VFD current ratings

Status
Not open for further replies.

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Putting some motors on VFD's. I know the VFD input wiring and supply OCP should be sized for the input current rating of the VFD. I expected that current rating to be a little higher than the motor rating, due to the inefficiency of the VFD, and due to some voltage drop across the VFD line reactor.

The VFD cut sheets list the exact opposite. The VFD input current is less than the motor current.

Example; 25HP motor - NEC rated current at 460V is 34 amps. VFD with a 3% line reactor is about 28.8 amps, with a 5% reactor its 27.9 amps.
Another one: 60 HP motor - rated current 77 amps. VFD with 3% = 67.4 amps, VFD with 5% is 66.9 amps.

Why would the VFD have a lower rated input current than the motor its supposed to drive? I thought it might be because the motor is rated at 460V, and the VFD is 480 volts. But that's only a 5% difference in voltage, and I'm seeing a 10% + drop in current for the VFD.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
Putting some motors on VFD's. I know the VFD input wiring and supply OCP should be sized for the input current rating of the VFD. I expected that current rating to be a little higher than the motor rating, due to the inefficiency of the VFD, and due to some voltage drop across the VFD line reactor.

The VFD cut sheets list the exact opposite. The VFD input current is less than the motor current.

Example; 25HP motor - NEC rated current at 460V is 34 amps. VFD with a 3% line reactor is about 28.8 amps, with a 5% reactor its 27.9 amps.
Another one: 60 HP motor - rated current 77 amps. VFD with 3% = 67.4 amps, VFD with 5% is 66.9 amps.

Why would the VFD have a lower rated input current than the motor its supposed to drive? I thought it might be because the motor is rated at 460V, and the VFD is 480 volts. But that's only a 5% difference in voltage, and I'm seeing a 10% + drop in current for the VFD.
You must have started this thread as I was posting to the one that you originally asked this on, so I'll repeat my answer here for anyone who sees this thread later in a search. One thing though, the original question came up regarding conductor sizing per the NEC article 430.122, which is not specifically referenced in this thread.

MOTOR full load current as shown on the nameplate is based on a measured power factor at full load, usually also stated on the nameplate. But with a VFD, the Power Factor, as it relates to the SUPPLY, is corrected, essentially because the VFD caps are storing the reactive power for the motor, basically like having PFC caps for it. So the ACTUAL current drawn by the VFD is always going to be lower than the current drawn by the motor. So that might make one think that 430.122 then is not going to be a problem.

The problem is actually in the reality of the drives business. Most VFDs, by aggregate volume, come from Asia made by Asian companies as part of a product that is designed to be sold all over the world. But for our North American market, they tend to have a difficult time interpreting the way UL requires testing and listing, plus they do NOT want to pay for multiple tests. So when they design a VFD for "400V Class" and then extend the input voltage range to 480V so they can sell them here, they test and rate them at 400V motor currents, which are higher than 480V currents for the same given power rating. That means when you buy a 250HP Japanese drive, it is going to be rated at 250HP 400V, which more than covers 250HP at 460V current capability. But it means that the MAXIMUM current rating on the VFD nameplate is higher than it could otherwise be, which then gets you in trouble with 430.122. Companies that design their drives in North America for our market, regardless of where they are actually manufactured, will size the current ratings based on the NEC motor HP charts, the same ones used for sizing the conductors. So that's where you tend to see that the VFD maximum input amps are the same or lower than the NEC HP chart amps.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Power factor - of course.

But one more question: I thought line reactors caused a voltage drop (in addition to doing its job of filtering out harmonics).

I read somewhere that a 3% reactor basically caused a 3% voltage drop across itself. And a 5% reactor caused a 5% drop. Based on this, I would of expected the input current ratings to be 3% higher and 5% higher respectively. But its just the opposite - a 5% reactor has the least input current.

I'm guessing the reason is that with fewer harmonics, there is less harmonic input current, and harmonic input current wouldn't be doing any useful work anyway. It would just be a wasted circulating current much like a low power factor causes a wasted current to flow back and forth from the source to the load.

Is that a reasonable guess?
 

kwired

Electron manager
Location
NE Nebraska
Power factor - of course.

But one more question: I thought line reactors caused a voltage drop (in addition to doing its job of filtering out harmonics).

I read somewhere that a 3% reactor basically caused a 3% voltage drop across itself. And a 5% reactor caused a 5% drop. Based on this, I would of expected the input current ratings to be 3% higher and 5% higher respectively. But its just the opposite - a 5% reactor has the least input current.

I'm guessing the reason is that with fewer harmonics, there is less harmonic input current, and harmonic input current wouldn't be doing any useful work anyway. It would just be a wasted circulating current much like a low power factor causes a wasted current to flow back and forth from the source to the load.

Is that a reasonable guess?
I can't say I entirely know the answer to the question, but will say that 3 or 5% is probably the impedance when pulling full load rating through it. Power factor may still have some impact as well.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
Power factor - of course.

But one more question: I thought line reactors caused a voltage drop (in addition to doing its job of filtering out harmonics).

I read somewhere that a 3% reactor basically caused a 3% voltage drop across itself. And a 5% reactor caused a 5% drop. Based on this, I would of expected the input current ratings to be 3% higher and 5% higher respectively. But its just the opposite - a 5% reactor has the least input current.

I'm guessing the reason is that with fewer harmonics, there is less harmonic input current, and harmonic input current wouldn't be doing any useful work anyway. It would just be a wasted circulating current much like a low power factor causes a wasted current to flow back and forth from the source to the load.

Is that a reasonable guess?
You are on the right track. When you have a non-linear load like a rectifier (the front end of a VFD), it draws current from the line in "gulps" only at the peaks of each sine wave. That is why you end up with ripple on the DC side, which is why you need capacitance to filter it. So if you add an inductor (reactor) on the line side, the inductive time constant of it acts to slow down the rise time of the "gulps", which is what reduces the effects they have on the line, the "harmonics". Harmonics also represent a power factor issue, called "distortion power factor", whereas the motor creates "displacement power factor". Most VAR metering systems are not fast enough to catch distortion power factor, so in essence we see only the improvement of the displacement power factor when it comes to billing issues. But that does not mean the distortion PF is not there, it is. That distortion PF then is what improves slightly when adding a reactor, and that improves the current draw relationship MORE than the voltage drop across the reactor will increase it. At 3% is about a wash, at 5% it can actually appear to be lower. But that also has a lot to do with other elements of the total circuit, so it's not always what it sometimes appears to be. As to the motor side, the caps on the DC bus still charge the bus up to what it needs to be in order to get the right RMS voltage to the motor, so the VD losses of the reactor on the motor itself are not showing up, only the small amount of losses on the input side of the VFD.
 

mike_kilroy

Senior Member
Location
United States
I read somewhere that a 3% reactor basically caused a 3% voltage drop across itself. And a 5% reactor caused a 5% drop.

Correct. when we designed them we used % voltage drop at given current & voltage rating (normally listed this as HP rating too) based on 60hz frequency. Nothing to do with real impedance.
 

Open Neutral

Senior Member
Location
Inside the Beltway
Occupation
Engineer
Correct. when we designed them we used % voltage drop at given current & voltage rating (normally listed this as HP rating too) based on 60hz frequency. Nothing to do with real impedance.

Thanks; I read up on this topic a few years ago, and no one ever explained it so succinctly.
 
Last edited:
Status
Not open for further replies.
Top