I am looking for an 80mm spindle but only have 110v. I see alot of spindle have 2.2kw motors that are offering 110v. But i also see people saying that isnt proper according to the amp draw. Is anyone using a 2.2kw with 110v, hard to think for all i see for sale that no one uses it like this.
Hey Kenneth,
people seem to buy those nonetheless but these are usually chinese crap. VFDs with only stupid V/f control but no vector control, thatâs stone age. The âmanufacturersâ of such VFDs willingly omit the input current draw on nameplate and on the manual to make you believe that there exist 110 V domestic supply circuits in North America that offer enough current for such a 2.2 kW spindle. But there is none at 110 V. So what happens to people if they nonetheless buy it? Well they will never be able to bring the spindle to its mechanical load limit, because it will blow their fuses long before they draw the rated power. Of course, if you make a few signs for the thanksgiving table, it could very well be that you will not hit the limit. But unlike a Makita hand trim router, a spindle is neither a hand tool, nor is it a machine, but it is a component intended to be the part of a machine. Therefore if you marry a Onefinity CNC machine and a VFD & spindle, itâs you who are the machine builder and therefore responsible to obey to the safety regulations. And that can never be the case if you connect a machine to a too weak supply circuit, or if you use a VFD whose input current rating is willingly omitted and kept secret by the manufacturer.
Look at the serious VFD manufacturers like Hitachi: The only VFDs they offer for 110 V are for 250 W (0.25 kW) spindle and 750 W (0.75 kW) spindle, but nothing bigger at 110 V. The bigger power VFDs are for 220 V or 400 V. Thatâs because they take into account what your 110 V supply circuits actually can deliver.
110 V is not for machines, it is for electric light and for small electrical appliances like phone chargers or radios. It is not for power motors. But usually in North America, you have 220 V (240 V in fact) anyway because you usually have split-phase electricity. You just need someone who can draw a cable to a NEMA 14 socket from your wall power box and install a circuit breaker fuse (MCB) for it.
With a VFD for a spindle of 1.5 kW rated mechanical power, at 110 V, you need approx. 30 A, but only 15 A max. when using a 220 V VFD/spindle for 1.5 kW. Reducing the current this way (by using a higher voltage) is the PURPOSE of the 220 V supply circuits, for heaters, ovens, air conditioners, table saws, router tables, planing machines, big dust collection systems â and for spindles which are induction motors.
Note that a 80 mm 1.5 kW spindle has more efficiency so delivers more mechanical power than a 65 mm 1.5 kW spindle.
My 220 V VFD for 2.2 kW spindle has a max input rating of 24 A, with 110 V it would be 48 A(!).
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Hey Aiph5u, would you mind elaborating on the supply requirements for a 1.5 kW spindle?
My actual design experience mostly is with 3-phase brushless motor drives, but Iâve worked with induction motors and VFDâs enough to be somewhat familiar with them. For the components Iâve analyzed, a system efficiency of 70% (mechanical output power / electrical input power) at full load appeared to be a reasonably conservative rule of thumb.
The following assumes that the VFD has a decent power-factor correcting input supply, so that itâs not pulling in huge current surges on every half-cycle of the supply voltage â and that may be too much to expect in a cheap VFD. But if you use a good VFD and that 70% efficiency assumption, 1500W of mechanical output power would require 1500 / 0.7 = 2143W of electrical input power. Which is too much for a 120V 15A circuit, but should be OK to run on a dedicated 120V 20A circuit.
But for sure, if the VFD manufacturer doesnât specify the full-load RMS input current, I wouldnât make any assumptions about how much mechanical power you can actually get out of the spindle system.
Hey Dennis,
you have trhee-phase electricity on the output of the spindle, each wave shifted by 120° at the same time. Also the 1.5 kW spindle that the VFD is for is by no means the electrical power draw but the mechanical power that is delivered at the shaft. The electrical power draw is much higher. I have shown in an example how to compute the power draw of a spindlle in in this posting for a VFD for 2.2 kW spindle. If you follow the computation example, what you have in the end is the electrical current draw at the spindle input = the output of the VFD, which is much higher than 1.5 kVA. THEN, for the current draw at the input of the VFD you still have to add the effficiency loss of the VFD itself. A VFD is an AC-to-DC-to-three-phase-AC inverter. It recitifies all AC at its input and stores it in a large capacitor, and then with six IGBTs, it switches the three phases on and off in a quasi-sine waves on three phases on three wires, with a variable frequency. If it has a single phase input, all the current comes in one wire, while at the output, the power is spread over three phases simultanously, so the current rating of the spindle (what is printed on the spindle), i.e. the current in one wire is usually a lower number. See also this posting here.
Also the formula for three-phase current is not P = V ¡ I, but P = V ¡ â3 ¡ I.
Hey Aiph5u, thanks for the reply.
I understand your calculations in the examples perfectly. In fact, we used almost exactly the same rule-of-thumb numbers for power factor and motor efficiency. In this example, you used a Power Factor of 0.82, and a motor efficiency of 85%. Your terminology was much more precise than mine, to be sure, but if you multiply your 0.82 by 85%, you get 0.697 â almost exactly my 70% âsystem efficencyâ (loosely speaking).
I didnât bother with calculating the motor phase current, because itâs not relevant to the system efficiency computation. But I do agree with your calculations in the example.
I think the difference in our conclusions about running a 1.5 kW (continuous mechanical output) spindle on 120V may come down to the efficiency of the VFD, which I had implicitly approximated as 100%.
So I think we agree that for a spindle to make 1500W of mechanical output power, it needs about 2143 W of electrical input power. A 120V, 20A supply can provide up to 2400 W continuous. The three-phase IGBT brushless servomotor drives that I once designed were about 96% efficient, IIRC, and so would need about 2143 / 0.96 = 2232 W of electrical input power to deliver 2143 W of power to the motor. That leaves just enough overhead for a 2232 / 2400 = 0.93 power factor at the input of the VFD. I honestly donât know if thatâs achievable, and it almost certainly doesnât leave room for even a 5% supply-voltage sag.
So bottom line, I would have to agree that a 1.5kW spindle / VFD system that runs on 120 V does need a supply current of more than 20A in order to deliver its rated continuous output power. Or else maybe a few drops of snake-oil lubrication in the bearings. 
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Hey Dennis,
anyway, you should always rely on the input current rating on the nameplate and the manual of the VFD. And if the VFDâs input current rating is willingly omitted and kept secret by the manufacturer, like in these examples here, I would never buy such a VFD. Itâs not serious.
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Hey Kenneth,
Daniel @PwnCNC does offer different voltage and power variations of their VFD/spindle but not 2.2 kW @ 110 V because there is no power supply circuit strong enough to expect in North American domestic areas. So if such a friendly VFD vendor says this, I would believe it. Although he seems wrong with the actual current draw of their VFDs. In every case, since you can say that a 110 V VFD for 2.2 kW spindle has a capacity of about 3.9â4.5 kVA plus VFDâs own efficiency loss, which is usually to expect, you need a supply that can deliver more than 35 A â 40 (plus reserve). So if you have this in your house (which I doubt) you surely have 240 V too.
Hey Dennis,
I forgot to mention that when the VFD manufacturers dimensionate the VFDâs capacity, they donât know how efficient the spindle is you will drive with it. In the computation example, I took my spindle which draws 8 A in continuous torque (CT) mode to deliver 2.2 kW mechanical power at the end of its shaft. But the operator could use a spindle with much worse efficiency, so VFDs are designed with some reserve. My VFD, the Omron MX2 for 2.2 kW spindle (aka Hitachi WJ200) can therefore deliver up to 11 A to a spindle with bad efficiency so that it still can deliver 2.2 kW mechanical power at its shaft. So this is the number that they take to put the rated capacity on the VFD datasheet. For my VFD, this is 3.800 kVA @ 200 V â 4.500 kVA @ 240V in constant torque mode. This is what it can really provide. This gives 19 A â 22.5 A. Then you add the efficiency loss of the VFD itself so you finally get what it can really draw, which is the 24 A on the nameplate (see posting above). Next fuse size then is 30 A.
With 120 V, it would simply be double the current, so 48 A VFD max. input rating, 60 A fuse to recommend.
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Thanks Aiph5u, thatâs good to know. And itâs good to see that these manufacturers are rating their equipment conservatively.
Since I was originally asking you about running a 1.5 kW spindle on 120 V, I tried scaling the numbers from your system to estimate the required VFD input current for that scenario. Assuming that the 1.5 kW spindle efficiency is the same as yours, the motor input current would be 8A * 1500 / 2200 = 5.45A at rated load. With a 240V input, the VFD input current would then be 5.45A * 22 / 11 = 10.91A. But with a 120V input, the required VFD input current would be 10.91A * 240 / 120 = 21.82A.
Which would seem to verify our earlier conclusion that a 120V, 20A supply is a little too small to generate a continuous 1.5 kW of mechanical output power, even with an efficient spindle and VFD.
I presume that you could program the VFD to limit its output power or current to a lower value. But then it would no longer be a 1.5 kW spindle system.
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Hey Dennis,
thatâs always possible e.g. if you want to buy a stronger spindle only later, you could start with a weaker spindle but already buy the VFD for the stronger spindle and limit the spindle current this way. Itâs just the voltage that youâll have to choose right away.
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