# Anyone using Huanyang GT series VFD? **URL:** \ **Category:** Spindle/VFD (Aftermarket/Unsupported)\ **Created:** [May 14, 2022, 7:57pm UTC](https://forum.onefinitycnc.com/t/anyone-using-huanyang-gt-series-vfd/13083 "2022-05-14T19:57:40Z")\ **Posts on this page:** 1\ **Showing post:** 12
**Author:** ![349iq2345](https://avatars.discourse-cdn.com/v4/letter/3/bb73d2/32.png) [@349iq2345](https://forum.onefinitycnc.com/u/349iq2345)\ **Post date:** [July 31, 2022, 4:19am UTC](https://forum.onefinitycnc.com/t/anyone-using-huanyang-gt-series-vfd/13083/12 "2022-07-31T04:19:38Z")
Hey Martin, > [@MVall](#): > > It’s a JGD-80/2.2R24. > It’s rated at 6A and Jianken told me to set VFD to 6A, which apparently is low for a 2.2kw spindle. They gave me the following explanation for their power rating: Power=volt \* current \* √3. However, they also said overcurrent could go up to 9A. The formula P=U·I·√3 is correct to calculate the so-called [apparent electric power](https://en.wikipedia.org/wiki/AC_power#Apparent_power). This is the information that is needed to dimension the power supply and the cables. But if they sold you a spindle as 2.2 kW spindle based on this formula, or if that is the power that is marked on the nameplate, then they fooled you. 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). For a [three-phase](https://en.wikipedia.org/wiki/Three-phase_electric_power) [induction motor](https://en.wikipedia.org/wiki/Induction_motor), this calculates this way: η = Pmech / Pelectr = Pshaft / (UN · IN · √3 · cos φ) where η (eta) is the [efficiency](https://en.wikipedia.org/wiki/Premium_efficiency), cos φ (cosinus phi) is the [power factor](https://en.wikipedia.org/wiki/Power_factor), U is the rated [voltage](https://en.wikipedia.org/wiki/Voltage), and I is the rated [current](https://en.wikipedia.org/wiki/Electric_current). So the rated mechanical output power on the nameplate of an induction motor is calculated this way: Pshaft = UN · IN · √3 · cos φ · η Let’s say for calculating an example let’s use [my spindle](https://www.spindel-shop.de/en/High-Frequency-Spindle-standard-Series/High-Frequency-Spindle-HFS-series/Water-Cooled-HF-Spindles--HF-Spindle--HF-Spindles--HF-Spindle--HF-Spindles--Motorspindle--Milling-Spindle--Grinding-Spindle-4.html). According to its [datasheet](https://www.mechatron-gmbh.de/fileadmin/user_upload/Datenblatt_HFS-8022-24-ER20.pdf), in duty type S1 (continuous running duty), the rated voltage and current are: UN = 230 V IN = 8 A I don’t have power factor and efficiency at hand, but let’s use the typical values: cos φ = 0.82 η = 85% This spindle is sold as a 2.2 kW spindle, so for the mechanical power available at the shaft, which shall be marked on the nameplate, you calculate: Pshaft = 230 · 8 · √3 · 0.82 · 85/100 = 2221.32 W Okay, this is true, it says 2.2 kW on the nameplate. Now for this spindle, the so-called electrical active (real) power is: Preal = 230 · 8 · √3 · 0.82 = 2613.31 W And still for this spindle, the so-called electrical apparent power is: Pappar = 230 · 8 · √3 = 3186.97 VA Okay, you see, a spindle rated as 2.2 kW spindle in fact sucks an apparent power of 3.2 kVA. The VFDs reflect that: If you have a look at the [datasheet](https://assets.omron.eu/downloads/datasheet/en/v11/i113e_mx2-series_variable_frequency_drives_datasheet_en.pdf) of my 2.2 kW VFD, it says, for a typical 2.2 kW induction motor, it provides 3.8 – 4.5 kVA output power (dependent on voltage, 200-240 V) (this VFD is rated for up to 11 A output). And it is sold as 2.2 kW VFD, which means, it is for a spindle that makes 2.2 kW mechanical power available at its shaft. So now to [your spindle](https://www.jian-ken.com/high-speed-spindle-motor-with-constant-troque/jgd-80-750w-1-5kw-2-2kw-2-5kw-24000rpm.html): So if they told you they calculate your 2.2 kW power rating by the formula P=U·I·√3, which is the formula for the so-called electrical apparent power, the current calculates this way: I = P / ( U · √3 ) 2200 / ( 220 · √3 ) = 5.77 Which corresponds approx. to the 6 A they told you. So let’s say the power factor cos φ = 0.82 (a typical value), then the active (real) power is: Preal = 220 · 5.77 · √3 · 0.82 = 1802.90 W and let’s say the efficiency is η = 85 % (a typical value) then the rated (mechanically available at the shaft) power is: Pshaft = 220 · 5.77 · √3 · 0.82 · 85/100 = 1532.47 W So in fact, to comply to IEC 60034-1, it should have marked “1.5 kW” on the nameplate (and not “2.2 kW”) --- _[View the full topic](https://forum.onefinitycnc.com/t/anyone-using-huanyang-gt-series-vfd/13083)._