Amperage difference on VFD

[jaykool]

When I measure the amps at the power source (breaker inside a MCC bucket) I get a reading of 32 amps. When I measure the amps on the load side of VFD it is around 27 amps.

Any explanation for the drop in amps?

I have a Allen-Bradley powerflex 700 series vfd feeding a 30HP motor.

 

[charlie b]

Measure the voltage and the frequency, and you will find differences there as well. So there is no reason to expect the amps to be the same. Calculate the power sent out from the MCC and the power sent out by the VFD, and the numbers will be closer, with the difference being due to power lost within the VFD.

 

[jaykool]

I agree that there is power loss due to internal components in the vfd. What are the internal components that would cause this drop?

 

[kingpb]

I'd say your VFD is operating at about 84.4% efficiency given your system parameters, and assuming input and output voltage are the same at the time you measured the current.

From AB literature the published efficiency is 97.5% at rated amps, nominal line volts, which BTW the listed "nominal" voltage for the 480V drive is 648Vac.

So, I doubt you will ever operate under the optimal rating; good luck ever reaching 97.5%, your 84.4% is much more reasonable.

 

[jaykool]

yep, makes sense. thanks for the input.

 

[Jraef]

It should also be noted that when measuring current on the output of a VFD it's difficult to get accurate readings with generally used metering equipment. It takes the best filtering capabilities to accurately interpret the PWM signals, something that most people not in the VFD servicing business usually pay for.

 

[gar]

101130-1403 EST

jaykool:

Why would you expect that amps in would equal amps out? What is inside a VFD drive? How does the output voltage of a VFD vary with motor speed? Is VFD output voltage equal to input voltage?

.

 

[Besoeker]

It should also be noted that when measuring current on the output of a VFD it's difficult to get accurate readings with generally used metering equipment. It takes the best filtering capabilities to accurately interpret the PWM signals, something that most people not in the VFD servicing business usually pay for.

Usually the relatively high carrier frequency of the the PWM and the leakage inductance of the motor results in a relatively good sinusoidal current. I use a simple 1000:1 split core CT with a 10 ohm burden resistor.
So no, it really doesn't need filtering or specialist kit.
As it happens, recently one of our drive partners borrowed this plus one of my digital scopes to prove that the readings on his drive modules were correct.

 

[bjcouche]

Kingpb is way off if he thinks 84.4% is a reasonable number. If you came up with 95% and it was rated as 97% I'd say you were within reason.

Here's a little background on VFD's that by your question, you didn't know. The VFD must vary the voltage and frequency in order to vary the motor's speed. For example, at full rated motor torque, operating at 50% rated speed, the motor voltage and frequency will also be 50%, so a 460V motor would be operating at 230V and 30Hz. For example a 1750rpm 30hp 35A motor operating at 100% torque but 50% speed. You would measure 230V to the motor at 30Hz and 35A. What would you expect the input of the drive to be? Remember the saying watts is watts. 230V at 35A 3 phase is 13943W. Lets calculate the drive input current, 13943W and 460V is 17.5A exactly half. There are some losses in the drive due to semiconductor IR drop which will account for a couple % more or so power on the input than the output. Just remember that the drive is a "power" converter type device.

How does it vary the voltage and frequency? This is a short explination and more in depth research may be required for a full answer but in short: The drive first creates DC out of the 3 phase AC by using a 6 diode bridge rectifier in the example of your PF700. The DC voltage is around the 650-680VDC range. Capacitors on the DC bus are used to smooth out the ripples. Next six IGBT's (big transistors) are used to pulse width modulate the DC to create a 3 phase sinusoidal current at the proper voltage and frequency based on the commanded speed. As others have said, measuring the current is fairly easy with normal instrumentation because the motor inductance smooths out the current waveform and the current is fairly sinusoidal. Measuring the voltage however requires special equipment as most meter's can't handle the non sinusoidal PWM voltage and will measure a voltage way off from the actual. Simply put, without specialized meters specially designed to measure PWM drive voltage, you are better off simply reading the output voltage from the drive display, as it will be more accurate on voltage measurement than your meter.

There are several operating modes FOC (Field Oriented Control) V/Hz, fixed boost, fan / pump curves, etc that can skew this linear voltage frequency speed curve a little to optimize specific performance, IE torque control or energy saving.

Here's another one to ponder. Most people know that an AC motor will draw roughly 1/3 of rated current at full speed with no load. What not everyone realizes is that not all that 33% is going up in wasted heat. Two words: Power factor. That current is purely inductive. OK, using your motor operating it unloaded draws 1/3 of 35= 11.7A at the output of the drive. What's the input current to the drive? Neglecting all the drive and motor losses the answer would be nearly 0A. This is because the drive also sort of corrects for power factor since it converts ac to dc and then back to ac again. If you are into phasors and power factor, then you would understand the math. Just remember that the current observed on the line side of the drive correlates to the actual work being done by the motor plus motor and inverter losses.

For example, For test purposes, I have seen a 200HP drive and motor be tested and operated up to full speed unloaded off of a 50A breaker. The only load on the breaker is the motor windage torque and the motor and inverter losses.

Brian
PS, I'm a past Rockwell drives service engineer so ask away...

 

[mull982]

Here's another one to ponder. Most people know that an AC motor will draw roughly 1/3 of rated current at full speed with no load. What not everyone realizes is that not all that 33% is going up in wasted heat. Two words: Power factor. That current is purely inductive. OK, using your motor operating it unloaded draws 1/3 of 35= 11.7A at the output of the drive. What's the input current to the drive? Neglecting all the drive and motor losses the answer would be nearly 0A. This is because the drive also sort of corrects for power factor since it converts ac to dc and then back to ac again. If you are into phasors and power factor, then you would understand the math. Just remember that the current observed on the line side of the drive correlates to the actual work being done by the motor plus motor and inverter losses.

Is it the internal capacitor in the drive which supplies the reactive power/current for the motor? If not what inside the drive provides the reactive current?

So you are saying then if you measured the current on the line side of the drive the current would all be completely in phase with the voltage? Or are there some reactive components in the drive prior to the rectifier bridge that may lead to some reactive current?

 

[kingpb]

OK, so, ex-service tech, why does AB say its 97.5% efficient at nominal voltage, then state the nominal rated voltage is 648V? Little deceiving isn't it since everybody is going to be using it at 480V. So what is the real efficiency at 480V? Humm, unpublished you say......so tell me again why 84.4% is so far off given no operating voltage efficiency is provided?

The published drive efficiency I would bet also applies to the drive being fully loaded or at some certain percentage.

Manufacturer's like to overstate their efficiency, sorta like car manufacturers telling you the car gets 38mi/gal hwy, but that number is not achievable since the actual use conditions will never be able to match the test conditions.

BTW: Last I saw efficiency is equal to power out divided by power in.

Call me a skeptic, but I bet you'd never see over 92% on the best day.

 

[Jraef]

OK, so, ex-service tech, why does AB say its 97.5% efficient at nominal voltage, then state the nominal rated voltage is 648V? Little deceiving isn't it since everybody is going to be using it at 480V. So what is the real efficiency at 480V? Humm, unpublished you say......so tell me again why 84.4% is so far off given no operating voltage efficiency is provided?

The published drive efficiency I would bet also applies to the drive being fully loaded or at some certain percentage.

Manufacturer's like to overstate their efficiency, sorta like car manufacturers telling you the car gets 38mi/gal hwy, but that number is not achievable since the actual use conditions will never be able to match the test conditions.

BTW: Last I saw efficiency is equal to power out divided by power in.

Call me a skeptic, but I bet you'd never see over 92% on the best day.

Nominal voltage is 648??? You must be reading the DC link voltage. They would be referring to nominal INPUT AC voltage if anything, because if you have a higher input voltage, your switching losses may increase, and if lower, the motor cannot get full voltage at full speed so it too looses efficiency if at full load.

Typically, peak efficiency of a VFD is rated at peak loading, not peak voltage. The efficiency drops as the speed drops mainly because the switching losses remain relatively constant, but the power flowing through the VFD is dropping with speed, so power in the switching losses represents a higher percentage of total power at that moment. You are still typically using less power than if you temporarily reduced speed in any other way.

 

[kingpb]

Nominal voltage is 648???

http://www.ab.com/support/abdrives/powerflex700/datasheet.html

 

[Jraef]

http://www.ab.com/support/abdrives/powerflex700/datasheet.html

Nominal Bus Voltage 281V DC 324V AC 540V AC 648V AC 810V DC

Huh... you're right in that it says that but that's undoubtedly a typo. Bus voltage would be DC, not AC.

 

[jaykool]

if i change the motor nameplate voltage parameter from 460V to 480V...how much of an effect on output current? I cant imagine the change will amount too much. any thoughts?

the reason i ask, this is a critical motor...in order to make parameter change...i have to shut down motor.

 

[Jraef]

if i change the motor nameplate voltage parameter from 460V to 480V...how much of an effect on output current? I cant imagine the change will amount too much. any thoughts?

the reason i ask, this is a critical motor...in order to make parameter change...i have to shut down motor.

The motor nameplate voltage rating is what the motor is designed for. If you change the parameter in the VFD to allow the VFD to deliver 480V to the motor at full speed (assuming you have 480V or more at the line terminals), it may drop the current a little bit, but it may not. Motors have a +-10% voltage tolerance anyway, so it should not do any immediate harm, but you are technically saturating the windings. So although the working current may drop, the heating effect from the increased winding saturation will increase the net current, so will not drop as much as you may think. But the motor will be running warmer, which is unnecessarily stressing it a little more. I see no net benefit to doing that.

Look at the chart below. At 480V you are going to be almost 5% higher than the nameplate rated voltage. Notice that at that point the current is still OK but is trending up, not down, and the efficiency is trending down as well, which means more losses in the motor.

Are you concerned about the motor current? Is the 27A you are reading on the VFD above the nameplate FLA of the motor? If not, you may be overly concerned.

 

[weressl]

http://www.ab.com/support/abdrives/powerflex700/datasheet.html

PowerFlex 700 Drive 200-208V Drive_|_240V Drive_|_380/400 Drive_|_480V Drive_|_ 600V Drive
Nominal Bus Voltage 281V DC_______|_ 324V AC__|_ 540V AC____|_648V AC__|_810V DC
Efficiency 97.5% at rated amps, nominal line volts.


So it is not clear to you that the AC voltage in the Nominal Bus Voltage is a printing error since it the only bus in a drive is the DC bus first and that the efficiency refers to line, not bus voltage?

Are the AB manuals are getting just as 'good' as the translated Japanese drive manuals? Not to mention Chinese.......

 

[Jraef]

... Are the AB manuals are getting just as 'good' as the translated Japanese drive manuals? Not to mention Chinese.......

A lot of instruction manuals are now being printed in China. Printers here in North America have been suffering for the last 5 years or so. Compared to local printers the Chinese are doing it for pennies on the dollar, including the shipping cost. Looks like A-B may have had the proof reading done there too.