What is your largest inverter size?

Example:

Busbar = 200A

Main = 200A

What is your largest inverter size?

                                 200A X 1.2 = 240A

                                 240A – 200A = 40A

                                  40A / 1.25 = 32A

Largest inverter is 32A

You can multiply by the ac voltage to get the power!

705.12(D)(2)(3)(C): SOLAR + LOAD BREAKER METHOD

Load and supply breakers do not exceed the main breaker (this also works for subpanels).

New in the 2014 NEC, 705.12(D)(2)(3)(c) allows us to add up the load and supply breakers and as long as they don’t exceed the rating of the main supply breaker (or the busbar rating) then we are good.

                                  Solar + Load ≤ Main

Note that for the solar + load breaker accounting method, we do not use

125% of the inverter current, we use the size of the breaker. This is easy and convenient when combining inverters with an ac combiner subpanel.

Ten-foot inverter tap rule

240.21(B)(1), the ten-foot tap rule (for taps of ten feet or less):

• Tap ampacity > tap load

• Ampacity of tap conductor has to be greater than the load.

Pa

Example:

• ten-foot tap rule

• 200A source breaker

• 16A inverter.

The tap conductor must be at least 10% of (source breaker + 125% inverter

current):

                                             16A inverter 3 1.25 = 20A

                                       20A + 200A source breaker = 220A

                                                   0.1 x 220A = 22A

Tap conductor must be at least 22A

Additionally, the inverter breaker must be less than 22A and for a 16A inverter;

we would use a 20A breaker (16A 3 1.25 = 20A breaker).

One more thing to check is the 230A feeder, which needs to be enough to carry

currents from the source and 125% of the inverter.

DC DISCONNECT LABEL CALCULATION

1. Operating current:

               A. Imp X number of source circuits.

2. Operating voltage:

               A. Vmp X number of modules in series per source circuit.

3. Max system voltage:

               A.Voc X Max number of modules in series per 

                source circuit X Correction Factor of Ambient temp.

4. Short circuit current:

               A. Isc X 1.25 X number of source circuits 

MARKING OF MODULES

•   Open-circuit voltage                                       Voc
•   Operating voltage                                           Vmp
•   Maximum permissible system voltage           Maximum system voltage
•   Operating current                                            Imp
•   Short-circuit current                                        Isc
•   Maximum power                                             Power of module (Vmp 3Imp)
•   Polarity                                                            + or –
•   Max series fuse rating                                     Max fuse size

How to find out Voltage drop ?

Voltage drop example question

You have a PV array in a field and an inverter at a house 250 feet away. You are

using 10AWG stranded copper wire. There is just one string and the voltage of

the string operates at Vmp of 200V and an Imp of 5A. What is the percentage

power loss due to dc voltage drop?

Answer

Voltage drop is calculated with Ohm’s Law, which states:

                                  Vdrop = I X R

We know that current is I = 5A

We will solve for resistance by multiplying:

ohm / kFT X kFT = ohm

The distance from the PV to the inverter is 250 feet, so there will be two wires

to complete the circuit, therefore the wire length is:

                        250 feet X 2 directions = 500 feet

                      500 feet / 1000 feet per kFT = 0.5kFT

                                  Distance = 0.5kFT

From looking up the properties of 10AWG stranded copper wire, we get

1.24 ohms/kFT, so:

                              ohm / kFT X kFT = ohm

                      1.24 ohms / kFT X 0.5 kFT = 0.62 ohm

                               Resistance = 0.62 ohm

Now back to Ohm’s Law:

                                     Vdrop = I X R

                             Vdrop = 5A X 0.62 ohm

                                 Vdrop = 3.1 volts

We now know that during peak sun conditions, we are losing 3.1 volts on the

wire and if we measured the voltage at the array, it would be 3.1 volts higher

than at the inverter.

To figure out our losses as a percentage we will divide voltage drop by system

voltage and then turn that into a percentage.

          (Vdrop / system voltage) X 100% = voltage drop percentage

                       (3.1V / 200V) X 100% = 1.55% voltage drop

Since voltage multiplied by current is power, then voltage drop percentage is the

same as power loss, so we are losing 1.55% of our power in this case.

           Answer to voltage drop question: 1.55% power loss at STC

                                          table use for ohm/KFT on wire.

Article Raceway or cable covered

320 Armored Cable: AC

330 Metal-Clad Cable: MC

334 Non-metallic-Sheathed Cable: NM, NMC and NMS (romex)

338 Service Entrance Cable: SE and USE (USE-2)

340 Underground Feeder and Branch-Circuit Cable: UF

342 Intermediate Metal Conduit: IMC

344 Rigid Metal Conduit: RMC

348 Flexible Metal Conduit: FMC

350 Liquidtight Flexible Metal Conduit: LFMC

352 Rigid Polyvinyl Chloride Conduit: PVC

356 Liquidtight Flexible Nonmetallic Conduit: LFNC

358 Electrical Metallic Tubing: EMT

NEC PV ARTICLES & SECTION

690 PV systems705 Interconnections

110.14(C) Terminal temperatures

110.21(B) Hazard marking

110.26(A)(1) Working spaces

110.28 Enclosure selection

200.6 6AWG and smaller can be marked white for PV

230 Services

240 OCPD

240.4(D) Small conductor rule

250 Grounding and bonding

250.52 Electrodes

250.66 AC GEC

250.166 DC GEC

250.122 ECG (ac and dc)

300.5 Underground installations (how deep to bury conduit)

300.7 Raceways exposed to different temperatures

310 Wire sizing

310.15(B)(2)(a) Ambient temperature correction factors

310.15(B)(3)(a) Adjustment for >3 current carrying conductors in conduit

310.15(B)(3)(c) Temperature adjustment in conduit in sun on roof

310.16 Conductor ampacity in raceway cable or buried

310.17 Conductor ampacity in free air

314 Junction boxes, enclosures, outlets

320–362 Raceways (conduit) and cables

330 Metal-Clad: MC cable

334 Non-metallic sheathed cable: NM, NMC and NMS

338 Service entrance cable: SE, USE (USE-2)

342 Intermediate metal conduit: IMC

344 Rigid metal conduit: RMC

352 Rigid PVC

356 Liquid tight flexible non-metallic conduit: LFNC

358 Electrical metallic tubing: EMT

480 Batteries (also in 690.71–690.74)

690.2 Definitions

690.5 Ground-fault protection

690.7 Maximum voltage

690.8 Currents and circuit sizing

690.9 Overcurrent protection

690.10 Stand-alone systems

690.11 Arc-fault protection (dc)

690.12 Rapid shutdown

690.13–.18 Part III Disconnecting means

690.31–.35 Part IV Wiring methods

690.41–.50 Part V Grounding

690.51–.56 Part VI Marking

690.71–.74 Part VIII Batteries

705.10 Directory

705.12 Point of connection

705.12(A) Supply-sideMemorize and familiarize

705.12(D) Load side

705.12(D)(2)(1) Feeders

705.12(D)(2)(2) Feeder taps

705.12(D)(2)(3) Busbars

705.12(D)(2)(3)(a) 100% rule

705.12(D)(2)(3)(b) 120% rule

705.12(D)(2)(3)(c) Sum rule (loads + breakers ≤ busbar)

705.31 Location of supply-side connection disconnect < 10 ft

705.32 Connect to supply-side of ground-fault protection