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Topic: Questions Regarding the minimum size of conductor for the inverter output circuit

User: Frank6172, Location: Maryland

Asked on: Thursday, March 6, 2025 at 10:17 pm EST
Answered on: Saturday, March 8, 2025 at 12:09 am EST

Question: Hi All,Here is the scenario: 2017 NEC, 208V, 3-phase, 4W, WYE-WYE, 35 degree C 2% high ambient temperature in ASHRAE, 3-current carrying conductors(L1,2 and 3) in the conduit. All lugs and breakers the conductors connected to are rated at 75 degree C. Continuous output current is 279A and we are using CU THWN-2 wire. We know the OCPD size should be 350A since it's the next size up of (279x1.25=348.75A). What is the minimum size of conductors? The AC run is very short so please don't consider voltage drop here. I got 500 kcmil but my colleague is saying 400 kcmil is fine? The 690.8(B) is telling me the conductor ampacity should be greater than (279x1.25=348.75A) and I looked up the 75 degree C column of 310.15(B)(16) due to the article 110.14 (C) (1) (b).Thanks very much in advance!

You are correct, and your reasoning is sound. Here's a breakdown:1. Continuous Load Calculation

279A 1.25 = 348.75A2.

OCPD Sizing

You correctly identify that the next standard overcurrent protection device (OCPD) size,
which is allowed, is 350A. (240.6(A))3.

Conductor Ampacity Requirement (690.8(B))

The conductors
must have an ampacity of at least 348.75A.
4.


Termination Temperature Limitations (110.14(C)(1)(b))

The terminations are rated for 75°C. Therefore, we must use the 75°C ampacity column in Table 310.15(B)(16) as the starting point.
5.


Initial Ampacity Check (75°C column)

400 kcmil CU THWN-2 has an ampacity of 335A in the 75°C column. This is not sufficient because it's less than 348.75A. 500 kcmil CU THWN-2 has an ampacity of 380A in the 75°C column.6.

Ambient Temperature Correction (310.15(B)(16))

You have a 2% increase for the high ambient. To calculate this, you need the correction factor from Table 310.15(B)(16). For conductors rated 90°C and operating at 35°C, the correction factor is 0.
96.
500 kcmil after correction: 380A 0.96 = 364.8A. Still greater than 348.75A7.

Number of Conductors Adjustment (310.15(B)(3)(a))

You have 3 current-carrying conductors in a raceway. According to Table 310.15(C)(1), this requires a 100% adjustment factor, meaning no change in ampacity.
8.


Final Decision

400 kcmil after correction: 335 0.96= 321.6A which is < 348.75A. Thus it does not work! 500 kcmil provides a corrected ampacity of 364.8A which is > 348.75A. This meets the requirements.

Therefore, your conclusion of 500 kcmil is correct.
400 kcmil is not sufficient, even after accounting for the ambient temperature correction. It doesn't have enough capacity to carry the necessary current after the temperature adjustment.

Key takeaways



Start with the Termination Temperature

Always base your initial conductor sizing on the weakest link – the temperature rating of your terminations.


Apply Adjustments Carefully

Remember to adjust for both ambient temperature and number of conductors after finding the initial ampacity from the table.*

Don't Skip Steps:
Each step in the NEC calculation is important for a safe and compliant installation. Flag for review

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