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

[Besoeker]

Labels R as in Red, B as in Blue and G as in Green.
No reference to "ground" except perhaps implied in the fact that the three AC lines are the voltages with respect to some common fixed point.

And when the voltages plotted are line to line voltages, even that implicit reference is gone.

Exactly that. With respect to each other. Line to line as you say. No common ground, fixed point, neutral, or reference required, inferred or implied.
It would have been red, yellow (not green) and blue but yellow didn't show up too well so I picked another colour. Just so happened.........water under the bridge.......no pun intended.

 

[GoldDigger]

... Just so happened.........water under the bridge.......no pun intended.

That's alright. None taken.



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[Besoeker]

That's alright. None taken.

Rectified then.............

 

[gar]

170522-0907 EDT

Smart $:

G is just one of the three input lines along with R and B. The letters correspond with the colors of the plotted curves. Others have pointed this out.

For convenience I would pick -DC output as the reference point, but note the AC input line that connects to -DC changes three times per cycle, and corresponds to the plotted waveform that is most negative.

If you go to 12 diodes, 6 transformers, and appropriately connected primaries, then you may need virtually no output capacitor, and ripple will be small. About 1/2 of the 6 diode bridge.

.

 

[Besoeker]

170522-0907 EDT

Smart $:

G is just one of the three input lines along with R and B. The letters correspond with the colors of the plotted curves. Others have pointed this out.

For convenience I would pick -DC output as the reference point, but note the AC input line that connects to -DC changes three times per cycle, and corresponds to the plotted waveform that is most negative.

If you go to 12 diodes, 6 transformers, and appropriately connected primaries, then you may need virtually no output capacitor, and ripple will be small. About 1/2 of the 6 diode bridge.

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This is something we have quite often done but with one transformer. Delta primary and two secondaries, one star (wye), the other being delta. Ddyn11. So called 12-pulse.
It was for supply harmonic mitigation the lowest being 11th. But yes, it does reduce the ripple also.

 

[junkhound]

'And as an exercise for the student' -
- take a look at the use of 126 transformers used for some megawatt high power high accuracy radar systems (where final DC ripple is only a few MICROvolts after added regulators) and the delta taps needed to feed the diodes.

 

[mivey]

It would appear all got happy with the diode refresher.

Am I to assume all got happy with the diode current not necessarily having a sqrt(3) factor for the single-phase input case?

 

[mike_kilroy]

It would appear all got happy with the diode refresher.

Am I to assume all got happy with the diode current not necessarily having a sqrt(3) factor for the single-phase input case?

Nope

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[topgone]

It would appear all got happy with the diode refresher.

Am I to assume all got happy with the diode current not necessarily having a sqrt(3) factor for the single-phase input case?

The input current depends on the load, hence not linear. Using 1.732 assumes a linear current draw. Does it approximate the current draw in that situation? Perhaps, but not always.

 

[mike_kilroy]

The input current depends on the load, hence not linear. Using 1.732 assumes a linear current draw. Does it approximate the current draw in that situation? Perhaps, but not always.

We will have to agree to disagree as this too is wrong.

Irms is Irms and Irms single phase is exactly sq rt 3 more than Irms three phase.

Just because one wants to discuss unlinear heating in a diode junction, it ain't gonna change physics.

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[GoldDigger]

We will have to agree to disagree as this too is wrong.

Irms is Irms and Irms single phase is exactly sq rt 3 more than Irms three phase.

Just because one wants to discuss unlinear heating in a diode junction, it ain't gonna change physics.

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Implicit in that statement, I think, is the assumption that Vrms single phase and Vrms three phase have a corresponding sqrt(3) relationship.

When charging a DC capacitor bank through the diode bridge and without inductors in the circuit to smooth out current flow that may not be a valid assumption.

And having a 120/240 single phase supply available to drive the VFD may not compare directly with having a 208Y/120 or a 240V delta available.

 

[Smart $]

...
When charging a DC capacitor bank through the diode bridge and without inductors in the circuit to smooth out current flow that may not be a valid assumption.
...

It seems everyone wants to make inconclusive statements for some reason...

While your assessment is accurate, it does not encompass the load as a whole. Technically, the motor is the load, not the VFD. The motor is a linear load. Other than losses, the current that goes through the VFD also goes through the motor. Modern VFD's are highly efficent. Losses are minimal if not outright insignificant. The problem I see with saying 1Ø VFD current in equals 1.732 × 3Ø VFD Irms out (in general) is that the VFD handles reactive current locally.

 

[mike_kilroy]

It seems everyone wants to make inconclusive statements for some reason...

While your assessment is accurate, it does not encompass the load as a whole. Technically, the motor is the load, not the VFD. The motor is a linear load. Other than losses, the current that goes through the VFD also goes through the motor. Modern VFD's are highly efficent. Losses are minimal if not outright insignificant. The problem I see with saying 1Ø VFD current in equals 1.732 × 3Ø VFD Irms out (in general) is that the VFD handles reactive current locally.

It is assumed we are talking about the real torque producing current, not the magnetizing current at all.

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[gar]

170526-2129 EDT

mike_kilroy:

Irms is Irms and Irms single phase is exactly sq rt 3 more than Irms three phase.

For what circuit, and assumptions? As a general statement and out of context it really means absolutely nothing. I have no idea what your sentence means. It is just some words stuck together.

What does "Irms is Irms" mean?

What is the relationship "Irms single phase is exactly sq rt 3 more than Irms three phase"? What assumptions, and where are these currents measured and what are the loads?

Something has to be defined or there is no possibility of your sentence meaning anything.

.

 

[GoldDigger]

Since the VFD is acting as a switching power supply, I am not completely sure that you can even say that Imotor must equal Iin even for a simple three phase VFD (some current could flow through snubbing diodes or equivalent rather than being sourced entirely from the DC bus.) I will have to think some more about that part.

I am not yet convinced that you can assert that for three phase input to VFD and a three phase motor the input current to the VFD (with a bunch of ways to potentially define that) will have the same ratio to the current for a three phase motor as the ratio for a single phase source and a single phase motor.

If the VFD were a black box that obeyed the rule you proposed, and you compare the three phase motor to the single phase motor, then I am leaning toward accepting the sqrt(3).

 

[Smart $]

It is assumed we are talking about the real torque producing current, not the magnetizing current at all.

The reactive component of the motor current just goes back and forth between VFD and motor. It's included in measured Irms out. We do not exclude the reactive component until we do power calculations V×(I×pf)×eff. The VFD input only has to provide the reactive component of the motor current at startup (assuming full load).

 

[gar]

170426-2341 EDT

If we assume the VFD contains a very large DC bus capacitor, or it could be a battery, then, if the parts of the VFD black box that are the circuit that produces the AC to the output motor, the output motor, and the mechanical load on the motor are constant, then for constant DC bus voltage the average pulsed DC input current to the capacitor must equal the average DC output current.

This does not imply what the specific AC RMS current input to the VFD will be for different sources. But total average current from the source or sources has to equal that required into the capacitor.

For a constant Iave the ratio of Irms to Iave will be a function of the waveforms.

.

 

[Smart $]

170426-2341 EDT

If we assume the VFD contains a very large DC bus capacitor, or it could be a battery, then, if the parts of the VFD black box that are the circuit that produces the AC to the output motor, the output motor, and the mechanical load on the motor are constant, then for constant DC bus voltage the average pulsed DC input current to the capacitor must equal the average DC output current.

This does not imply what the specific AC RMS current input to the VFD will be for different sources. But total average current from the source or sources has to equal that required into the capacitor.

For a constant Iave the ratio of Irms to Iave will be a function of the waveforms.

.

Well evaluating each stage will get us there, so consider the broader scope of the motor being a linear load coupled with a fairly-efficient electronic interface—say 95%. Only the interface can introduce a non-linear waveform on the AC source. Just how much impact can 5% of the input current make on linearity?

 

[gar]

170527-0812 EDT

Smart $:

The difference in the input to the VFD between single phase input and three phase input is solely a function of the circuitry between the AC input and the DC bus filter capacitor for a constant motor load.

Everything beyond the filter capacitor remains unchanged for a constant motor load if the filter capacitor voltage is held constant. Ideally we can expect that connecting to an AC waveform of the same peak to peak voltage will produce the same DC bus voltage independent of whether the input is three or single phase.

What the input current waveform is for three or single phase will be dependent upon that circuitry.

.

 

[Besoeker]

It seems everyone wants to make inconclusive statements for some reason...

Because you can't without knowing component values. There is no fixed fudge factor.