Single Phase current draw for a 3 phase output VFD: technical discussion

[mivey]

Rules of thumb and such are great but as I noted before, one must be aware of the accompanying constraints. When those constraints are ignored then incorrect statements can occur.

For example, we typically say power in equals power out even though that is not always true. There are rules for the rules. Learn them lest ye be tripped up.

Bes and gar have stated the issues quite plainly but still some readers let that go over their heads. I don't get that.

 

[Ingenieur]

We are talking about a vfd and using 1 ph in / 3 ph out
Pin = Pout + losses
all mfgs say use 1.732 or 2
my model confirms this, post 172
as do simple calculations
we have made this overly complicated for the original query

for basic electrician instruction on selecting a vfd for use w/1 ph in and a 3 ph motor
what sizing factor should be used?
Perfect world 1.732 minimum or with losses use 2, next std size
verify input i < rated i

 

[Smart $]

170609-2434 EDT

jpeer:

If I have a black box device that is 100 % efficient (meaning no power loss in the black box), it produces 3 phase power to an unbalanced load, even asume the load is resistive, then what is the load current I use for you to apply the 1.732 factor to? And further does this even imply that the input current to the black box is sinusoidal?

You just can not arbitrarily apply equations and constants to a circuit without knowing details of the circuit.

.

gar

I get your point, but I think you aren't seeing the forest for the trees. If you have a black box that is 100% efficient, power in equals power out. The waveform being sinusoidal does not matter if we are using RMS values as opposed to instantaneous values... and if the box is 100% efficient, it does not have a power factor of its own.

The only power factor that could be exhibited on the input is the power factor of the output. However, the method and means of conversion actually compensate for the output power factor. It is not passed upstream to the input side.

 

[gar]

170610-2359 EDT

Smart $:

I am talking about power factor.

The power factor of the load on the output of the black box does not of necessity reflect back thru the black box to be the power factor at the input to the black box.

Consider a black box that consists of an input bridge rectifier connected to a large capacitor input filter with a converter load that generates a single phase sine wave output voltage to a resistive load. The ouput load has a power factor of 1. The black box input is not a power factor of 1, but something less. Yet assuming no losses in the black box the input power equals the output power.

.

 

[Smart $]

170610-2359 EDT

Smart $:

I am talking about power factor.

The power factor of the load on the output of the black box does not of necessity reflect back thru the black box to be the power factor at the input to the black box.

Consider a black box that consists of an input bridge rectifier connected to a large capacitor input filter with a converter load that generates a single phase sine wave output voltage to a resistive load. The ouput load has a power factor of 1. The black box input is not a power factor of 1, but something less. Yet assuming no losses in the black box the input power equals the output power.

.

The highlighted statements are a contradiction for a 100% efficient black box considering there are no reactive components in the input stage. If any of the reactiveness of the capacitor is leaked back to the input, then the black box is not 100% efficient. Keep in mind, when I say power in equals power out, I am talking real power (W), not apparent power (VA).

 

[Besoeker]

The highlighted statements are a contradiction for a 100% efficient black box considering there are no reactive components in the input stage.

There are non-linear components therefore distortion power factor.

 

[Smart $]

There are non-linear components therefore distortion power factor.

Yes, but now we are talking something that is less than 100% efficient.

In a "real box", what is the power factor and efficiency?

 

[gar]

170611-0854EDT

Smart $:

In post numbered #3 I showed a plot of the DC pulse current from a full wave bridge rectifier to a capacitor input filter with a long time constant. That means essentially a constant voltage across the capacitor.

Flip one of the two pulses shown and you have the AC input current. The power factor of a sine wave input voltage and the input current is not 1.

Put a simple resistive load on the output of the DC supply and it has a power factor of 1. Thus, input power factor does not equal load power factor.

However, input power minus the small losses in the power supply will equal load power.

Have you ever made scope, volt, current, and power measurements on a simple capacitor input DC power supply to see the results? If not, then try the experiment. Otherwise repeat the experiment and check your results.

.

 

[Besoeker]

Yes, but now we are talking something that is less than 100% efficient.

That's not really significant. A diode drop is in the order of 1V on load.For a full wave single-phase bridge two in series conduct at any one time.

In a "real box", what is the power factor and efficiency?[/QUOTE]
For efficiency, see above.
For power factor we are talking about distortion power factor, not displacement PF.
The distortion depends on the external component values. Both gar and I have submitted waverforms and they show just how non-linear the current is. I don't know how gar arrived at his. Mine were calculated from known component values. Whatever, they are remarkably similar in shape. But change a few values.......

In short, there is no "one size fits all" solution.

 

[mike_kilroy]

It appears the confusion is over real work producing input current vs. reactive input current to the vfd.

Everyone should agree that 1ph real input current is 1.73x higher than the real output 3ph current into the motor.

Everyone should agree that the output reactive current and power stay on the motor/bus of the vfd and do not effect the ac input to the vfd.

Everyone should agree the input rms current is true indication of diode heating and accurate representation of total input power (real and reactive).

Since we all say the vfd has .95 displacement PF, we sometimes get misled and assume the input PF is effectively 1, ignoring the distortion PF. I think this discussion would have gone much better if typical PFdist values were given for typical vfd inputs. This is typically .75 and often goes down to .30. THIS is why that true RMS current value is so much higher than expected sometimes. Remember ALL this extra rms current above 1.73 is from distortion PF.

Check out this CLEAR description with real world rms input values:

https://www.motioncontrolonline.org...ng-Power-Factor-and-Harmonics/content_id/1545

So those who say the input rms current to produce motor output power is 1.73x higher, they are right.

So those who say the input rms apparent current (total) is higher than 1.73x, they too can be right.

 

[Besoeker]

It appears the confusion is over real work producing input current vs. reactive input current to the vfd.

There is little reactive input current for a diode bridge.

Everyone should agree that 1ph real input current is 1.73x higher than the real output 3ph current into the motor.

Components have to be rated for actual current, not just one component of it.

Everyone should agree the input rms current is true indication of diode heating

Not really. The current can be anything from a square wave to something very peaky as shown earlier. Diode dissipation from nearly level continuous current to peaky discontinuous is quite different.

Since we all say the vfd has .95 displacement PF, we sometimes get misled and assume the input PF is effectively 1, ignoring the distortion PF. I think this discussion would have gone much better if typical PFdist values were given for typical vfd inputs. This is typically .75 and often goes down to .30.

There's the crunch and why you cannot specify a fixed multiplying factor.

 

[Ingenieur]

As background, for a given power (kW/hp) and voltage, the ratio of current for a single phase circuit will be 3 (1.732) times that of a three-phase circuit. This means that theinput rectifier will see 1.732 times the current of the output devices. When powered bythree-phase, these currents are nearly the same. This higher current would destroy theinput of the drive if an oversized inverter were not used. Furthermore, full-wave rectifiedsingle-phase power has a much higher harmonic content than full-wave rectified three-phase power. This would introduce large ripple into the DC bus of the inverter,potentially causing other malfunctions. Larger size inverters have larger bus capacitors,thus more inherent filtering. So upsizing the drive ameliorates the ripple problem as well.

The rule of thumb Hitachi recommends is to start with the 3-phase motor’s nameplate fullload amperage (FLA) rating and double it. Then select an inverter with this doubledcontinuous current rating. This will give adequate margin in the input rectifier bridge andbus capacitors to provide reliable performance. NOTE: Fusing or Circuit Breakers shouldbe sized to match the INVERTER input current rating, NOT the motor current rating!

http://www.hitachi-america.us/supportingdocs/forbus/inverters/Support/AN032404-1_Rev_A_Sizing_for_Single-Phase.pdf

 

[Besoeker]

When powered bythree-phase, these currents are nearly the same.

That's just not true.
With decent filtering, the Idc is approximately level. Distortion PF is about 0.958.
The last motor I dealt with (225kW 400V) had a PF of 0.7. The currents can't be nearly the same.

With minimal filtering the input PF could be as low as 0.3. The output power factor would still have been 0.7. Again, the currents can't be nearly the same.
Like I said, "there is no one size fits all" factor.

 

[Ingenieur]

That's just not true.
With decent filtering, the Idc is approximately level. Distortion PF is about 0.958.
The last motor I dealt with (225kW 400V) had a PF of 0.7. The currents can't be nearly the same.

With minimal filtering the input PF could be as low as 0.3. The output power factor would still have been 0.7. Again, the currents can't be nearly the same.
Like I said, "there is no one size fits all" factor.

you need to take that up with Hitachi's vfd design engineers (that is an excerpt from the doc)
I agree with them, though

Hitachi says next size up from a factor of 2

pf does not factor into S = sqrt3 x v x i

 

[Besoeker]

pf does not factor into S = sqrt3 x v x i

It factors into both input current and motor current.

 

[mivey]

you need to take that up with Hitachi's vfd design engineers (that is an excerpt from the doc)

I doubt there is much to "take up". Besoeker is a vfd design engineer. I would think two vfd design engineers could talk beyond a layman's term write-up and agree on more precise facts. Others however...

 

[Besoeker]

I doubt there is much to "take up". Besoeker is a vfd design engineer. I would think two vfd design engineers could talk beyond a layman's term write-up and agree on more precise facts. Others however...

Thanks mivey. The voice of reason.
There is a point though. I think it was one I made in response to a comment made by Jaref about ratings."Ask the manufacturer" was my response.
Hitachi are probably the best source for an answer about the performance of their products.

And I think I'm done and dusted with this topic. I've tried to explain but horse and water come to mind.
Thank you for your patience.

 

[Ingenieur]

I doubt there is much to "take up". Besoeker is a vfd design engineer. I would think two vfd design engineers could talk beyond a layman's term write-up and agree on more precise facts. Others however...

I disagree
I put my money on Hitachi's 40 years experience and team of PhD's
they would not use 1.732 as a basis and 2 (plus next std size up) as a recommendation if it was blowing drives up

I hardly qualify as a 'layman'
grad degree in EE with several courses on power electronics
working with drives since the mid 80's

 

[gar]

170611-1705 EDT

If I have a circuit consisting of a DC battery of constant voltage with a resistive load of constant resistance, then what is the power factor of that load.

Next add some inductance in series with the resistor. What is the power factor of the series resistor and inductor under steady state conditions?

.

 

[Ingenieur]

VFD Input Current > Motor Current Rating * 1.73
The VFD input current must be equal to or greater than the Motor Current Rating * 1.73
http://www.dartcontrols.com/wp-content/uploads/2012/01/De-Rating-VFD-for-Single-Phase-Power.pdf