Detached Shop Feeder: Conductor Size, Grounding, and Load
A homeowner wants a 100 A subpanel in a detached 30 × 40 ft shop about 165 ft from the house. The house has a 200 A, 120/240 V single-phase service with an exterior main panel. The shop will contain a 5 hp, 240 V air compressor, a 240 V welder rated 48 A at 40% duty cycle, LED lighting totaling 1,500 VA, and general-use receptacles totaling 3,000 VA. An existing 1-1/4 in. PVC conduit runs between the buildings at 18 in. of cover. Utility voltage at the house measures 242 V under light load. The owner has a coil of 2-2-2-4 aluminum SER left over from another job and wants to pull it in, drive a ground rod at the shop, and skip the equipment ground to save copper.
1. Evaluate the proposed 2-2-2-4 aluminum SER for a 100 A feeder at this distance. What conductor do you actually need?
Read the worked answer
Two independent failures.
1. Ampacity. 2 AWG aluminum at 75 °C is rated 90 A. A 100 A feeder requires a conductor with ampacity of at least 100 A (the next-size-up rule does not apply to feeder conductors for a load that is actually calculated at 100 A). Minimum compliant aluminum is 1 AWG (100 A at 75 °C); 1/0 aluminum (120 A) gives margin.
2. Voltage drop at 165 ft, 242 V, 100 A using VD = 2 × K × I × L ÷ CM (K ≈ 21.2 for aluminum):
- 2 AWG (66,360 cmil): 2 × 21.2 × 100 × 165 ÷ 66,360 = 10.5 V = 4.4%
- 1/0 (105,600 cmil): 2 × 21.2 × 100 × 165 ÷ 105,600 = 6.6 V = 2.7%
4.4% exceeds the 3% feeder recommendation before you add any branch-circuit drop at the compressor, which will sag noticeably on start.
3. A material issue the owner did not mention. SER cable is not listed for wet locations. Inside a buried conduit it is a wet location, so the cable would need to be USE/RHH-2 rated or a wet-location-listed jacketed assembly. Leftover SER from an indoor job is often not.
Recommendation: pull 1/0 aluminum XHHW-2/USE or 3 AWG copper in the existing 1-1/4 in. PVC - both fit, both meet ampacity and the 3% target.
2. The owner wants to drive a ground rod at the shop and use the earth as the return path instead of pulling an equipment grounding conductor. Is that permitted? What grounding is actually required at the detached building?
Read the worked answer
No - the earth may never serve as the equipment grounding conductor. The feeder to a separate structure must include an equipment grounding conductor run with the circuit conductors, sized per the feeder overcurrent device (100 A → 8 AWG copper or 6 AWG aluminum). Removing it creates a path through earth and the house's grounding electrode system, which is high-impedance, unreliable, and will not clear a fault.
What the shop actually needs:
- An equipment grounding conductor in the feeder, terminated on the subpanel's equipment grounding bar.
- A grounding electrode system at the separate structure, which typically means a driven rod (8 ft, 1/2 in. or larger, 8 ft of cover) connected to the subpanel's grounding bar with a grounding electrode conductor. The conductor to a single rod is not required to be larger than 6 AWG copper.
- The neutral must remain isolated from the grounding bar at the subpanel. There is exactly one neutral-to-ground bond in the system, at the service. Bonding at the shop panel puts neutral current on the equipment grounding conductor and on any metallic path between buildings.
- Disconnecting means at the shop, rated at least 100 A, marked to indicate the location of the house service disconnect.
The rod is a supplement to the equipment ground, not a substitute. Driving it is required; relying on it alone is a red tag.
3. Using the shop loads given, does a 100 A feeder actually cover the shop? Show the calculation.
Read the worked answer
Convert the motor first. Per the motor tables, a 5 hp, 240 V single-phase motor draws 28 A. The largest motor gets 125%: 28 × 1.25 = 35 A.
Welder. At 40% duty cycle the welder demand factor is 0.63, so 48 A × 0.63 = 30.2 A. (Do not apply 125% to the welder as a continuous load - it is not one.)
Lighting. 1,500 VA ÷ 240 V = 6.25 A.
General receptacles. 3,000 VA ÷ 240 V = 12.5 A.
Total calculated demand: 35 + 30.2 + 6.25 + 12.5 = 84 A
Conclusion: a 100 A feeder is adequate at 84% loading, with roughly 16 A of headroom. Two cautions before you commit:
- The compressor is the starting load. At 165 ft, even a compliant conductor will produce a noticeable dip; confirm the compressor has a listed starter or that the feeder's voltage drop holds under locked-rotor current.
- If the owner later adds a second welder or a 240 V heater, the 84 A becomes 95 A+ quickly. Quote 1/0 aluminum now; the difference between 1/0 and 1 AWG in this size is a few dollars a foot and avoids a re-pull.