Tool Enable - M15 - IOT Relay - & Breakout board (auto turn off/on router)

Hey Julian,

But here you misunderstand what 8 A means on three-phase electricity. This means the VFD’s output current, and it is 8 A per phase on a three-phase current, which is what a VFD produces. What you need is to know the input current of the VFD. With a VFD for a 220 V spindle that is rated 2.2 kW, this is usually 24 A.

On three-phase electricity, the so-called apparent power (this is the power that really flows and which is the base of dimensionating the wire gauges and the fuse), is calculated this way:

Pappar = 230 · 8 · √3 = 3186.97 VA

So you see, a VFD for your 2.2 kW 230 V spindle draws up to 3.2 kVA of current! To this, you still have to add more current because of the loss because the efficiency of the VFD itself. A VFD is a AC-to-DC-to-three-phase inverter. Of course there is additional loss.

Also according to IEC 60034-1 the power rating on spindles does not mean the electrical power they draw, but the mechanical power they are able to deliver at their shaft (see here for details).

The electrical power is higher because of loss (efficiency and because of the so-called power factor). A spindle’s current draw and power is calculated → this way.

Here you can see the nameplate of my VFD and Michael’s, both are VFDs for 2.2 kW 220 V spindle:

Look at the input current, and at the output current. The output current refers to the current per phase!

Also you have to know that despite the fact that it is recommend to connect a VFD to the supply circuit with a magnetic contactor, you NEVER switch the spindle off by cutting the current to the VFD! NEVER cut the power to a VFD while the spindle is in RUN mode!

The only allowed way to stop a spindle is to use the STOP (or disable RUN or disable REV) command, either by VFD keypad, Modbus command, or programmed input terminal. Otherwise you may damage the VFD and the spindle or both.

There is enough knowledge in this forum. Please use the search function.

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