single phase to three phase w/ VFD
- Thread starter keelv
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skeshesh
Senior Member
- Location
- Los Angeles, Ca
I think you've calculated the maximum allowed OCPD correctly but don't think you need to go that far with a VFD unless you have a line-voltage bypass, which you don't because you need the VFD to convert the phases. The VFDs do have internal losses - I would contact the VFD vendor and get more details.
- Location
- San Francisco Bay Area, CA, USA
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- Electrical Engineer
First off you have to double the size of the VFD for single phase input (assuming it will allow 1 phase input, not all will) compared to the motor FLA. So in your case the VFD must be rated for 84A minimum. Then per the NEC, you must size the OCPD at 125% of the VFD Max Amp Rating, not the motor rating. So find a VFD with a rating of at least 84A and whatever it says is the MAX input amp rating of the VFD is what you size the OCPD at +25%, rounded to the nearest size.
So for example If you find a VFD that is rated for 78A and another rated for 93A, you must use the 93A rated drive, then the minimum OCPD would be 116A, so a 125A CB or fuses (and conductors).
So for example If you find a VFD that is rated for 78A and another rated for 93A, you must use the 93A rated drive, then the minimum OCPD would be 116A, so a 125A CB or fuses (and conductors).
Thanks for your replies.
Don't have the NEC with me right now...I know, per NEC, the conductors would be sized at 125% of the rated input of the VFD, but can you give me the article for sizing the OCPD at 125% for VFD.
- Location
- Massachusetts
I don't think there is such a requirement, I think as long as the conductors are sized as Jraef instructed that the OCPD could be as small as you want but no larger than the ampacity of the conductors.
That said, I can't see running a set of 100 amp conductors to a 50 amp breaker for a 75 amp input rated VFD.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
110129-1353 EST
keelv:
My answer is not related to code but to provide you with a fundamental concept of how the system works.
Assume there is no false advertising on the output rating. Then a continuous duty load of 15 HP = 15*746 = 11,190 W. Assume 85% efficiency, then the input power required is 13,165 W.
If this were supplied from a DC source, batteries for example, at 240 V, then the DC input current would be about 55 A. This assumes bypassing the input rectifier diodes in the VFD if they are not large enough.
A VFD drive supplied from a 3 phase system will have a bank of rectifiers at the input feeding a capacitor bank. The capacitor bank serves two purposes --- to filter some of the input ripple from the rectifiers, and to absorb some energy during deceleration.
Rectifier-capacitor input filters produce non-sinusoidal pulses of current in the input lines. This means the RMS input current is higher than if the same energy was transferred from a DC source.
If you take a VFD drive designed for a 3 phase input and use it on a single phase system, then the diodes and the filter bank may be under sized. This is why Jaref suggested you may need a 3 times larger VFD. It is not because of the DC to AC conversion electronics, but because of the input diodes and capacitors.
If you were to remove the input diodes and hang a large battery bank in parallel with the input capacitor bank, and add a charger for the batteries, then, depending upon load characteristics, you might actually operate at an input current level much less than the above 55 A DC referenced to the AC input. On a typical usage of a CNC machine this might work on a 20 A circuit.
This does not answer your direct question, but it may give you an insight to the system.
In your application, given that you have a VFD system that will provide the 15 HP output from a 240 V single phase input, then you need to know the RMS line current under these conditions, and apply NEC rules for this current level.
.
keelv:
My answer is not related to code but to provide you with a fundamental concept of how the system works.
Assume there is no false advertising on the output rating. Then a continuous duty load of 15 HP = 15*746 = 11,190 W. Assume 85% efficiency, then the input power required is 13,165 W.
If this were supplied from a DC source, batteries for example, at 240 V, then the DC input current would be about 55 A. This assumes bypassing the input rectifier diodes in the VFD if they are not large enough.
A VFD drive supplied from a 3 phase system will have a bank of rectifiers at the input feeding a capacitor bank. The capacitor bank serves two purposes --- to filter some of the input ripple from the rectifiers, and to absorb some energy during deceleration.
Rectifier-capacitor input filters produce non-sinusoidal pulses of current in the input lines. This means the RMS input current is higher than if the same energy was transferred from a DC source.
If you take a VFD drive designed for a 3 phase input and use it on a single phase system, then the diodes and the filter bank may be under sized. This is why Jaref suggested you may need a 3 times larger VFD. It is not because of the DC to AC conversion electronics, but because of the input diodes and capacitors.
If you were to remove the input diodes and hang a large battery bank in parallel with the input capacitor bank, and add a charger for the batteries, then, depending upon load characteristics, you might actually operate at an input current level much less than the above 55 A DC referenced to the AC input. On a typical usage of a CNC machine this might work on a 20 A circuit.
This does not answer your direct question, but it may give you an insight to the system.
In your application, given that you have a VFD system that will provide the 15 HP output from a 240 V single phase input, then you need to know the RMS line current under these conditions, and apply NEC rules for this current level.
.
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