I have a 220v 1.5K Longwang VFD and 1.5K 65 mm spindle. I’ve just ordered the 80 mm Z20 slider from 1F and am getting a sleeve printed so I can continue using the 65mm spindle. However, I’d like to upgrade to an 80mm spindle with an ER20 collet.
Will my 7.5 amp 1.5k VFD drive an 8 amp 2.2k spindle
It’ll work until your spindle needs more than 2HP, then it will overload and ruin your carve. Upsize the VFD when you get a larger spindle.
Hey Brian,
yes, but the maximum current flowing through the spindle will be less than what the spindle can take, and provide less maximum power than the spindle is rated, but I assume you intend to put up with that. The rated spindle current (in Ampères) is the most important setting in a VFD. If you have a VFD for 1.5 kW motor, you would probably set the rated motor current to the maximum that the VFD will allow which will probably be less than the rated current of the 2.2 kW spindle. The other way around is also possible, you can run a 1.5 kW spindle on a 2.2 kW VFD since the current is always limited by the VFD, because the rated motor current setting is a mandatory setting that has to be set in every VFD anyway.
By the way, that is also a difference between using a spindle/VFD and using a hand trim router: If the latter gets overloaded, it will overheat and may burn your workshop down, while with the first, if it is exposed to a too high mechanical load (which will make it go beyond the motor current limit) the VFD will stop the spindle, send an alarm, and if your emergency stop circuit is wired correctly, will halt the g-code program running in the CNC controller.
Thanks Aiph (??). Here’s the info on the spindle I’m looking at:
And here’s the info on my current spindle and vfd. It seems that, other than size, the ratings are the same - esp. the 8 amp rating.
Hey Brian,
this looks like a Huanyang HY Series VFD. Unfortunately, it is not capable of Sensorless vector control (SVC) and I consider this to belong to “cheap chinese” VFDs, even if “Huanyang” seems to be a brand.
Regarding the general mandatory VFD settings, you have to set:
Specifally the number of magnetic poles (value to set PD143) is not mentioned in the image you posted above, and since it is “4” by default, your spindle would run at half the speed. On a spindle that runs with 24 000 rpm at 400 Hz, you know that it has two magnetic poles, not four.
Pole pairs on three-phase asynchronuous induction motors:
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| 2 magnetic poles (1 pole pair) | 4 magnetic poles (2 pole pairs) |
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Also I would set PD011 (frequency lower limit) to what the spindle manufacturer says in the spindle manual. Usually spindle manufacturers require spindles with 2 magnetic poles that can do 24 000 rpm @ 400 Hz to set this limit to 6 000 rpm (100 Hz) or 8 000 rpm (133 Hz).
Since the spinde speed is calculated with:
this would mean:
100 Hertz × 2 × 60 seconds / 2 poles = 6,000 rpm
133 Hertz × 2 × 60 seconds / 2 poles = 7,980 rpm (≈ 8,000 rpm)
What is mentioned in your image is the setting PD144 (rated motor revolution) whose meaning is often not clear because of the lousy manuals of cheap chinese VFDs. It means “rotations per minute with which the spindle would run if provided with a frequency of 50 Hz”.
This would mean:
50 Hertz × 2 × 60 seconds / 2 poles = 3,000 rpm
So here you got to set PD144 to “3000”, this is correct.
PD070 (analog input), PD072 (lower analog frequency) and PD073 (higher analog frequency) are only necessary if you want to control spindle speed with a potentiometer through the analog voltage or analog current input (or if you have the Masso G3 controller which controls the spindle this way too because it does not support Modbus control). Since I can assume you have the buildbotics-derived Onefinity Controller, you would let the CNC controller control spindle speed and STOP/RUN command via Modbus via its RS-485 serial communication port instead. In this case you would need to set the communication settings PD163, PD164 and PD165 to the same values as set in the Onefinity CNC Controller on the “I/O” page, and also you’ll have to set:
Since your VFD is not capable of vector control, you can only use the (more stupid) U/f (voltage/frequency) control. The settings PD005–PD010 are used to fine-tune the U/f mode:
– Source: Huanyang Inverter HY Series Instruction Manual
Follow the instruction of each setting in the manual to choose the correct value.
If I can give you an advice: Never set a setting without knowing what it is for, which also implies, never run a device with default settings (even if the designers of a popular graphical operating system tries to teach people since 1985 that this is a good idea. In fact, it is not.).
Default, n.: The vain attempt to avoid errors by inactivity. Possibly from Black English “De fault wid dis system is you, mon.” “Nothing will come of nothing: speak again.” – King Lear.
– Ambrose Bierce, The devil’s dictionary.
Please do not ever use the word default in a program designed for
humans. Default is something the mortgage went into right before the
evil banker stole the Widow Parson’s house. There is an exhaustive list
of substitutes (previous, automatic, standard, etc.) in the Appendix to
How to Write a Manual.Defaults should be declared, not assumed . Undeclared (not displayed)
defaults such as pressing RETURN for Yes (or for No?) will cause
confusion and anger.– Source: Apple IIe Design Guidelines, 1982
Hey Brian,
I forgot to answer to this.
It may very well be that a spindle specified with 220 V / 8 A delivers a different amount of power (in kiloWatts) than another one that is also specified with 220 V / 8 A. You need to know that the values for voltage and current on the spindle nameplate are electrical values that occur during current consumption, but that the power specification “1.5 kW” or “2.2 kW” for a motor does not refer to the electrical power, but to the mechanical power at the end of the shaft.
According to IEC 60034-1, an electric motor shall be provided with a rating plate on which shall be marked the rated output. According to IEC 60034-1, the rated output is the mechanical power available at the shaft and shall be expressed in watts (W).
It is calculated this way.
So in this case it seems that the larger diameter of the rotor and stator on an 80 mm spindle already results in a higher leverage force acting on the axle than on a 65 mm spindle. Therefore, it may well be that the spindle delivers more mechanical power than a narrow spindle simply due to its size. In this case, it appears that this is the case with identical electrical current, i.e. for this 80 mm spindle you would not change the values for voltage and current in the VFD, but it has greater efficiency than the narrow spindle and delivers more mechanical power.
There are also some other parameters that can influence a spindle’s efficiency, like the amount, type and size of the bearings used (less friction), and others.
Thanks … what is your name? I feel odd calling you Aiph, unless that IS your name.
I’m trying to translate your message. It seems that what I’m proposing might be OK… or are you saying that my rig will blow up if I connect a 2.2k spindle?
