De-rating for reliability is perhaps a slightly old fashioned idea. Especially in the home workshop where things are rarely (deliberately?) driven flat out.

I'd say buy oversize and derate a bargain low end model as these are generally not engineered to run full power for long periods. As always its price / performance ratio. Why pay extra to be able to run flat out for hours on end when full oomph is needed for maybe 5 minutes a month. If there is an r in it.

Buy the right size if getting a good brand name vector drive unit. Vector drives do a good job of self derating by their very nature as they largely only deliver the power needed. Getting the right size means that all the factory default safety settings will be set correctly. No need to delve into the manual and parameter lists to set the maximum current et al.

At those sizes there is little objective difference in actual £ between a larger, inexpensive, version to be run derated and a decent brand vector unit of the right size.

Clive

Edited By Clive Foster on 14/07/2020 17:42:46

I bought a new 0.75KW motor and an 0.75KW Schneider VFD. They work perfectly together, I heeded some advise not to waste money on a larger VFD as the safety features built in were designed to be matched to a similar power motor.

In the factory I worked in we had many hundreds of VFDs but the specification only called up a few sizes of drive. Many were over rated for the motors but they were sophisticated drives and all the motor parameters were part of the setup so the drive would be optimised for the motor to be controlled. Personally I would buy a decent branded drive to match the motor and I would expect it to be able to deliver full performance for a long working life. The component most likely to fail are the capacitors and companies that repair and refurbish drives will replace them as a matter of course, long before terminal failure they will suffer a reduction in performance.

Mike

Oh dear, I hope Bob doesn't think I'm out to make his life miserable by questioning two posts on the same day?

I believe 'A motor will take pretty much the same current irrespective of its load, the power factor alters to produce the output power' is wrong.

• The current drawn by a motor does vary with load. More work = more amps.
• Power Factor doesn't alter to produce output power. As I understand it PF is a measure of distribution efficiency as distorted by inductive or capacitive loads. Inductive electric motors cause current to lag behind the volts, breaking the simple W=VI relationship on which Electricity Bills are based. Plus other side-effects.

I'm not clear on Conduction Angles either. Bob may be assuming a particular type of controller, perhaps SCR or Triac based? Are they common in welders?

My lathe's VFD charges a bank of big capacitors with high-voltage DC derived from rectifiers (about 170° conduction angle) . Then the electronics simulate 3-phase AC by chopping up the DC supply. I don't think limited conduction angles are an issue in this design.

I agree about buying VFDs sized to match the motor, rather than over egging. Electronic capacity for work is mainly determined the devices ability to get rid of heat. In that sense bigger heat-sinks on the next model up wouldn't be a bad thing. But the money's probably wasted. Even cheap VFD's are able to apply current limits and shutdown when they get too hot. And lathe motors are often pretty idle in amateur service. They aren't thrashed for long periods and get plenty of time to cool off while the operator gets ready for the next stage. It's a vacuum cleaner motor in the hands of a house-proud housewife that deserves our sympathy!

I tested my 1.5kW rated hobby lathe with a power meter and even brutal high-speed 6 mm deep cuts into grotty steel didn't consume more than 1.2kW. In normal use I've never managed to get the motor warm to touch. Others might be less happy with the same machine. It's a largish lathe for the sort of work I do and I rarely push it. A busy workshop frequently needing to remove a lot of metal in a hurry might well warm her up! However, I suggest not worth blowing lots of cash on an oversized VFD unless the machine works much harder than average. In which case, the lathe should be sized for industrial work too!

Dave

Oh dear, on some points I disagree with Bob and SoD.

For an induction motor at no load the current is low and so is the power factor (around 0.1 to 0.3), ie, the motor looks pretty much like an inductor. As the load increases the current increases and the power factor improves to around 0.7 to 0.9 (still inductive) at full load.

With a VFD all the above is isolated from the mains input. The classic rectifier/capacitor circuit has a short conduction angle. Once the capacitors are charged the rectifier diodes only conduct when the mains input voltage is higher than the voltage on the capacitors. So current only flows for a short period near the peak voltage of the input. This is bad for several reasons. One, the rectifier diodes need to carry a much larger current (for a short period) to supply the nominal power rating of the device. That cause more heating and the need for bigger heatsinks. Two, current is taken from the mains as a series of short spikes, which have a high harmonic content. That upsets the energy suppliers. To comply with regulations a cheaper VFD (with a simple front end rectifier) should be fed from a filter so that the assembly looks like a resistive load from the mains.

Larger and/or better quality (more expensive ) VFDs may have a power factor corrector (PFC) at the front end. The PFC is basically a DC-DC boost converter, but with a control loop that forces the current draw to be proportional to the input voltage, ie, resistive.

Andrew

YES! But doesn't need to be that much bigger so 0.5 kW should be fine.

That may be because there is no straight answer. As an electrician taking on this job for a customer I would want to see the pillar drill and check for my self before fitting an inverter. To insure the correct size was fitted. Too small risks repeated tripping and even burning out the inverter; too big costs more than it needs to and far too big the inverter may not work correctly and may damage both the inverter and the motor.

Tom

Is a 0.5kW VFD OK for a 0.37kW motor? Russell said yes, and no-one disagreed with him. 0.45kW would do too.

The discussion moving on more generally into the principles behind VFD sizing may have caused confusion, but may help others.

Rule of thumb: double the size is a bit much. In practice, I'd fit a spare 1.5kW VFD to a 350W motor to save a few quid. Not best practice though, and deliberately buying a new 1.5kW VFD for the same job would be daft.

I'm worrying about what Andrew said. Turns out a 22A dc load is an average 11A per diode, no problem, but a 20% ripple means a peak current of 164A. What makes my head hurt is whether it matters! If I buy a 20A Bridge Rectifier, its rated for 20A average, and – on average – the sums work out. Provided the rectifier's average rating isn't exceeded, it should be OK. And a bog-standard 1000piv 50A rectifier costing £3.37 has a surge rating of 400A, which is reassuring. Rectifiers do die, but not as often as the big amp number suggests.

What rectifier spikes do to the mains is another question. Can't help thinking of the good old days when the nation's millions watched television on valve sets drawing between 80W and 600W for big screen colour. Their PSUs ignored the negative and drank deep of the positive. I don't recall lack of positivity being a problem, but maybe the horror broke strong men at the power station.

Dave

Just don't go too OTT. if the VFD is way too big, its safety features might not work on a smaller motor. for ex if you have a 3HP VFD, the internal safety, overload protection might not perform correctly and you risk not protecting the motor.

also I suggest 1 VFD per 1 motor.

Greg

Let us have some basics.

All motors are rated on their output power, nothing else.

A VFD is rated on its output power, plus about 2000 other not important factors.

An induction motor is an amazingly complex electromagnetic machine. Pick up any book on motors from any charity shop to convince yourself. An induction motor is essentially a transformer where power is transferred from primary, stator, to secondary, rotor. With the huge air gap between the power factor of the motor just rotating, nothing whatever attached to the spindle, will be about 0.2. At full load the power factor of a 2.2kW motor will be 0.85. If you measure the current, note current, taken at these two points it will be near identical, about 5%, possibly 10%. Read up about circle diagrams.

To determine the current taken by an induction motor you take the rating plate Watts divided by voltage equals current, then divide by 0.8 for efficiency and again by 0.8 for power factor. This is pretty worst case. Note that efficiency and power factor improve as the motor gets larger. If you have a large machine with multiple kW motors then divide by 0.85.

This leads onto a little known fact about induction motors, they are very fussy about their voltage. The power output, that is torque, is proportional to the voltage squared. So if your mains is 216 volts, and your motor is old and rated at 240V then this is 0.9 down, so squared is 81% power!

Conduction angle. Oh, this has hurt many people! The VFD has a bridge rectifier stuck across the mains with the electrolytic caps after, the idea being to charge the caps. But the caps are also being discharged by the motor as it is driven. Now, the maximum voltage the caps can be charged to is the RMS line voltage times 1.41, root 2, the difference between the RMS and peak voltages. The conduction angle is the actual part of the sinewave that the mains is more positive than the cap, so current flows through the rectifier to charge the cap. If the cap is at 300V then it can be seen that this charging period is not 180 degrees, more like 40 degrees. The problem here is that all the power needed to run the motor, 370W, has to be passed through the rectifier in these 40 degrees. So whilst the motor is happily delivering 370W with a current of 1.1/2A for 180 degrees, the poor rectifier has to pass that power in a quarter of the time, so 6A nominally flows. We then have the problem that the current is actually peaky, more like 10A peak for very few degrees. This is why they are so infernally noisy. Need very very fast switching diodes to do it with minimum loaa and noise. Now factor in the usual case where the workshop is run from some lighting flex and it is obvious that current drop due both to the wire resistance and impedance start to become important.

Just to keep you all happy there were some posts about single phase in to 400V three phase out VFDs. Now these really are pushing the limits because they use a voltage doubling circuit to generate the 400V or so for the caps. So what is in the single phase VFD a full wave bride, in these it is half wave. One half wave charges the cap to 240V, the next half wave pushes the 240V on the cap to 450V. Not at all nice. Now you see what three phase is used, and wired to create 6 phases.

For SoD, more load means more REAL amps, not inductive amps. Power factor REALLY does vary widely as the current taken shifts from inductive to real amps. I don't mind at all, I make as many mistakes as anyone else!

Foe AJ, same as SoD, at no load the motor is a large leaky inductor with non0existant [power factor.

I said buy a VFD same size as the motor.

Buying a larger VFD really does nothing to improve the conduction angle or charge current.

The bottom line is that the electronics, particularly the rectifiers and caps, are seriously stressed. One 4kW VFD I have actually has a programme option to measure the impedance of the caps.

@ Bob Worsley

While your conclusion ".. buy a VFD same size as the motor. " is correct, your explanation is seriously flawed.
It falls over in the first 3 lines. If motors and VFDs were both rated on output power picking a VFD the same size (rating) would result in an undersized VFD.

The VFD power rating (output) should match, or be slightly greater than, the electrical power rating (input) of the motor.

I'm not even going to bother addressing the rest of your explanations flaws.

Robert G8RPI.