I am about to buy a spindle from pwnc’ the 2.2kw 220 volt for my onefinity journeyman with the buildbotics controler. my electrician wants to know what amp breaker to install in my panel i will be getting two separate lines of 110 and a dedicated 220 for the spindle a simple answer to this question with ouut all the gobbledeegook of tech this and code that that I would not understand is all i require. Is a dedicated 220 line with 30 amp breaker enough?
yes, it’s enough. Your HY motor might have a stamp on the side, mine shows 11 amps for a 220v, 2.2KW. Math should be 2200 watts @ 220v = 10 amps.
Jim
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Hey Robert,
30 A for a 220 V VFD for 2.2 kW spindle is correct. It’s always the rated input current on the nameplate and in the manual of the VFD that tells you the answer. On my 220 V VFD for 2.2 kW spindle (Omron MX2 aka Hitachi WJ200), the nameplate says 24 A max. input current and the manual says to use a fuse of 30 A.
I recommend to never buy a VFD that has the rated input current value omitted. It’s no serious equipment then.
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HY VFDs usually have the rated input current omitted (missing on nameplate and in the manual). Don’t confuse input current with output current. Output current is the spindle current and it’s three of them (over three wires, three-phase electricity). The Ampère value is per phase. The formula for the three-phase current that goes to the spindle is not 220 V × 10 A = 2200 W but 220 V × √3 × 10 A = 3810.5 W. See here for example calculation of spindle power.
What you need to size your supply circuit and breaker is the VFD’s input current rating, not the spindle input current rating. Here you can see it on the nameplates of these 220 V for 2.2 kW spindle VFDs:
Also Huanyang HY Series VFDs don’t even have sensorless vector control, only dumb v/f control.
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thank you for this information.
If you have the same 2.2 kW PwnCNC system that I do (VFD model CDI-EM61G2R2S2), the VFD’s rated input current is 27A. Here’s the table from the VFD manual with that spec. It doesn’t say if that’s a peak or continuous rating.
HOWEVER, the power entry module (where the power cord plugs in) for the VFD enclosure is rated for only 10A at 250V. (The 15A printed at the bottom of the label is its rating for a 120V input.) On top of that, the supplied power cord is only 14 AWG, which shouldn’t be used for more than 15A continuous, and the twist-lock connector on it is rated for only 15A at 250V.
Theoretically, 10 amps at 240V single phase could supply 2.4kW of continuous input power (under ideal resistive-load conditions). But there’s no way, in the real world, that an inexpensive VFD and spindle could convert that into 2.2 kW of continuous mechanical output power. So my guess is that 2.2 kW is a peak output power rating, not a continuous rating. Which is still a lot more than a trim router, mind you. Ditto for the 27A input current rating.
I’m assuming that the VFD’s firmware limits its continuous input current. Otherwise, that input module could smoke with a breaker larger than 10A.
FWIW, I ran a 240V 20A circuit for my spindle. I haven’t yet run it under load.
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30Amp 240 for mine. Been running for years, no issue, even though it is a Huanyang HY Series VFD.
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Hey Dennis,
damn, you are right. Is this the line filter of the Delixi PwnCNC VFD?
It’s already the dimensional size of the line filter that seems much too small to me. A line filter should be rated with at least the input current rating and input voltage of the VFD plus a reserve, so at least 240 V 30 A in this case. This 120 V 15 A / 250 V 10 A line filter could fail and this is very dangerous. A line filter has safety capacitors both continously in line and in series with full voltage and current. They must be rated accordingly.
I bought the matching EMI filter with my VFD (specified in the manual). I can say it’s already the size that is much different, it has the same width and height than the VFD itself(!). I can show a photo later if I find the time.
Meanwhile see this photo that shows the matching EMI filter (VFD control cabinet from mechatron-gmbh.de / spindel-shop.de):
Power cable for 1-phase input at 30 A should be 10 AWG (I took 4 mm²)
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Yes, it’s the AC input module in PwnCNC’s enclosure for the Delixi VFD. You can see it in the upper right corner of the pic below. I had partially disassembled the enclosure when I was trying to determine my own AC supply circuit requirements.
15 amps at 240 V is potentially 3.6 kW, which is possibly enough for 2.2 kW of output power, if the VFD’s DC supply has a power-factor correcting front end. But you’re right about that 10A input module potentially being a safety hazard, if any firmware overload protection in the VFD is assuming a 15A input supply. I should probably reach out to PwnCNC about that, in case they misinterpreted the device specs. The thing is, I bought the pre-integrated spindle system from them specifically because I didn’t want to spend my time on integration engineering of a one-off system! The Delixi VFD manual alone runs 193 pages.
Hey Dennis,
this is often misunderstood: A “2.2 kW VFD” does not mean that it provides 2.2 kW output power, but the 2.2 kW mean “A VFD for a spindle that delivers 2.2 kW mechanical power at the shaft”. Such a VFD usually can draw up to 5 kVA. This is how you come to the rated 24 A (VT mode) 22 A (CT mode). See here for example calculation of spindle power. See also the “capacity” in VFD datasheet: The 2.2 kW model is the AB022 and it delivers 3.8 – 4.5 kVA in continuous duty mode, add the efficiency loss of the VFD itself so you come to 5 kVA input power so this explains why you need 22 A / 24 A (and the EMI filter should be rated this value plus reserve)
According to my VFD manual, the EMI filter that matches the 2.2 kW version is:
I have the Schaffner brand. Here you can see the EMI filter datasheet
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Yes, I should have said “… possibly enough for 2.2 kW of mechanical output power.”
In your power-calculation post that you linked (with which I have no quibbles), you used an assumed motor efficiency of 85%. The military and industrial 3-phase motor drives that I’ve designed have an efficiency of about 96%, which with your motor efficiency assumption would give a total system efficiency of 0.96 * 0.85 ~= 82%. So with 3.6 kW of available VFD input power, the mechanical output power could conceivably be as high as 3.6 kW * 82% = 2.95 kW.
Which is not to say I believe that an inexpensive system can do that well, but it is at least possible.
I have to assume that Omron simply rates it’s drives very conservatively, so as to ensure that they can generate the rated mechanical output power even when driving very inefficient motors. After all, they have no control over the efficiency of the motor that a customer or integrator may choose. But given control of the entire system design, such generous safety margins may be needlessly expensive.
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Hey Dennis,
it is true that for dimensioning of a VFD “for 2.2 kW mechanical power” spindle, the manufacturer takes inefficient spindles into account. But they have to do, they have no choice. That’s what a user can attach to the VFD anytime. You simply can’t dimensionate a VFD for best cases (best spindles) only. And no electrician will tell you it’s okay to attach an electric device to a supply circuit that does not fulfill the rated values. Also don’t forget apparent vs. real power, the apparent current is one that really flows! Even if you don’t pay for it so has to be taken into account when sizing wires, fuses, and supply circuits.
So all VFDs “for 2.2 kW mechanical power spindle” are in the range of being able to provide 4–5 kW electric power and thus to draw 22 – 27 A @200–250 V, and are therefore rated this way.
And those VFD “manufacturers” which produce VFDs that are not able to do this, well, they simply omit the input current rating and fool you this way. And offer VFDs at extremely low prices. That is not serious equipment, because you can have no knowledge of what they provide and what they draw, and when they will fail, making the price paid a bad decision, no matter how low it was.
I like to use hardware that I can rely on and sleep well then. I don’t like crap hardware whose “manufacturer” fools you with its (absent) rating values. I think if I pay €380 for a good VFD of industrial quality, I get something that I can rely on and sleep well. Look at the manual, 400 pages of dense information and leaves no questions open. In contrast, the VFDs of many cheap VFDs are just a joke. Not surprising that buyers search for the parameter settings at youtube – the VFD manual does not tell them in a good, didactical way, and the translation from chinese is often like a joke. My industrial VFD offers its 400-page manual in EN, DE, FR, IT, ES, CS, PL, RU: and this manual allows me to do everything from unpacking, connecting, sizing circuit, performing a first spindle test run, to sophisticated use cases. Such a manual is worth a lot, and takes part in justifying the price of the VFD.
Just an FYI. I’m feeding everything on my 1F from 1 30a 240v line. That’s a 2.2kW water cooled spindle, the 1hp dust collector (120v), a water pump (120v), and the controller + monitor. On occasion I’ll also feed a power supply for my rotary’s stepper. Only one cord is needed. I split up the power into 120v at the 1F. You’ll need to balance the load (I have the dust collector on one leg) and everything else on the other. I haven’t had any issues with tripping a breaker nor EMI.
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thank you all for your input. No offence to any of you but with a few exceptions this is exactly what happens when I and many other newbies get when asking for a simple answer. An over technical long winded, kamala Harris type answer that has the OP wondering what the heck people are saying. I am a bit surprised that one of the answers wasn’t “I grew up in a middle class family”! My question was is a 230 volt 30 amp circuit enough to run a pwnc 2.2kw spindle. Yes Or No? LOL God bless your hearts one and all.
Reach out to your spindle manufacturer. They should give you that information.
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