Group Welder Feeder Demand Calculator
Size the feeder that supplies a group of arc welders under NEC 2020 630.11(B). Enter each welder's rated primary current and duty cycle, and the tool computes each machine's 630.11(A) current, applies the 100/85/70/60 percent diversity, returns the smallest Table 310.16 conductor that meets the demand, and reports the 630.12(B) feeder overcurrent ceiling.
How to use this calculator
- Add a row for each welder on the feeder, with its rated primary current and duty cycle from the nameplate.
- Set each welder's type: nonmotor-generator (transformer or inverter machines) or motor-generator.
- Choose the feeder conductor material and the termination temperature column (75 degrees C is the usual default).
- Adjust ambient temperature or conductor count if the feeder is not at the base Table 310.16 conditions.
- Read the 630.11(B) feeder demand, the smallest qualifying conductor, and the 630.12(B) maximum overcurrent device.
NEC reference
This calculator follows NEC 2020 Article 630. Section 630.11(B) sets the feeder or service conductor ampacity at 100 percent of the two largest welders' 630.11(A) currents, 85 percent of the third, 70 percent of the fourth, and 60 percent of the rest; Section 630.11(A) and Table 630.11(A) give each welder's duty-cycle-adjusted current. The conductor comes from Table 310.16, with Table 310.15(B)(1) ambient correction and Table 310.15(C)(1) adjustment when you enter non-base conditions, and the 110.14(C) termination cap. Section 630.12(B) sets the feeder overcurrent at not more than 200 percent of the conductor ampacity. Article 630 is unchanged in these sections across the 2017, 2020, and 2023 editions.
Results are for reference only. Verify against the applicable adopted edition of the NEC and consult a licensed electrician for code compliance.
The 630.11(B) diversity method
A shop with several welders on one feeder never runs every machine at full output at the same instant. NEC 630.11(B) builds that diversity into the feeder size. First find each welder's 630.11(A) current, which is its rated primary current times the Table 630.11(A) duty-cycle multiplier. Then rank the welders by that current and take 100 percent of the two largest, 85 percent of the third, 70 percent of the fourth, and 60 percent of every welder after that. The sum is the minimum feeder conductor ampacity.
Because the discount grows as the group grows, the feeder is always smaller than the arithmetic sum of the individual branch conductors, and a single large welder dominates the result. The factors apply to the duty-cycle currents, so a bank of low-duty-cycle machines lands on a much smaller feeder than their nameplate amps suggest.
Worked example: three 60 A welders
- Three nonmotor-generator welders, each 60 A rated primary at 100 percent duty, so each 630.11(A) current is 60 x 1.00 = 60 A.
- Rank them: the two largest at 100 percent (60 A and 60 A) and the third at 85 percent (0.85 x 60 = 51 A).
- The feeder demand is 60 + 60 + 51 = 171 A.
- In the 75 degrees C column of Table 310.16, 1/0 copper is 150 A (short) and 2/0 copper is 175 A, so the feeder is 2/0 copper.
- The 630.12(B) feeder device may not exceed 200 percent of the 175 A conductor ampacity, or 350 A, which is a standard rating.
One welder is a branch circuit, not a feeder
The 630.11(B) diversity is a feeder rule. A single welder on its own circuit is sized under 630.11(A) for the conductor and 630.12(A) and (B) for the overcurrent device, both keyed on that one machine. Use the Welder Circuit Calculator for a single welder's branch circuit; this tool takes over once two or more welders share a feeder. Resistance welders are a separate method again, under 630.31 and Table 630.31, and are outside this tool's scope.
Frequently asked questions
How is a feeder to several welders sized?
Take 100 percent of the two largest welders' 630.11(A) currents, 85 percent of the third, 70 percent of the fourth, and 60 percent of the rest, then pick the smallest Table 310.16 conductor that meets the sum. Three 60 A welders come to 171 A, which is 2/0 copper at 75 degrees C.
What are the 630.11(B) demand factors?
100 percent for the two largest welders, 85 percent for the third, 70 percent for the fourth, and 60 percent for every remaining welder, applied to the Table 630.11(A) duty-cycle currents rather than the rated primary currents.
Why is a welder feeder smaller than the sum of the welders?
Because the welders do not all strike an arc at once. 630.11(B) discounts every welder past the two largest, so the feeder carries less than the sum of the individual branch conductors.
What overcurrent device protects a welder feeder?
630.12(B) caps it at 200 percent of the feeder conductor ampacity; the next higher standard 240.6(A) rating is permitted where that value is not standard.
Does the 240.4(D) small-conductor rule apply?
No. Article 630 permits welder-circuit overcurrent up to 200 percent of the conductor ampacity, above the 240.4(D) caps, so this tool sizes the feeder on the 630.11(B) demand alone.
Related calculators
- Welder Circuit Calculator: a single arc welder's branch circuit under NEC 630.11(A) and 630.12.
- Wire Ampacity Calculator: the Table 310.16 ampacity of a chosen conductor, with ambient and fill derating.
- Several Motors on One Branch Circuit: the analogous multi-load feeder for motors under NEC 430.24.
- Breaker Size Calculator: the standard 240.6(A) breaker for a general-use load with the 125 percent continuous rule.
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