Solar PV breaker and 120% rule calculator
Enter the module short-circuit current and number of parallel strings. You get the minimum conductor ampacity and fuse size. For a load-side connection, add the inverter output and panel ratings to check the 120% rule.
Built on the 2026 NEC · Free, no sign-up · Updated October 2026
Result
Conductor ≥ 35.9 A · OCPD ≥ 36 A · 120% rule allows 40 A (✓ compliant)
How it was calculated
- Max current per 690.8(A): 23 × 1.25 = 28.75 A.
- Conductor & OCPD per 690.8(B): 28.75 × 1.25 = 35.94 A (use next standard OCPD size at or above this).
- 705.12(B)(2)(a) 120% rule: 200 × 1.2 − 200 = 40.00 A available. 125% × 32 A inverter output = 40.00 A fits ✓.
Assumptions
- Combined Isc = 11.5 A × 2 strings = 23 A.
- Two 125% multipliers stack: 690.8(A)(1)(a)(1) maximum current + 690.8(B)(1) conductor sizing → 156.25% of Isc. The OCPD is at least 125% of the maximum current per 690.9(B)(1).
- Load-side interconnection — 705.12(B)(2)(a) 120% rule evaluated with 125% of the 32 A inverter output current, a 200 A busbar, and a 200 A busbar OCPD. Assumes the power source and main OCPD are at opposite ends of the busbar.
Sections used
These are the sections the calculator applies. Look them up in your code book before you rely on the result, and check the code your AHJ has adopted.
Sizing PV circuits from Isc
Under 690.8(A)(1), the maximum current of a PV circuit is the sum of the parallel modules' short-circuit current (Isc) × 125%. That covers the higher output you get in strong sunlight.
Under 690.8(B), conductors and overcurrent devices are then sized at 125% of that maximum current, unless an engineered or listed method applies. Together that's 156% of Isc. Round the overcurrent device up to the next standard size.
The 120% rule for load-side connections
When the inverter backfeeds a breaker in a panel, the busbar can see current from the utility and the inverter at the same time. The most common check, in 705.12(B)(2), limits that: 125% of the inverter's output current plus the main breaker rating can't be more than 120% of the busbar rating, with the inverter breaker at the opposite end of the bus from the main.
For a 200 A bus with a 200 A main, that leaves 40 A, which allows an inverter with up to 32 A of continuous output.
This is 705.12(B)(2)(a), and it also requires a warning label next to the backfed breaker. If it fails, 705.12(B) has other methods, such as the sum of all the breakers in 705.12(B)(3). You can also install a smaller main breaker, or connect on the supply side under 705.11.
2023 to 2026: our list has 12 changes in Article 690. See the changes
Worked example
Two parallel strings of 11.5 A Isc modules, a 32 A inverter, and a 200 A main on a 200 A busbar:
- Module Isc (amps): 11.5
- Parallel strings: 2
- Interconnection method: Load-side (120% rule)
- Inverter output current (amps): 32
- Main OCPD (amps): 200
- Busbar rating (amps): 200
Result
Conductor ≥ 35.9 A · OCPD ≥ 36 A · 120% rule allows 40 A (✓ compliant)
How it was calculated
- Max current per 690.8(A): 23 × 1.25 = 28.75 A.
- Conductor & OCPD per 690.8(B): 28.75 × 1.25 = 35.94 A (use next standard OCPD size at or above this).
- 705.12(B)(2)(a) 120% rule: 200 × 1.2 − 200 = 40.00 A available. 125% × 32 A inverter output = 40.00 A fits ✓.
Assumptions
- Combined Isc = 11.5 A × 2 strings = 23 A.
- Two 125% multipliers stack: 690.8(A)(1)(a)(1) maximum current + 690.8(B)(1) conductor sizing → 156.25% of Isc. The OCPD is at least 125% of the maximum current per 690.9(B)(1).
- Load-side interconnection — 705.12(B)(2)(a) 120% rule evaluated with 125% of the 32 A inverter output current, a 200 A busbar, and a 200 A busbar OCPD. Assumes the power source and main OCPD are at opposite ends of the busbar.
Sections used
These are the sections the calculator applies. Look them up in your code book before you rely on the result, and check the code your AHJ has adopted.
Questions electricians ask
What is the 120% rule for solar?
An option in 705.12(B)(2): 125% of the inverter output current plus the main breaker rating can't be more than 120% of the busbar rating, with the backfed breaker at the far end of the bus from the main.
How big a solar breaker can I put on a 200 A panel?
With a 200 A main on a 200 A bus, the 120% rule allows 40 A. That fits an inverter with up to 32 A of rated output, since 32 × 1.25 = 40.
Why multiply Isc by 1.56?
Two separate 125% factors. One in 690.8(A) for irradiance higher than test conditions, and one in 690.8(B) for continuous current. 1.25 × 1.25 = 1.5625.
Teaching this in a class?
Build your own cited lesson, quiz, and practice on any part of the NEC, in your order.
How this calculator was built
The calculator runs the same code as the Ask Shorty iPhone app. Automated tests check it against hand-worked answers, and every section it names is checked against the 2026 NEC text. It shows section numbers, not code text. Read the sections in your own code book and confirm which edition your AHJ enforces.
How we test the calculators. See something wrong? Tell us and we will check it against the code.
Related calculators
- Residential service loadStandard-method dwelling load calculation, with EV charger.
- Ampacity deratingApply ambient temperature and bundling adjustments to a conductor.
- Voltage dropVolts lost and percent drop on a copper or aluminum run, single- or three-phase.
Have a code question the calculator doesn’t cover?
Ask Shorty in plain English. You get a straight answer with the NEC 2026 section it came from.