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I'm building a CNC router table that has a three-phase 220 V spindle motor, and 120 V computer (and accessories). The spindle's nameplate:

HSD ES330 Nameplate

As you can see it tops out at 20.3 A @ 220 V.

The VFD’s (Delta Electronics VFD25AM23ANSHA) nameplate says it’s rated for 5.5 kW, 25 A output, 30.0 A @ 200 - 240 V input.

I have three-phase 208 V power. I would like to have a single plug and power cord for the spindle motor and related 120 V accessories (PC, monitor, drive electronics, solenoid valves, etc). The total of these accessory loads is probably less than 500 W.

Can I use 10/5 SJOOW power cord (not sure I can get that, but I know I can get SOOW 10/5 cord), and a 5-conductor locking plug to provide the three phases, neutral, and ground?

The NFPA 2017 NEC Handbook tables 400.5(A)(1) and 400.5(A)(3) seem so suggest I'm limited to 20 A with 10/5 cable.

NEC Table 400.5(A)(1) NEC Table 400.5(A)(3)

400.4 also says:

A neutral conductor that carries only the unbalanced current from other conductors of the same circuit shall not be required to meet the requirements of a current-carrying conductor.

That could allow me to consider my cord as having three conductors, and thus not needing derating to 20 A. But it also refers me to b.310.15(b)(2)(11) for derating for load diversity.

What's my best bet here?

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    \$\begingroup\$ so what's the efficiency of your VFD? it seems to me that you're looking at over 25A from your 208V supply if you want to run that motor at full rated power \$\endgroup\$ Commented Nov 11, 2018 at 6:19
  • \$\begingroup\$ @Jasen I updated the question with information about the VFD. \$\endgroup\$
    – Rick
    Commented Nov 11, 2018 at 6:26

1 Answer 1

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The nameplate appears to be for the motor, but the motor must be supplied by a VFD electronic speed controller. You need to consider the current at the VFD input. The VFD input will not reflect the motor power factor, but will have harmonic current in the input current. Also, the torque rating of the motor will need to be reduced somewhat due to the input voltage being 208 volts rather than the 230 or 240 volts expected at the input of the VFD.

The neutral will certainly carry much less current than the other conductors. It certainly doesn't need to be considered as carrying the same current. It might be reasonable to treat the cable as having only 3 current-carrying conductors, but an electrical inspector might not see it that way.

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  • \$\begingroup\$ I’ve edited the question with information about the VFD. It is input-rated for 30.0 A @ 200 - 240 V, 25 A output (5.5 kW). I assume I can derate the input requirement based on the maximum draw from the spindle. \$\endgroup\$
    – Rick
    Commented Nov 11, 2018 at 6:24
  • \$\begingroup\$ I believe you may need to wire based on the VFD nameplate input regardless of the motor rating. Read the VFD section of NEC. \$\endgroup\$
    – user80875
    Commented Nov 11, 2018 at 6:28
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    \$\begingroup\$ 430.122 Conductors - Minimum Size and Ampacity. (A) BranchlFeeder Circuit Conductors. Circuit conductors supplying power conversion equipment included as part of an adjustable-speed drive system shall have an ampacity not less than 125 percent of the rated input current to the power conversion equipment. \$\endgroup\$
    – user80875
    Commented Nov 11, 2018 at 6:43
  • \$\begingroup\$ Thanks for that. I'm not sure what they mean by “rated.” If one considers the entire system, vs. the nameplate specification (since it’s not likely the spindle will be changed), then one would think I could specify the rating for the system. Assuming 30 A, that really increases the wire gauge; it would have to be 5 ga wire, which seems ridiculous (125% + the 80% derating for more than three conductors). \$\endgroup\$
    – Rick
    Commented Nov 11, 2018 at 6:48
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    \$\begingroup\$ I would use a separate 120V supply for the control equipment, though. \$\endgroup\$ Commented Nov 11, 2018 at 6:55

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