EE – 4420 Electric Machine Analysis
Fall 2005
THREE-PHASE INDUCTION (ASYNCHRONOUS) MACHINES
1. CONSTRUCTION AND PRINCIPLE OF OPERATION
There are two types of induction machines: slip-ring and squirrel cage machine.
1.1 Induction machine with slip-ring rotor (wound-rotor type)
The motor structure is shown in Fig. 1. A stator has a 3-phase winding supplied from
a 3-phase source. A rotor winding is of the same form as the stator winding and is
embedded in the slots. The stator and the rotor cores are composed of laminations of
high-grade sheet steel. The wire is insulated from the lamination. The rotor winding is
connected in star and its left 3 free terminals are conducted through the hollow shaft and
connected to the 3 copper rings (insulated from the rotor shaft). Using stationary brushes
pressing against the slip rings, the rotor terminals can be connected to an external circuit
e.g. external three-phase resistor. The three-phase windings on the stator and on the rotor
are distributed windings. The winding of each phase can be distributed over several slots.
When the current flows through the distributed winding it produces an essentially
sinusoidal space distribution of mmf.
3-phase stator winding
3-phase rotor winding
A1
X
X
X
variable
resistance
B2
brushes
X
X
C2
X
X
C1
slip rings
B1
X
X
A2
circuit diagram of 3-phase
rotor winding connected in star
Fig.1 Construction scheme of a slip-ring machine
1
1.2 Machine with squirrel-cage rotor
There is no any difference in the stator construction of both types of machines. The
difference is in the rotor winding. The squirrel-cage winding consists of aluminum or
copper bars embedded in the rotor slots and short-circuited at both ends by aluminum or
copper end rings as shown in Fig.2.
Fig.2 Construction scheme of rotor squirrel-cage winding
1.3 Principle of Operation
If the three-phase stator winding is connected to a three-phase supply, a rotating
magnetic field will be produced in the air gap. The field rotates at synchronous speed ns =
n1 (given by equation (13) – notes: Rotating Magnetic Field). The rotating field when
moving with respect to the copper bars of stator and rotor windings induces voltages. If
the rotor circuit is closed, the induced voltages (in the rotor winding) cause the currents to
flow. Their interaction with the air gap magnetic field produces a torque. The rotor if free
to do so will then starts rotating. According to Lentz’s law, the rotor rotates in the
direction of the rotating field such that the relative speed between the rotating field and
the rotor winding decreases. The rotor will eventually reach the steady-state speed n that
is less than the synchronous speed ns at which the stator rotating field rotates in the air
gap. At rotor speed n = ns there will be no induced voltage and currents in the rotor
circuit and hence no torque.
The difference between the rotor speed n and the synchronous speed ns of the rotating
field is called slip-speed. The rotor slip s is defined as:
s=
ns − n
ns
(1)
If you were sitting on the rotor you would find that the rotor was slipping behind the
rotating field by the slip-speed:
s ⋅ ns = ns − n = n2r
(2)
2
The frequency of the induced voltage and current in the rotor circuit will correspond
with this slip speed, because this is the relative speed between the rotor winding and the
rotating field. Due to the slip-speed the voltages and the currents are induced in the rotor
winding. They also produce a rotating field moving with the speed n2r with respect to the
rotor. The induced rotor magnetic field rotates with respect to the stator at speed:
a f
n2r + n = ns s + ns 1 − s = ns
(3)
Therefore, both, the stator field and the induced rotor field rotate in the air gap at the
same synchronous speed ns. The stator magnetic field and the rotor magnetic field are
therefore stationary with respect to each other. This is the one of conditions to produce
the average torque (see section Introduction to Electric Motors).
The frequency f2 of the rotor currents will change according to the variation of the
rotor slip s. This frequency can be expressed in terms of the supply (stator) frequency f1
and the rotor slip s. Similar as for a stator rotating field (see equation (13) – Rotating
Magnetic Field) this speed is given by the following equation:
n2r =
120 f2
= ns − n
p
(4)
p
s ⋅ ns
120
(5)
The frequency
f2 =
The synchronous speed ns =
f2 =
120 f1
, thus:
p
120 f1
p
s⋅
120
p
(6)
f2 = s ⋅ f1
(7)
and finally:
This rotor frequency f2 is called a slip frequency.
The voltage induced in the rotor circuit at slip s is:
E2 = 4.44 ⋅ f2 N2 Kw 2Φ
= 4.44 ⋅ sf1 N2 Kw 2Φ
(8)
= sE2 o
where:
E2 o = 4.44 ⋅ f1 N2 Kw 2Φ
(9)
is the voltage induced in the rotor winding at rotor speed n = 0.
3
2. THREE MODES OF OPERATION
Theoretically, the rotor speed can change in the range −∞ < n < +∞ and thus the rotor
slip s. The speed axis and the corresponding slip axis are shown in Fig.3. Within this
range the induction machine operates in three modes:
- motoring 0 < n < ns , 1 > s > 0 , f1 > f2 > 0 , torque T>0
- generating (regenerative braking) ns < n < ∞ , 0 < s < −∞ , 0 < f2 < ∞ , T<0
- plugging (braking) −∞ < n < 0 , +∞ > s > 1 , ∞ > f2 > f1 , T>0
There are two particular points of operation:
- at n = 0, s = 1, f2 = f1, T>0 – transformer operation,
- at n = ns, s = 0, f2 = 0, T = 0 – ideal no-load operation.
An illustration of the above description is shown in Fig.3.
Transformer
Plugging
s
Ideal no-load
operation
Motoring
Generating
0
n1= ns
1
0
n
Fig.3 Three modes of operation of an induction machine
2.1 Motoring
In this natural mode of operation the motor develops the torque in the direction of
rotating field and the rotor moves in the same direction.
2.2 Generating
In this mode of operation the rotor is driven with speed grater than synchronous speed
of rotating field. Therefore it brakes the rotor and the power of the driving machine is
converted by the induction machine to supply. Thus, the induction machine becomes the
induction generator. It brakes the driving machine. The process is known as regenerative
braking.
2.3 Plugging
If the rotor of the induction machine is driven in an opposite direction to the stator
rotating magnetic field the torque that acts on the rotor will oppose its motion. Thus, this
torque is a braking torque.
4
2.4 Transformer operation
If the rotor is blocked and the stator winding is supplied from the three-phase source
the induction machine behaves as a three-phase transformer with the short-circuited
secondary (rotor winding). The frequency at both sides (stator and the rotor winding) is
the same.
2.5 Ideal no-load operation
The rotor rotates with synchronous speed. In practice, when the rotor is not loaded it
experiences braking by the friction in bearings and windage loss, so its speed drops
below synchronous speed. Theoretically, if there is no friction at all the rotor (when the
machine is supplied) moves synchronously with the stator magnetic field.
3. EQUIVALENT CIRCUIT OF THE MOTOR
The machine analysis will be based on the equivalent circuit. To derive it the
equivalent circuit of the transformer will be used. Since there are different frequencies in
stator and rotor windings we draw first the equivalent circuit separately for both windings
with the magnetic coupling between them as shown in Fig.4.
I1
R1
Xl1
I2
Xl 2(f )
2
R2
Φ
V1
RFe
E1
Xµ
E2
f 2 = f1
Fig.4 Equivalent circuit of the motor with magnetic coupling of the stator and rotor
windings
The leakage reactance of the rotor that depends on the varying frequency f2 can be
written as follows:
X l 2( f2 ) = 2π f 2 Ll 2 = 2π f1s ⋅ Ll 2 = sX l 2
(10)
where Xs2 – is the leakage reactance of the rotor at standstill when the frequency is equal
to that in the stator. Thus, this reactance does not depend on the rotor slip. To make the
rotor circuit parameters independent on the rotor speed let us write the rotor current as:
5
I2 =
E2
R +X
2
2
2
l 2( f 2 )
=
sE2 o
R + ( sX l 2 )
2
2
2
=
E2 o
2
⎛ R2 ⎞
2
⎜ ⎟ + Xl2
⎝ s ⎠
(11)
According to expression (11) the rotor circuit can be redrawn as in Fig.5 where the
frequency is constant and equal to f1.
I2
E2o
Xl2
R2 / s
f 2 = f1
Fig.5 Rotor equivalent circuit with f2 = f1
To avoid the magnetic connection between the two circuits the rotor circuit
parameters should be transferred to the stator side as it was done for a transformer. After
this the induction machine equivalent circuit is as in Fig.6. The rotor parameters are
expressed as follows:
Xs' 2 = a 2 Xs 2
R2' = a 2 R2
(12)
E2' o = aE2 o
I2' =
I2
a
where:
a=
m1 Kw1 N1
m2 Kw 2 N2
(13)
and m1 and m2 – are the number of phases in the stator and the rotor winding
respectively.
6
I1
R1
,
,
Xl1
X l2
I2
,
a R2 / s
Ie
IFe
Iµ
,
V1
E1= E2o
RFe
Xµ
b
Fig.6 Equivalent circuit of the induction machine
The power that is transferred to the rotor through the air gap
Pag = m1
R2' '
I2
s
d i
2
(14)
should be the sum of power loss in the rotor resistance
d i
∆Pw2 = m1 R2' I2'
2
(15)
and the mechanical power Pm. It means that
Pm = Pag − ∆Pw 2 = m1
R2' '
I2
s
d i
2
d i
− m1 R2' I2'
2
(16)
After transformation:
Pm = m1 R2'
1− s '
I2
s
d i
2
(17)
According to equation (17) the mechanical power is “dissipated” in the resistance
R2'
1− s
. The equivalent circuit of the rotor with the rotor resistance splits into two
s
components is shown in Fig.7.
Looking at the rotor equivalent circuit, keeping in mind the equations (14) – (17), the
following relations between the rotor powers can be derived:
a fd i
R2'
1 − s I2'
s
= 1 − s Pag
Pm = m1
a f
Pm = m1 R2'
2
(18)
a1 − sf d I i
s
1− s
=
∆Pw 2
s
' 2
2
(19)
7
,
a I2
,
R2
, 1- s
R2 s
b
Fig.7 Rotor equivalent circuit with the rotor resistance split into “mechanical” resistance
and winding resistance
From equations (18) and (19):
∆Pw 2 = sPag
(20)
Thus:
a f
Pag : ∆Pw 2 : Pm = 1: s: 1 − s
(21)
Equation (21) indicates that, of the total power input to the rotor (i.e., power crossing
the air gap, Pag), a fraction s is dissipated in the resistance of the rotor circuit (rotor
winding loss, ∆Pw2) and the fraction (1-s) is converted into mechanical power. Therefore,
for efficient operation of the induction machine, it should operate at a low slip so that
more of the air gap power is converted into mechanical power. Part of the mechanical
power will be lost to overcome the windage and friction. The remainder of the
mechanical power will be available as output shaft power.
The complete equivalent circuit is shown in Fig.8.
I1
R1
,
Xl1
I2
,
X l2
,
R2
Ie
IFe
V1
Iµ
,
RFe
E1= E2o
Xµ
,
R2
1- s
s
Fig.8 A complete equivalent circuit of the induction machine
8
4. ELECTROMAGNETIC TORQUE
The electromagnetic torque developed by the induction machine can be derived from
the following equation:
Tem =
Pm
ωm
(22)
where mechanical angular speed:
ωm =
2πn
60
(23)
It is related to the synchronous speed by equation:
a f
n
=
2π a1 − sf
60
ωm = ωs 1− s
s
(24)
After substitution of above formulae to equation (22) we obtain
Tem =
=
Pm
ωs 1− s
a f
Pag
(25)
ωs
or
Tem = 9.55
Pag
ns
(26)
The power crossing the air gap
Pag = m
R2' ' 2
( I2 )
s
(27)
Since
E
d i = F R Id i
GH s JK + d X i
2
I2'
' 2
2o
'
2
2
(28)
' 2
s2
The electromagnetic torque:
9
Tem =
m
(E )
2
'
2o
ωs ⎛ R2' ⎞ 2
R2'
⋅
s
'
⎜ ⎟ + ( Xl2 )
s
⎝ ⎠
,
(29)
,
(30)
2
When we assume that E2' o ≅ V1
Tem ≅
m
(V1 )
ωs ⎛ R2' ⎞ 2
2
R2'
⋅
s
'
⎜ ⎟ + ( Xl2 )
s
⎝ ⎠
2
The torque-speed (slip) characteristic drawn from the above formula is shown in
Fig.9.
R2'
At low value of slip
>> X l' 2 , and the torque
s
Tem ≅
b g
2
m V1
s,
ω s R2'
(31)
T
Tmax
0
1
Plugging
sb
Motoring
ns
0
n
s
Generating
Fig.9 Torque-speed characteristic at V1 = const and f1 = const.
10
At larger values of slip
R2'
<< X l' 2 and the torque
s
Tem ≅
m
(V1 )
2
ωs ( X l' 2 )2
R2'
s
(32)
To determine the maximum torque or breakdown torque we differentiate the torque
with respect to slip and setting it equal to zero:
dT
=0
ds
(33)
The above equation is fulfilled when
R2'
X l' 2
If the breakdown slip sb formula is placed to Eq. (30) we obtain
sb ≅ ±
m (V1 )
≅
ωs 2 X l' 2
(34)
2
Tmax
(35)
In the calculation of motor performance very often the ratio of torque to its maximum
value is applied
2 R2' ⋅ X l' 2
T
2
=
=
2
sb s
Tmax ( R ' )
+
2
' 2
+ s ( Xl2 )
s sb
s
(36)
5. PERFORMANCE CHARACTERISTICS
When the induction motor is loaded by an external machine (Fig.10) the point of
operation P of the machine set at steady-state is at the intersection of characteristics of
both motor and the load, where the torque developed by the motor is equal and opposite
to the torque of the loading machine (Fig.11).
The torque depends on the number of variables and the torque-speed characteristic
can be modified in different ways. Let examine the influence of the rotor resistance on
torque-speed characteristic (Fig.12). The change of R2' does not influence the maximum
torque (Eq.35) but affects the breakdown-slip sb (Eq.34). The proper value of rotor
resistance (by adding to slip-rings an external resistance) can be chosen to get the
maximum torque at start (n = 0). This is applied in motor starting.
11
n,T
Induction
machine
Load
TL
Fig.10 Machine set: induction machine + load
T
Tmax
Load characteristic (T)L
P
0
1
ns
0
sb
n
s
Fig.11 Mechanical characteristics of induction motor and the load; P – point of operation
at steady state
T
Tmax
,,
R2
0
,,
1= s b
>
,
R2
,
sb
>
R2
sb
ns
0
n
s
Fig.12. Family of mechanical characteristics at different rotor resistances
12
The supply voltage and frequency also influences the torque-speed characteristic.
This will be discussed later on at speed control.
The input power before is converted to the mechanical power on the rotor shaft is
partially lost in the stator and the rotor. The power flow diagram shown in Fig.13
illustrates the way the power is transmitted from the input to output.
∆Pm
Pin = Pelec
Pout = Pshaft
Pm
∆P1
Pag
∆P2
Fig.13 Power flow diagram for motoring: ∆P1 = mR1 I12 , ∆P2 = mR2 I 22 , ∆PFe ≅ 0
The efficiency of the induction motor is
Eff =
Pout
Pin
(37)
If we neglect all losses except those in the resistance of the rotor circuit
Pag = Pin
(38)
∆P2 = sPag
(39)
a f
(40)
Rotor power loss
and output power
Pout = Pm = Pag 1 − s
The ideal efficiency is
Effideal =
Pout Pag (1 − s )
=
= (1 − s )
Pin
Pag
(41)
The above formula indicates that an induction machine must operate near its
synchronous speed if high efficiency is desired. This is why the slip is very low for
normal operation of the induction machine.
If other losses are included, the actual efficiency is lower than the ideal efficiency of
Eq.41. The full-load efficiency of a large induction motor may be as high as 95 %.
13
The induction machine when operates in other than motoring modes has the negative
mechanical power. This is also seen when (s < 0) or (s > 1) is put to the formula of
resistance that symbolizes the mechanical power:
1− s
1− s ' 2
< 0 and Pm = mR2'
I2 < 0
s
s
2
R'
Pag = m 2 I2' < 0
s
' 1− s
' 1− s ' 2
< 0 and Pm = mR2
I2 < 0
At plugging mode: s > 1 , R2
s
s
2
R'
Pag = m 2 I2' > 0
s
d i
d i
d i
d i
At generating mode: s < 0 , R2'
At generating mode the power “flows” from rotor to stator and next to supply. At
plugging mode the power “flows” from both sides, and as a result the loss in the rotor
circuit, ∆P2 , is enormously increased.
The power flow for these two modes is shown in Fig.14 and 15.
∆Pm
Pout = Pelec
P in = Pshaft
∆P1
Pag
∆P2
Pm
Fig.14. Power flow diagram for generating mode
∆Pm
Pin= Pelec
P in = Pshaft
∆P1
Pag
Pm
Pout= ∆P2
Fig.15. Power flow diagram for plugging mode
14
6. STARTING OF INDUCTION MOTORS
To start the motor special attention should be pay to avoid to large starting current
that may flow in the line. There are few methods that are applied, different for slip-ring
and squirrel-cage motors.
6.1 Slip-ring motor
An increase of torque at n = 0 caused by the increase of rotor resistance is used to
start the slip-ring motor. Fig.16 illustrates the starting process when the external
resistance connected to the rotor is being changed. An increase of rotor resistance
contributes not only to the increase of torque but also a decrease of stator current, what is
highly appreciated behavior.
I1
R ex3> R ex2 > R ex1
R ex = 0
I st
T
Tst
n
s
ns
0
0
1
R ex3> R ex2 > R ex1 R ex = 0
TL
Point of operation
0
1
ns
0
n
s
Fig.16 Starting process of slip-ring motor
15
6.2 Squirrel-cage motor
Since there is no access to the rotor resistance of squirrel-cage motor different
methods should be used to start the large power motors. There are two methods:
1) reduction of the supplied voltage,
2) application of deep-bar cage or double-cage to the rotor
The reduction of supplied voltage can be achieved by application of:
a) A three-phase step-down autotransformer,
b) A solid-state voltage controller,
c) A star-to-delta switch.
Two methods will be discussed.
6.2.1 Star-to-delta (Y/∆) switch method
The stator winding is supplied through the switch, which connects it first in star, then
in delta during the starting (Fig.17). This allows reducing the phase voltage and consequently – the starting current. An explanation of this starting process is shown in
Fig.18.
A
B
C
Vp
∆
Stator
Y
Fig.17 Connection of stator winding to Y - ∆ switch
16
∆
Y - connection
IY
A
V
B
I∆
A
C
I pY
- connection
V
C
B
Ip∆
VpY
Vp ∆
V pY =
V
3
V p∆ = V
Ip =
Vp
I pY =
I p∆
Zp
3
I p∆ =
I pY = IY
I∆
3
I∆
3
T
TY = ∆
3
IY =
Fig.18 Explanation to an application of Y/∆ switch
Applying the same line-to-line voltage to the winding, first connected in star, then in
delta the phase voltage in star is 3 times lower than the one at delta connection. The
same is with the phase currents. The line currents in both cases differ 3 times from one
another. Unfortunately the same is with the torque. At star connection the torque is 3times lower than that at delta connection. The whole starting process is illustrated in
Fig.19 with the aim of torque-speed and current-speed characteristics. The stator current
does not exceed the desire current Id during the starting process.
17
I1
I∆
Id
I st
0
1
IY
ns
0
n
s
ns
0
n
s
T
T∆
TY
Tst
0
1 1
Fig.19 Current-speed and torque-speed characteristics at starting by using Y/∆ switch
6.2.2 Deep-bar squirrel cage rotor
To modify the torque-speed characteristic and in consequence – the starting torque
there are applied the deeper cage bars (Fig.20). At speed equal to 0 the rotor current
frequency is equal to supply frequency. Due to the slot opening the lower layers of the
bar experience higher impedance than the upper layers. It causes the current flow
distribution in the whole cross-section of the bar non-uniform with the higher current
18
density close to the air-gap (see Fig.20). This in fact makes the resistance of the bar R2
higher what contributes to an increase of the starting torque and decrease of starting
current, similar to that what was done for starting of slip-ring motor. When the rotor
speed goes up the rotor frequency lowers. This causes the impedance of the lower part of
the bar becomes lower too. The current can flow now across the entire cross-section of
the bar with the same density. The effective resistance of the cage decreases, what
contributes to an increase of the current and the torque. Fig.21 illustrates the change of
the torque-speed characteristics during the starting process.
(a)
h
i2
Φs2
(b)
R2eq
0
60 Hz
f2
Fig.20.a Deep-bar of the rotor and characteristics of the current density i2 in the bar, 20.b
Equivalent rotor resistance as the function of rotor current frequency.
19
T
Tst
ns
0
0
1
n
s
Fig.21 Torque-speed characteristic modified by the increase of the equivalent impedance
of the bar
Similar process takes place for the rotor with double-cage. The upper cage is called
start cage, since the current at start flows mainly through this cage. During the normal
operation most of the current flows in the inner cage, called work cage (Fig.22).
start cage
work cage
Fig.22 Double-cage rotor bars
20
7. SPEED CONTROL
From the rotor slip: s =
ns − n
we can derive the formula for rotor speed:
ns
n = n1 (1 − s)
=
(42)
120 f
(1 − s)
p
Looking at Eq. 42, the rotor speed can be varied by changing:
• frequency
• number of poles
• slip, which depends on: - supply voltage
- rotor resistance
7.1 Line frequency control
The change of frequency influences not only the rotating field speed but the
maximum torque according to equation 35 where:
Tmax
⎛V ⎞
V12
m V12
m
=
=
= KT' ⎜ 1 ⎟
'
ω1 2 X l 2 2π f1 2 ⋅ 2π f1 Ll 2
⎝ f ⎠
2
(43)
. Due to that two techniques are applied to control the speed:
a) constant voltage technique: f = var., V1 = const
b) constant torque (flux) technique: f = var., V1 = var.
ad a) constant voltage
Torque-speed characteristics in Fig.23 illustrate how the rotor speed changes at
constant load torque TL.
when f ↓ , then T ↑, n ↓ at constant load torque.
ad b) constant flux
The supply voltage
V1 ≅ E1 = 4.44 fNKw Φ
Hence
Φ = KΦ
V1
f
21
Fig.24 shows how the speed varies when the voltage and frequency change at the
V
V
ratio Φ ≈ 1 = const . When f1 ↓ and V1 ↓ at 1 = const , then T = const , while n ↓ at
f
f1
const load torque.
T
f (4) < f (3) <
f (2) < f (1)
TL
0
n(4) < n (3) < n (2) < n (1)
n
s
Fig.23 Variation of rotor speed according to the change of line frequency at constant
supply voltage
T
f (4) < f (3) < f (2) < f (1)
TL
n
0
n(4) < n (3) < n (2) < n (1)
Fig.24 Variation of rotor speed according to the change of line voltage frequency at
constant magnetic flux
22
7.2 Line voltage control
A set of T-n characteristics for various terminal voltages is shown in Fig.25. The
maximum torque varies according to equation:
2
Tmax =
m V1
= KV V12
'
ω1 2 X l 2
while the slip at maximum torque remains unchanged as indicates the equation below:
R2'
sb ≅ ± '
Xl2
The terminal voltage can be varied by using a three-phase autotransformer or a solid-state
voltage controller.
T
V(1) > V(2) > V(3)
TL
ns
0
n
n b n (3) n (1)
n (2)
Fig.25 Speed control by changing of line voltage
When V1 ↓ - then n ↓ at const load torque. This type of control is applied to small
power motors, e.g. motors, which drive the fans.
23
7.3 Rotor resistance control
According to Eqs.35 and 34, which are quoted below:
R2'
sb ≅ ± '
Xl2
2
Tmax =
m V1
ω1 2 X l' 2
the change of rotor resistance influences the slip at maximum torque, while the
breakdown torque remains unchanged. This illustrates Fig.26, which shows the set of T-n
characteristics drawn for various rotor resistances.
T
R ex3> R ex2 > R ex1
R ex = 0
TL
ns
0
n
n (2)
n(4)
n (3)
n (1)
Fig.26 Control of rotor speed by rotor resistance
When R21 ↑ then n ↓ at constant load torque. The major disadvantage of the rotor
resistance control method is that the efficiency is low at reduced speed because of higher
slips. However, this control method is often employed because of its simplicity. It is
applied only to slip-ring motors.
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7.4 Pole changing
A change of number of poles of the machine causes a change of synchronous speed. This
change can be done by changing the coil connections of the stator winding as shown in
Fig.27. This method provides two synchronous speeds.
N
S
p=2
I
A1
S
A2
N
S
S
N
p=4
I
A1
A2
Fig.27 Pole changing method for rotor speed control
This method can be applied only to squirrel-cage motors because the change of pole
number must be applied to both stator and rotor windings and the cage-rotor can operates
with any number of stator poles. It is obvious, however, that the speed can be changed
only in discrete steps and that the elaborate stator winding makes the motor expensive.
The motor with the possibility of pole change is called multiple-speed motor.
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8. SINGLE-PHASE INDUCTION MOTORS
The motor has cage rotor and single-phase distributed or concentrated stator winding.
The motor scheme with the latter-type of winding is shown in Fig.28.
Φ
stator
rotor
stator winding
rotor cage
V1
Fig.28. Scheme of the single-phase induction motor.
The single-phase winding produces the alternating magnetomotive force that
contributes to alternating magnetic flux. As was stated before (see section Introduction
to Electric Motors), such a magnetic flux may be represented by two flux components:
one rotating forwards and another - backwards with the same absolute speeds (Fig.29).
Each component develops its own torque as showed in Fig.30. The resultant torque at
standstill is equal to 0. It means, that the single-phase motor does not develop the starting
torque. When the motor begins to run it develops the torque directed towards the speed it
currently operates. In order to make the motor start rotating, some arrangement is
required so that the motor produces a starting torque.
ΣΦ
+
Φ
+
ω
Φ
_
ω−
Fig.29 The alternating magnetic flux ΣΦ represented by two rotating fluxes
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T
T = T ++ T
-n s
-
0
T
T+
ns
n
-
Fig.30. Torque-speed characteristics developed by the flux components of the singlephase motor
In general there are two methods for producing the starting torque:
• motor with auxiliary winding, and
• shaded-pole motor
8.1 Motor with auxiliary winding
A schematic diagram of the motor is shown in Fig.31. The auxiliary winding is placed
perpendicularly to the main winding. To get the rotating magnetic field from both
windings their currents must be shifted in time too, preferably by the same angle. It can
be obtained by connecting in series with auxiliary winding either:
• capacitor (capacitor-start motor), or
• resistor (split-phase motor).
In Fig.31 the first solution is applied. The capacitor changes the phase angle between
the voltage and current (Fig.32). The value of the capacitance should be chosen to get the
shift angle equal to 90o between two fluxes in order to get a circle-rotating field, which
gives the best motor performance (Fig.33). After the motor starts the auxiliary winding is
switched off at approximately 0.8ns by means of centrifugal switch (Fig.31).
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Fig.31 Scheme of the circuit of the single-phase motor with an auxiliary winding
Fig.32 Phasor diagrams of the motor currents and magnetic fluxes produced by the main
winding and auxiliary winding
T
T = Ta + Tm
Tm
TL
0
n ns
n
Fig.33 Torque-speed characteristics, which illustrate the starting process of single-phase
motor
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8.2 Shaded-pole motor
The motor has a salient pole construction (Fig.34). A shaded band consisting of a
short-circuited copper turn, known as a shading coil, is used on one portion of each pole.
The main single-phase winding is wound on one “leg” of the core, opposite to the “leg”
with rotor.
Φ
stator
shading coil
stator winding
n
rotor cage
V1
rotor
Fig.34 Scheme of shaded-pole single-phase motor
The role, which plays the shading coil, is explained in Fig.35. The flux that links the
coil induces in it the current (voltage) which next generates its own flux, which is added
to (or subtracted from) the main one. The result is that the flux in the shaded portion of
the pole lags the flux in the unshaded portion of the pole. Therefore the flux in the shaded
portion reaches its maximum after the flux in the unshaded portion reaches its maximum.
This is equivalent to a progressive shift of the flux from the unshaded to the shaded
portion of the pole. It is similar to a rotating field moving from the unshaded to the
shaded portion of the pole. As a result, the motor produces a starting torque.
Fig.35 Phasor diagram that illustrates the generation of rotating flux in shaded-pole motor
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The typical torque-speed characteristic is shown in Fig.36. Shaded-pole motors are
the least expensive motors of the fractional horsepower motors and are generally built for
low horsepower rating.
T
Tm
TL
P
0
n ns
n
Fig.36 A typical torque-speed characteristic of shaded-pole motor
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