TEMPUS ENERGY:
BRUSHLESS DOUBLY FED INDUCTION MACHINE
1: Introduction
Figure 1: Wind turbine with brushless doubly fed induction machine
Reducing the maintenance cost of a wind turbine is important. Indeed, performing maintenance and repair
taks in the nacelle is time consuming since a lot of time is needed to reach the nacelle which is mounted
on a high tower. Especially when considering offshore wind energy, reaching the nacelle is very time
consuming and expensive implying a further reduction of maintenance and repair tasks is desirable.
In order to reduce these maintenance costs, it is useful to use a brushless synchronous generator instead of
a classical synchronous generator (with slip rings and brushes) or it is useful to use a brushless doubly fed
induction generator instead of a classical doubly fed induction generator.
Figure 1 (H. Serhoud, D. Benattous) visualizes a wind turbine with a brushless doubly fed induction
machine. The rotor blades of the wind turbine drive the induction machine using a gearbox.
The stator of the induction machine contains two three phase windings. The so called power winding is
directly fed by the 50 Hz grid whereas the so called control winding is fed by a frequency converter.
These windings differ in pole number i.e. the power winding generates a magnetic rotating field having p P
pole pairs and the control winding generates a magnetic rotating field having p C pole pairs. In low speed
drives like wind turbines, generally a pP larger than pC is chosen.
Due to the frequency converter, the control winding is fed by a frequency f C which generally differs from
the 50 Hz grid frequency fP applied to the power winding. The amount of active power processed through
the control winding is only a fraction of the total electrical power of the machine. This implies the
frequency converter has a reduced size, rating, losses and cost compared with the frequency converter
needed in a wind turbine equipped with a synchronous generator (where the full power is processed by the
frequency converter).
By adjusting the frequency applied to the control winding, the desired speed of rotation of the rotor (and
the rotor blades) can be obtained. By adapting the speed of rotation of the rotor blades to the wind speed,
a larger power can be extracted from the wind (a higher rotor efficiency is obtained).
Similar with the classical doubly fed induction machine, the power winding is able to generate reactive
power. This situation occurs by an appropriate use of the frequency converter when feeding the control
winding (also the frequency converter itself can consume or generate reactive power).
2: Operating modes of the brushless doubly fed induction machine
A brushless doubly fed induction machine has three important operating modes:
-
Simple induction mode,
Cascade induction mode,
Synchronous operation mode.
Figure 2: Brushless doubly fed induction machine
Figure 2 (B.V. Gorti, G.C. Alexander, R. Spée) visualizes a brushless doubly fed induction machine.
Although for the brushless doubly fed induction machine, operation in synchronous mode is the most
important mode of operation, the present text will discuss all three modes of operation. Simple induction
mode and cascade induction mode do not require a frequency converter as it is the case for operation in
synchronous mode (visualized in Figure 2). This implies in case of a failure of the frequency converter
power generation is still possible using simple or cascade induction mode.
3: Simple induction mode
Figure 3: Torque speed characteristic in simple induction mode (2 pole pairs)
Figure 4: Torque speed characteristic in simple induction mode (4 pole pairs)
In case only one stator winding is used and the other one is left open, the ‘classical’ behavior of an
induction machine is obtained implying the speed depends on the mechanical load. Figure 3 shows torque
speed characteristics of a brushless doubly fed induction machine in case a first stator winding, having two
pole pairs, has been fed by a 50 Hz grid and the second stator winding is left open (P.C. Roberts, R.A.
McMahon, P.J. Tavner, J. M. Maciejowski, T.J. Flack).
Figure 4 shows torque speed characteristics of the same brushless doubly fed induction machine in case
the second stator winding, having four pole pairs, has been fed by a 50 Hz grid and the first stator winding
is left open. In Figure 3 and Figure 4 these characteristics have been measured with different types of
rotors, but always the appropriate synchronous speed is noticed. Below this synchronous speed the
machine behaves as a motor and above this synchronous speed the machine behaves as a generator.
A discussion of all these rotor types is beyond the scope of the present text but details are available in
(paper Roberts) and (thesis Roberts). Rotor 4 is a conventional squirrel cage rotor performing very well.
Notice Rotor 1 (see Figure 8) has a specialized cage structure containing pP + pC nests as mentioned in
Figure 2 and does not perform well in case of simple induction mode.
4: Cascade induction mode using a classical cascade connection
In case only one stator winding is fed by a three phase voltage and the other stator winding is short
circuited, the machine behaves as an induction machine having (pP + pC) or (pP - pC) pole pairs. In order to
understand this cascade induction mode, it is useful the study the classical ‘cascade connection’.
Consider a first induction machine where the three phase stator winding is fed by a grid frequency f 1= 50
Hz. The machine has p1 pole pairs and the axis is rotating having a slip s1 (suppose s1 > 0: motoring). A
speed
𝑛=
𝑓1
(1 − 𝑠1 )
𝑝1
is obtained. The induction machine contains a wound rotor and in this rotor voltages having a frequency
𝑠1 𝑓1 are induced. This three phase rotor voltage is used to supply the stator windings of a second induction
machine (by means of slip rings and brushes) as visualized in Figure 5.
Figure 5: The classical cascade connection
The second machine has a speed
𝑛=
𝑠1 𝑓1
(1 − 𝑠2 )
𝑝2
and the frequency of the induced rotor voltage equals 𝑠1 𝑠2 𝑓1. By a mechanical coupling of the two motor
axes, twice the same speed of rotation is obtained giving
𝑛=
𝑓1
𝑠1 𝑓1
(1 − 𝑠1 ) =
(1 − 𝑠2 ).
𝑝1
𝑝2
Since the rotor windings of the second machine are short circuited, 𝑠2 is small (since the rotor windings of
the first machine are connected with the stator of the second machine, 𝑠1 is not negligible small). This
implies
𝑓1 𝑠1 𝑓1
𝑠1 𝑓1
𝑓1
𝑠1 𝑓1
−
=
− 𝑠1 𝑠2
≅
𝑝1
𝑝1
𝑝2
𝑝2
𝑝2
giving
𝑠1 ≅
𝑝2
.
𝑝1 + 𝑝2
The speed of the motor axes equal
𝑛=
𝑓1
𝑓1
(1 − 𝑠1 ) ≅
.
𝑝1
𝑝1 + 𝑝2
Due to 𝑠2 which is small but not zero, a speed of rotation close to the synchronous speed 𝑓1⁄(𝑝1 + 𝑝2 ) is
obtained. As already mentioned, the cascade connection of Figure 5 behaves as an induction machine
having 𝑝1 + 𝑝2 pole pairs.
4.1: Calculating the slip of the classical cascade connection
When considering machine 1 in the cascade connection visualized in Figure 5, there is a stator rotating
magnetic field having a speed 𝑓1⁄𝑝1 (suppose it rotates in a clockwise direction). The rotor has a speed
𝑛=
𝑓1
(1 − 𝑠1 )
𝑝1
in the same clockwise direction. The rotor currents having a frequency 𝑠1 𝑓1 generate a rotor rotating field
having a speed 𝑠1 𝑓1 ⁄𝑝1 in clockwise direction (machine 1 operates as a motor) with respect to the rotor.
With respect to a stationary observer, the rotor rotating field has the same speed 𝑓1⁄𝑝1 in clockwise
direction as the stator rotating field.
When considering machine 2 in Figure 5, suppose there is a stator rotating magnetic field having a speed
𝑠1 𝑓1 ⁄𝑝2 in clockwise direction. The rotor has a speed
𝑛=
𝑓1
𝑠1 𝑓1
(1 − 𝑠1 ) =
(1 − 𝑠2 )
𝑝1
𝑝2
in the same clockwise direction. Due to rotor currents having a frequency 𝑠2 𝑠1 𝑓1, a rotor rotating field
having a speed 𝑠2 𝑠1 𝑓1 ⁄𝑝2 in clockwise direction (also machine 2 operates as a motor) with respect to the
rotor is obtained.
The mechanical load determines the slip 𝑠1 and the slip 𝑠2. Using the two expressions of the rotor speed
allows to calculate 𝑠2 in case 𝑠1 is known. In case for instance 𝑝1 = 5, 𝑝2 = 2 and 𝑓1 = 50 𝐻𝑧 a slip 𝑠1 =
0.4 implies a slip 𝑠2 = 0.4 (other example: a slip 𝑠1 = 0.5 implies a slip 𝑠2 = 0.6). Verify that indeed
(1 − 𝑠1 )
𝑓1
𝑠2 𝑠1 𝑓1
𝑠1 𝑓1
+
=
.
𝑝1
𝑝2
𝑝2
In case (with 𝑝1 = 5, 𝑝2 = 2 and 𝑓1 = 50 𝐻𝑧) slip 𝑠1 = 0.3, a small slip 𝑠2 = 0.067 is obtained giving a
speed of rotation close to the synchronous speed 𝑠1 𝑓1⁄𝑝2 = 𝑓1⁄(𝑝1 + 𝑝2 ).
4.2: The number of pole pairs of the classical cascade connection
In the previous paragraph a number of pole pairs equal to 𝑝1 + 𝑝2 has been obtained. This situation is
obtained when the two rotating magnetic fields in the machines rotate in the same direction. By changing
the phase sequence when feeding the stator of the second machine, the speed of rotation becomes
𝑛=
𝑓1
𝑠1 𝑓1
(1 − 𝑠1 ) = −
(1 − 𝑠2 ).
𝑝1
𝑝2
Since 𝑠2 is negligible small, the slip
𝑠1 ≅
giving a speed of rotation
𝑝2
𝑝2 − 𝑝1
𝑛≅
𝑓1
.
𝑝1 − 𝑝2
This cascade connection behaves as an induction machine having 𝑝1 − 𝑝2 pole pairs. Combining the
results of the previous and the present paragraph, it is clear the cascade connected system can run at two
possible speeds.
When considering wind energy, due to the low speed of rotation of the rotor blades, a low speed of
rotation of the generator is desirable (in combination with a high torque). This implies the configuration
realizing 𝑝1 + 𝑝2 pole pairs is much more useful than the configuration realizing 𝑝1 − 𝑝2 pole pairs.
5: Connecting rotors to realize a cascade connection
Instead of connecting the rotor of the first induction machine to the stator of the second induction
machine, it is also possible to connect the two rotors electrically as visualized in Figure 6 (this avoids
electrical contacts between the rotating slip rings and the fixed brushes). Instead of short circuiting the
rotor windings of the second machine, the stator windings of the second machine must be short circuited.
Notice the induced voltages in the stator of the second machine have a frequency 𝑓2.
Figure 6: The cascade connection using the rotors
5.1: Calculating the slip and the frequency in the second stator
The rotor has a speed (suppose in clockwise direction)
𝑛=
𝑓1
(1 − 𝑠1 ).
𝑝1
The rotor voltages and currents have a frequency 𝑠1 𝑓1 generating in machine 2 a rotor rotating magnetic
field having a speed 𝑠1 𝑓1 ⁄𝑝2 (suppose in clockwise direction) with respect to the rotor. With respect to a
stationary observer this gives a rotor rotating magnetic field having a speed
(1 − 𝑠1 )
𝑓1
𝑠1 𝑓1
+
.
𝑝1
𝑝2
Due to the induced stator frequency 𝑓2, the stator rotating magnetic field has the same speed implying
𝑓2
𝑓1
𝑠1 𝑓1
= (1 − 𝑠1 ) +
.
𝑝2
𝑝1
𝑝2
By combining this expression with
𝑛=
𝑓1
𝑓2
(1 − 𝑠1 ) =
(1 − 𝑠2 ),
𝑝1
𝑝2
it is possible to calculate 𝑓2 and 𝑠2 in case 𝑓1, 𝑝1 , 𝑝2 and 𝑠1 (which depends on the mechanical load) are
given. In case for instance 𝑝1 = 5, 𝑝2 = 2 and 𝑓1 = 50 𝐻𝑧, a slip 𝑠1 = 0.3 implies a frequency 𝑓2 =
29 𝐻𝑧 and a slip 𝑠2 = 0.517 (other example: a slip 𝑠1 = 0.5 implies a frequency 𝑓2 = 35 𝐻𝑧 and a slip
𝑠2 = 0.714). Verify that
𝑠1 𝑓1 = 𝑠2 𝑓2
which means that the frequencies of the currents in rotor 1 and rotor 2 are the same. Indeed, the
relationship
𝑓2
𝑓1
𝑠1 𝑓1
𝑓2
𝑠1 𝑓1
= (1 − 𝑠1 ) +
= (1 − 𝑠2 ) +
𝑝2
𝑝1
𝑝2
𝑝2
𝑝2
shows 𝑠1 𝑓1 = 𝑠2 𝑓2 is valid in general.
6: The brushless doubly fed induction machine in cascade induction mode
Instead of using two physically separated machines and connecting the two rotors electrically and
mechanically as visualized in Figure 6, it is possible to realize an entire brushless doubly fed induction
machine as one single machine.
In the paper (P.C. Roberts, R.A. McMahon, P.J. Tavner, J. M. Maciejowski, T.J. Flack), a brushless
doubly fed induction machine is studied. The first stator winding has p1 = 2 pole pairs and the second
stator winding has p2 = 4 pole pairs (both windings have the same power rating). The torque speed
characteristics, when considering several rotor types, are visualized in Figure 7 (when operating in
cascade induction mode). In Figure 7, the winding having p1 = 2 pole pairs is fed by a 𝑓1 = 50 𝐻𝑧 voltage
and the winding having p2 = 4 pole pairs is short circuited.
First notice the cascade synchronous speed (the so called natural speed)
𝑛=
𝑓1
= 500 𝑟𝑝𝑚
𝑝1 + 𝑝2
where the torque equals zero. The torque also equals zero at the synchronous speed of 1500 rpm and an
intermediate frequency between the natural speed and the synchronous speed of 1500 rpm (depending on
the rotor type).
In Figure 3 and Figure 4 when the machine is used in simple induction mode, the specialized cage
structure Rotor 1 containing pP + pC (here 𝑝1 + 𝑝2 ) nests does not perform well. Notice however this
Rotor 1 behaves well when the machine is used in cascade induction mode as visualized in Figure 7.
Figure 7: Cascade induction mode
Although a brushless doubly fed induction machine is able to operate in simple induction mode and in
cascade induction mode, the major interest is the operation in synchronous mode. As visualized in Figure
1 and Figure 2, this synchronous operation mode needs a frequency converter connected with the control
winding.
7: Synchronous operation mode
In the cascade connection of Figure 6, the stator of the second induction machine is short circuited. When
connecting resistors with this stator winding, the speed of rotation can be changed. However, in order to
avoid heat dissipation in these resistors, it is useful to connect this stator winding of the second machine
with the output of a frequency converter giving a frequency 𝑓2 (instead of dissipating active power in the
resistors, the power can be injected into the grid increasing the energy efficiency).
By adjusting 𝑓2, the speed of rotation can be adjusted to the wind speed in order to maximize the rotor
efficiency of the wind turbine. Indeed, a synchronous speed of rotation depending on 𝑓1, 𝑓2, 𝑝1 = 𝑝𝑃 and
𝑝2 = 𝑝𝐶 is obtained. This speed of rotation does not depend on the mechanical power.
In case both stator windings (the power winding and the control winding) are mounted in one single
machine and there is one single (well constructed) rotor, a brushless double fed induction machine is
obtained as visualized in Figure 2. The frequency converter (generally with vector control) in Figure 2 is
bidirectional implying active power can be injected or extracted from the control winding. Also the power
winding can consume active power (motor) from the grid or inject active power (generator) into the grid.
7.1: Rotor types of a brushless doubly fed induction machine
Different rotor types are discussed in literature, but especially in (P.C. Roberts, R.A. McMahon, P.J.
Tavner, J. M. Maciejowski, T.J. Flack) and (dissertation of Roberts) attention goes to these rotor types.
Rotor 1 (see Figure 3, Figure 4 and Figure 7) has a specialized cage structure which is visualized in Figure
8. This Rotor 1 contains NR = 6 nested windings where NR = p1 + p2 with 𝑝1 = 2 and 𝑝2 = 4. Each nested
winding is composed of a set of concentric loops (paper Blazquez).
Figure 8: Rotor 1 containing nested windings
Although several other cage rotor types can be used (having different performances), it is also possible to
use a wound rotor as visualized in Figure 9. Since the first stator winding contains 4 poles also the first
rotor winding is a 4 pole winding. Since the second stator winding contains 8 poles also the second rotor
winding is a 8 pole winding. (bron: dissertation Roberts)
Figure 9: Wound rotor of a brushless doubly fed induction machine
7.2: Behavior in case of a high wind speed
Consider the straightforward rotor configuration visualized in Figure 9. The frequency of the rotor
currents due to the power winding (first winding) equals 𝑠1 𝑓1 and the frequency due to the control winding
(second winding) equals 𝑠2 𝑓2. By connecting the two rotor windings,
𝑠1 𝑓1 = 𝑠2 𝑓2 .
The connections inside the rotor imply the rotor currents generate two rotating magnetic fields having,
with respect to the rotor, a different direction of rotation (for instance the first one counterclockwise and
the second one clockwise).
Suppose the power winding injects active power into the grid i.e. active power is generated. Suppose the
speed of rotation 𝑛 of the rotor is larger than the speed of the stator magnetic field having 𝑝𝑃 = 𝑝1 pole
pairs. Suppose the rotor rotates in a clockwise direction. The stator rotating magnetic field also rotates in a
clockwise direction having a speed 𝑓1⁄𝑝1 . Finally, the rotor rotating magnetic field rotates in a
counterclockwise direction, with respect to the rotor, having a speed 𝑠1 𝑓1 ⁄𝑝1 . With respect to a stationary
observer, the rotor and the stator rotating magnetic fields have the same speed of rotation implying
𝑛−
𝑠1 𝑓1 𝑓1
= .
𝑝1
𝑝1
In this case, electrical power must be extracted from the rotor winding which is linked with the power
winding (instead of dissipating this power as it is the case when using rotor resistors). Via the rotor
winding linked with the control winding and the control winding, this power flows to the frequency
converter and this power is also injected into the grid.
The speed of rotation 𝑛 (clockwise) is smaller than the speed of the stator magnetic field having 𝑝𝐶 = 𝑝2
pole pairs. Indeed, the rotor rotating magnetic field rotates in a clockwise direction, with respect to the
rotor, having a speed 𝑠2 𝑓2 ⁄𝑝2 . Finally, the stator rotating magnetic field rotates in a clockwise direction
having a speed 𝑓2⁄𝑝2 . With respect to a stationary observer, the rotor and the stator magnetic fields have
the same speed of rotation implying
𝑛+
𝑠2 𝑓2 𝑓2
=
𝑝2
𝑝2
𝑛+
𝑠1 𝑓1 𝑓2
= .
𝑝2
𝑝2
or equivalently (by using 𝑠1 𝑓1 = 𝑠2 𝑓2)
Once by multiplying with 𝑝1 and once by multiplying with 𝑝2 , one obtains that 𝑝1 𝑛 − 𝑠1 𝑓1 = 𝑓1 and that
𝑝2 𝑛 + 𝑠1 𝑓1 = 𝑓2 . By adding these two expressions, (𝑝1 + 𝑝2 )𝑛 = 𝑓1 + 𝑓2 giving a speed of rotation
𝑛=
𝑓1 + 𝑓2
.
𝑝1 + 𝑝2
For a particular generator, the grid frequency 𝑓1 and the pole pair numbers 𝑝1 and 𝑝2 are fixed. In order to
have a speed of rotation 𝑛 adapted to the wind speed, a frequency
𝑓𝐶 = 𝑓2 = (𝑝1 + 𝑝2 )𝑛 − 𝑓1
must be applied to the control winding by the frequency converter.
This situation occurs in case of a high wind speed. The maximum rotor efficiency is obtained when the
speed of rotation of the rotor of the generator is larger than 𝑓1⁄𝑝1 . Due to the high wind speed a lot of
mechanical power is available which allows to inject active power into the grid by the power winding and
by the control winding (via the frequency converter). The total generated active power equals the sum of
the power of the power winding and the power of the control winding.
7.3: Behavior in case of a low wind speed
Also in this situation, 𝑠1 𝑓1 = 𝑠2 𝑓2 and the connections inside the rotor imply the rotor currents generate
two rotating magnetic fields having, with respect to the rotor, a different direction of rotation.
Suppose the power winding injects active power into the grid i.e. active power is generated. Suppose the
speed of rotation 𝑛 of the rotor is smaller than the speed of the stator magnetic field having 𝑝𝑃 = 𝑝1 pole
pairs. Suppose the rotor rotates in a clockwise direction. The stator rotating magnetic field also rotates in a
clockwise direction having a speed 𝑓1⁄𝑝1 . Finally, the rotor rotating magnetic field rotates in a clockwise
direction, with respect to the rotor, having a speed 𝑠1 𝑓1 ⁄𝑝1 . With respect to a stationary observer, the rotor
and the stator rotating magnetic fields have the same speed of rotation implying
𝑛+
𝑠1 𝑓1 𝑓1
= .
𝑝1
𝑝1
In this case, electrical power must be injected into the rotor winding which is linked with the power
winding. This power is provided by the rotor winding linked with the control winding and the control
winding. This power is provided by the frequency converter and extracted from the grid.
The speed of rotation 𝑛 (clockwise) is larger than the speed of the stator magnetic field having 𝑝𝐶 = 𝑝2
pole pairs. Indeed, the rotor rotating magnetic field rotates in a counterclockwise direction, with respect to
the rotor, having a speed 𝑠2 𝑓2 ⁄𝑝2 . Finally, the stator rotating magnetic field rotates in a clockwise
direction having a speed 𝑓2⁄𝑝2 . With respect to a stationary observer, the rotor and the stator magnetic
fields have the same speed of rotation implying
𝑛−
𝑠2 𝑓2 𝑓2
=
𝑝2
𝑝2
𝑛−
𝑠1 𝑓1 𝑓2
= .
𝑝2
𝑝2
or equivalently (by using 𝑠1 𝑓1 = 𝑠2 𝑓2)
Once by multiplying with 𝑝1 and once by multiplying with 𝑝2 , one obtains that 𝑝1 𝑛 + 𝑠1 𝑓1 = 𝑓1 and that
𝑝2 𝑛 − 𝑠1 𝑓1 = 𝑓2 . By adding these two expressions, (𝑝1 + 𝑝2 )𝑛 = 𝑓1 + 𝑓2 giving a speed of rotation
𝑛=
𝑓1 + 𝑓2
.
𝑝1 + 𝑝2
This situation occurs in case of a low wind speed. The maximum rotor efficiency is obtained when the
speed of rotation of the rotor of the generator is smaller than 𝑓1⁄𝑝1 . Due to the low wind only a limited
mechanical power is available implying only a limited total active power can be generated and injected
into the grid. Indeed, still active power is injected into the grid by the power winding but the control
winding extracts (consumes) active power (via the frequency converter). The total generated active power
equals the power of the power winding minus the power of the control winding.
7.4: Practical realization of the synchronous operation mode
In the paper (Carlston et al), a brushless doubly fed induction machine is connected with a 60 Hz grid. The
power winding has 𝑝𝑃 = 𝑝1 = 6 pole pairs and the control winding has 𝑝𝐶 = 𝑝2 = 4 pole pairs. The wind
speed varies linearly from 6.5 m/s up to 12 m/s as visualized in Figure 10. The speed of rotation of the
generator varies also linearly from 252 rpm to 468 rpm.
Figure 10: Variation of the wind speed and the speed of rotation of the generator
As the wind speed increases, the available mechanical torque and mechanical power increases as
visualized in Figure 11 (due to the increased speed of rotation, the increase of the power is larger than the
increase of the torque).
Figure 11: Evolution of the mechanical torque and the mechanical power
The stator rotating magnetic field, due to the power winding, has a speed of 𝑓1⁄𝑝1 = 10 𝑟𝑝𝑠. In case of a
rotor speed of 468 rpm = 7.8 rps, using
𝑛=
𝑓1 + 𝑓2
,
𝑝1 + 𝑝2
a frequency 𝑓2 = 18 𝐻𝑧 must be applied to the control winding by the frequency converter. In case of a
rotor speed of 252 rpm = 4.2 rps, also a frequency 𝑓2 = 18 𝐻𝑧 must be applied to the control winding by
the frequency converter. However when comparing the 468 rpm and the 252 rpm speeds, the three phase
18 𝐻𝑧 voltages applied by the frequency converter must have a different phase sequence. By changing the
phase sequency the sign of the 𝑓2 frequency changes in the formula.
Figure 12: Current in the control winding
In case of a speed of 360 rpm = 6 rps, a frequency 𝑓2 = 0 𝐻𝑧 must be applied (a DC voltage). Indeed, the
generator rotates at its natural speed (the cascade synchronous speed). Figure 12 visualizes the 𝑓2
frequency must increase as the rotor speed 𝑛 deviates from 360 rpm = 6 rps.
7.5: Remark 1
As already explained, in case the power winding of the brushless doubly fed induction machine injects
active power into grid (generating active power), there are two possibilities.
-
When the rotor speed 𝑛 is larger than 𝑓1⁄𝑝1 , power must be extracted from the rotor winding
coupled with the power winding.
When the rotor speed 𝑛 is smaller than 𝑓1⁄𝑝1 , power must be injected into the rotor winding
coupled with the power winding.
However, the power winding of the brushless doubly fed induction machine is also able to consume active
power from the grid (motoring which implies the rotor axis is driven by the machine). Using the brushless
doubly fed induction machine, the rotor of the wind turbine can indeed be electrically driven (for instance
to start up or to rotate in case there is no wind during a demonstration day). Also here, there are two
possibilities.
-
When the rotor speed 𝑛 is smaller than 𝑓1⁄𝑝1 , power must be extracted from the rotor winding
coupled with the power winding.
When the rotor speed 𝑛 is larger than 𝑓1⁄𝑝1 , power must be injected into the rotor winding
coupled with the power winding.
Notice these results (generating is possible with speeds smaller and larger than 𝑓1⁄𝑝1 , motoring is possible
with speeds smaller and larger than 𝑓1⁄𝑝1 ) are also obtained for the classical doubly fed induction motor
with brushes.
7.6: Remark 2
In case the connections inside the rotor imply the rotor currents generate two rotating magnetic fields
having (with respect to the rotor) a different direction of rotation, the speed of rotation equals
𝑛=
𝑓1 + 𝑓2
.
𝑝1 + 𝑝2
In case the internal connections inside the rotor are different implying the rotor currents generate two
rotating magnetic fields having (with respect to the rotor) the same direction of rotation, it is possible to
prove the speed of rotation equals
𝑛=
𝑓1 + 𝑓2
.
𝑝1 − 𝑝2
Since the user is generally not able to change the internal connections in the rotor and since the generators
in wind turbines generally have a low speed of rotation, mainly
𝑛=
𝑓1 + 𝑓2
𝑝1 + 𝑝2
has practical relevance.
References
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