How do large power supplies work? (VFD's etc.)

Question

I've looked at the designs of switching power supplies (buck/boost/buck-boost) and I basically understand how they work. But is this the same thing used in large power supplies such as those in the VFD of a tower crane (etc.)? Like the larger of the two hoist winches offered for the Liebherr 710 HCL 25/50 tower crane, which uses a 220 kW (294 H.P.) AC motor. How does the power supply maintain DC output while the motor is bogged out to ~18% of full speed?** Do large VFD's and whatnot still use a conventional buck (boost) switching converter to get the required "rail voltage"?

Note that 480 Vrms coming in gets you approx. 680 VDC through the rectifier and reservoir capacitors.

**Here's part of the performance data if anyone wants to know:

0 - 3450 kg*** at 0 - 238 m/min (0 - 6900 kg at 0 - 119 m/min)

0 - 25000 kg at 0 - 43 m/min (0 - 50000 kg at 0 - 22 m/min)

***238 m/min only goes up to 1800 kg on 1-part line if the reduced speed curve is used.

Answer 1

You don't want to control a "rail voltage". You want to control current. Why? Because the fundamental connections between electrical and mechanical world are

$$P_\mathrm{el} = I \cdot U = M \cdot \omega = P_\mathrm{mech}$$

and, in addition:

$$U \sim \omega$$

$$M \sim I$$

It's as great as it is simple.

For your crane example this means, depending on the load on the hook, your motor needs a certain torque – a certain current – to keep that load at an equilibrium of constant speed.

That's why you want to measure and control the current. The current depends directly on the load applied. Speed doesn't, voltage doesn't.