# AC DRIVES

```AC MOTOR DRIVES
AC MOTOR DRIVES
AC motor Drives are used in many industrial and domestic
application, such as in conveyer, lift, mixer, escalator etc.
The AC motor have a number of advantages :
• Lightweight (20% to 40% lighter than equivalent DC motor)
• Inexpensive
• Low maintenance
• 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 – DC – AC CONVERSION
control
THE INDUCTION MOTOR BASICS
This type of drives are commonly called variable-speed drives (VSD).
The basic layout of a three-phase induction motor consists of:
Three-phase
windings
Rotor windings
• Stator
Threephase
supply
• Rotor
• Air gap

Stator
Air gap
s
Rotor
m
T
Three-phase
windings
Rotor windings
Threephase
supply

Stator
Air gap
s
Rotor
m
T
The stator windings are supplied with balanced three-phase AC voltage,
which produce induced voltage in the rotor windings. It is possible to
arrange the distribution of stator winding so that there is an effect of
multiple poles, producing several cycles of magnetomotive force (mmf) or
field around the air gap.
The speed of rotation of field is called the synchronous speed s ,
which is defined by :
2
Ns is syncronous speed [rpm]
s 
or
p is numbers of poles
p
ω is the supply frequency [rad/sec]
f is the supply frequency [Hz]
120 f
Ns 
Nm is motor speed
p
THE INDUCTION MOTOR SPEED
The rotor speed or motor speed is :
Where S is the slip defined as :
m  s (1 S )
S
 S  m
S
Three-phase
windings
S
or
Rotor windings
Threephase
supply

Stator
Air gap
s
Rotor
m
T
NS  Nm
NS
EQUIVALENT CIRCUIT OF AN INDUCTION MOTOR
Three-phase
windings
Rotor windings
Threephase
supply
Where :
Rs is resistance per-phase of
stator winding

Stator
Air gap
Rotor
m
s
Is
Xs
T
Xs is leakage reactance perphase of the winding stator
Xr’
Rs
Ir’
Im
Rr’/s
Vs
Xm
Rm
Rr is resistance per-phase of rotor
winding
Xs is leakage reactance perphase of the winding rotor
Xm is magnetizing reactance
Stator
Air gap
motor
Rm is Core losses as a reactance
INDUCTION MOTOR LOSSES
Is
Xs
Xr’
Rs
Im
Ir’
Rr’/s
Vs
Xm
Stator
Rm
Air gap
motor
Stator copper loss :
Ps cu  3 I s Rs
Rotor copper loss :
Pr cu  3 ( I r ) 2 Rr
2
'
2
Core losses :
2
V
V
Pc  3 m  3 s
Rm
Rm
'
DEVELOPED POWER AND TORQUE OF AN INDUCTION
MOTOR
- Power developed on air gap (Power from stator to
'
rotor through the air gap) :
R
'
Pg  3 ( I r ) 2
- Power developed by motor :
or
Pd  Pg (1  S )
- Torque of motor :
or

Td 
Pg (1  S )
S (1  S )
Pd
m

Pg
s
r
S
'
R
Pd  Pg  Pr cu  3 ( I r ) r (1  S )
S
' 2
or
Pd 60
Td 
2 N m
EFFICIENCY OF AN INDUCTION MOTOR
Input power of motor :
Output power of motor :
Efficiency :
If
Pg  ( Pc  Ps cu )
and
Pi  3Vs I s cos m Pc  Ps cu  Pg
Po
 
Pi Pc  Ps cu  Pg
so, the efficiency can be calculated as :
Pd Pg (1  S )
 
1  S
Pg
Pg
SIMPLIFIED EQUIVALENT CIRCUIT OF AN INDUCTION
MOTOR
Generally, value of reactance magnetization Xm >> value Rm (core
losses) and also
X m  ( Rs  X s )
2
2
2
So, the magnetizing voltage same with the input voltage :
Vm  Vs
Therefore, the equivalent circuit is :
Is
Xs
Ii
Xr’
Rs
Im
Xm
Stator
Rm
Air gap
Xr’
Rs
Rr’/s
Vs
Xm
Po
Pi
motor
Is=Ir’
Ir’
Im
Ir’
Rr’/s
Vs
Xs
Stator
Air gap
rotor
Ii
Xs
Xr’
Rs
Is=Ir’
Ir’
Im
The equivalent circuit is :
Rr’/s
Vs
Xm
Po
Pi
Stator
Air gap
rotor
'
R
 X m ( X s  X r )  jX m ( Rs  r )
S
Zi 
'
Rr
'
Rs 
 j( X m  X s  X r )
S
'
Total Impedance of this circuit is :
Ir 
Vs
'
The rotor current is :

Rr' 
  X s  X r '
 Rs 
S 

2


2



1
2
INDUCTION MOTOR TORQUE vs SPEED CHARACTERISTIC
Ii
Xs
Xr’
Rs
Is=Ir’
Ir’
Im
Rr’/s
Vs
Po
Pi
Stator
Tmax
Air gap
Td 
3 Rr' Vs2
' 2

R 
S s  Rs  r   X s  X r'
S 


rotor
Td
Tst
TL
Tm=TL
Operating point
S=1
Nm =0
Smax S=Sm S=0
m  s
Nm N s

2



THREE - REGION OPERATION
1. Motoring :
0 S  1
2. Regenerating :
S 0
3. Plugging :
1 S  2
Torque
Forward
regeneration
Forward
motoring
Reverse
plugging
Tmax
m  s
m s
m s
Tst
S=Sm
-Smax
s
S=-1
s Smax
S=0
Ns
-Tmax
s =0
S=1
Nm =0
 s
S=2
STARTING TORQUE OF AN INDUCTION MOTOR
Starting speed of motor is m = 0 or S = 1,
Starting torque of motor is : Tst 
3 Rr' Vs2
' 2

Rr 
s  Rs    X s  X r'
S 


Slip for the maximum torque Smax can be found by setting :
So, the slip on maximum torque is :
S max  

2



d Td
0
dS
Rr'
R   X  X  
2
s
s
1
2
'
2
r
Speed-Torque Characteristic :
Td 
2

Rr' 
S s  Rs    X s  X r'
S 

X
For the high Slip S. (starting)
So, the torque of motor is :
3 Rr' Vs2

s
Td 
And starting torque (slip S=1) is :
X

' 2
r

R 
  Rs  
S 

'
r
3 Rr' Vs2

S s X s  X
Tst 

2

' 2
r
'
r
2
s
3R V
 s X s  X

' 2
r
2



For low slip S region, the motor speed near unity or synchronous
speed, in this region the impedance motor is :
X
So, the motor torque is :
s
X

' 2
r
R'r

 Rs
S
3Vs2 S
Td 
 s R 'r
And the slip at maximum torque is :
S max  
Rr'
R   X  X  
1
2
'
2
r
2
s
The maximum motor torque is :
Td 
s
3 Rr' Vs2
' 2

R 
S s  Rs  r   X s  X r'
S 



2



STATOR VOLTAGE CONTROL
AC
Variable
Voltage
Sources
Controlling Induction Motor Speed by
Td 

Rr' 
S s  Rs    X s  X r'
S

Ii
Xs

Xr’
Rs

Vs
Td
3 Rr' Vs2
2
IM

2



Is=Ir’
Ir’
Im
Vs > Vs1 > Vs2
Tmax
TL
Tst
Tst1
Tst2
Rr’/s
Vs
Po
Pi
Stator
air
gap
rotor
S=1
Nm =0
2 1 
S=0
s
Ns
Td
STATOR FREQUENCY CONTROL
Td
fs2 < fs1 < fs
Ii
Xs
Xr ’
Rs
Ir’
Im
Vs
Tmax
Is=Ir’
Tst2
Rr’/s
f
Po
Pi
Stator
Air
gap
Tst1
Tst
rotor
S=1
 m =0
Since
Eag = k f ag
TL
1
S=0
2
S=0
fs1
fs2
𝝓𝒂 =
𝑬𝒂
≈
s
S=0
fs
𝑽
Speed is adjusted by varying f - maintaining V/f constant to avoid flux saturation
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:
s   b
And :
The motor torque :
S
b  m

1  m
b
b
Td 
3 Rr' Vs2
2

Rr' 
S s  Rs    X s  X r'
S 

Td 


2



3 Rr' Vs2

Rr'
S b  Rs 
S

2

  X s  X r'



2



If Rs is negligible, the maximum torque at the base speed as :
Tmb
3 Vs2

2S b X s  X r'


2
Vs
3
And the maximum torque at any other frequency is : Tm 
2S b X s  X r'  2

Sm 
At this maximum torque, slip S is :
Rr '
 X s  X r'


2
Vs
3
Tm 
2S b X s  X r'  2


Normalizing :
Tmb
3 Vs2

2S b X s  X r'

Tm
1
 2
Tmb 

Tm   Tmb
2
And

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 :
Base frequency :
460
 265 volt
3
b  2  f  2 x 3.14 x 60  377 rad / s
Base Torque : Tmb 
Motor Torque :
Vs 
60 Po
60 x11200

 61.11 Nm
2  N m 2 x 3.14 x1750
Tm  35 Nm
a) the frequency of supply voltage :
Tm
1
 2
Tmb 
Tmb
61.11


 1.321
Tm
35
Synchronous speed at this frequency is :
s 1.321 x 377  498.01rad / s
s   b
or
60 x 498.01
 4755.65 rpm
N s   Nb 
2
So, the supply frequency is :
p NS
4 x 4755.65
fs 

158.52 Hz
120 b
120
b) speed of motor at the maximum torque :
At this maximum torque, slip Sm is :
Sm 
Rr '
 X s  X r'


Rr’ = 0.38W, Xs = 1.14W, Xr’ = 1.71W and   1.31
So,
Sm 
0.38
or,
 0.101
1.3211.14 1.71
N m  N S (1  S )  4755.65 (1  0.101)  4275 rpm
TYPES OF AC DRIVES
• Static frequency changers
• Static voltage controllers
• Rectifier – inverter systems with line commutation
• Rectifier – inverter systems with self commutation
• Pulse Width Modulation systems
STATIC FREQUENCY CHANGER
SQUIRREL-CAGE INDUCTION MOTOR WITH
CYCLOCONVERTER
Single phase
waveform
STATIC VOLTAGE CONTROLLER
Significant reduction of
the torque is noted
RECTIFIER – INVERTER SYSTEMS WITH LINE COMMUTATION
Synchronous motor drive
Induction motor with
wound rotor drive
RECTIFIER – INVERTER SYSTEMS WITH SELF COMMUTATION
PULSE WIDTH MODULATION (PWM) SYSTEMS
PULSE WIDTH MODULATION CONVERTERS
PULSE WIDTH MODULATION CONVERTERS
PWM wave
formation
Effect of
narrower
pulses for
voltage
regulation
CONTROLLING INDUCTION MOTOR SPEED
USING ROTOR RESISTANCE
(Rotor Voltage Control)
Equation of Speed-Torque :
In a wound rotor induction motor, an external
three-phase resistor may be connected to its
slip rings,
Td 
3 Rr' Vs2

Rr'
S s  Rs 
S

2

  X s  X r'


3Vs2 S
Td 
 s R 'r
RX
Stator
RX
Rotor
Three-phase
supply
RX

2



The 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.
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)
THREE-PHASE DIODE RECTIFIER AND A
DC CHOPPER
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:
Re  R(1  k )
Where k is duty cycle of DC chopper
The speed can controlled by varying the duty cycle k, (slip power)
Id
Ld
Stator
D1
D3
Vd
Rotor
Three-phase
supply
D5
D4
D6
D2
GTO
R
Vdc
POWER RECOVERY IN A WOUND-ROTOR
INDUCTION MOTOR
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
Transformer
Id
Ld
Stator
D1
D3
Vd
Rotor
Slip Power
D5
D4
D6
Diode rectifier
D2
T1
T3
T5
T2
T4
T6
Vdc
Controlled rectifier/
inverter
Na:Nb
Example:
A three-phase induction motor, 460V, 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 W and
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.
Id
Ld
Stator
D1
D3
Vd
Rotor
Three-phase
supply
Vs 
D5
D4
D6
460
 265.58 volt
3
p 6
  2 x 60  377 rad / s
s  2 x 377 / 6  125.66 rad / s
D2
GTO
R
Vdc
The dc voltage at the rectifier output is :
Vd  I d Re  I d R (1  k )
and
Nr
Er  S Vs
 S Vs nm
Ns
For a three-phase rectifier, relates Er and Vd as :
Vd 1.65 x 2 Er  2.3394 Er
Using :
Nr
Er  S Vs
 S Vs nm
Ns
Vd  2.3394 S Vs n m
Pr
If Pr is the slip power, air gap power is : Pg 
S
Pr
3Pr (1  S )
 S) 
Developed power is : Pd  3( Pg  Pr )  3(
S
S
Because the total slip power is 3Pr = Vd Id and
Pd  TL m
(1  S )Vd I d
Pd 
 TLm  TLm (1  S )
S
Substituting Vd from
Vd  2.3394 S Vs n m
Solving for Id gives :
TLs
Id 
2.3394Vs nm
In equation
Pd above, so :
Which indicates that the inductor current is independent of the speed.
From equation :
So,
Which gives :
Vd  I d Re  I d R (1  k )
and equation :
I d R(1  k )  2.3394 S Vs nm
I d R(1  k )
S
2.3394 S Vs n m
Vd  2.3394 S Vs n m
The speed can be found from equation :
I d R(1  k )
S
2.3394 S Vs n m
as :

I d R(1  k ) 
m  s (1  S )  s 1 

2
.
3394
V
n
s m


TLs R(1  k ) 
m  s 1 
2
(
2
.
3394
V
n
)
s m


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 s
m  180  / 30  83.77 rad / s
From torque equation : TL
 K vm
2
2
 800 
 750 x 
  347.67 Nm
 1175 
From equation :
TLs
Id 
2.3394Vs nm
The corresponding inductor current is :
Id 
347.67 x 125.66
 78.13 A
2.3394 x 265.58 x 0.9
The speed is minimum when the duty-cycle k is zero and equation :

I d R(1  k ) 
m  s (1  S )  s 1 

2
.
3394
V
n
s m

78.13 R
)
83.77  125.66(1 
2.3394 x 265.58 x 0.9
And :
R  2.3856 W
SOFT STARTERS
SOFT STARTER MODES
SOFT STARTER MODES
```
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