 degrees keeps the wire below 90 degrees to protect the insulation. The CU @ 75 degrees derating of the conductor keeps it below 75 degrees to protect the terminals. You run both calcs and the lesser of the two is your "real" ampacity.
I think i see what you are saying first determine how much current it would take #10 cable terminal to get to 75C which is 35x0.8=28A. Compare that to what you get after derating. If derated ampacity is less than terminal ampacity then ok. If greater then you take terminal ampacity.

Make sense where is all of this in NEC 2014? Which code section?

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Last edited by hhsting; 06-25-19 at 08:26 PM.

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Originally Posted by hhsting
Ok here is my question:
18A is Isc of the optimizer.
If i do 18x1.25x1.25= 28A then i need 28A.

Now without any temperature correction or any derating since my circuit is less than 100A I would take 60C but 60C does not have THWN-2 column so what would be the size if THWN-2 is provided?

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The rule on 60C terminals is usually more academic than practical. It is a rule that applies in theory, for when equipment/terminals are not specifically listed for any temperature (which is rare). If you are taking an exam, you can't assume 75C by default like you can in practice. The exam problem would have to state it, or the exam would have to give it in a general statement for all problems. Most equipment you use, will be listed and labeled otherwise for 75C.

Don't expect to find a 90C rated wire in the 75C column. It is allowed to be sized for either a 60C or 75C rating, if other factors govern. THWN-2 being in the 90C column means that it is 90C rated and rated to everything lower than 90C. The breakdown of the term "THWN-2" means thermoplastic, heat and water resistant, with a nylon jacket and dual rating for 75/90C. One H = 75C, two H's = 90C, "-2" means dual rated for 90C in addition to the 75C rating it would have without it. THWN without "-2" is 75C rated in wet locations, and usually carries THHN rating allowing a 90C rating in dry locations. Most modern wire in this family is THWN-2 fully rated for both high heat and water, but check the datasheet if you depend on this being true. Some datasheets have fine print that don't carry the full rating in all sizes. You often will see THWN-2's print legend marked THHN/THWN/THWN-2, because it carries the legacy ratings of its predecessor products.

The rule about using both 125% factors, only applies to uncontrolled outputs of PV modules. It does not apply to inverters or DC/DC converters, which are current limited. The first 125% is for the possibility of more than a "full sun" illuminating the panel, such as reflection from the surroundings (glass building, snow, etc) making the irradiance more than 1000 W/m^2. The second 125% factor is the continuous load factor. It applies to continuous loads in general, to limit the risk of nuisance tripping the fuse or breaker. For inverter outputs and DC/DC converter outputs, the device is current limited. Your 18A value is used in place of 125%*Isc.

Back to your 18A optimizer output question. 18A*125% = 22.5A. #12 Cu wire carries a 25A rating at 75C. If it weren't for the "small conductor rule" in 240.4(D), #12 Cu would be our answer. If fusing is in place, you'd need #10 Cu to use either a 25A or 30A fuse, since #12 is limited to 20A fuses/breakers. What if fusing isn't in place? I hope some other members can assist on this one, because I rarely specify smaller than #10 on string wiring (other factors usually govern larger sizes), so I haven't researched how to rule out the need to apply 240.4(D).
Last edited by Carultch; 06-25-19 at 08:59 PM.

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