Doubly-fed induction generator (DFIG): how variable-speed wind turbines work

Why wind turbines need variable speed

The power in the wind is P = ½ × ρ × A × v³, and a turbine captures a fraction Cp of it. Cp depends on the tip-speed ratio λ, the blade-tip speed divided by the wind speed, and has a single peak (about 0.45 to 0.5 for modern three-bladed rotors, near λ ≈ 8). To stay at that peak as the wind changes, the rotor speed has to change in proportion to the wind speed.

A squirrel-cage induction generator connected to the grid runs at almost constant speed, so it cannot do this. The options are a full-power converter between the generator and the grid, which works but must carry 100% of the power, or a doubly-fed induction generator, which makes the same variable-speed operation with a converter that is much smaller. Wind speed also fluctuates, so allowing the rotor to speed up and slow down stores gusts as rotor kinetic energy and reduces the stress on the gearbox.

How a DFIG is built

The machine has a three-phase stator winding and a three-phase wound rotor, with the rotor terminals brought out through slip rings. The turbine hub drives the rotor through a gearbox (for example about 90:1, stepping a hub speed of roughly 10 to 20 rpm up to the generator speed). The two windings are connected like this:

• The stator is connected directly to the grid at the grid frequency.
• The rotor is connected to the grid through two converters sharing a DC link: a rotor-side converter that applies a controlled AC voltage to the rotor at slip frequency, and a grid-side converter that exchanges power with the grid.

The name doubly-fed refers to the two electrical connections to the grid, stator and rotor. The rotor-side converter regulates the rotor current, which sets both the torque and the reactive power of the stator. The grid-side converter keeps the DC link voltage steady.

Sub- and super-synchronous operation

The synchronous speed of the stator field is n_s = 120 × f / p, 1500 rpm for four poles at 50 Hz. The slip is s = (n_s − n) / n_s, positive when the rotor is slower than the field and negative when it is faster. The frequency of the rotor currents is |s| × f, so a rotor at 1200 rpm sees 10 Hz and one at 1800 rpm sees 10 Hz in the opposite phase sequence.

The converter can generate this slip frequency, and the direction of rotor power depends on the sign of the slip.

• Sub-synchronous (n < n_s, s > 0): the rotor absorbs power from the grid through the converter. The stator delivers more than the shaft power.
• Super-synchronous (n > n_s, s < 0): the rotor delivers power to the grid through the converter. The grid receives stator power plus rotor power.
• At synchronous speed the rotor current is DC and the rotor power is just copper loss.

Neglecting losses, the stator delivers P_s and the rotor power is

P_rotor ≈ −s × P_s

with the sign convention that the rotor power is positive when it flows to the grid. The total power at the grid is P ≈ (1 − s) × P_s, and the shaft power is the same. Because the rotor voltage needed is proportional to |s|, the converter only needs a voltage and a power rating proportional to the slip range.

A worked example

A 2 MW, 690 V, 50 Hz, four-pole DFIG has a ±30% speed range, so the rotor speed runs from 1050 to 1950 rpm.

1. Synchronous speed: 120 × 50 / 4 = 1500 rpm. 2. At the top of the range, 1950 rpm, s = (1500 − 1950) / 1500 = −0.30. At the bottom, 1050 rpm, s = +0.30. 3. If the machine delivers 2000 kW in total at s = −0.30, then P_s = 2000 / 1.3 ≈ 1540 kW and the rotor power is 0.30 × 1540 ≈ 460 kW. The converter handles about a quarter of the total power. 4. At 1050 rpm the machine at low wind delivers, say, 500 kW to the grid: P_s = 500 / 0.7 ≈ 714 kW, and the rotor absorbs about 214 kW.

A full-converter system would put all 2000 kW through its converter. This is the economic argument for the DFIG: a smaller converter means lower cost and lower converter losses, and it is why a 30% converter became so common.

Reactive power and the grid

Besides controlling the active power through the rotor current, the rotor-side converter can magnetise the machine from the rotor. A set-point for the stator reactive power can therefore be held at unity power factor, or made to deliver or absorb reactive power to support the grid voltage, without using capacitors. The range is limited by the converter voltage and current ratings, and by the stator current limit. The grid-side converter can also exchange reactive power on its own.

Compared with a full-converter turbine (which uses a synchronous or induction generator and a converter rated at the full power), the DFIG has these trade-offs:

• Smaller, cheaper converter and lower converter losses.
• The stator is connected directly, so it feels grid disturbances more: a voltage dip causes high rotor currents, which requires a crowbar or other fault ride-through protection.
• It needs slip rings and brushes, which are a maintenance item, and a gearbox.
• Its speed range is limited to roughly ±30%, while a full converter can allow any speed.

How the simulator models the DFIG

Curvo Studio's doubly-fed induction generator simulator uses a steady-state, per-phase T equivalent circuit referred to the stator, in motor convention, with the rotor supplied at slip frequency. The turbine uses the Heier/Slootweg Cp(λ, β) curve (peak Cp about 0.48 at λ ≈ 8.1), a gearbox with 97% efficiency, MPPT speed control limited to ±30% slip, and blade pitching above the rated wind. Cut-in is 4 m/s and cut-out 25 m/s. The converter loses 1.5% of the rotor apparent power, and the grid side is assumed to work at unity power factor.

The model is quasi-steady. It does not simulate converter control loops, grid faults or ride-through. The default machine is 2000 kW, 690 V, four-pole, 90:1 gearbox, 11 m/s rated wind, a 30% converter and a unity-power-factor set-point.

At the default 10 m/s wind, the tracker holds the rotor at about 1676 rpm (slip −11.8%, super-synchronous). The stator supplies about 1344 kW, the rotor adds about 155 kW through the converter, and the grid receives about 1496 kW.

Try it in the DFIG simulator

Open the doubly-fed induction generator simulator and try the following.

• Leave Speed control on Wind (MPPT) and sweep the Wind speed. At 10 m/s the Operating region is super-synchronous and the Rotor power is positive, at about 155 kW. At 7 m/s the speed is about 1173 rpm, the region is sub-synchronous, the slip is about +22%, and the Rotor power is about −148 kW (power is taken from the grid by the rotor).
• At 5 m/s the rotor speed is held at the limit of 1050 rpm (slip 30%), and at 12 m/s it is held at 1950 rpm while the Blade pitch rises to about 3° and the Turbine power is held near its rated value.
• Below 4 m/s the machine is parked and the readouts show Below cut-in wind.
• Switch Speed control to Manual speed and set the Rotor speed to 1500 rpm. The slip is zero and the Rotor current frequency is 0 Hz.
• Set the Stator reactive power to +500 or −500 kvar (positive means delivered to the grid) and watch the Converter loading and the Power factor at the grid. At +1000 kvar the loading reaches about 100%.
• Reduce the Converter rating to 10% with the manual speed at 1700 rpm and the Converter loading rises above 100%: the converter is overloaded and an operating warning appears.

The Power flow against rotor speed chart shows the stator, rotor and total power along the MPPT curve, with the rotor power changing sign at synchronous speed.

Try it in the simulator