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Old December 15th, 2013, 07:19 AM   #1
Tharon
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Question about VFD Input current draw.

I'm using a 40HP variable frequency drive to power a 30HP motor. The manufacturer specs list 60A as the input current of the drive.

Is that 60A during full load operation? Is full current drawn during initial power up when the drive is fully discharged? If I'm only running a 30HP motor, will the input current be lower? Can I, or should I, size the input fusing to match the designed max load, plus dissipation, of the VFD?

Thanks in advanced.
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Old December 15th, 2013, 11:20 AM   #2
Tom Jenkins
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Those are some good questions, and the answer to any good question is "It depends".

The law of conservation of energy dictates that the power input must equal the power output plus the losses to heat. VFDs have a fairly constant power factor and losses so the power input will be proportional to the power demand of the load. If you are running a 40 hp load the power and current in will be less than a 60 hp load.

The way the demand varies with speed is determined by the load. Centrifugal pumps, for example, have the power vary as the cube of the speed. Conveyors have power draw that varies linearly with speed.
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Old December 15th, 2013, 12:02 PM   #3
Lancie1
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Assuming a 480 volt 3-phase supply voltage, your 30 HP motor should need about 40 Amps full-load current. Because your VFD will limit the start-up current to slightly above the motor FLA, you should use 45 or 50-Amp fuses on the input side of the VFD.

Only use 60-Amp fuses if you plan to very soon replace the 30 HP with a 40 HP. Fuses should be sized for actual load, not for the size of the largest electrical device (VFD) in the circuit. For example if you had your 30 HP motor connected to a motor starter, and the starter happened to be larger than needed, say 100 HP (124 Amp load) then you would not use 150 amp fuses just because your starter happened to be much larger than needed.

If your electrical device is larger than needed, remember that smaller fuses will also protect the device (as well as the motor) from short-circuits. You can go larger for your electrical controller, but you should not go smaller!

Also you should set the motor overload trip point inside your VFD for a 30 HP motor, not 40 HP. Most VFDs have a section of parameters where you can enter the motor nameplate information, such as HP, Amps, Volts, and RPM. Make sure you enter those values to get the best control.

Last edited by Lancie1; December 15th, 2013 at 12:12 PM.
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Old December 15th, 2013, 07:34 PM   #4
Tharon
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That's what I had thought. Thanks for the explanation.
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Old December 17th, 2013, 07:55 AM   #5
DickDV
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Just as Lancie1 said, if you program the VFD properly, it will protect the smaller motor correctly. I would, however, size the larger drive feeder conductors and fusing as the manual suggests for the actual rating of the drive. Undersizing the feed could lead to blown fuses when powering up due to the capacitor charging inrush currents.
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Old December 17th, 2013, 11:46 AM   #6
jraef
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Here in the US, the NEC dictates this, very specifically because of this exact issue. You are REQUIRED, per Article 430.122, to size the feeder circuit for 125% of the VFD MAXIMUM INPUT amps, regardless of the size of the motor connected behind it. The motor will still only draw as much current as the load needs to do the work, but the intent of the NEC rule change is to cover what the VFD might be used for by someone else in the future, who reads the nameplate and decides LATER to use a larger motor.

Then because of that change, which originally took place in the 2002 code under a different clause, UL has changed their rules on getting a VFD UL listed. In the past, VFDs were not actually required to provide motor protection, because they were classified as "Power Converters". Some mfrs started including it anyway, but UL was not actually requiring it. Now UL says if they are "Motor Controllers", they must provide both the running overload protection (OL) AND the motor Branch Short Circuit and Ground Fault protection. This must now be based upon the connected motor, NOT the incoming power. So when you program the VFD with the motor nameplate data, you are at the same time programming the OL and the "Electronic Circuit Breaker" protection functions that are built-in to the VFD.

Word of caution here: Not all VFDs are created equal, and a lot of the cheap junk showing up on places like FleaBay are either not UL listed at all, which means they do NOT provide this, or they are using a trick of getting them UL listed as "Power Converters" still, which allows them to ignore this requirement. They cannot lie about it though, so look at what the UL label says; if the label says something to the effect of it being a "Power Converter", it may not be providing this level of connected motor protection (it also may, it's just not required to, so you have to check). If it says "Motor Controller" in any form, it must then provide this.
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