A question has come up on a design for some panels with multiple Variable Frequency Drives and I was hoping you could provide a clarification. I am using the 2002 code as a reference. 1999 is still in effect in our area but they are basically the same. I have not gotten the 2005 version yet.

430.22 deals with sizing branch circuit conductors that supply a single motor. Normally the size of the conductor is based upon 125% of the full load current rating of the motor. Exception No. 2 states that when you are using and adjustable speed drive system such as a VFD you shall use 125% of the rated input to the power conversion equipment rather than the full load current of the motor.

I do not fully understand why they use the maximum rating of the drive rather than the actual load which is the motor but I can live with this section without problem. The issue is an interpretation of 430.24 which deals with several motors or motors and other loads. This would be the guidelines for sizing the feeder.

430.24 states that you use 125% of the full load current rating of the highest rated motor plus the sun of the full load current ratings of all the other motors in the group. It refers back to 430.6 A. Neither 430-24 or 430.6A make any mention of power conversion devices or adjustable frequency drives. (Although they are covered in 430.6C)

Here is where I get confused.

In order to minimize inventory we have standardized on 5 HP Variable Frequency Drives to be used for all motors from 1 HP to 5 HP. The rated input amps at 480 VAC for this drive is 9 amps. Lets take a worse case example. I have a panel with 10 VFDs all serving 1 HP motors. If I look at the full load amps of the motors I need to size the feeder for approximately 19 amps (10.25 x 1.8). If I need to be concerned with the input power to the conversion equipment then I need to size the feeder for roughly 92 amps (10.25 x 9). This is a HUGE difference.

My interpretation was that 430.24 which determines the feeder wiring does not mention power conversion devices so I did not need to use that information in sizing the feeder although I would need to use it in sizing the branch circuit. A design firm I have working for me feels that the feeder needs to be sized using the rated input power to the VFD. This results in much larger equipment. In this case we have existing motors utilizing contactors that are being replaced with VFDs. I would need to repull the feeder to the panels, replace the main disconnect on the panel, replace the feed to the panel in the MCC, etc. even though the actual load at the equipment has not changed at all.

The next step would deal with the overcurrent protection. I have been sizing the overcurrent protection for the motor and VFD based upon the FLAs of the motor and table 430.52. This has worked fine for years. There is a section 430.52-5 which allows you to use different fuses for power electronic devices. Sometimes some manufacturers recommend a quicker fuse to protect the electronics rather than the dual element time delay. Using the Bussman SPD Motor Circuit Protection Tables the optimal branch circuit protection for that 1 HP motor would be a 3 2/10 LPJ_SP fuse. This would never allow the 9 amps for the rated input power of the VFD to happen.

Finally, what I need is an answer in sizing the feeders. Do I use the FLA of the motor or the rated input power of the VFD? If it is the input power of the VFD, could you explain the logic behind this reasoning.