Induction and synchronous Motor drives (AC motor drives)

AC motor drives are used in many industrial and domestic application, such as in conveyer, lift, mixer, escalator etc,...

1. AC DRIVES
The AC motor have a number of advantages :
•Lightweight (20% to 40% lighter than equivalent DC motor)
•Inexpensive
•Low maintenance
The Disadvantages AC motor :
* The power control relatively complex and more expensive
There are two type of AC motor Drives :
1.Induction Motor Drives
2.Synchronous Motor Drives
AC motor Drives are used in many industrial and domestic
application, such as in conveyer, lift, mixer, escalator etc.

2. INDUCTION MOTOR DRIVES
Three-phase induction motor are commonly used in adjustable-speed
drives (ASD).
Basic part of three-phase induction motor :
•Stator
•Rotor
•Air gapStator
Three-phase
windings
RotorAir gap
 T
Rotor windings
Three-
phase
supply
m
s


3. Thestatorwindingaresuppliedwithbalancedthree-phaseACvoltage,
whichproduceinducedvoltageintherotorwindings.Itispossibleto
arrangethedistributionofstatorwindingsothatthereisaneffectof
multiplepoles,producingseveralcycleofmagnetomotiveforce(mmf)or
fieldaroundtheairgap.
The speed of rotation of field is called the synchronous speed
s , which
is defined by :p
s


2

ω
sissyncronousspeed[rad/sec]
N
sissyncronousspeed[rpm]
pisnumbersofpoles
ωisthesupplyfrequency[rad/sec]
fisthesupplyfrequency[Hz]
N
mismotorspeedStator
Three-phase
windings
RotorAir gap
 T
Rotor windings
Three-
phase
supply
m
s
 p
f
N
s
120

or

4. Stator
Three-phase
windings
RotorAir gap
 T
Rotor windings
Three-
phase
supply
m
s
 The rotor speed or motor speed is :)1(S
sm

Where S is slip, as defined as : S
mS
S


 Or S
mS
N
NN
S


The motor speed

5. Equivalent Circuit Of Induction MotorStator
Three-phase
windings
RotorAir gap
 T
Rotor windings
Three-
phase
supply
m
s
 Stator Air gap motor
Vs
Is
Im Ir’
Xs
Xr’Rs
RmXm
Rr’/s
Where :
Rs is resistance per-phase of stator winding
Rr is resistance per-phase of rotor winding
Xs is leakage reactance per-phase of the
winding stator
Xs is leakage reactance per-phase of the
winding rotor
X
mis magnetizing reactance
R
m is Core losses as a reactance

6. Performance Characteristic of
Induction Motor
Stator copper loss :Stator Air gap motor
Vs
Is
Im Ir’
Xs
Xr’Rs
RmXm
Rr’/s sscus
RIP
2
3 '2'
)(3
rrcur
RIP
Rotor copper loss :m
s
m
m
c
R
V
R
V
P
22
33
Core losses :

7. S
R
IP
r
rg
'
2'
)(3 )1()(3
'
2'
S
S
R
IPPP
r
rcurgd
 )1(SPP
gd
 -Powerdevelopedonairgap(Powerfropmstatorto
rotorthroughairgap):
Performance Characteristic of
Induction Motor
-Powerdevelopedbymotor:
or
-Torqueofmotor:m
d
d
P
T

 s
g
S
g P
S
SP





)1(
)1(
orm
d
d
N
P
T
2
60

or

8. mssi IVP cos3 gcusc
PPP  Inputpowerofmotor:
Performance Characteristic of
Induction Motorloadnodo
PPP  gcusc
loadnod
i
o
PPP
PP
P
P



Outputpowerofmotor:
Efficiency:

9. )(
cuscg
PPP  loadnod
PP S
P
SP
P
P
g
g
g
d


 1
)1(
 If
and
so,theefficiencycancalculatedas:
Performance Characteristic of
Induction Motor

10. )(
222
ssm
XRX  Generally,valueofreactancemagnetizationX
m
>>valueRm(core
losses)andalso
So,themagnetizingvoltagesamewiththeinputvoltage:smVV
Therefore, the equivalent circuit is ;Stator Air gap motor
Vs
Is
Im Ir’
Xs
Xr’Rs
RmXm
Rr’/s Stator Air gaprotor
Vs
Ii
Im Ir’
Xs Xr’Rs
Rr’/s
Pi
Po
Is=Ir’
Xm
Performance Characteristic of
Induction Motor

11. )(
)()(
'
'
'
'
rsm
r
s
r
smrsm
i
XXXj
S
R
R
S
R
RjXXXX
Z


 Total Impedance of this circuit is :
Performance Characteristic of
Induction MotorStator Air gaprotor
Vs
Ii
Im Ir’
Xs Xr’Rs
Rr’/s
Pi
Po
Is=Ir’
Xm
The rotor current is : 
2
1
2
'
2
'
'



















rs
r
s
s
r
XX
S
R
R
V
I

12.  



















2
'
2
'
2'
3
rs
r
ss
sr
d
XX
S
R
RS
VR
T
 Stator Air gaprotor
Vs
Ii
Im Ir’
Xs Xr’Rs
Rr’/s
Pi
Po
Is=Ir’ Tmax
Smax
Tst
Td
S=0
s
Ns
S=1
TL
S=Sm
m
NmNm =0
Tm=TL
Operating point Torque –speed Characteristic

13. Tmax
Smax
Tst
S=0
s
Ns
S=1
S=Sm
Nm =0
ss =0
s
-Tmax
-Smax
S=-1
S=2
Forward
regeneration
Forward
motoring
Reverse
plugging
s
m
sm
sm
Torque Three region operation :
1.Motoring :
2.Regenerating :
3.Plugging :10S 0S 21S

14. Starting speed of motor is m= 0 or S = 1,
Performance Characteristic of
Induction Motor
Starting torque of motor is : 



















2
'
2
'
2'
3
rs
r
ss
sr
st
XX
S
R
R
VR
T

Slip for the maximum torque S
maxcan be found by setting :0
dS
Td
d
So, the slip on maximum torque is :  
2
1
2
'2
'
max
rss
r
XXR
R
S



15.  






2'2
2
max
2
3
rssss
s
XXRR
V
T
 Performance Characteristic of
Induction Motor
Torque maximum is :
And the maximum regenerative torque can be found as : 






2'2
2
max
2
3
rssss
s
XXRR
V
T

Where the slip of motor s = -S
m

16.  



















2
'
2
'
2'
3
rs
r
ss
sr
d
XX
S
R
RS
VR
T
 Speed-Torque Characteristic : 
2
'
2
'









S
R
RXX
r
srs  
2
'
2'
3
rss
sr
d
XXS
VR
T


  
2
'
2'
3
rss
sr
st
XX
VR
T



ForthehighSlipS.(starting)
So,thetorqueofmotoris:
Andstartingtorque(slipS=1)is:

17.  
s
r
rs R
S
R
XX 
'2
' rs
s
d
R
SV
T
'
3
2

 ForlowslipSregion,themotorspeednearunityorsynchronous
speed,inthisregiontheimpedancemotoris:
So, the motor torque is :  
2
1
2
'2
'
max
rss
r
XXR
R
S


And the slip at maximum torque is :
The maximum motor torque is : 



















2
'
2
'
2'
3
rs
r
ss
sr
d
XX
S
R
RS
VR
T


18. Tmax
S=0
s
Ns
S=1
TL
Nm=0
Td
Vs1Vs Vs2>>
12
Tst
Tst1
Tst2 Stator VoltageControl
Controlling Induction Motor Speed by
Adjusting The Stator Voltage Td
IM
AC
Variable
Voltage
Sources
Vs Stator
air
gap
rotor
Ii
Im Ir’
Xs Xr’Rs
Rr’/s
Pi
Po
Is=Ir’
Vs  



















2
'
2
'
2'
3
rs
r
ss
sr
d
XX
S
R
RS
VR
T


19. Frequency VoltageControl
Controlling Induction Motor Speed by
Adjusting The Frequency Stator Voltage Td
IM
AC
Variable
Voltage
Sources
Vs
f Stator
Air
gap
rotor
Ii
Im Ir’
Xs Xr’Rs
Rr’/s
Pi
Po
Is=Ir’
Vs
f T
max
S=0
S=1
T
L
m=0
T
d
<<
12
T
st
T
st1
T
st2
s
fs
S=0
fs1fs2
S=0
fs2 fsfs1
  



















2
'
2
'
2'
3
rs
r
ss
sr
d
XX
S
R
RS
VR
T


20. If the frequency is increased above its rated value, the flux and torque
would decrease. If the synchronous speed corresponding to the rated
frequency is call the base speed 
b,
the synchronous speed at any other
frequency becomes:bs
And : b
m
b
mb
S






 1
The motor torque :  



















2
'
2
'
2'
3
rs
r
ss
sr
d
XX
S
R
RS
VR
T
  



















2
'
2
'
2'
3
rs
r
sb
sr
d
XX
S
R
RS
VR
T


21. If R
sis negligible, the maximum torque at the base speed as :  
'
2
2
3
rsb
s
mb
XXS
V
T



And the maximum torque at any other frequency is :  
2
2
'
2
3

s
rsb
m
V
XXS
T


At this maximum torque, slip S is :  
'
'
rs
r
m
XX
R
S



Normalizing :  
'
2
2
3
rsb
s
mb
XXS
V
T


  
2
2
'
2
3

s
rsb
m
V
XXS
T

 2
1


mb
m
T
T
And mbm TT 
2


22. Example :
A three-phase , 11.2 kW, 1750 rpm, 460 V, 60 Hz, four pole, Y-connected
induction motor has the following parameters : Rs = 0.1W,Rr’ = 0.38W,Xs =
1.14W,Xr’ = 1.71W,and Xm = 33.2W.If the breakdown torque requiretment is
35 Nm, Calculate : a) the frequency of supply voltage, b) speed of motor at
the maximum torque
Solution :
Input voltage per-phase : voltV
s 265
3
460
 sradxxf
b /3776014.322 
Base frequency :Nm
xx
x
N
P
T
m
o
mb 11.61
175014.32
1120060
2
60

 NmT
m35
Base Torque :
Motor Torque :
a) the frequency of supply voltage :2
1


mb
m
T
T 321.1
35
11.61

m
mb
T
T


23. Synchronous speed at this frequency is :bs sradx
s /01.498377321.1 
orrpm
x
x
NN
bs 65.4755
2
01.49860



So, the supply frequency is :Hz
xNp
f
b
S
s
52.158
120
65.47554
120

b) speed of motor at the maximum torque :
At this maximum torque, slip S
mis :  
'
'
rs
r
m
XX
R
S



Rr’ = 0.38W, Xs = 1.14W, Xr’ = 1.71Wand 1.31 
101.0
71.114.1321.1
38.0



mS
So,rpmSNN
Sm 4275)101.01(65.4755)1( 
or,

24. CONTROLLING INDUCTION MOTOR SPEED USING
ROTOR RESISTANCE
(Rotor Voltage Control)

25. Wound rotor induction motor applications
cranes

26. CONTROLLING INDUCTION MOTOR SPEED USING
ROTOR RESISTANCE
(Rotor Voltage Control)
Equation of Speed-Torque : 



















2
'
2
'
2'
3
rs
r
ss
sr
d
XX
S
R
RS
VR
T
 rs
s
d
R
SV
T
'
3
2


In a wound rotor induction motor, an external
three-phase resistor may be connected to its
slip rings, Three-phase
supply
Rotor
Stator
RX
RX
RX

27. These resistors Rx are used to control motor starting and stopping
anywhere from reduced voltage motors of low horsepower up to
large motor applications such as materials handling, mine hoists,
cranes etc.
The most common applications are:
AC Wound Rotor Induction Motors –where the resistor is wired into the
motor secondary slip rings and provides a soft start as resistance is
removed in steps.
AC Squirrel Cage Motors –where the resistor is used as a ballast for soft
starting also known as reduced voltage starting.
DC Series Wound Motors –where the current limiting resistor is wired to
the field to control motor current, since torque is directly proportional to
current, for starting and stopping.

28. The developed torque may be varying the resistance Rx
The torque-speed characteristic for variations in rotor resistance
This method increase the starting torque while limiting the starting current.
The wound rotor induction motor are widely used in applications requiring
frequent starting and braking with large motor torque (crane, hoists, etc)

29. The three-phase resistor may be replaced by a three-phase diode rectifier
and a DC chopper. The inductor Ld acts as a current source Id and the DC
chopper varies the effective resistance: )1(kRR
e 
Where k is duty cycle of DC chopper
The speed can controlled by varying the duty cycle k, (slip power) Three-phase
supply
Rotor
Stator
D1
D2
D3
D6D4
D5
GTO
R Vdc
Id
Ld
Vd

30. The slip power in the rotor circuit may be returned to the supply by
replacing the DC converter and resistance R with a three-phase full
converter (inverter)Three-phase
supply
Rotor
Stator
D1
D2
D3
D6D4
D5
Id
Ld
Vd
T2
T5
T4
T3T1
T6
Vdc
Transformer Na:Nb
Diode rectifier
Controlled rectifier/
inverter
Slip Power

31. Example:
A three-phase induction motor, 460, 60Hz, six-pole, Y connected, wound rotor
that speed is controlled by slip power such as shown in Figure below. The
motor parameters are Rs=0.041 W, Rr’=0.044 W, Xs=0.29 W, Xr’=0.44 Wand
Xm=6.1 W. The turn ratio of the rotor to stator winding is nm=Nr/Ns=0.9. The
inductance Ld is very large and its current Id has negligible ripple.
The value of Rs, Rr’, Xs and Xr’ for equivalent circuit can be considered
negligible compared with the effective impedance of Ld. The no-load of motor is
negligible. The losses of rectifier and Dc chopper are also negligible.
The load torque, which is proportional to speed square is 750 Nm at 1175 rpm.
(a) If the motor has to operate with a minimum speed of 800 rpm, determine
the resistance R, if the desired speed is 1050 rpm,
(b) Calculate the inductor current Id.
(c) The duty cycle k of the DC chopper.
(d)The voltage Vd.
(e)The efficiency.
(f)The power factor of input line of the motor.

32. voltV
s 58.265
3
460
 6p sradx /377602  sradx
s /66.1256/3772  The equivalent circuit :

33. The dc voltage at the rectifier output is : )1(kRIRIV
dedd  ms
s
r
sr nVS
N
N
VSE 
and
For a three-phase rectifier, relates Er and Vd as :rrd
EExV 3394.2265.1 
Using :ms
s
r
sr nVS
N
N
VSE  msd nVSV 3394.2
If Pr is the slip power, air gap power is :S
P
P
r
g
Developed power is : S
SP
S
S
P
PPP
rr
rgd
)1(3
)(3)(3



34. Because the total slip power is 3Pr = Vd Id and mLdTP
So, )1(
)1(
STT
S
IVS
P
mLmL
dd
d 

 
Substituting Vd from msd nVSV 3394.2 In equation Pd above, so
:
Solving for Id gives :ms
sL
d
nV
T
I
3394.2


Which indicates that the inductor current is independent of the speed.
From equation : )1(kRIRIV
dedd  and equation :msd nVSV 3394.2
So,msd nVSkRI 3394.2)1( 
Which gives :ms
d
nVS
kRI
S
3394.2
)1(


35. The speed can be found from equation :ms
d
nVS
kRI
S
3394.2
)1(
 as :




 

ms
d
ssm
nV
kRI
S
3394.2
)1(
1)1(  




 

2
)3394.2(
)1(
1
ms
sL
sm
nV
kRT

Which shows that for a fixed duty cycle, the speed decrease with load
torque. By varying k from 0 to 1, the speed can be varied from minimum
value to ssrad
m /77.8330/180 
From torque equation :2
mvL
KT Nmx 67.347
1175
800
750
2








36. From equation :ms
sL
d
nV
T
I
3394.2

 The corresponding inductor current is :A
xx
x
I
d 13.78
9.058.2653394.2
66.12567.347

The speed is minimum when the duty-cycle k is zero and equation :




 

ms
d
ssm
nV
kRI
S
3394.2
)1(
1)1(  )
9.058.2653394.2
13.78
1(66.12577.83
xx
R

And :W3856.2R