Larger VFD/Motors

[Carl FarringtonParticipant @carlfarrington53722]

It's probably not a bad idea to adhere to what the datasheet of your VFD says, is it. If you can anyway.

If I have to run a VFD off of a 13 amp plug or something, I'll probably buy a smaller motor than 2.2kW.

 

There's a 41 page discussion on the subject here: https://forums.mikeholt.com/showthread.php?t=184232

 

They can't agree a number, but at a glance the consensus seems to be that 1ph to 3ph VFD will put a 1.5 – 2x current increase on the 1ph input side vs the 3ph output side.

Edited By Carl Farrington on 06/05/2019 16:29:38

[Anonymous]

Posted by SillyOldDuffer on 06/05/2019 13:45:45:

I waffled about this in another recent thread. Briefly, the maximum currents quoted in the specs are peak, and they mainly occur during start-up.

……………….

I am not an expert! If a professional disagrees with any of the above listen to him.

I couldn't possibly comment on the last statement.

The first statement is correct but for the wrong reason. Agreed that when starting a motor direct on line there is a brief time when the motor currents are much higher than the motor rated current. This is while the motor speeds up and starts producing a back emf that opposes the applied voltage. When running via a VFD the start up currents will not be large. The VFD is programmed to increase the applied voltage over time, keeping the currents to a sensible value during start up. I would expect the basic efficiency of the output stage of a VFD to be on the order of 85-95%. So why the large input currents? Two basic reasons, one the appalling power factor as illustrated by Ian Parkin. Two, most small, and cheaper, VFDs do not have a power factor corrector as a first stage, but simply a rectifier. Once the DC link capacitors are charged the input current is anything but sinsoidal. Current only flows near the peak of the voltage waveform when the input voltage exceeds that of the DC link. So the current is taken in short bursts twice a cycle. During each burst the input current is much higher than one might calculate from simple V x I requirements. The bursts of current are also bad as they create harmonics that are reflected back into the power network. To prevent these most VFD manuals recommend using an external power line filter.

Andrew.

[Martin KyteParticipant @martinkyte99762]

. . . . and most external power line filters will trip your RCCB's.

regards Martin

[SillyOldDufferModerator @sillyoldduffer]

Posted by Andrew Johnston on 06/05/2019 22:20:25:

Posted by SillyOldDuffer on 06/05/2019 13:45:45:

I waffled about this in another recent thread. Briefly, the maximum currents quoted in the specs are peak, and they mainly occur during start-up.

……………….

I am not an expert! If a professional disagrees with any of the above listen to him.

I couldn't possibly comment on the last statement.

The first statement is correct but for the wrong reason.

…

Andrew.

Thanks for that Andrew. I find it hard to get my head round what the effects of starting and loading a 3-phase motor powered by a VFD might be on current flowing on the input supply. On average – I think! – a motor working at a steady rate is a good fit to this sort of analysis:

1. The motor draws the same average current on each artificial phase.
2. The current required by each phase is provided by the VFD. It does this by switching DC power stored in a capacitor such that each of three coils on the motor gets an approximate sine wave 120° out-of-phase with the others. But the DC supply simply 'sees' the total average load created by the sum of 3 averages.
3. The DC power supply is sized to deliver the total average current. The capacitors need only be big enough to store enough charge to fill the gaps left between rectified half-cycles, ie the time the single-phase mains isn't providing current. Again simple enough to derive the average current.
4. On the mains side, the supply 'sees' an average current that can be used to size the fuses.

If, god forbid, I was ordered to design such a set-up, I'd size everything relative to the power of the motor, working in watts rather than amps, and allowing for conversion inefficiencies. Thus a 1kW motor would get, guesstimated, three 400W switches and a 1.4kW DC power supply. The currents involved can then be calculated from whatever voltages are involved. For example, a 1.4kW supply on a 240V main would draw about 6A. I'd fit a 10A equipment fuse, and wire the whole to a 13A plug.

Trouble with my semi-educated approach is everything revolves around assumptions about averages. I don't believe this to be completely unreasonable, but the design is liable to come unstuck because it doesn't consider current spikes. These require deeper consideration of when current flows and why. Time matters in addition to volts and amps and this stretches my imagination and maths to breaking point. Examples of non-average current bursts include:

1. Motor and other start-up surges
2. Capacitor charging and discharging
3. Harmonics and switching transients caused by the switches
4. Harmonics and switching transients caused by the power supply diodes

I have severe difficulty imagining the effect of these non-sinusoidal demands on current flowing in a sinusoidal single-phase supply. In the same way I sort of understand the effect on power factor of an inductive load like a motor but I've no idea what putting a DC power supply between motor and mains does.

Now I'm thinking of experimenting with an oscilloscope in hope of capturing the current waveform when my VFD powered lathe is first switched on. If I put small resistor (20mm of Nichrome wire) in the Live, I can measure the voltage on both sides with a 2-channel oscilloscope. Never used it in anger, but my scope has a 'Math' function that displays the voltage difference between 2 channels, which I think reveals current. If so, the oscilloscope might catch what happens to mains current over several 50Hz cycles.

Might not be practical. The filtering on my lathe is basic and transients may mask the effect. This is assuming I don't make any accidental smoke wiring it up…

Dave

[Anonymous]

Posted by SillyOldDuffer on 07/05/2019 10:36:40:

I find it hard to get my head round what the effects of starting and loading a 3-phase motor powered by a VFD might be on current flowing on the input supply. On average – I think! – a motor working at a steady rate is a good fit to this sort of analysis:

The problem is that one still needs to account for phase angles, even with averages. So ideally one needs to use complex numbers, or as I prefer to call them two dimensional numbers.

If you're going to do some measurements, and I'd be interested to see the results, you'd be better off (and safer) using a current transformer, like these:

**LINK**

The datasheets are worse than useless and the frequency range is a bit limited. But they'll give an indication of the current waveforms, and it's a lot safer than playing with high side current sense resistors. You may want to check the maximum input voltage wrt earth before you go connecting to the mains.

Andrew

[Robert Atkinson 2Participant @robertatkinson2]

Another reason for the input current of 23.2A specified for the Eaton DE1 2.2kW drive linked to by Carl **LINK** is the overload rating and input voltage range. It says 2.2kW output so about 95% efficiency gives 2.31kW. Minimum input voltage is 200V 2.31 /200 = 11.55A. The overload rating is 200% for 1.875s 2 x 11.55 = 23.1A no missing power or incorrect data

If running at nominal 230 V this becomes 10.05A full load and 20.1 at 200% overload. Thus a "13A" outlet would be adequate.

Power factor is less of an issue with the better modern drives as the input may have PF correction.

Robert G8RPI.

[Carl FarringtonParticipant @carlfarrington53722]

Posted by Robert Atkinson 2 on 07/05/2019 12:51:51:

Another reason for the input current of 23.2A specified for the Eaton DE1 2.2kW drive linked to by Carl **LINK** is the overload rating and input voltage range. It says 2.2kW output so about 95% efficiency gives 2.31kW. Minimum input voltage is 200V 2.31 /200 = 11.55A. The overload rating is 200% for 1.875s 2 x 11.55 = 23.1A no missing power or incorrect data

If running at nominal 230 V this becomes 10.05A full load and 20.1 at 200% overload. Thus a "13A" outlet would be adequate.

Power factor is less of an issue with the better modern drives as the input may have PF correction.

Robert G8RPI.

I'm not sure it's the 200% overload. I think it's the single phase to 3 phase conversion.

Have a look at the number on this one instead: **LINK**

or this one: **LINK**

or this one: **LINK**

[Robert Atkinson 2Participant @robertatkinson2]

Posted by Carl Farrington on 07/05/2019 12:59:09:

Posted by Robert Atkinson 2 on 07/05/2019 12:51:51:

Another reason for the input current of 23.2A specified for the Eaton DE1 2.2kW drive linked to by Carl **LINK** is the overload rating and input voltage range. It says 2.2kW output so about 95% efficiency gives 2.31kW. Minimum input voltage is 200V 2.31 /200 = 11.55A. The overload rating is 200% for 1.875s 2 x 11.55 = 23.1A no missing power or incorrect data

If running at nominal 230 V this becomes 10.05A full load and 20.1 at 200% overload. Thus a "13A" outlet would be adequate.

Power factor is less of an issue with the better modern drives as the input may have PF correction.

Robert G8RPI.

I'm not sure it's the 200% overload. I think it's the single phase to 3 phase conversion.

Have a look at the number on this one instead: **LINK**

or this one: **LINK**

or this one: **LINK**

Well your first example makes no sense at all with "Input Current: 19.6A (Single Phase), 13.3A (Three Phase)" so at a nominal 230V the input power is 4.5kW on single phase and 9.2kW on 3 phase! At 200V it's 3.9kW 1Ph & 8kW 3Ph. (single phase nearly matches the overload figure).

At least the Eaton figures make sense. Without qualifying what load and input voltage the input current apples to it's impossible to be sure what it means. For practical purposes the motor rating (in kW) / 200 is a good estimate for full load input on single phase. so for 2.2kW you are looking at about 11A so a "13A" outlet should be OK unless you are overloading the motor.(the 200 is based on 230V, 0.95 power factor, 90% efficiency and rounded up).

Robert G8RPI.

[Carl FarringtonParticipant @carlfarrington53722]

Posted by Robert Atkinson 2 on 08/05/2019 07:49:43:

Posted by Carl Farrington on 07/05/2019 12:59:09:

Posted by Robert Atkinson 2 on 07/05/2019 12:51:51:

Another reason for the input current of 23.2A specified for the Eaton DE1 2.2kW drive linked to by Carl **LINK** is the overload rating and input voltage range. It says 2.2kW output so about 95% efficiency gives 2.31kW. Minimum input voltage is 200V 2.31 /200 = 11.55A. The overload rating is 200% for 1.875s 2 x 11.55 = 23.1A no missing power or incorrect data

If running at nominal 230 V this becomes 10.05A full load and 20.1 at 200% overload. Thus a "13A" outlet would be adequate.

Power factor is less of an issue with the better modern drives as the input may have PF correction.

Robert G8RPI.

I'm not sure it's the 200% overload. I think it's the single phase to 3 phase conversion.

Have a look at the number on this one instead: **LINK**

or this one: **LINK**

or this one: **LINK**

Well your first example makes no sense at all with "Input Current: 19.6A (Single Phase), 13.3A (Three Phase)" so at a nominal 230V the input power is 4.5kW on single phase and 9.2kW on 3 phase! At 200V it's 3.9kW 1Ph & 8kW 3Ph. (single phase nearly matches the overload figure).

At least the Eaton figures make sense. Without qualifying what load and input voltage the input current apples to it's impossible to be sure what it means. For practical purposes the motor rating (in kW) / 200 is a good estimate for full load input on single phase. so for 2.2kW you are looking at about 11A so a "13A" outlet should be OK unless you are overloading the motor.(the 200 is based on 230V, 0.95 power factor, 90% efficiency and rounded up).

Robert G8RPI.

How do you arrive at those ~9kW figures ?

All those examples I linked to are 230v input, and you have a choice of feeding them 230V 3 phase or 230V single phase. The point being, by looking at those examples rather than just single-phase examples such as the Eaton and trying to guess whether they're accounting for overload current or something else (e.g. DOL surge which was previously suggested by someone else but is not relevant for VFDs), it makes it very clear that single phase input increases the current requirements dramatically, because they all give both single phase and 3 phase input options, for the same voltage, and the single phase is always in the region of 50% higher than 3 phase input current, and certainly usually around, or greater than twice the output 3 phase motor nameplate rating (which is obviously a combination of efficiency losses and what not, but also the single phase to 3 phase conversion that nobody so far seems to accept even exists. Did anybody look at that 51 page discussion on the electrical forum I linked to earlier ?)

In the example that you queried above (the one you said makes no sense), have a look at page 167 of the installation manual here: **LINK**

I'm not suggesting this really matters too much in the real world, for you and I in our garages. I'm just after clear understanding.

Edited By Carl Farrington on 08/05/2019 08:46:03

[Robert Atkinson 2Participant @robertatkinson2]

Hi,

230V x 13.3A x 3ph = 9.177kW. I assume that the 13.3A they quote is for all three phases added together (which would give a more reasonable 3kW) but this is NOT the normal convention which is to specify the current for 1 phase and assumes the 3 are equal.
1 to 3 phase conversion is of course possible. As virtually all the modern drives convert the input AC to DC before converting back to AC it doesn't matter what the input phase is. In fact as long as you down-rate the power so the input rectifiers is not overloaded and the assuming the DC bus capacitor is up to it you can jus stick single phase into most drives and they will run fine. If like most, it has a connection to the DC bus available you could run it off about 20 car batteries and a bunch of solar cells. Or more realistically use an external single phase rectified and capacitor to run a 3 phase unit at full rating. Obviously if the VFD has an input transformer or AC fan this approach won't work.

Robert G8RPI

[Carl FarringtonParticipant @carlfarrington53722]

Posted by Robert Atkinson 2 on 08/05/2019 14:45:35:

Hi,

230V x 13.3A x 3ph = 9.177kW. I assume that the 13.3A they quote is for all three phases added together (which would give a more reasonable 3kW) but this is NOT the normal convention which is to specify the current for 1 phase and assumes the 3 are equal.
1 to 3 phase conversion is of course possible. As virtually all the modern drives convert the input AC to DC before converting back to AC it doesn't matter what the input phase is. In fact as long as you down-rate the power so the input rectifiers is not overloaded and the assuming the DC bus capacitor is up to it you can jus stick single phase into most drives and they will run fine. If like most, it has a connection to the DC bus available you could run it off about 20 car batteries and a bunch of solar cells. Or more realistically use an external single phase rectified and capacitor to run a 3 phase unit at full rating. Obviously if the VFD has an input transformer or AC fan this approach won't work.

Robert G8RPI

I don't think you just multiply by 3 like that. Pretty sure that's not how it works at all.

[Carl FarringtonParticipant @carlfarrington53722]

You're supposed to multiply by the square root of 3 (1.732), for starters.

**LINK**

[Bill DawesParticipant @billdawes]

Sorry if this has already been answered in the voluminous calcs. but I think a possible issue is not just fuse rating but the fuse type. Motors driving a high starting torque machine need a motor rated fuse. If you look at the time/current graph of a domestic fuse to a motor rated one you will see that the motor rated one blows after a much longer period, it can be milliseconds v several seconds.

I know this from my job in industrial fans where acceleration times can be many seconds, a non motor rated fuse would be useless.

Having said that for a normal domestic situation, even a machine tool of a kw or so, I would not have expected a domestic fuse to be an issue.

Bill D

[Alan VosParticipant @alanvos39612]

It isn't just motors with brief high initial currents. Last year I used a current clamp and an oscilloscope to assess the inrush current of a decent quality 19V 65W switch mode power supply (Intel NUC). Nominal current, say 1/3 amp. Peak observed, nearly 90 amps. But only for about 250 microseconds.

[Robert Atkinson 2Participant @robertatkinson2]

Posted by Carl Farrington on 08/05/2019 14:48:33:

Posted by Robert Atkinson 2 on 08/05/2019 14:45:35:

Hi,

230V x 13.3A x 3ph = 9.177kW. I assume that the 13.3A they quote is for all three phases added together (which would give a more reasonable 3kW) but this is NOT the normal convention which is to specify the current for 1 phase and assumes the 3 are equal.
<SNIP>

Robert G8RPI

I don't think you just multiply by 3 like that. Pretty sure that's not how it works at all.

Yes you do, there are 3 phases, they should have quoted the per phase current. see next post for root 3.

Robert.

[Carl FarringtonParticipant @carlfarrington53722]

Posted by Robert Atkinson 2 on 08/05/2019 19:38:26:

Posted by Carl Farrington on 08/05/2019 14:48:33:

Posted by Robert Atkinson 2 on 08/05/2019 14:45:35:

Hi,

230V x 13.3A x 3ph = 9.177kW. I assume that the 13.3A they quote is for all three phases added together (which would give a more reasonable 3kW) but this is NOT the normal convention which is to specify the current for 1 phase and assumes the 3 are equal.
<SNIP>

Robert G8RPI

I don't think you just multiply by 3 like that. Pretty sure that's not how it works at all.

Yes you do, there are 3 phases, they should have quoted the per phase current. see next post for root 3.

Robert.

I suppose that would make sense, if that's how they usually specify 3 phase current..

[Robert Atkinson 2Participant @robertatkinson2]

Posted by Carl Farrington on 08/05/2019 14:55:52:

You're supposed to multiply by the square root of 3 (1.732), for starters.

**LINK**

 

No you don't in this case. It is all referenced to 240V Phase to Neutral voltage this includes the root 3 factor from the Phase to Phase voltage (415V). If I multiplied by 1.72 again the power would go even higher.

I do know about these things, it's part of my day job designing aircraft electrical systems. I'm currently working on a system that is the inverse of a VFD. It takes variable frequency from a engine driven generator and produces fixed 400Hz at more than 10kW.

Robert G8RPI.

Edited By Robert Atkinson 2 on 08/05/2019 19:55:33

[Robert Atkinson 2Participant @robertatkinson2]

Posted by Alan Vos on 08/05/2019 18:14:54:

It isn't just motors with brief high initial currents. Last year I used a current clamp and an oscilloscope to assess the inrush current of a decent quality 19V 65W switch mode power supply (Intel NUC). Nominal current, say 1/3 amp. Peak observed, nearly 90 amps. But only for about 250 microseconds.

Indeed, I've seen similar issues. In your case it was almost certainly EMI fliter capacitors charging. The main supply capacitor charging would be at least 10s of milliseconds. It's the main supply capacitor charging that drives the need for slower circuits breakers, typically type C.

Robert G8RPI.

[Carl FarringtonParticipant @carlfarrington53722]

Posted by Robert Atkinson 2 on 08/05/2019 19:48:36:

Posted by Carl Farrington on 08/05/2019 14:55:52:

You're supposed to multiply by the square root of 3 (1.732), for starters.

**LINK**

No you don't in this case. it is all referenced to 240V Phase to Neutral voltage this includes the root 3 factor from the Phase to Phase voltage (415V). If I multiplied by 1.72 again the power would go even higher.

I do know about these things, it's part of my day job designing aircraft electrical systems. I'm currently workin on a sytem that is the inverse of a VFD. It takes variable frequency from a engine driven generator and produces fixed 400Hz at more than 10kW.

Robert G8RPI.

I think I finally get it..

If the 3 phase current requirements on the datasheets are per phase like you said, then that makes sense, although it puts it quite high for the 3 phase input on most of them.

I must admit, I'm still a bit confused, but it's been a long day and I've just had a beer.

If 3 phase ratings are always given per phase, then that means our 2.2kW motors which say something like 8A @ 230v at full load, well is that 8A x 230V x 3 ? or 8A x 230V x 1.732 ?

[Robert Atkinson 2Participant @robertatkinson2]

Posted by Carl Farrington on 08/05/2019 19:55:11:

Posted by Robert Atkinson 2 on 08/05/2019 19:48:36:

Posted by Carl Farrington on 08/05/2019 14:55:52:

You're supposed to multiply by the square root of 3 (1.732), for starters.

**LINK**

 

No you don't in this case. it is all referenced to 240V Phase to Neutral voltage this includes the root 3 factor from the Phase to Phase voltage (415V). If I multiplied by 1.72 again the power would go even higher.

I do know about these things, it's part of my day job designing aircraft electrical systems. I'm currently workin on a sytem that is the inverse of a VFD. It takes variable frequency from a engine driven generator and produces fixed 400Hz at more than 10kW.

Robert G8RPI.

I think I finally get it..

If the 3 phase current requirements on the datasheets are per phase like you said, then that makes sense, although it puts it quite high for the 3 phase input on most of them.

 

I must admit, I'm still a bit confused, but it's been a long day and I've just had a beer.

If 3 phase ratings are always given per phase, then that means our 2.2kW motors which say something like 8A @ 230v at full load, well is that 8A x 230V x 3 ? or 8A x 230V x 1.732 ?

HI,

8A x 230V x 3

= 5.5kW seems high but they are taking the worst case of high motor load and low input voltage (230/240V is nominal voltage). If the voltage was specified as 398V (415 for 240V phase to phase) then you would multiply by root 3 instead of 3. This gives the same answer 5.5kW

Don't worry about it I keep having to explain this to supposed qualified electronics engineers.

Robert G8RPI.

Edited By Robert Atkinson 2 on 08/05/2019 20:06:04

[Simon Williams 3Participant @simonwilliams3]

About this root 3 business, it depends on whether the line voltage (230V) is a line to line or a line to neutral figure.

To calculate the power drawn by a three phase motor load supplied with three phase 400 volts (line to line inferred) you need the root three factor. If the 230 volts figure is the same – line to line – then it follows that the root three factor is still needed.

If the supply is specified as 230 volts line to neutral then the root three factor has already been applied by dint of getting from 400 volts to 230 volts.

In each case the convention is that the line current in one phase is specified, and the other two phases carry nominally equal currents.

The confusion arises because while it is obvious that a single phase input/three phase output VSD is supplied with 230 volts line to neutral, to develop full power at the motor it generates three phase 230 volts line to line, not line to neutral. The motor rating plate assumes that the delta connected motor will be supplied with three phase 230 volts, just as the star connected motor requires 400 volts line to line to develop full power.

Of course the arithmetic changes with phase coverters that can generate 400 volts line to line three phase from 230 volts line to neutral single phase input.

HTH Simon

[Carl FarringtonParticipant @carlfarrington53722]

It does say 'of each phase' in that 3 phase watts law thing I came across before doesn't it.. http://c03.apogee.net/contentplayer/?coursetype=foe&utilityid=mp&id=4576

I'm just going to wire the thing up with the 25A breaker and >25A switch and have done with it

Edited By Carl Farrington on 08/05/2019 20:13:45

[Carl FarringtonParticipant @carlfarrington53722]

Posted by Robert Atkinson 2 on 08/05/2019 20:05:22:

Posted by Carl Farrington on 08/05/2019 19:55:11:

I must admit, I'm still a bit confused, but it's been a long day and I've just had a beer.

If 3 phase ratings are always given per phase, then that means our 2.2kW motors which say something like 8A @ 230v at full load, well is that 8A x 230V x 3 ? or 8A x 230V x 1.732 ?

HI,

8A x 230V x 3

= 5.5kW seems high but they are taking the worst case of high motor load and low input voltage (230/240V is nominal voltage). If the voltage was specified as 398V (415 for 240V phase to phase) then you would multiply by root 3 instead of 3. This gives the same answer 5.5kW

Don't worry about it I keep having to explain this to supposed qualified electronics engineers.

Robert G8RPI.

It does seem weird that a motor rated at 2.2kW (and its VFD) would pull over double that. Seems very high.

[Robert Atkinson 2Participant @robertatkinson2]

Posted by Simon Williams 3 on 08/05/2019 20:06:35:

About this root 3 business, it depends on whether the line voltage (230V) is a line to line or a line to neutral figure.

To calculate the power drawn by a three phase motor load supplied with three phase 400 volts (line to line inferred) you need the root three factor. If the 230 volts figure is the same – line to line – then it follows that the root three factor is still needed.

If the supply is specified as 230 volts line to neutral then the root three factor has already been applied by dint of getting from 400 volts to 230 volts.

In each case the convention is that the line current in one phase is specified, and the other two phases carry nominally equal currents.

The confusion arises because while it is obvious that a single phase input/three phase output VSD is supplied with 230 volts line to neutral, to develop full power at the motor it generates three phase 230 volts line to line, not line to neutral. The motor rating plate assumes that the delta connected motor will be supplied with three phase 230 volts, just as the star connected motor requires 400 volts line to line to develop full power.

Of course the arithmetic changes with phase coverters that can generate 400 volts line to line three phase from 230 volts line to neutral single phase input.

HTH Simon

While this is generally correct, you would not normally see 230V or 240V line to line quoted in the UK. The output voltage of a VFD is not normally fixed, it varies with the speed with the voltage dropping with the speed reducing. This is to stop excessive current flow. This called V/Hz control and maintains a constant ratio between the two. for a 230V 50Hz motor the ratio is 4.6V/Hz so if running at 3/4 speed the frequency is 37.5Hz and voltage 172V. Note that while this imples that running at 1.25x speed the voltage would be 287V the voltage is not normally incresaed above nominal unless the motor is specified for inverter use at higher speed. This is becuse the higher voltage (and induced transients) can cause insulation breakdown.

In any case the power delivered should not exceed the motor rating (average and overload).

Robert G8RPI.

Edited By Robert Atkinson 2 on 08/05/2019 22:30:25

[Carl FarringtonParticipant @carlfarrington53722]

Posted by Robert Atkinson 2 on 08/05/2019 22:28:40:

Posted by Simon Williams 3 on 08/05/2019 20:06:35:

About this root 3 business, it depends on whether the line voltage (230V) is a line to line or a line to neutral figure.

To calculate the power drawn by a three phase motor load supplied with three phase 400 volts (line to line inferred) you need the root three factor. If the 230 volts figure is the same – line to line – then it follows that the root three factor is still needed.

If the supply is specified as 230 volts line to neutral then the root three factor has already been applied by dint of getting from 400 volts to 230 volts.

In each case the convention is that the line current in one phase is specified, and the other two phases carry nominally equal currents.

The confusion arises because while it is obvious that a single phase input/three phase output VSD is supplied with 230 volts line to neutral, to develop full power at the motor it generates three phase 230 volts line to line, not line to neutral. The motor rating plate assumes that the delta connected motor will be supplied with three phase 230 volts, just as the star connected motor requires 400 volts line to line to develop full power.

Of course the arithmetic changes with phase coverters that can generate 400 volts line to line three phase from 230 volts line to neutral single phase input.

HTH Simon

While this is generally correct, you would not normally see 230V or 240V line to line quoted in the UK. The output voltage of a VFD is not normally fixed, it varies with the speed wit the voltage dropping with the speed setting This is to stop excessive current flow. This called V/Hz control and maintains a constant ratio between the two. for a 230V 50Hz motor the ratio is 4.6V/Hz so if running at 3/4 speed the frequency is 37.5Hz and voltage 172V. Note that while this imples tht running at 1.25x speed the voltage would be 287V the voltage is not normally incresaed above nominal unless the motor is specified for inverter use at higher speed. This is becuse the higher voltage (and induced transients) can cause insulation breakdown.

In any case the power delivered should not exceed the motor rating (average and overload).

What about sensorless vector control & similar?

[Author]

Posts