I'm working on a robotics project where we intend to deliver ~500W of power over a 1000 ft tether. The current plan is to run ~400VDC down the tether on a pair of 16 AWG wires. These power wires will be bundled with other control wires as part of a cable assembly.

We are intending to include basic safety features like fuses and overcurrent/overvoltage/temperature sensors. We're considering including some type of DC Ground Fault Interrupter. Additionally, this is not a consumer product, it's an industrial tool for infrastructure maintenance work.

I have seen in this post that many of these standards are handled by UL standards, but I am not sure if complying with UL standards is a legal requirement, or only necessary if we are pursuing a UL certification.

What sort of regulatory standards do we need to comply with to distribute power to our robot in this manner?

  • \$\begingroup\$ You need a regulatory engineer from your region, who is familiar with this. Also the service will probably be paid. The wires are too small according to me. You will have a lot of losses too for this device. \$\endgroup\$
    – CFCBazar
    Commented May 13, 2020 at 19:02
  • \$\begingroup\$ Sending 380-500vAC down the cable would be much more practical, much less lossy, and is directly applicable to a wide range of industrial facilities worldwide. \$\endgroup\$
    – rdtsc
    Commented May 13, 2020 at 19:03
  • \$\begingroup\$ @rdtsc Why would that be much less lossy? And why more practical? (also, the wiring OP has in mind seems a bit flimsy to be directly connected to three-phase mains power) \$\endgroup\$
    – marcelm
    Commented May 13, 2020 at 19:10
  • 1
    \$\begingroup\$ If I want 400V/500W at the robot, I'm getting a current of 1.25A. I'm calculating a resistance of 8.01 ohms for 2000 ft of 16 AWG wire, giving a power loss of 12.5 W. 500W delivered / 512.5W supplied = 97.5% efficiency. Sufficient for our application. \$\endgroup\$ Commented May 13, 2020 at 20:03
  • \$\begingroup\$ I agree. In fact I'd say you should lower the voltage because you don't need voltage that high, and further, you should change it to AC because poor Nipsy the elephant. Seriously, HVDC is a VERY nasty customer, and if you don't have a fearsome appreciation of that, well... you soon will. I'd much rather shake hands with 277 VAC than 120 VDC. \$\endgroup\$ Commented May 14, 2020 at 5:01

3 Answers 3


You need to comply with the safety codes where the equipment will be used. That means the state, county and perhaps city electrical code. Those codes are based on the National Electrical Code (NEC) but contain some additions and may contain some exceptions. You need to study the codes carefully for industrial use exemptions. NEC requires equipment to be acceptable to the authority having jurisdiction(AHJ). The NEC has provisions calling for listing and labeling by a nationally recognized testing laboratory (NRTL). The Occupational Safety and Health Administration (OSHA) maintains a list of NRTL's. UL and ETL are the most well known electrical NRTL's on that list. Although there may be several NRTL'S to choose from, most of the applicable standards are published by UL. Regardless of which NRTL you may want to use, you probably need to get advice from someone who is familiar with UL standards. You can probably find a consultant to advise you. It is best to get advice during design rather than after the design is substantially complete. For an industrial establishment, OSHA requirements may be the primary concern. The local AHJ may defer to OSHA. You may also want to talk to your insurance underwriter.


Your proposed Power Cable is DC not AC .Sure this has some advantages But a disadvantage is Arcing .DC Arcs much worse than AC generally.You should incorperate Arc Protection.


I would as a minimum add a portable earth leakage to the AC part of the circuit - before the 300 V rectifier. This would protect against any sensitive earth fault on the tether. This can also be done by electronics.

I would also not include any storage or ripple at the source side - use a diode to block back-feed of the stored voltage into a fault on the tether.


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