Power electronics: the drive and its brake

A variable-frequency drive (VFD, variateur) changes an AC motor's speed by changing the supply frequency, since $n_s \propto f$. It works in three stages.

1. Rectifier. Diodes turn the three-phase AC into pulsating DC.
2. DC link. A large capacitor smooths it into a steady bus voltage of about $V_L\sqrt{2} \approx 565$ V on a 400 V supply.
3. Inverter. Six fast switches (IGBTs) chop the DC bus into pulses thousands of times a second.

Pulse-width modulation

The inverter can only connect each motor terminal to the top or the bottom of the DC bus. By changing how long it stays on each side (the pulse width), it makes the average voltage follow a sine wave of any frequency. The motor's inductance smooths the current into a near-sine.

Modulation index m

0.80

Carrier frequency

15 × f

Output frequency

50 Hz

Fundamental peak

226 V

Switching frequency

750 Hz

V/Hz command

400 V @ 50 Hz

V / f = 400 V / 50 Hz = 8 V/Hz → 50 Hz : 400 V

Raise the carrier ratio and watch the filtered curve become cleaner. Drop the modulation index and the fundamental shrinks.

FoundationStart here if this is new to you

Think of a dimmer switch that flicks a lamp on and off very fast. If it is on 80 % of the time, the lamp looks 80 % bright. PWM does the same with voltage, and changes the "on" fraction smoothly over each cycle to draw a sine wave.

V/Hz control

To keep the magnetic flux constant, and so the torque capability, the voltage is reduced in proportion to the frequency: 400 V / 50 Hz = 8 V/Hz, so 200 V at 25 Hz. At very low frequency the stator resistance matters, so drives add a small voltage boost. Above 50 Hz the voltage cannot rise any further: that is field weakening, where the drive gives constant power and torque falls with speed.

Vector control (field-oriented control) is the high-performance mode. The drive splits the stator current into a flux part and a torque part and controls each separately, as in a DC motor. With a shaft encoder it gives full torque at zero speed, which every hoist and elevator needs.

Where does braking energy go?

Predict first

A hoist lowers its load and the motor becomes a generator. The drive has a plain diode rectifier. What happens to the returned energy?

It flows back to the grid through the diodesIt charges the DC-link capacitor and the bus voltage risesIt disappears in the motor

There are two fixes:

• Dynamic braking. A chopper transistor connects a resistor across the bus above a threshold (for example 750 V) and burns the energy as heat: $P = V_\text{dc}^2 / R$. Cheap and simple.
• Active front end (AFE). An IGBT rectifier replaces the diodes and feeds the energy back to the grid. It pays off on machines that generate often: hoists, mine winders, rolling mills.

A soft starter is simpler than a VFD. Two thyristors per phase, back to back, switch on late in each half-cycle, which lowers the RMS voltage. It ramps the voltage up gently to limit inrush, then a bypass contactor takes over. It cannot control speed: once at full voltage, speed is set by the grid frequency.

ExplorerGo deeper: derivations and open questions

Why $V_L\sqrt{2}$? A diode bridge charges the capacitor to the peak of the line-to-line voltage: $400\sqrt{2} = 566$ V. Under load the average drops towards $1.35 V_L = 540$ V.

Switching frequency trade-off. A higher carrier frequency gives a cleaner current and less motor noise, but more switching loss in the IGBTs and steeper voltage edges on the cable. Chapter 6 shows how those edges double the voltage at the motor terminals on long cables.