The Role of Inspectors<\/h2>\n
Before we begin, it\u2019s important to consider the role of inspectors when examining PV systems. I spent most of my career as an installer (25 years). Although I am now an AHJ, I don\u2019t see it as a major difference because, in either case, correct installation is the goal.<\/p>\n
Correct installation is essential for any electrical system but can be more challenging when dealing with new technologies. Constantly evolving technology results in new products, installation methods and code requirements. Confusion regarding the requirements can also result.<\/p>\n
PV Systems Basics<\/h2>\n
What is a PV system<\/strong>? The National Electrical Code<\/em> (NEC<\/em>) defines a photovoltaic (PV) system in Article 100 as \u201cthe total components and subsystems that, in combination, convert solar energy into electric energy for connection to a utilization load.\u201d<\/p>\n A PV system is not just the modules up on the roof, but it\u2019s the total components and subsystems that, in combination, convert solar energy into electric energy. If the system is not interconnected with a utility or other primary source, then the requirements of NEC<\/em> Chapters 1-4 and Article 690 are applicable. If interconnected with the utility grid or other primary source , the interconnection requirements will be covered by Article 705 of the NEC<\/em>.\u00a0The systems discussed in this article today will be central inverter, utility-interactive, and interconnected with a primary source.<\/p>\n An off-grid system that is not interconnected with any other primary sources of electricity would only be subject to NEC<\/em> Chapters 1-4 and Article 690. Some requirements are duplicated in NEC<\/em> Articles 690 and 705 because application is dependent upon the interconnection.<\/p>\n What does a PV inverter<\/strong> do?\u00a0The inverter is an electronic power converter that converts the direct current (DC)<\/em> output from an array of modules into alternating current (AC).<\/em> The alternating current output of the inverter can be used to supply a premises electrical system, feedback to the utility grid, or both. The AC output of a utility-interactive inverter will be synchronized with the utility grid or other primary source.<\/p>\n Figure 1 is a block diagram that illustrates the concept. PV source circuits are indicated by the red box on the far left. The box represents the array of modules on the roof which produce the direct current. The PV output circuits route the DC to the inverter input circuit. The inverter converts the DC to AC synchronized with utility or other primary source.\u00a0 The inverter AC output is used to supply the grid and\/or the premises wiring system.<\/p>\n <\/p>\n Figure 2 provides an illustration identifying the specific components.\u00a0Please take note of the red line located in the middle of the inverter. We\u2019re going to look at the system conceptually in two parts: the DC side and the AC side. I like to approach PV system design review in the middle, by starting with the inverter and the AC output.<\/p>\n <\/p>\n Why start with the AC side? My experience has been that most plan review comments result from design problems on the AC half of the system. That doesn\u2019t mean there aren\u2019t important design considerations for the DC portion, but we\u2019ll leave those for another article.<\/p>\n Photo 1 is a picture of an actual system like the block diagram in Figure 2.<\/p>\n <\/p>\n The basic elements of a PV system array are illustrated by Figure 3. The individual photovoltaic cells are combined in a manufactured unit known as a module<\/strong>. The modules are connected in series strings and often connected with other strings in a series-parallel arrangement. The output of the combined PV source circuits is known as the PV output circuit.<\/p>\n <\/p>\n The Figure 3 illustration indicates three source circuit fuses. Be aware that fuses are not always required.\u00a0Based on the rules and exception of NEC<\/em> 690.9, fuses are generally not necessary for smaller residential systems.<\/p>\n Photo 2 is a picture of an array I took during an inspection at an apartment complex.\u00a0It\u2019s difficult to see the string conductors because the modules are close to the roof, but the DC\u00a0output circuit conductors are located in the conduit on the lower right side of the picture.<\/p>\n <\/p>\n There are three basic design elements of the AC side that require simple math to validate: the inverter output overcurrent protective device (OCPD) rating, the output conductor ampacity, and the permitted PV contribution to the premises wiring system. These are also the elements that generate most plan review comments for me.<\/p>\n <\/p>\n Please refer to Figure 4. Three questions need to be answered:<\/p>\n The inverter size (the KW rating) in Figure 4 is not identified, but a 20-ampere circuit breaker is indicated for the inverter. Is this correct?<\/p>\n Overcurrent protection is required for the protection of conductors and equipment. Current must be limited in accordance with the ampacity rating of equipment and conductors to prevent overheating and\/or failure. An overcurrent protective device (OCPD) is used to limit current flow to a safe level.<\/p>\n All PV system currents (DC or AC) are continuous. Why? Because the sun generally will be shining for more than three hours a day. If it didn\u2019t, you probably wouldn\u2019t be installing a PV system! Anytime maximum current is expected to continue for three hours or more, it\u2019s a continuous current, and there are requirements that must be considered.<\/p>\n The minimum rating for the PV inverter AC overcurrent device is 125% of the rated inverter continuous output current unless the overcurrent device is listed for continuous operation at 100% (see NEC <\/em>705.60). The circuit breaker in our sample system is not listed for continuous operation therefore we must apply the 125% factor.<\/p>\n <\/p>\n Figure 5 provides an example of a typical inverter datasheet. A key element of the datasheet is the inverter rating (kW). The normal output power is indicated as 3000 watts (3 kW) and the inverter can be interconnected to a 208, 240, or 277-volt premises distribution system. The output current may differ based on the system voltage, so be sure to reference the correct column.<\/p>\n The bottom row provides the maximum<\/em> current output <\/em>of the inverter. This system is interconnected at 240-volts.\u00a0See the center column. Notice that if this were a 208-volt system, the maximum current rating would still be 14.5 amperes.\u00a0But for a 277-volt interconnection the output would drop to 12.0 amperes. Our system is for a residential dwelling unit with a 240-volt supply, and therefore, a 14.5 ampere current output.<\/p>\n Per the Code <\/em>reference, the minimum rating for the PV inverter (AC) overcurrent device is 125% of the rated inverter continuous output.<\/p>\n The datasheet in Figure 5 states that the maximum output current is 14.5 amperes at 240-volts.<\/p>\n Minimum OCPD is 14.5 A x 1.25 = 18.125 A<\/p>\n I am not aware of an 18-ampere circuit breaker. The Code <\/em>permits use of the next larger (800 ampere or less per NEC<\/em> 240.4) standard overcurrent device rating; 20 amperes in this example. That\u2019s the minimum rating. The OCPD rating can\u2019t be rated less than 20 amperes. Is it possible to use an OCPD with a higher current rating?<\/p>\n If the inverter were tested with larger overcurrent protective device rating during the certification process, then a larger OCPD may be allowed. In that case, the installation instructions should provide the information. I recommend that plan reviewers or inspectors consider the minimum required OCPD rating as the maximum unless someone can show you differently. If the designer can provide documentation that a larger ODPD rating is permitted by the manufacturer in accordance with the listing, then approval is possible.<\/p>\n In our jurisdiction, the field inspector cannot approve an installation where the installed inverter circuit breaker rating does not match what is required by the approved plan.\u00a0 The correct inverter OCPD rating is validated at the plan review stage.<\/p>\n Standard overcurrent device (fuses and inverse time circuit breakers) ratings are provided by NEC<\/em> Table 240.6(A). <\/em><\/p>\n It may be desirable for you to create a chart like that in Table 1 for the common inverter ratings. If you perform a lot of PV system plan reviews, you will soon memorize the OCPD ratings based on common inverter kW sizes.<\/p>\n <\/p>\n Table 1 does not contain every possible inverter rating. Always confirm the rated output through the manufacturer\u2019s datasheet.<\/p>\n A common question involves systems with multiple central inverters.\u00a0 How do we do the math? Do we just add up the minimum OCPD ratings for each inverter, and use the sum for the single interconnection circuit breaker?<\/p>\n <\/p>\n <\/p>\n Figure 6<\/strong> provides an example of a system with multiple central inverters. There are two inverters on the left \u2014 3 KW on the top and a 6 KW on the bottom. Each of those is required to be protected individually by an overcurrent protective device. The requirements differ for microinverter and AC module systems.<\/p>\n In this example, the two inverter outputs are combined in a panelboard. A panelboard used in this manner might be described as an \u201cAC combiner panel\u201d. A standard panelboard or a specially designed unit could be utilized. But in either case, individual OCPD is required for each central inverter. A 20-ampere circuit breaker for the 3 KW inverter and a 35-ampere circuit breaker for the 6 KW inverter is indicated.<\/p>\n What would be the minimum required rating for the panelboard feeder circuit breaker?\u00a0Is it the sum of the 20 ampere and 35 ampere individual inverter circuit breaker ratings (55 amperes)? Is a 55 -ampere circuit breaker available? So, would a 60-ampere circuit breaker be required? Is that the minimum required rating? The answer is no. Let\u2019s look at the correct method.<\/p>\n <\/p>\n <\/p>\n Figure 7<\/strong> provides an inverter manufacturer datasheet. The 3-kW inverter provides a maximum continuous output current of 12.5 amperes, and the 6 KW has an output of 25 amperes.<\/p>\n For the 3 KW inverter<\/strong> the continuous output current is 12.5 amperes.<\/p>\n 12.5 amperes x 1.25 = 15.6 A<\/p>\n Therefore a 20-ampere circuit breaker is required.<\/p>\n For the 6 KW<\/strong> inverter<\/strong> the continuous output current is 25 amperes.<\/p>\n 25 A x 1.25 = 31 A<\/p>\n Therefore a 35-ampere circuit breaker is required.<\/p>\n We have established the minimum circuit breaker ratings for each inverter. But how is the circuit breaker rating for the feeder that supplies the panelboard determined? In the same manner as the individual circuits.<\/p>\n Sum of both inverter outputs:\u00a0\u00a0\u00a0 25 + 12.5 = 37.5 A<\/p>\n Minimum required circuit breaker rating: 37.5 A x 1.25 = 46.9 A<\/p>\n Therefore, the minimum circuit breaker rating is 50 amperes<\/p>\n <\/p>\n Simply adding the two individual circuit breaker ratings (20 + 35) results in a 55-ampere rating. But you come up with a different answer when the correct method is used.\u00a0That\u2019s critical because you wouldn\u2019t want to turn down a plan if you think they need a 60-ampere circuit breaker based on adding 20 + 35. The actual minimum required rating is 50 amperes in this example.<\/p>\n The minimum feeder OCPD rating could be greater than 50 amperes if desired, or if necessary for another reason (such as supplying loads).<\/p>\n The method is the same regardless of the quantity of inverters.\u00a0 The numbers may be larger but the same process is used.<\/p>\n Next are calculations for the inverter output conductor ampacity.<\/strong> After establishment of the minimum overcurrent device rating, the corresponding conductor ampacity must be validated. The order of validation is important because the conductor ampacity will be directly related to the overcurrent protective device rating. The minimum required ampacity for the conductors is established by NEC<\/em> 705.60 and 690.8. The minimum ampacity for the inverter output conductor must be larger of 1) or 2) below:<\/p>\n For simplicity, in our example here, we will assume that the OCPD is not rated for continuous operation and that 1) above will provide the larger result. For this calculation we will also reference the datasheet of Figure 5. The inverter continuous output rating was used to establish the minimum (and perhaps maximum) overcurrent protective device rating of 20 amperes. \u00a0The designer specified #10 THWN-2 copper conductors. Will the conductors have the sufficient ampacity?<\/p>\n It may seem redundant, but the process for determining conductor ampacity begins the same way as for minimum overcurrent protection.\u00a0The inverter datasheet of Figure 5 identifies the continuous current output as 14.5 amperes at 240 volts for the 3 kW unit.<\/p>\n Output Current = 14.5 amperes<\/p>\n Minimum Conductor Ampacity\/OCPD<\/p>\n Rating = 14.5 A x 1.25 = 18.125 A<\/p>\n <\/p>\n The conductor and OCPD must have an ampacity of at least 18 amperes. Because a 20 ampere OCPD is necessary (18 ampere is not a standard OCPD rating), the conductor will also require an ampacity of 20 amperes.<\/p>\n What is the ampacity of the conductor? See NEC<\/em> Table 310.15(B)(16). The ampacity will be based on the conductor insulation temperature rating of 60\u00b0C, 75\u00b0C, or 90\u00b0C.<\/p>\n Be sure to consider the asterisks (*) of Table 310.15(B)(16) for conductor AWG sizes 18, 16, 14, 12, and 10. The asterisks refer the user to 240.4(D) for conductor overcurrent protection limitations. Regardless of the temperature rating of the conductor, and any conditions of use, there are OCPD limitations for small conductors.<\/p>\n Figure 4 identifies the inverter output conductor type and size as #10 THWN-2. Table 310.15(B)(16), for Types THWN-2 provides an ampacity rating of 40 amperes before application of any conditions of use.<\/p>\n There are maximum OCPD limitations for small conductors (except in special conditions). Three common conductor sizes and associated limitations include:<\/p>\n The most common design violation I have encountered with residential PV systems is use of a #10 conductor for a 35 or 40-ampere inverter output circuit. Table 1 indicates that a minimum 35-ampere device is required for a 6 kW inverter and a 40-ampere device is necessary for a 7 kW system. These are very common residential PV system and\/or inverter ratings.<\/p>\n Unfortunately, what\u2019s also very common is that designers will specify a #10 conductor for the 6- and 7-KW inverter output circuit. The #10 conductor size cannot be approved based on the rules of 240.4(D) even if otherwise permitted by the ampacity tables and conditions of use.<\/p>\n The allowable PV system contribution is based on the ampacity rating of the panelboard bus and overcurrent protective device. Are there limitations? The answer is\u00a0 yes, there are limitations, but also several options for compliance.<\/p>\n Busbars are conductors, and all conductors must be protected at their ampacity. The panelboard bus rating determines the main circuit breaker or feeder circuit breaker rating. Can a 100-ampere panelboard bus be protected\u00a0 by a 200-ampere main breaker circuit breaker? The answer is no; you must protect all conductors, including the bus, at rated ampacity.<\/p>\n Where a panelboard is supplied by a utility or other primary source, the PV system will be an additional source of electricity.\u00a0A PV system is a supply of electricity for the panelboard.<\/p>\n Source currents are only limited by the source overcurrent device. Some might say \u201cwell, yeah, but if the PV system is supplying current, then that means the utility is not, so why does it matter?\u201d It matters because the available current, from all sources, cannot exceed the rating of the busbars except as permitted by NEC<\/em> 705.12.<\/p>\n The sum of branch circuit breaker ratings may exceed the panelboard busbar ampacity. The bus is protected at rated ampacity by either the feeder or the main circuit breaker.<\/p>\n <\/p>\n Figure 8 provides an illustration of the concept. The PV system inverter is a supply to the panelboard along with the utility.\u00a0The utility supply is limited by the 100-ampere main circuit breaker at the top and the PV system supply is limited by the 20-ampere circuit breaker at the bottom. Both the PV input and the utility connection are supplies. What limits ampacity on this 100-ampere rated bus bar? The utility and PV system overcurrent devices provide the necessary limitation.<\/p>\n The rules for load-side source interconnections are found in NEC<\/em> 705.12. The basic concept is that the panelboard must be equipped with bus sufficient for the PV contribution.<\/p>\n There are specific limitations where panelboards are supplied simultaneously by a primary source of electricity and one or more other load side power sources and are also capable of supplying multiple branch circuits and\/or feeders. We will consider two options. The \u201c120% rule\u201d of 705.12(B(2)(3)(b) and the \u201climited load rule\u201d of 705.12(B)(2)(3)(c).<\/p>\n <\/p>\n \u00a0<\/em><\/p>\n NEC<\/em> 705.12(B)(2)(3)(b) states that the sum of the main circuit breaker rating (or the feeder breaker) and 125% of the inverter output are not to exceed 120% of the panelboard bus rating. There are requirements that must be met to use this allowance. First, the PV system circuit breaker must be located at the opposite end of the panelboard bus from the primary supply connection. With all of the load connections located between the two supplies there cannot be an oversupply at any one location on the bus bars. A warning label is also required to prohibit relocation of the PV system circuit breaker.<\/p>\n Common panelboard and main circuit breaker ratings are indicated in Table 2. The top row (row 1) contains common panelboard bus ratings. Under the 120% rule, a contribution from all sources up to 120% of the busbar rating is permitted. That equates to 120 amperes for a 100-ampere panel bus (row 2) from all sources. The next variable is the main circuit breaker or feeder circuit breaker rating (row 3).\u00a0 A 100 ampere-panelboard equipped with a 100-ampere main circuit breaker is permitted up to 20 amperes of PV supply. A 125 ampere-panelboard equipped with a 100-ampere main is allowed up to 50 amperes of PV system contribution.<\/p>\n But a 125-ampere panelboard equipped with a 125-ampere main is only permitted up to 25 amperes from the PV system.\u00a0 There\u2019s an interaction between the main circuit breaker rating and panel bus that determines the permitted PV supply under this rule.<\/p>\n Remember that 125% of the inverter output is used for these calculations, not the inverter circuit breaker rating.<\/p>\n![\"Figure](\"https:\/\/iaeimagazine.org\/wp-content\/uploads\/2022\/02\/2022-03-Jackson-FIG1.jpg\")
<\/a><\/a><\/a><\/a><\/a><\/a>\n
Inverter Output OCPD Rating<\/h2>\n
<\/a>Standard Ampacity Ratings<\/h2>\n
<\/a>What if there are multiple inverters?<\/h2>\n
<\/a><\/a><\/a><\/a>Inverter Output Conductor Ampacity<\/h2>\n
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PV System Contribution<\/h2>\n
<\/a><\/a>120% rule<\/h3>\n
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