Huanyang VFD current draw measurements.

[magicniner]

Well, for instance, the current at the input is relevant to me as I want to have some idea of how much current the whole system is likely to draw in total when at full blast

Motor Stall Current multiplied by one over the VFD efficiency or the maximum rated current draw for the VFD will give you a far better idea of the maximum current under heavy working conditions than spinning up the motor and sticking a meter on the mains input.
Maths and manufacturers data are your friends here, not a meter and a finger in the air ;-)

- Nick

[Edward]

If you took manufacturer's info from certain devices, my Chinese servo for instance, you wouldn't even want to plug it in your 13A fused socket:) Steppers are another example. While some people spend their precious time solving complicated equations to do with inductance, back EFM and speed graphs, all I need to know - not being of a brainy disposition - is, can I plug it in with a reasonable margin of safety?

BTW, do you happen to know the Motor stall current x 1/VFD efficiency of the typical Huanyang/2.2Kw combo? I just can't find it anywhere in the instructions:)

Edward

[magicniner]

BTW, do you happen to know the Motor stall current x 1/VFD efficiency of the typical Huanyang/2.2Kw combo? I just can't find it anywhere in the instructions:)

Motor stall current should be on the manufacturer's spec sheet for the motor, VFD efficiency should be on the manufacturer's spec sheet for the VFD but you do to some extent get what you pay for.

If these are not available then ball park figures can be used from a device of similar specification from a manufacturer that does publish technical data and you'll still be closer to the number you are trying to get than you would be with a clamp meter, spindle running with no/moderate load.

- Nick

[Edward]

Hmmm, unfortunately there are too many variants that affect the actual current draw of a particular device, for instance the forces applied at the cutting end. And often practical experience will tell you more than a myriad of calculations.

And when I do some heavy cutting, I will observe the current to see where I am in relation to the max current set for the device, it will be interesting to know, that's all.

I only measured the input current with no load just out of sheer curiosity, nothing else, I am not trying to defy science;)

Edward

[magicniner]

Hmmm, unfortunately there are too many variants that affect the actual current draw of a particular device, for instance the forces applied at the cutting end.

Incorrect
0/10

[Edward]

"Hmmm, unfortunately there are too many variants that affect the actual current draw of a particular device, for instance the forces applied at the cutting end."

Incorrect
0/10

Puerile as puerile goes. But whatever:)

[magicniner]

"Hmmm, unfortunately there are too many variants that affect the actual current draw of a particular device, for instance the forces applied at the cutting end."

That's quite obviously utter bunkum, otherwise nothing electrical other than simple resistive load systems in industry could ever be built to a specification without years of iterative testing, fettling and re-testing.

0/10 seems extremely apt as it happens ;-)

You're saying that something calculated every day by professional engineers cannot be calculated - you're full of it and don't like that you are demonstrably incorrect, I suggest you get used to it as you seem hell bent on continuing the pattern :D

[m_c]

The problem with monitoring the line in current, is it's relation to motor current draw varies depending on the motor speed.

The motor will only see the full mains voltage when running at the rated speed. Any speed below the rated speed, the voltage applied is proportional to the speed i.e. 50% speed, 50% voltage.
What this means, if using a 2.2kw spindle rated at 220V with a 220V supply for example, is if you're running at maximum torque, at rated speed you'll be pulling 10A with the spindle ideally seeing 220V, and using 2.2KW. Now if you half the speed, the voltage drops to 110V, but the current remains constant at 10A, and you're only using 1.1KW. The supply voltage is still 220V, and due to the wonders of the switching in VFDs, the VFD will only draw 1.1KW/5A from the supply (that's assuming 100% efficiency - there will be some power loss).

Now from a spindle monitoring point, the supply current draw is useless, unless you're always going to combine it with the motor speed to get some kind of meaningful figure.


As for manufacturer's recommendations for supply fusing. They are based on worse case scenarios. Nearly all VFDs (and that includes servo drives, as they are essentially a VFD) have huge instantaneous switch on current surges. When first powered up, they appear as a short on the supply until their internal capacitor bank charges.
Then due to the switching frequency, you can get very high currents (why noise filters should be fitted), that can cause fuses to fail. Even though the average current will never exceed the rated power, there could be spikes many times the average current, which can be worsened if multiple drives share the same unfiltered supply, due to noise/frequency harmonics. In extreme cases, harmonics can cause wiring to burn out due to overheating, yet not trip any protection system.

[Edward]

Thank you m_c, for the detailed explanation.

I am aware of the current spikes on switch on, but if we leave that aside for the moment, I thought that the power line current was also proportional to the current drawn on the motor side, for instance, I thought that if the VFD panel showed an increase of 30% in current, then the inline current would also increase by the same rate. I have not had a chance to observe that. Obviously if that isn't the case, in terms of monitoring the current, the VFD measurement is the one to monitor. Thanks

Edward

[m_c]

It's a common thing that people don't understand the fundamentals of motor current and voltage.
Current = Torque. Voltage = Speed.
As motor speed increases, back EMF increases, so you need more voltage to push/drive the required current through the windings.
As load increases, due to the motor poles dragging further behind the magnetic fields, it takes more current to pull/push the rotor along.

If you were to run a motor at say 10% rated speed, but still apply rated voltage, you would burn out the windings, as due to the lack of back EMF to limit current, excess current would flow through the windings.
That's why VFDs reduce the voltage in proportion to commanded speed.

Then combine that with the glorified switch mode power supply that is a VFD, and the lines can get very blurred, but in nutshell, Power out = Power In.
So supply current = (motor voltage * motor current) / supply voltage.
That obviously doesn't allow for any inefficiency, but I'd guess that even cheap VFDs will still have efficiency ratings in the high 90s, as they don't generate any major amount of heat.


You could use the supply current to calculate spindle power, however without knowing the speed, it's relatively meaningless figure, as max spindle power is proportional to speed, so you end up with a percentage of a proportional power. And that's before you consider any inefficiency in the VFD.
That's why the VFD motor current figure is far more useful, as it corresponds directly to motor torque, and is measured after any VFD inefficiency, so you can quite easily convert to a percentage of available torque.

With the KFlop, you could pull that data from the VFD using Modbus, then show it in KMotionCNC.