How to Size an Electrical Feeder: Step by Step

Four checks, in a fixed order

Sizing a feeder is not one calculation. It is four, and they fail differently: the conductor can be too small to carry the current, too small to hold the voltage up at the far end, too large to fit the raceway already in the wall, or correct on all three counts and still wrong because the load was counted as non-continuous. Running them in a fixed order is what stops you redoing the arithmetic after the pipe is installed.

The order is: load to amps, continuous-load factor, ampacity table, voltage drop, then raceway and box. Every figure in the worked examples comes from the same arithmetic the calculators on this site use.

Step 1 — Turn the load into amps

Everything downstream needs amperes, and loads arrive as kVA, kW, horsepower or plain amps. The three-phase conversion is the one people get wrong, because of the square root of three.

The √3 is not a fudge factor. In a balanced three-phase system the phases sit 120 degrees apart, so line-to-line voltage is √3 times line-to-neutral — 208 V line-to-line is 120 V line-to-neutral, and 480 V is 277 V. The kVA calculator runs the conversion in both directions.

Step 2 — Apply 125 percent for continuous load

A continuous load is one expected to run three hours or more. For those, feeder conductors are sized at 125 percent of the load current — NEC 215.2(A)(1) for feeders, 210.19(A)(1) for branch circuits. The margin is not arbitrary: a conductor held at its table ampacity for hours is at its thermal limit, with nothing left for a hot day or a slightly derated installation.

The practical consequence is that a load computing to 83 A does not get a conductor rated 85 A. It needs 104 A, often two sizes larger. Skipping this step is the most common reason a feeder that checks out on load alone gets flagged on inspection.

Step 3 — Read one column of the ampacity table

Ampacity for a given size appears three times in NEC Table 310.16, once per insulation rating: 60, 75 and 90 degrees Celsius. The column is set by the weakest link, and the weakest link is usually the terminal rather than the wire. Most breakers and lugs are rated 75 degrees, so a 90-degree THHN conductor still gets sized from the 75-degree column unless the terminations are also rated 90 degrees.

Two habits prevent almost every mistake. Write down which column you are using before you look anything up, and stay in it for the whole job. And do not treat the 90-degree column as a free upgrade — mixing columns inside one calculation, taking 6 AWG from 75 degrees and 2 AWG from 90 degrees, produces a size that matches no table in the code book.

Step 4 — Check voltage drop before you accept it

An ampacity-sized conductor can still be the wrong conductor. Voltage drop is a separate check, and on long runs it decides the answer. For three-phase:

VD = √3 × K × I × L ÷ CM

K is the resistivity constant in ohm-circular-mil per foot — 12.9 for copper at 75 degrees Celsius, 21.2 for aluminium. I is load current, L is one-way length in feet, CM is the conductor's circular-mil area. For single-phase, drop the √3 and count length twice, because the return path is a real conductor.

The usual targets come from NEC informational notes rather than a mandatory rule: 3 percent on a branch circuit, 5 percent for feeder and branch combined. Those are guidance, but equipment nameplates assume them. The voltage drop calculator uses the same K and the same circular-mil table, so hand and tool should agree to the last decimal.

Step 5 — Confirm the raceway and box will hold it

The last check has nothing to do with electricity. Four 1/0 conductors plus a ground occupy real volume, and a raceway limits how much of it they may fill: no more than 40 percent of internal cross-section for three or more conductors. A conduit sized for the old feeder often will not take the new one, and learning that after the pull is expensive.

Add each conductor's area from NEC Chapter 9 Table 5, divide by the raceway's total internal area from Table 4, compare with 0.40. The conduit fill calculator does the whole table at once; the box fill calculator answers the same question at the enclosure, where the limit is counted in conductor volumes instead of a percentage.

Worked example: 30 kVA at 208Y/120, 160 ft

A 30 kVA transformer feeding a panel 160 ft away on 208Y/120, power factor 0.80, treated as continuous.

Amps. I = 30,000 ÷ (√3 × 208) = 83.27 A.

Continuous. 83.27 × 1.25 = 104.09 A, so ampacity must reach 104.09 A. From the 75-degree column that is 2 AWG at 115 A. One size smaller, 3 AWG at 100 A, carries the 83.27 A load current comfortably and still fails — this is the trap in Step 2.

Voltage drop. With K = 12.9 and 66,360 circular mils: VD = 1.732 × 12.9 × 83.27 × 160 ÷ 66,360 = 4.49 V, or 2.16 percent of 208 V. Under 3 percent, so 2 AWG stands.

What doubling the distance does

Move the panel from 160 ft to 320 ft and nothing about the load changes — same 30 kVA, same 83.27 A, same 104.09 A continuous requirement. The ampacity answer stays 2 AWG. The drop answer does not: 2 AWG at 320 ft gives 8.97 V, or 4.31 percent, straight through the 3 percent target.

Back under 3 percent at that distance needs 1/0 AWG, at 5.64 V and 2.71 percent. The conductor grew four sizes for a reason unrelated to current capacity — 2 AWG was already more than adequate for 104 A. This is why a feeder size quoted without a distance cannot be evaluated, and why distance rather than load is what makes feeders expensive.

Example: conductor and raceway in one pass

The 1/0 AWG answer needs a raceway. Three phases, a neutral and a 6 AWG equipment grounding conductor, all THHN. From Chapter 9 Table 5 each 1/0 THHN is 0.1855 square inches, so four of them are 0.742; the 6 AWG ground adds 0.0507. Total 0.793 square inches.

Divided by 0.40, the raceway needs 1.982 square inches of internal area. A 1-1/2 inch EMT offers 1.784, which puts fill at 44.4 percent — over the limit. A 2 inch EMT offers 2.316, giving 34.2 percent, which passes. That difference is the difference between a pull that goes in and one that stops halfway.

Example: power factor moves the whole chain

Take 24 kW of real load on the same 208 V system, 160 ft of copper, and change only power factor from 0.80 to 0.60. Real power is identical. Apparent power is not: 24 ÷ 0.80 is 30 kVA, while 24 ÷ 0.60 is 40 kVA.

Current rises by a third, the conductor goes up several sizes, and voltage drop improves — the larger conductor's lower resistance more than offsets the higher current. That last cell is worth pausing on: sized first and measured later, the installation would have looked fine. It was fine for the load you assumed, not the load that was there. The power factor calculator works the relationship from the other direction.

What this guide does not decide

Five things sit outside the arithmetic above, and each can change the answer. Ambient temperature correction and the adjustment for more than three current-carrying conductors both reduce usable ampacity below the table value — a 2 AWG that passes here can fail in a hot mechanical room with six conductors in one pipe. Parallel conductors, short-circuit withstand and equipment grounding conductor sizing are separate articles with their own tables. And many jurisdictions amend the code locally.

Treat this as the arithmetic, not as the approval. A feeder is a code-compliance question as much as a calculation, and the person signing the drawing is the one who has to satisfy the authority having jurisdiction.

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