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VFDs Current
What is the relation between input current and output current in a VFD?
By nvrsrinivas on 4 August, 2016 - 12:31 pm

I want to know the relation between input current and output current in a VFD. What is the difference in the current drawn by motor at 50 Hz and at 40, i.e., whether we can use this current to calculate motor power? How to distinguish current at 40 Hz and 50Hz in a VFD fed to motor? Whether both currents can be measured by clampmeter? Whether these currents can be used to calculate motor power?

Please clarify.

What could be powerloss/efficiency in a VFD?
Thanks.nvrsrinivas100@rediffmail.com

By W.L. Mostia on 4 August, 2016 - 4:15 pm
1 out of 1 members thought this post was helpful...

Yes, it is possible to measure the current on the output side of a VFD if you have the correct meter or equipment to do it and also the voltage. Your problem is that the output waveform of a VFD is typically non-sinusoidal and the voltage also varies with the speed, which makes calculating power not a trivial matter. An example of such a meter can be found at:

http://www.testequipmentdepot.com/application-notes/pdf/multimeter/multimeter-measurements-on-variable-frequency-drives-using-the-new-fluke-289-dmm_an.pdf

I suggest that you talk to the people at Load Controls Incorporated about their Hall Effect power monitors for variable speed drives. An example instrument can be found at:

http://loadcontrols.com/wp-content/uploads/2015/02/ph3a-1000-datasheet.pdf

Also, please see this article:

http://www.plantengineering.com/home/single-article/electrical-power-measurement-on-3-phase-motors/0a092fdb11ff7ee850e42cba66428cb2.html

I have no connection to the companies mentioned here.

William (Bill) L. Mostia, Jr. PE
ISA Fellow, SIS-TECH Fellow,
FS Eng. (TUV Rheinland)
SIS-TECH Solutions, LP

"No trees were killed to send this message, but a large number of electrons were terribly inconvenienced." Neil deGrasse Tyson

Any information is provided on a Caveat Emptor basis.

By Bob Peterson on 4 August, 2016 - 5:55 pm
1 out of 1 members thought this post was helpful...

*I want to know the relation between input current and output current in a VFD. *

Output current is always higher than input current.

*What is the difference in the current drawn by motor at 50 Hz and at 40, *

Depends on the load.

*i.e., whether we can use this current to calculate motor power? *

Not readily.Current is not directly related to power in a motor.

*How to distinguish current at 40 Hz and 50Hz in a VFD fed to motor? *

My guess is there is some instrument that can do it, but why would it matter any? The VFD will tell you what the output frequency and the output current is already.

*Whether both currents can be measured by clampmeter? *

If it is a true RMS, yes.

*Whether these currents can be used to calculate motor power?*

No. Current is not directly related to power in a motor.

*What could be powerloss/efficiency in a VFD?*

This would be in the spec from the manufacturer.

Current is directly related to torque, but only indirectly related to load. So you can determine a RELATIVE DIFFERENCE in load by looking at current in two different scenarios, but it's not possible to accurately determine actual load without knowing the power factor of the motor at any given moment, and that is very difficult to measure on the output of a VFD. But as said, VFDs will tell you the current, and many will even tell you the calculated load in kW, because the VFD already has what it needs to do the accurate measurements.

So between 40Hz and 50Hz there WILL be a difference in kW, in fact it's likely to be 4/5ths at 40Hz, as long as it is not a centrifugal load, like a pump or fan. In a centrifugal load the power varies at the cube of the speed change, so 4/5ths speed is 80%, so kW will be 51%, plus whatever losses there are in the VFD.

Losses within the VFD are variable as well, based on output and also carrier frequency. But at the most efficient, meaning lowest carrier frequency, full speed, the losses can be estimated at 3%. Some mfrs claim 2%, but I think that's more of a marketing gimmick.

By Curt Wuollet on 5 August, 2016 - 11:52 pm
1 out of 1 members thought this post was helpful...

Lots of questions, Generally power out = power in minus 5 or 10 % to account for efficiency, but that's a really rough rule.

One thing I will say is that yes, you can measure the currents with a clamp meter, but not your average clamp meter. We keep around a Fluke meter that can handle high crest factors and give you a reasonable answer. Your average clamp meters give you readings whose only value is that they are obviously wrong, so you don't believe them. None of the current waveforms around most VFDs look like the sine waves.that most clamp meters are calibrated for.

The input is generally a bridge with a capacitive filter, it draws current in narrow spikes aligned with the peaks of the input voltage. The output voltage of most VFDs is something you have to see to believe. The PWM converters make vary high amplitude hash, the "sine" wave inverters really aren't much better. But this works because the motor is inductive and responds to the average current, But the current waveform, while much closer to a sine, has a lot of high frequency detail. The high crest factor basically means that the electronics can handle the high frequency components and weigh them into the total properly. I have seen a 50:1 relationship between the utility clamp meters for general use and the Fluke. By the way this isn't just a problem with VFDs. Almost every device now draws non-linear current and can fool the clamp ons that were fine not so many years ago. I think every facility should have at least one admittedly high buck current clamp with a specified high crest factor specifically to handle these jobs. The guys now know that if the readings look bogus, to check things with the Fluke. I'm sure there are others that have this capability, but Fluke is my go to when I need readings I can believe.

By the way, _their_ utiity grades get fooled too.

Hope this helps,

Regards
cww