Battery-to-Inverter DC Conductor and OCPD (NEC 690.9 / 706)

ScopeDC input current = AC power / (battery voltage x inverter efficiency), then NEC 690.8(B) / 706 / 240.4 size the conductor and overcurrent device at 125% of that current, with the OCPD rounded up to the next standard size (240.6). A 4 kW inverter on a 48 V bank at 90% efficiency draws about 92.6 A, so the conductor is rated at least 115.7 A (a #1 Cu at 75 C, 130 A) on a 125 A DC fuse; a 24 V bank would double the current. Use a listed DC-rated (often Class T, for a battery's high available fault current) fuse and switch, keep the run short and heavy, and terminate at the battery's rated torque. A sizing estimate; the inverter and battery datasheets, the fault-current rating, the AHJ, and the adopted NEC edition govern.

Formula or tabledc_current_a = inverter_power_w / (battery_voltage_v x efficiency_pct/100); min_conductor_ampacity_a = 1.25 x dc_current_a; ocpd_a = next standard size (NEC 240.6) >= 1.25 x dc_current_a.

Source and editionNEC 690.8(B) (continuous current), 706 (energy storage systems), 240.4 (conductor protection), and 240.6 (standard OCPD sizes), by name; the inverter and battery datasheets and the AHJ govern.

Free accessThe DC-current relation is Ohm's/power law (public); the 125% continuous factor and the standard OCPD sizes are public NEC, and the inverter power, efficiency, and battery voltage come from the datasheets.

What governsEstimate. AHJ and licensed professional govern.

API field namesinverter_power_w, battery_voltage_v, efficiency_pct, dc_current_a, min_conductor_ampacity_a, ocpd_a

• 125% continuous conductor and OCPD at 1.25 x the DC input current; OCPD to the next standard size (240.6)NEC 690.8(B) / 240.4 / 240.6
• DC-rated / Class T a battery's high fault current needs a listed DC-rated fuse (often Class T) and disconnect with adequate interrupting ratingNEC 706 / practice