Code Corner: PV Circuit Sizing & Current Calculations: Page 2 of 2


The second calculation and comparison is outlined in 690.8(B)(2)—690.8(B)(2)(b) in the 2011 version. Here, after conditions of use have been applied, we compare the conductor’s ampacity to the maximum current as calculated in 690.8(A). This section does not require multiplying by 125% a second time.

Let’s examine these calculations with an example three-string batteryless grid-tied PV system in Sacramento, California. The source circuits run in EMT from a rooftop junction box and down to the inverter. To begin, we consider the type of conductor used: THWN-2, which is rated at 90°C and for use in wet or dry environments. Use the ampacity values in the 90°C column of Table 310.15(B)(16) when adjusting the conductor’s ampacity.

The temperature limitation of the conductor’s terminals must also be considered. Although the direct reference to 110.14(C), which states that the conductor’s ampacity rating needs to correlate with the terminal’s rating, has been removed in the 2014 NEC, that Code section is still necessary. If the terminal is rated for 75°C (common for terminals in PV junction boxes), the ampacity of the conductor at the terminal is considered to be that of a conductor with the 75°C rating, as long as the actual conductor rating is 75°C or more.

To start sizing the conductor, first consider 690.8(B)(1) from the 2014 NEC. The source circuit’s maximum current is 8.8 A × 1.25 = 11 A. The conductor must have an ampacity of 11 A × 1.25 = 13.75 A. Referring to Table 310.15(B)(16), in the 75°C column (because of the terminal limitations), 14 AWG is the smallest conductor listed with an ampacity greater than 13.75 A.

For 690.8(B)(2), the effects of temperature and the number of conductors in a conduit are considered. The 90°C rating for the conductors can be used. We need to know the local ambient temperature, the number of current-carrying conductors in the conduit or raceway, and whether any additional conditions apply. In this system, the EMT is installed 4 inches above the roof and runs 12 feet before going down to the inverter. Therefore, the correction factors in Table 310.15(B)(3)(c) need to be applied to determine the temperature imposed on the conductors. Looking at our correction factors:

• ASHRAE 2% ambient temperature: 38°C. The ASHRAE data is referenced in an Informational Note to Table 310.15(B)(3)(c). (See

• For a raceway that lies 4 inches above the rooftop and is exposed to sunlight, add 17°C to the ambient temperature for correction-factor calculations:

38°C + 17°C = 55°C

Temperature correction factor for conductors rated at 90°C in 55°C environment:

0.76 – 310.15(B)(2)(a)

• With three source circuits, there are six current-carrying conductors in the raceway.

Table 310.15(B)(3)(a) correction factor is 80%

• To compare the 14 AWG conductor found in 690.8(B)(1), use the ampacity of 14 AWG from the 90°C column and apply conditions of use:

25 A × 0.76 × 0.8 = 15.2 A

This ampacity must be greater than the maximum current calculated in 690.8(A): 15.2 A > 11 A, so the 14 AWG fits for this calculation.

To complete the conductor sizing, verify that the conductors are properly protected by any OCPD in the circuit (refer to 690.9 in the 2014 NEC or 690.8(B)(1) in the 2011 version for the OCPD requirements).


Ryan Mayfield is the principal at Renewable Energy Associates, a design, consulting, and educational firm in Corvallis, Oregon, with a focus on PV systems.

Comments (7)

Ryan Mayfield_2's picture

Thanks for the comment. The way Code reads in 690.8(B), we have to look at the ampacity two different ways. The first is to apply two 125% factors to the Isc (8.8A x 1.25 x 1.25 = 13.75A in my example) and verify the conductor chosen has enough ampacity based on that value without applying conditions of use. This is how I got to a 14AWG conductor in my example. The second part is to apply conditions of use to the conductor's ampacity (from Table 310.15(B)(16)) and make sure that corrected factor is greater than the Isc multiplied by a single 125% factor.

So in the example I used in the article, I corrected the 14AWG conductor down to having 15.2A of ampacity because of the conditions of use. Per 690.8(B)(2) that corrected value needs to be greater than Isc x 125% (8.8A x 1.25 = 11A). That is the defined maximum current in 690.8(A).

I know this section of Code can be confusing and difficult to follow. I hope this helps.

Ryan's picture

Thank you for detailed explanation.
If corrected ampacity because of condition of use lies between 11A & 13.75A. Say if its 12A Can we still use this wire because we normally protect circuit with Fuse rating greater than 13.75A thinking that Fuse operates at 80% of its rated value. For OCPD we normally use Iscx1.25x1.25. If i use 15A Fuse & wire with corrected ampacity between 11 & 13.75, will it damage a wire before fuse blow?

Oz Baeza's picture

Not quite sure about this: 690.9(B) mentions OCPD ratings no less than 125 percent of max current rating calculated in 690.8(A). This is the 2014 cycle.
"For OCPD we normally use Isc x 1.25 X11.25."
Should that be only Isc x 1.25, so 8.8 x 1.25 = 11a and a 15a fuse x 0.80 = 12a and would be in accordance to 240?

Ryan Mayfield_2's picture

Oz, for OCPD sizing, your are correct: Minimum OCPD rating = Isc x 1.25 x 1.25 (or Isc x 1.56). The reason for the way I described it is, I like to break out all the individual sections and 690.8(A) is where we apply the first 1.25 factor. Then 690.9(B) says the OCPD needs to be no less than 125% of that value from 690.8(A). As an industry, we are quick to multiply Isc by 1.56 (1.25 x 1.25 = 1.56) and run with that number for everything. The reason I advocate for not doing that is back in 690.8(B), there are two checks we do for conductor sizing, one uses Isc times both 1.25 factors and one only uses a single 1.25 factor. If you apply both factors for all scenarios, you will end up over sizing conductors in some cases. So while that isn't a safety issue or Code violation, it may cost you more money.

Now for the example, the OCPD for that module (or string) would be 8.8 x 1.25 x 1.25 = 13.75A. We can use a 15A fuse here because of 240.4 allowances.

Ryan Mayfield_2's picture

Great question, one that comes up quite often and I should cover in an upcoming Code Corner since I couldn't get to this point in my last column. In 690.9, Code instructs us to use Article 240 to properly protect the conductors. In 240.4(B), we are told we can protect a conductor with the next standard overcurrent device, if the conductor's ampacity doesn't correspond to a standard size. Standard size overvcurrent devices can be found in NEC 240.6.

So if the conductor's ampacity after conditions of use was 12A like you said, we could still use this conductor, protected by a 15A fuse. This is because we need to protect the circuit with a 15A OCPD (as required by 690.9) and Code allows a conductor with 12A of ampacity to be protected by a 15A OCPD.

I know this is counter-intuitive, and the reason it is OK is based on what is called the "I squared T" values of OCPDs and conductors. In short, conductors can withstand higher current values than we are allowed to use and as long as we protect them by no more than the next standard OCPD size, we will not damage the conductors before the OCPDs have an opportunity to clear the overcurrent situation.

Ryan's picture

Interesting..Thanks a lot for clarification's picture

This ampacity must be greater than the maximum current calculated in 690.8(A): 15.2 A > 11 A, so the 14 AWG fits for this calculation.

If i am not wrong the above statement in the article should be 15.2A>13.75A

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