AC 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 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
3 phase induction motor
drives
STATOR FREQUNCY
CONTROL(cont
CONTROL(
cont….)
….)

V=2Π
V=2
ΠfT
fTφφKW

φ V/F
Low frequency operation at constant voltage:
V constant; f decrease ; φ increases

High frequency operation at constant voltage:
V constant ; f increase ; φ decrease
V/f control(cont
control(cont….)
….)

f increase ; N increase ; Tmax decrease

V increase; ; Tmax increase
(cont
cont….)
….)
(cont
cont….)
….)









Applications:
·
heating,
·
ventilation,
·
air conditioning systems,
·
waste water treatment plants,
·
blowers,
·
fans,
·
textile mills,
·
rolling mills, etc
variable voltage and variable
frequency(v/f) (cont
(cont….)
….)
The variable frequency and variable voltage can be
obtained by,


Voltage source inverter(VSI)
Cycloconverter control
VSI(cont
VSI(
cont….)
….)
VSI
.
Different schemes of VSI

If regeneration is necessary, the phase controlled bridge
rectifier is replaced by dual converter
Different schemes of VSI

Due to chopper ,the harmonic injection into the ac
supply is reduced
Variable current and variable
frequency (CSI) (cont
(cont….)
….)
current source inverter(CSI)
(cont
cont….)
….)

Input voltage kept constant output current
depends upon the nature of load
Different schemes of CSI(cont
CSI(cont….)
….)
Different schemes of CSI(cont
CSI(cont….)
….)
Closed loop control of CSI fed IM
Rotor resistance control(cont
control(cont….)
….)
T
R1< R2< R3
R3
R1
R2
nr1< nr2< nr3
T
nr3
nr2 nr1
ns~nNL
n

Static rotor resistance
control(cont
control(
cont….)
….)
An electronic chopper implementation is also
possible as shown below but is equally
inefficient.
Closed loop control static rotor
resistance(contn
resistance(
contn….)
….)
Slip power recovery

Instead of wasting the slip power in the rotor circuit
resistance, a better approach is to convert it to ac line
power and return it back to the line. Two types of
converter provide this approach:
1) Static Kramer Drive - only allows
operation at subsub-synchronous speed.
2) Static Scherbius Drive - allows
operation above and below
synchronous speed
Conventional kramer
sysytem((contn
sysytem
contn….)
….)
Improved version of kramer
system(Ns to half of Ns) (contn
(contn….)
….)
Improved version of kramer
system(zero to Ns) (contn
(contn….)
….)
Static kramer system(
system(contn
contn….)
….)
Appliaction((contn
Appliaction
contn….)
….)


Large pumps
Fan type loads
Closed loop control of static kramer
system(contn
system(
contn….)
….)
(contn
contn….)
….)
Modified kramer system(
system(contn
contn….)
….)
Scherbius system(conventional
scherbius system) (contn
(contn….)
….)
Static scherbius system(
system(contn
contn….)
….)
slip power –rectifier
rectifier--inveter
inveter-transformer--supply
transformer
Supply--transformer
Supply
transformer--rectifier
rectifier--inverter
inverter-rotor circuit(contn
circuit(contn….)
….)
Closed loop control of scherbius
sytem((contn
sytem
contn….)
….)
Cycloconverter static
scherbius((contn
scherbius
contn….)
….)
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 gap
The stator winding 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 cycle of magnetomotive force (mmf) or
field around the air gap.
The speed of rotation of field is called the synchronous speed ws , which
is defined by :
ωs is syncronous speed [rad/sec]
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 motor speed
The rotor speed or motor speed is :
Where S is slip, as defined as :
m  s (1 S )
  m
S S
S
Or
S
NS  Nm
NS
Equivalent Circuit Of Induction Motor
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
Xm is magnetizing reactance
Rm is Core losses as a reactance
Performance Characteristic of
Induction Motor
2
Stator copper loss :
Ps cu  3 I s Rs
Rotor copper loss :
Pr cu  3 ( I r ) 2 Rr
'
2
Core losses :
2
V
V
Pc  3 m  3 s
Rm
Rm
'
Performance Characteristic of
Induction Motor
- Power developed on air gap (Power fropm stator to
'
rotor through air gap) :
' 2 Rr
Pg  3 ( I r )
S
'
'
2
- Power developed by motor : Pd  Pg  Pr cu  3 ( I r )
Pd  Pg (1  S )
or
- Torque of motor :
or
Pd
Td 
m

Pd 60
Td 
2 N m
or
Pg (1  S )
 S (1  S )

Pg
s
Rr
(1  S )
S
Performance Characteristic of
Induction Motor
Input power of motor :
Pi  3Vs I s cos m
 Pc  Ps cu  Pg
Output power of motor :
Po  Pd  Pno load
Pd  Pno load
Po
Efficiency :   
Pi Pc  Ps cu  Pg
Performance Characteristic of
Induction Motor
If
Pg  ( Pc  Ps cu )
and
Pd  Pno load
so, the efficiency can calculated as :
Pd Pg (1  S )
 
1 S
Pg
Pg
Performance Characteristic of
Induction Motor
Generally, value of reactance magnetization Xm >> value Rm (core
losses) and also X m 2  ( Rs 2  X s 2 )
So, the magnetizing voltage same with the input voltage :
Therefore, the equivalent circuit is ;
Xm
Vm  Vs
Performance Characteristic of
Induction Motor
Total Impedance of this circuit is :
'
R
 X m ( X s  X r )  jX m ( Rs  r )
S
Zi 
'
Rr
'
Rs 
 j( X m  X s  X r )
S
'
Xm
'
The rotor current is :
Ir 
Vs
2

Rr' 
'
 Rs    X s  X r
S 



2



1
2
Td 
3 Rr' Vs2
' 2

R 
S s  Rs  r   X s  X r'
S 


Torque – speed Characteristic

2



Three region operation :
1. Motoring :
0 S  1
2. Regenerating :
S 0
3. Plugging :
1 S  2
Performance Characteristic of
Induction Motor
Starting speed of motor is wm = 0 or S = 1,
Starting torque of motor is : Tst 
3 Rr' Vs2
' 2

R 
s  Rs  r   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'
1
2
'
2
r
R   X  X  
2
s
s
Performance Characteristic of
Induction Motor
Torque maximum is :
Tmax 
3 Vs2


2


2 s  Rs  Rs  X s  X r'


2
2 s  Rs  Rs  X s  X r'

2
And the maximum regenerative torque can be found as :
Tmax 
3 Vs2
Where the slip of motor s = - Sm
2



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

R 
  Rs  
S 


2
3 Rr' Vs2

S s X s  X
Tst 

' 2
r
3 Rr' Vs2
 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 :
2
R'
X
So, the motor torque is :
s
 X r'

r

S
3Vs2 S
Td 
 s R 'r
And the slip at maximum torque is :
S max  
Rr'
1
2
'
2
r
R   X  X  
2
s
The maximum motor torque is :
 Rs
Td 
s
3 Rr' Vs2
' 2

Rr 
S s  Rs    X s  X r'
S 



2



Stator Voltage Control
Controlling Induction Motor Speed by
Adjusting The Stator Voltage
Td 
3 Rr' Vs2
2

Rr' 
S s  Rs    X s  X r'
S 



2



Frequency Voltage Control
Controlling Induction Motor Speed by
Adjusting The Frequency Stator Voltage
Td 
3 Rr' Vs2

Rr'
S s  Rs 
S

2

  X s  X r'



2



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 wb, the synchronous speed at any other
frequency becomes:
 s   b
And :
S
b  m

1  m
b
b
The motor torque :
Td 
3 Rr' Vs2
' 2

R 
S  s  Rs  r   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

2 S b X s  X r'


And the maximum torque at any other frequency is :
2
Vs
3
Tm 
2 S b X s  X r'  2


Sm 
At this maximum torque, slip S is :
Rr '
 X s  X r'


2
Normalizing :
Vs
3
Tm 
2 S b X s  X r'  2

Tmb

3 Vs2

2 S b X s  X r'

Tm
1
 2
Tmb 

And
Tm  2  Tmb
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 : Vs 
460
 265 volt
3
Base frequency : b  2  f  2 x 3.14 x 60  377 rad / s
Base Torque : Tmb 
Motor Torque :
60 Po
60 x 11200

 61.11 Nm
2  N m 2 x 3.14 x 1750
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   b
 s 1.321 x 377  498.01 rad / s
or
60 x 498.01
N s   Nb 
 4755.65 rpm
2 x
p NS
4 x 4755.65
So, the supply frequency is : f s 

 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  = 1.321
So,
Sm 
0.38
 0.101
1.3211.14  1.71
or,
N m  N S (1  S )  4755.65 (1  0.101)  4275 rpm
CONTROLLING INDUCTION MOTOR SPEED USING
ROTOR RESISTANCE
(Rotor Voltage Control)
Wound rotor induction motor applications
cranes
CONTROLLING INDUCTION MOTOR SPEED USING
ROTOR RESISTANCE
(Rotor Voltage Control)
Equation of Speed-Torque :
Td 
3 Rr' Vs2
' 2

R 
S  s  Rs  r   X s  X r'
S 


In a wound rotor induction motor, an external
three-phase resistor may be connected to its
slip rings,

2
3Vs2 S
Td 
 s R 'r



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.
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)
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)
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)
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 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.
460
Vs 
 265.58 volt
3
p 6
  2 x 60  377 rad / s
 s  2 x 377 / 6  125.66 rad / s
The equivalent circuit :
The dc voltage at the rectifier output is :
Vd  I d Re  I d R (1  k )
Nr
Er  S Vs
 S Vs nm
Ns
and
For a three-phase rectifier, relates
Er and Vd as :
Vd  1.65 x 2 Er  2.3394 Er
Nr
E

S
V
 S Vs nm
Using :
r
s
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
So,
Pd  TL m
(1  S )Vd I d
Pd 
 TLm  TLm (1  S )
S
Substituting Vd from
Solving for Id gives :
Vd  2.3394 S Vs n m
TLs
Id 
2.3394Vs nm
In equation
:
Pd above, so
Which indicates that the inductor current is independent of the speed.
Vd  I d Re  I d R (1  k )
I d R(1  k )  2.3394 S Vs n m
From equation :
So,
Which gives :
I d R(1  k )
S
2.3394 S Vs n m
and equation : Vd
 2.3394 S Vs n m
I d R(1  k )
S
2.3394 S Vs n m

I d R(1  k ) 
m  s (1  S )  s 1 

2
.
3394
V
n
s m

The speed can be found from equation :
as :

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 ws
m  180  / 30  83.77 rad / s
2
 K v m
2
 800 
 750 x 
  347.67 Nm
 1175 
From torque equation : TL
From equation :
TLs
Id 
The corresponding inductor current is :
2.3394Vs nm
347.67 x 125.66
Id 
 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