50 A welder at 30% duty cycle
The classic hobby-to-light-fabrication machine.
27.5 A of conductor — 10 AWG — behind a 100 A device
Conductor and breaker size for an arc welder — the one load the code lets you size below its nameplate current, because duty cycle means the conductors never see it continuously.
Input is what the circuit sees. Output is welding amps and is far larger.
From the nameplate — the INPUT figure.
Percent of a ten-minute period the machine can weld. 30% is 3 minutes in 10.
For a feeder serving several machines. Leave at 0 for a single welder.
630.11(A): 50.0 A rated × 0.55 = 27.5 A of conductor, so 10 AWG copper at 75 °C. The multiplier is the square root of the duty cycle, because conductor heating goes as I²t — half the duty cycle is not half the heat.
This is the one load the code lets you size below its nameplate. A 50 A welder at 30% duty cycle needs conductors for 28 A — a saving of 22 A, and entirely legitimate.
630.12: overcurrent protection at not more than 200% of the RATED current — 100 A, so a 100 A device. Note this is 200% of the rated current, not of the derated conductor ampacity: the device has to survive striking an arc, the conductors only the average.
So a 100 A device ahead of 10 AWG conductors rated 28 A. Like a motor circuit, this looks wrong under 240.4 and is exactly what Article 630 requires.
Duty cycle is quoted at a stated output current and falls as you turn the machine up — a welder rated 60% at 200 A may be 20% at its 250 A maximum. Size on the setting you actually use, not the best number on the plate.
The classic hobby-to-light-fabrication machine.
27.5 A of conductor — 10 AWG — behind a 100 A device
Industrial machine, no reduction at all. Compare the conductor size.
Full 50 A of conductor — the multiplier is 1.00
630.11(B) diversity. Welders do not all strike an arc together.
97.6 A feeder against a naive 110 A
When the input figure is missing. Rough, and clearly labelled as such.
Estimates about 33 A input — check the plate
When to use this: wiring a welder outlet in a shop or garage, or sizing a feeder for several machines.
Almost every load in the code is sized at 100% of its rating or more. A welder is the exception: NEC 630.11 lets you size the conductors below the nameplate current, because a machine at 30% duty cycle welds three minutes in ten and the conductors never reach the temperature a continuous draw would produce.
A 50 A welder at 30% duty cycle needs conductors rated 27.5 A. That is a saving of 22.5 A, and it is entirely legitimate.
Conductor heating goes as I²t, so the multiplier is √(duty cycle) rather than the duty cycle itself. Half the duty cycle is not half the heat, and it is not half the conductor.
| Duty cycle | Multiplier | √(duty cycle) | 20 A welder | 30 A welder | 40 A welder | 50 A welder | 70 A welder |
|---|---|---|---|---|---|---|---|
| 100% | 1.00 | 1.0000 | 20.0 A | 30.0 A | 40.0 A | 50.0 A | 70.0 A |
| 90% | 0.95 | 0.9487 | 19.0 A | 28.5 A | 38.0 A | 47.5 A | 66.5 A |
| 80% | 0.89 | 0.8944 | 17.8 A | 26.7 A | 35.6 A | 44.5 A | 62.3 A |
| 70% | 0.84 | 0.8367 | 16.8 A | 25.2 A | 33.6 A | 42.0 A | 58.8 A |
| 60% | 0.78 | 0.7746 | 15.6 A | 23.4 A | 31.2 A | 39.0 A | 54.6 A |
| 50% | 0.71 | 0.7071 | 14.2 A | 21.3 A | 28.4 A | 35.5 A | 49.7 A |
| 40% | 0.63 | 0.6325 | 12.6 A | 18.9 A | 25.2 A | 31.5 A | 44.1 A |
| 30% | 0.55 | 0.5477 | 11.0 A | 16.5 A | 22.0 A | 27.5 A | 38.5 A |
| 20% | 0.45 | 0.4472 | 9.0 A | 13.5 A | 18.0 A | 22.5 A | 31.5 A |
The tabulated values are the square root rounded to two places. The 60% row is the one that does not round cleanly — √0.6 is 0.7746, which rounds to 0.77, and the code publishes 0.78. The table is the legal value, so that is what this calculator uses; the discrepancy is worth knowing about rather than assuming somebody mistyped it.
Because the table is a square root, unlisted duty cycles are not a problem: a machine rated 35% simply takes √0.35.
The most common error on this page's topic, and it is a factor of five. A welder nameplate carries two current ratings:
Everything in Article 630 uses the input figure. Sizing a circuit from a 180 A output rating gives conductors three times larger than needed and a breaker that will never trip.
Inverter machines complicate the intuition further: an inverter rated 200 A output may draw 25 A input where an old transformer machine would draw 50. Read the plate.
630.12 permits overcurrent protection at up to 200% of the rated primary current — not of the derated conductor ampacity. For our 50 A welder that is 100 A, so a 100 A device ahead of conductors rated 27.5 A.
Like a motor circuit, this looks like a serious violation under 240.4 and is exactly what the code requires. The device has to let the machine strike an arc without opening; the conductors only have to survive the average. Our motor breaker calculator covers the same idea from the Article 430 side.
630.11(B) applies diversity more generously than almost anywhere else in the code, because welders in a shop genuinely do not all strike an arc at the same instant.
| Position | Factor | Example, four 27.5 A machines |
|---|---|---|
| Largest | 100% | 27.50 A |
| Second | 100% | 27.50 A |
| Third | 85% | 23.38 A |
| Fourth | 70% | 19.25 A |
| Feeder | 97.6 A | |
| Adding them up instead | 110.0 A | |
Everything from the fifth machine onward counts at 60%, so a large shop's feeder grows much more slowly than the machine count suggests.
Manufacturers quote the best number, and duty cycle is stated at a specific output current. A machine rated 60% at 200 A may be 20% at its 250 A maximum — a multiplier of 0.78 against 0.45, which is most of a conductor size.
Size on the setting you actually weld at. If you routinely run the machine flat out, use the duty cycle at that output rather than the headline figure.
The nameplate carries both an output rating in welding amps and an input rating in primary amps, and they differ by a factor of five or more. The circuit only cares about the input.
The percentage of a ten-minute period the machine can weld continuously. Note the output current it is quoted at — duty cycle falls as you turn the machine up.
630.11 multiplies the rated current by the square root of the duty cycle. A 50 A welder at 30% needs 27.5 A of conductor.
630.12 permits up to 200% of the rated primary current — not of the derated conductor ampacity. The device has to survive striking an arc.
100% of the two largest, 85% of the third, 70% of the fourth, 60% of the rest.
Less than its nameplate current, which surprises people. NEC 630.11 multiplies the rated primary input current by the square root of the duty cycle: a 50 A welder at 30% duty cycle needs conductors rated 27.5 A, which is 10 AWG copper. At 100% duty cycle there is no reduction and you need the full 50 A.
Because it does not run continuously. A 30% duty cycle machine welds three minutes in ten, and conductor heating goes as current squared times time — so the average heating over the cycle is far below what a continuous draw would produce. The multiplier is the square root of the duty cycle for exactly that reason, which is why halving the duty cycle does not halve the conductor size.
Up to 200% of the rated primary current under 630.12, with the next standard size permitted where 200% is not one. A 50 A welder takes a 100 A device. That is deliberately much larger than the conductors, because the device has to let the machine strike an arc without opening while the conductors only have to survive the average.
No, and confusing the two is the most common error here. The output rating is welding amps at the electrode — 180 A, 250 A. The input rating is what the machine draws from the panel, typically 25 to 50 A at 240 V. Everything in Article 630 uses the input figure. If the plate shows only output, our estimate from output current, arc voltage and supply voltage is a starting point, not a substitute for the plate.
NEC 630.11(B): 100% of the two largest, 85% of the third, 70% of the fourth, and 60% of every one after that — after each has already been derated for its own duty cycle. Four identical 27.5 A machines give 97.6 A rather than 110. The code is unusually generous here because welders in a shop genuinely do not all strike an arc at the same instant.
Yes, substantially, and manufacturers quote the best number. A machine rated 60% at 200 A may be 20% at its 250 A maximum. Size on the setting you actually weld at rather than on the headline figure — and if you routinely run the machine flat out, size on the duty cycle at that output.
Article 630 covers arc welders, resistance welders and plasma cutting equipment, and the duty cycle approach applies to all of them. Inverter machines draw considerably less input current for the same output than an old transformer machine, so read the plate rather than assuming — an inverter rated 200 A output may draw only 25 A input where a transformer machine would draw 50.

Technical reviewer
Electrician · 10+ years of installation work in Bangladesh and the wider South Asian region
He reads the result the way an installer would: are the defaults values people actually meet, does the warning fire where you would stop and think, and is the answer something you could buy and fit? The code figures themselves come from the published standards cited below, not from him — that boundary is set out on his profile.
The full process is written up in the methodology and editorial policy. Results are engineering guidance, not a code sign-off — see the disclaimer. If a result looks wrong, tell us; corrections are answered before anything else.
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FLAFull load current from motor kW or hp, including efficiency — the term most calculators drop, which understates the current by 10–15%.
AWGWhat size wire you need, from the load, the run length and the conditions — sized against both NEC ampacity and voltage drop, with the terminal temperature rule that stops 90 °C wire giving 90 °C ampacity.
kWWhat size generator you actually need — running watts plus the largest single starting surge, derated for altitude and heat. Not the sum of every surge, which is how most calculators oversize you.