Sinamics V90 power supply

Hi,

I am reading the V90 manual and I don't quite understand some things. 

1FL6034-2AF21-1AG1 has a nominal torque of 1.27Nm, 3000 RPM, 0.4kW power. The servo driver is 6SL3210-5FB10-4UF1 single phase 200VAC, also 0.4kW.

Now, the nominal input current of the driver is 5A, which is equal to 1kVA power. Where the difference goes 1kVA - 400W = 600VA? Further, the manual says that power supply has to deliver 1.1kVA continuously. Why? 

I would expect that nominal current shall be no more than 3A, since the power can't evaporate. Where am I doing wrong? 

Well, the peak can last 0.3s for every 10s period, if you calculate this overall time it brings another 25W, not 600W.

You don't interpret the data correctly. The motor has 2.6A nominal current, thus a nominal voltage acording this formula:

P=sqrt(3)*U*I, is U=400W/(1.73*2.6A)= 89Vrms phase to phase. It can never rise to 200VAC as the motor would blow due to centrifigal force 6800rpm

Motor is also rated 400W, not 900W, the short peak is not part of nominal characteristics/current/voltage. So the driver will never deliver 900W continuous, you have to clearly distinguish between S1 and S6, as S6 is not continuous/nominal power.

However it true, that 2.6A nominal current on 200VAC output voltage will produce 900W of power, but there is no such combination motor-driver for V90, the only suitable motor is 400W with low Ke factor (beside there is no documented parameter list of the PMSM).

Similar packs from ESTUN or DELTA  (which to my opinion all of them are the same product, including V90) have a nominal input current of 3.2A, which makes sense.