PV Circuit Sizing & Current Calculations

Intermediate

Section 690.8 of the National Electrical Code (NEC)  deals with PV circuit sizing and current calculations, and defines how to calculate four maximum circuit current values. These maximum circuit currents are used in additional calculations in sections 690.8(B). But before jumping into calculations, a few NEC definitions will be helpful, since the rules for correction factors and overcurrent requirements can change based on the specific circuit. Working from the array to the inverter, we have:

PV source circuits are conductors between the modules, and from modules to a common point of connection, typically a junction box or combiner box. In industry terms, these are often called the “home runs” from the individual strings.

PV output circuits are conductors between the PV source circuits and the inverter or DC utilization equipment. These are the circuit conductors after a combiner box to the inverter or charge controller. 

Inverter input circuits, in a battery-based system, are the conductors between the inverter and the battery bank. In a grid-tied system, they are the conductors between the inverter and PV output circuits. Typically, these are the conductors between the inverter’s integrated DC disconnect and the inverter’s DC input connection.

Inverter output circuits are the AC conductors from the inverter to the ultimate connection to the AC distribution system for either stand-alone or utility-interactive systems.

Calculations

The first calculation, from 690.8(A)(1), results in the maximum PV source-circuit current. The rated short-circuit current (Isc) is multiplied by 125%. For example, if a PV module has an Isc of 8.8 amps, this calculation is: 8.8 A × 1.25 = 11 A.

Section 690.8(A)(2) covers the maximum current for PV output circuits. For output circuits, multiply the Isc by the number of circuits in parallel, and then by 125%. A common installation method is to keep the source circuits separate until they reach the inverter’s integrated DC combiner and disconnect. In that case, there are no output circuits to consider because the source circuits are not placed in parallel outside of the inverter.

Section 690.8(A)(3) defines the maximum current for the inverter’s output circuit. For utility-interactive inverters, there isn’t a calculation required, since the maximum current is defined as the inverter’s continuous output rating.

Section 690.8(A)(4) shows the calculation for the highest input current of a stand-alone inverter. This value helps determine the conductor size and overcurrent protection device (OCPD) rating between the batteries and the inverter. Divide the inverter’s continuous power output rating by its lowest DC operating voltage, and then multiply by the inverter’s rated efficiency under those conditions.

Part 5 of 690.8(A), added to the 2014 Code, defines the maximum output current of DC-to-DC converters as the rated output per the manufacturer’s specifications. No additional calculations are required.

In the 2014 NEC, 690.8(B), which outlines the rules for calculating minimum conductor sizes in PV circuits, is titled “Conductor Ampacity.” The OCPD section has been relocated to 690.9. The method for conductor sizing has not changed, although the 2014 sections incorporate some clarifications.

In 690.8(B)—690.8(B)(2) in the 2011 edition—two calculations must be run; the circuit conductor size must be based on the larger of the two values calculated. The first calculation is in 690.8(B)(1)—690.8(B)(2)(a) in the 2011 edition. Because PV system currents are considered continuous, the maximum currents calculated in 690.8(A) must be multiplied by 125% to calculate the minimum conductor size. This calculation ensures that the conductors do not carry more than 80% of the continuous current value (0.8 is the inverse of 1.25), a standard procedure in earlier Code articles. In the PV industry, the result of this calculation is commonly referred to as the “156% factor.” When this rule is applied, the module’s rated Isc has been multiplied by 156% (125% × 125% = 156%). However, don’t just multiply everything by 156%. Inverter output circuits were not multiplied by 125% originally, so the 156% factor doesn’t apply to them.  This calculation is done before applying any adjustment and correction factors, commonly referred to as “conditions of use,” which include corrections for conductors exposed to temperatures in excess of 30°C or more than three current-carrying conductors within a conduit. The ampacity of the conductor, at a minimum, then, needs to be greater than or equal to the maximum current in 690.8(A) × 1.25.

Comments (5)

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

prakash.nadagoudra@gmail.com'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?

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

prakash.nadagoudra@gmail.com's picture

Interesting..Thanks a lot for clarification

prakash.nadagoudra@gmail.com'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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