So where does extra current go with bad distortion PF (Standard VFD input)?

[steve66]

A recent thread examined displacement PF and VFDs. It was shown this displacement PF stayed inside and on the output of the VFD so did not effect the AC input to the VFD.

So many say since the input displacement PF to the VFD is nearly 1, there is no extra input current, like out of phase, or 'imaginary.'

Others point out the input RMS current can be 2,3,4,5 times HIGHER that to produce power in the motor at output; yet since displacement PF is effectively 1.0, how can this be? So where does this extra rms current GO? It is caused by distortion PF, even though it is effectively in phase current.

Example: Our 460vac 3 ph input VFD outputs 460Vac 3ph into a fully loaded motor pulling 6 amps RMS in phase current (forget the out of phase aka magnetizing aka imaginary current flowing back and forth from VFD to motor). So assuming near 100% efficiency of the VFD, we expect to see same 6 amps RMS at VFD input terminals. But the input RMS current is measured at 12 amps!

Where does the extra 6 amps go?? Is it real? Imaginary? Power Factor?

Understanding THIS will go a long way to helping eliminate confusion and helping some folks grasp the whole picture.

Where does the extra 6 amps go??

I think you stated this wrong. The input to the VFD basically has a 1.0 PF, so it might be 6 amps. However the current to the motor will have a lower power factor (maybe 0.8), so the current to the motor will actually be higher (probably not 12 amps, but maybe 8 amps.)

So I think you are actually asking "Where did the extra 2 amps come from?".

The answer is that reactive power (and current) is not dissipated in heat like real power is. It constantly flows back and forth between the source (the VFD) and the load (the motor).

So when the motor is started or stopped, there may be a transient current surge on the VFD input to get the reactive current flowing (or stopped), but in a steady state condition, the output reactive current doesn't need an equal current on the input to keep it going.

 

[GoldDigger]

There may have been some confusion also with the peak input current rather than the RMS, and with the current from a single phase source when driving a three phase output. And finally, perhaps, the Code requirement to size the input circuit by the VFD rated current rather than by the actual load.
Regardless of any of those points of potential confusion the input power is greater than the output power only by the inefficiency of the conversion circuit.
Finally some VFDs allow a higher output voltage than input voltage using voltage multiplier circuits to produce a higher DC bus voltage. In that case equal power means a higher input current than output current by the voltage conversion ratio.

Sent from my XT1585 using Tapatalk

 

[Smart $]

100% efficiency means there is no losses doesn't it?
...

Correct. 100% Efficiency rating is also impossible in the real world... but as you say, not by much.

 

[Ingenieur]

100% efficiency means there is no losses doesn't it?



Have we developed conductors with no resistance? AFAIK there is always some loss, we just have found some ways to keep them pretty minimal.

We are talking about the vfd losses
if we assume vfd to be 100% eff it has no losses
in reality on the order of 2%
for the purpose of this discussion it has none

 

[Ingenieur]

It doesn't go any where.

if it 'goes' nowhere it isn't moving, it is not flowing, ie static, therefore not current since current is charge/time, a sort of 'flow rate or Q' of charge

 

[Ingenieur]

I think you stated this wrong. The input to the VFD basically has a 1.0 PF, so it might be 6 amps. However the current to the motor will have a lower power factor (maybe 0.8), so the current to the motor will actually be higher (probably not 12 amps, but maybe 8 amps.)

So I think you are actually asking "Where did the extra 2 amps come from?".

The answer is that reactive power (and current) is not dissipated in heat like real power is. It constantly flows back and forth between the source (the VFD) and the load (the motor).

So when the motor is started or stopped, there may be a transient current surge on the VFD input to get the reactive current flowing (or stopped), but in a steady state condition, the output reactive current doesn't need an equal current on the input to keep it going.

This makes sense

 

[Smart $]

I think you stated this wrong. The input to the VFD basically has a 1.0 PF, so it might be 6 amps. However the current to the motor will have a lower power factor (maybe 0.8), so the current to the motor will actually be higher (probably not 12 amps, but maybe 8 amps.)

So I think you are actually asking "Where did the extra 2 amps come from?".

The answer is that reactive power (and current) is not dissipated in heat like real power is. It constantly flows back and forth between the source (the VFD) and the load (the motor).

So when the motor is started or stopped, there may be a transient current surge on the VFD input to get the reactive current flowing (or stopped), but in a steady state condition, the output reactive current doesn't need an equal current on the input to keep it going.

Depends on whether the power factor is displacement only or displacement and distortion.

https://mobile.yaskawa.com/delegate...D.10&cmd=documents&documentName=WP.AFD.10.pdf

 

[Ingenieur]

Depends on whether the power factor is displacement only or displacement and distortion.

https://mobile.yaskawa.com/delegate...D.10&cmd=documents&documentName=WP.AFD.10.pdf

good paper
it was posted in the other thread
but in both cases in the chart the line side is lower than the vfd output
the one vfd low load where util is 1-3 % (vs 25-35% for util) seems low?
motor draws the same low load 25-35% in both cases?

 

[Besoeker]

Ah, but it does.

Where?

 

[Smart $]

good paper
it was posted in the other thread
but in both cases in the chart the line side is lower than the vfd output

Yes... but the VFD has a dc link choke which is added to the VFD design intentionally to improve power factor. Not all VFD's are designed with power factor improvement in mind.

the one vfd low load where util is 1-3 % (vs 25-35% for util) seems low?
motor draws the same low load 25-35% in both cases?

Mostly reactive current? 0.2 pf is practically as low you can go.

VFD input won't see that, which is why it is 1-3%.

 

[Ingenieur]

Yes... but the VFD has a dc link choke which is added to the VFD design intentionally to improve power factor. Not all VFD's are designed with power factor improvement in mind.



Mostly reactive current? 0.2 pf is practically as low you can go.

VFD input won't see that, which is why it is 1-3%.

in the op's example motor is fully losded
so from the chart input is 95% vs motor 100%

to get input 2 times output, ie 0.5 pf tdh must be 150%, unrealistically high
assume util pf 0.9

Adjusted pf = 0.9 / (sqrt(1 + tdh))
tdh = 1.5 or 150%

 

[Smart $]

in the op's example motor is fully losded
so from the chart input is 95% vs motor 100%

to get input 2 times output, ie 0.5 pf tdh must be 150%, unrealistically high
assume util pf 0.9

Adjusted pf = 0.9 / (sqrt(1 + tdh))
tdh = 1.5 or 150%

Well, I agreed with you about the plausibility from the get go. :happyyes:

 

[GeorgeB]

Example: Our 460vac 3 ph input VFD outputs 460Vac 3ph into a fully loaded motor pulling 6 amps RMS in phase current (forget the out of phase aka magnetizing aka imaginary current flowing back and forth from VFD to motor). So assuming near 100% efficiency of the VFD, we expect to see same 6 amps RMS at VFD input terminals. But the input RMS current is measured at 12 amps!

My first guess would be that you have mislabeled input and output. My second, if that is not true, is that you have single-phase input; have you checked input current in all 3 legs?

 

[Ingenieur]

Well, I agreed with you about the plausibility from the get go. :happyyes:

:lol:

 

[Ingenieur]

My first guess would be that you have mislabeled input and output. My second, if that is not true, is that you have single-phase input; have you checked input current in all 3 legs?

1 ph to 3 ph was my thought too (also lower v in than out)
but he clearly stipulates 3 ph/460 in/out

 

[Besoeker]

if it 'goes' nowhere it isn't moving, it is not flowing, ie static, therefore not current since current is charge/time, a sort of 'flow rate or Q' of charge

So where does ir go?

 

[Ingenieur]

So where does ir go?

If we accept it as reactive it oscillates between L and C elements gradually dissapating and needing replacement if R elements exist in the path

I do not believe the op's scenario plausible as described

 

[Besoeker]

If we accept it as reactive it oscillates between L and C elements gradually dissapating and needing replacement if R elements exist in the path

I do not believe the op's scenario plausible as described

Which L and wich C.
?

 

[Ingenieur]

Which L and wich C.
?

I asked
he said 'irrelevant'

what happens if the 'motor' is over-driven, ie, generating?

 

[Besoeker]

I asked
he said 'irrelevant'

what happens if the 'motor' is over-driven, ie, generating?

To sustatin that you'd need the input bridge to be regererative. I don't think that is being discussed here.