Motor Drives
Dr. Nik Rumzi Nik Idris
Dept. of Energy Conversion, UTM
2013
Conventional electric drives (variable speed)
•
Bulky
•
Inefficient
•
inflexible
Modern electric drives (With power electronic converters)
•
Small
•
Efficient
•
Flexible
Electric Drives
DC Drives:
•
•
uses DC motors as prime movers
Easy to control  need variable
DC voltage
AC Drives:
•
•
uses AC motors as prime movers
More difficult to control  need
variable AC voltage (magnitude and
frequency)
DC motors and AC motors
DC motors: Regular maintenance, heavy, expensive, speed limit
Easy control, decouple control of torque and flux
AC motors: Less maintenance, light, less expensive, high speed
Coupling between torque and flux – variable
spatial angle between rotor and stator flux
Overview of AC and DC drives
Extracted from Boldea & Nasar
DC DRIVES
DC Machine:
• Can be operated as motor
• Can be operated as generator
Armature
terminals
Field
terminals
In basic construction, it consist
of 4 terminals:
• Armature terminals
• Field terminals
DC DRIVES
DC Motors - 2 pole
Rotor
Field flux
Field circuit produces field flux
Stator
DC DRIVES
DC Motors - 2 pole
• Current flowing in armature
circuit interact with field
winding to produce torque
X
X
X
X
X
DC DRIVES
DC Motors - 2 pole
• Current flowing in armature
circuit interact with field
winding to produce torque
• Torque produced is
proportional to the armature
current
Torque
X
X
X
• The torque will make the
rotor to rotate (clockwise)
X
X
Torque
• As the rotor rotates, voltage
will be induced in the rotor –
which is known as the back
emf
In order to look on how the speed is control on DC motor, we
need to model the DC motor with electric circuit
DC DRIVES
Ra
Lf
La
ia
+
+
Rf
if
+
Vt
ea
Vf
_
_
_
dia
vt = Raia + L
+ Ea
dt
Te = ktfia
Ea = kefw
f µ if
Electric torque
Armature back e.m.f.
The flux per pole is proportional to if
v f = R f if + L
di f
dt
DC DRIVES
• If the field flux comes from permanent magnet, the motor is called the
permanent magnet DC motor
• For permanent magnet DC motor, the field current CANNOT be
controlled, therefore the torque and back e.m.f. can be written as
Te = ktia
kt is the torque constant
• Most of the time, kt = ke
Ea = kew
ke is the back e.m.f. constant
DC DRIVES
• Armature circuit:
Vt = Raia + L
dia
+ Ea
dt
• In steady state current will not change with time (dia/dt =0),
Vt = R a Ia + Ea
• Therefore steady state speed is given by,
Vt
RaTe
=

ke kt ke  2
• For the case kt = ke,
Vt
RaTe
=

kt kt 2
DC DRIVES
Vt
RaTe
=

kt kt 2
• Three possible methods of speed control:
• Armature voltage Vt
• Field flux
• Armature resistance Ra
• Control using Ra is seldom used because it is inefficient due to
the losses in the external armature resistance
DC DRIVES
Controlling Vt and if using Controlled Rectifier
Controlled Rectifier
+
3-phase
supply
Vt
Controlled Rectifier
if
+
Vf
−
−
αt
Vt =
3Vl-l
p
αf
cosat
Vf =
3Vl-l
p
cosa f
Current if is changed by changing Vf
DC DRIVES
Controlling Vt using Controlled Rectifier
Controlling if using resistance
Controlled Rectifier
+
3-phase
supply
Vt
if
Un-controlled Rectifier
Rf
+
Vf
−
−
Vf =
αt
Vt =
3Vl-l
p
cosat
3Vl-l
p
Current if is changed by changing Rf
DC DRIVES
Example:
A wound stator DC motor which is fed by a 3-phase controlled rectifier
has the following parameters:
Ra = 1.2 Ω,
ktΦ = 0.5 Nm/amp,
keΦ = 0.5 V/rad/s
The amplitude of the line-line input voltage of the rectifier is 200V. If the
motor runs at 1500 RPM, and the armature current is 15 A, calculate the
delay angle of the rectifier.
What is the torque developed?
If the field is reversed and what would happen to the motor? If the current
is to be maintained at 15A, what should be the delay angle of the
rectifier?
DC DRIVES
Example:
Torque developed is Te= ktIa = 0.5 *15
= 0.5*15 = 7.5 Nm
Controlled Rectifier
+
Vt
From DC machine equation,
−
Vt = R a Ia + Ea
where Ra = 1.2 Ω, Ia = 15 A and Ea = keω .
αt
The given speed is in rotation per minute (RPM) and has to be converted to rad/s.
w=
N(2p ) 1500(2p )
=
=157 rad / s
60
60
Using
Vt =
3(200)
p
\Vt =1.2(15)+ (0.5)157 = 96.5V
æ p (96.5) ö
o
cosat , αt can be calculated as: at = cos ç
÷ = 59.6
è 3(200) ø
-1
DC DRIVES
Controlled Rectifier
Ra
+
+
Vt
-78.5V
−
−
αt
At the instant the field is reversed, the speed will remain at 1500rpm.
The back emf be -78.5V. The DC machine will be operated as a
generator. Hence to maintain the current at 15A,
\Vt =1.2(15)-(0.5)157 = -60.5V
\-60.5 =
3(200)
p
cos at Þ at = 108.46 o
DC DRIVES
Controlling Vt using a DC-DC converter
DC-DC converter
Un-controlled Rectifier
3-phase
supply
+
+
VDC
Vt
−
−
V DC =
3Vl l


Vt is change by changing the duty cyle 
DC DRIVES
Controlling Vt using a DC-DC converter
Ra
T1
+
+
D1
3-phase
supply
V DC =
3Vl l

VDC
Vt
−
-
ia
DC DRIVES
Controlling Vt using a DC-DC converter
When T1 ON, vt = VDC
Ra
T1
ia
When T1 OFF, D1 will ON and vt = 0
ton
+
VDC
D1
Vt
0
1
Vt =
T
If
ò
T
0
=
1
ton
ton
vt dt = [ t (VDC )] 0 = VDC
T
T
t on
T
= duty cycle, then,
T
Vt = VDC

The voltage Vt canbe controlled by changing
the duty cycle 
DC DRIVES
Example:
A permanent magnet DC motor which is fed by a DC-DC converter
(BUCK converter) has the following parameters:
Ra = 1.2 Ω,
kt = 0.5 Nm/amp,
ke = 0.5 V/rad/s
When the motor runs at a speed of 1200 RPM, the torque developed by
the motor is 12Nm. The DC input voltage of the converter is 200V.
(i) Calculate the duty cycle of the converter.
(ii) If the period of the waveform is 20 s, draw the output voltage
waveform of the converter.
DC DRIVES
Ra
T1
ia
+
D1
Vt
-
12Nm of torque is due to Te/kt = 12/0.5 = 24 A
of armature current. 1200 RPM is equivalent
to:
N (2 ) 1200 (2 )
=
=
= 125 .67 rad / s
60
60
Hence the back e.m.f.,
Ea = 125.67 *0.5 = 62.83V
From the motor equation, we can calculate
the terminal voltage
Vt = Ra I a + Ea = (1.2)24 + 62.83 = 91.63V
Since
Vt = VDC
 =
Vt
91.63
=
= 0.46
VDC
200
DC DRIVES
9.2s
200V
0
20 s
=
ton
= 0.46  ton = 0.46 (20  10 6 ) = 9.2s
T
AC DRIVES
• In modern AC drive systems, the speed of the AC motor is
controlled by controlling:
(i) Magnitude of the applied voltage
(ii) Frequency of the applied voltage
• Magnitude and frequency of the AC motor can be controlled by
using power electronic converters
• There are two most widely used inverter control technique:
(i) Six-step voltage source inverter
(ii) PWM voltage source inverter
AC DRIVES
(i) Six-step voltage source inverter (assuming available
source is 3-phase supply)
Controlled Rectifier
3-phase VSI
Variable DC
a
+
3-phase
supply
b
Vdc
_
Amplitude
control
A
c
Frequency
control
f
AC motor
AC DRIVES
(i) Six-step voltage source inverter (assuming available
source is 3-phase supply)
With the correct switching signals to the VSI, the following
voltages can be obtained:
Vdc
2
Vdc
3
line-line voltage
phase voltage
AC DRIVES
(i) Six-step voltage source inverter (assuming available
source is 3-phase supply)
A1
A2
Example of amplitude control
A1 < A2 < A3
A3
AC DRIVES
(i) Six-step voltage source inverter (assuming available
source is 3-phase supply)
1/f1
1/f2
1/f3
Example of frequency control
f1> f2 > f3
AC DRIVES
(ii) PWM voltage source inverter (assuming available
source is 3-phase supply)
Controlled Rectifier
3-phase VSI
Fixed DC
a
+
3-phase
supply
b
Vdc
_
c
Frequency
and
amplitude
control
A, f
AC motor
AC DRIVES
(ii) PWM voltage source inverter (assuming available
source is 3-phase supply)
With the correct switching signals to the VSI, the following
voltages can be obtained:
line-line voltage
phase voltage
AC DRIVES
(ii) PWM voltage source inverter (assuming available
source is 3-phase supply) Variable amplitude
AC DRIVES
(ii) PWM voltage source inverter (assuming available
source is 3-phase supply) Variable frequency
AC DRIVES
(ii) PWM voltage source inverter (assuming available
source is 3-phase supply) Variable frequency
INDUCTION MOTOR
DRIVES
a
120o
Construction of
induction machine
120o
c’
b’
c
b
a’
120o
Stator – 3-phase winding
Rotor – squirrel cage / wound
INDUCTION MOTOR
DRIVES
Single N turn coil carrying current i
Spans 180o elec
Permeability of iron >> o
→ all MMF drop appear in airgap
a

a
’
Ni / 2
-
-Ni / 2
-/2
/2


INDUCTION MOTOR
DRIVES
Distributed winding
– coils are distributed in several slots
Nc for each slot

(3Nci)/2
(Nci)/2
-
-/2
/2


INDUCTION MOTOR
DRIVES
Phase a – sinusoidal distributed winding

Air–gap mmf
F()

2

INDUCTION MOTOR
DRIVES
• Sinusoidal winding for each phase produces space sinusoidal
MMF and flux
• Sinusoidal current excitation (with frequency s) in a phase
produces space sinusoidal standing wave MMF
• Combination of 3 standing waves resulted in MMF wave rotating
at:
2
s = 2f
p
p – number of poles
f – supply frequency
INDUCTION MOTOR
DRIVES
INDUCTION MOTOR
DRIVES
• Rotating flux induced:
emf in stator winding (known as back emf)
Emf in rotor winding
Rotor flux rotating at synchronous frequency
•
Rotor current interact with flux producing torque
•
Rotor ALWAYS rotate at frequency less than synchronous, i.e. at slip speed (or slip
frequency):
sl = s – r
•
Ratio between slip speed and synchronous speed known as slip
s=
s  r
s
INDUCTION MOTOR
DRIVES
Stator voltage equation:
Vs = Rs Is + j(2f)LlsIs + Eag
Eag – airgap voltage or back emf
Eag = k f ag
Rotor voltage equation:
Er = Rr Ir + js(2f)Llr
Er – induced emf in rotor circuit
Er /s = (Rr / s) Ir + j(2f)Llr
INDUCTION MOTOR
DRIVES
Rs
+
Per–phase equivalent circuit
Llr
Lls
+
Is
Vs
–
Rs –
Rr –
Lls –
Llr –
Lm –
s–
Im
Lm Eag
–
stator winding resistance
rotor winding resistance
stator leakage inductance
rotor leakage inductance
mutual inductance
slip
Ir
+
Er/s
–
Rr/s
INDUCTION MOTOR
DRIVES
Per–phase equivalent circuit
We know Eg and Er related by
Er s
=
Eg a
Where a is the winding turn ratio
The rotor parameters referred to stator are:
Rr '
L lr '
Ir = a (Ir ' ) , R r = 2 , L lr = 2
a
a
 rotor voltage equation becomes
Eg = (Rr’ / s) Ir’ + j(2f)Llr’ Ir’
INDUCTION MOTOR
DRIVES
Rs
Per–phase equivalent circuit
Is
Llr’
Lls
+
+
Vs
–
Rs –
Rr’ –
Lls –
Llr’ –
Lm –
Ir ’ –
Ir’
Lm
Im
Eag
–
stator winding resistance
rotor winding resistance referred to stator
stator leakage inductance
rotor leakage inductance referred to stator
mutual inductance
rotor current referred to stator
Rr’/s
INDUCTION MOTOR
DRIVES
Power and Torque
Power is transferred from stator to rotor via air–gap,
known as airgap power
Pag = 3Ir' 2
Rr '
=
s
3Ir' 2R r ' +
Lost in rotor
winding
3Ir' 2
Rr '
1  s
s
Converted to mechanical
power = (1–s)Pag
INDUCTION MOTOR
DRIVES
Power and Torque
Mechanical power, Pm = Tem r
But, ss = s - r

r = (1-s)s
 Pag = Tem s
Pag
3Ir' 2R r '
Tem =
=
s
ss
Therefore torque is given by:
Vs2
3R r '
Tem =
2
ss 
Rr ' 
2
Rs +
 + X ls + X lr '
s 

INDUCTION MOTOR
DRIVES
Power and Torque
Tem
sm = 
Pull out
Torque
(Tmax)
Tmax
Trated
s
1
0
sm
rated s
0
Rr
R s + X ls + X lr 
2
2

Vs2
3 
=
ss  R  R 2 + X + X 2
s
ls
lr
 s
r




INDUCTION MOTOR
DRIVES
Speed Control of IM
Control of induction machine based on steady-state model (per phase steadystate equivalent circuit) is known as scalar control
Rs
Is
Llr’
Lls
+
+
Vs
–
Ir’
Lm
Im
Eag
–
Rr’/s
INDUCTION MOTOR
DRIVES
Speed Control of IM
Te
Pull out
Torque
(Tmax)
Te
Intersection point
(Te=TL) determines the
steady –state speed
TL
s
sm
rotor s
r
INDUCTION MOTOR
DRIVES
Speed Control of IM
Given a load T– characteristic, the steady-state speed can be changed by
altering the T– of the motor:
Pole changing
Synchronous speed change with
no. of poles
Discrete step change in speed
Variable voltage (amplitude),
variable frequency
Using power electronics converter
Operated at low slip frequency
Variable voltage (amplitude),
frequency fixed
E.g. using transformer or triac
Slip becomes high as voltage
reduced – low efficiency
INDUCTION MOTOR
DRIVES
Speed Control of IM
Variable voltage, fixed frequency
600
Low speed  high slip
500
Therefore, low efficiency at
low speed
Torque
400
300
200
100
0
0
20
40
60
80
w (rad/s)
100
120
140
160
INDUCTION MOTOR
DRIVES
Speed Control of IM
Variable voltage, variable frequency
T-ω characteristic of IM when air-gap is kept constant:
900
800
50Hz
700
30Hz
Torque
600
500
10Hz
400
300
200
100
0
0
20
40
60
80
100
120
140
160
INDUCTION MOTOR
DRIVES
Speed Control of IM
Variable voltage, variable frequency
How do we keep air-gap flux constant?
Eag = k f ag
At high speed, Eag ≈ Vs
\fag
=
E ag
f

V
= constant
f
• Speed is adjusted by varying f - maintaining V/f constant to
avoid flux saturation
• This method is known as Constant V/f (or V/Hz) method
INDUCTION MOTOR
DRIVES
Speed Control of IM
Variable voltage, variable frequency
Constant V/Hz – open-loop
Vs
Vrated
frated
f
INDUCTION MOTOR
DRIVES
Speed Control of IM
Variable voltage, variable frequency
Constant V/Hz – open-loop
Rectifier
3-phase
supply
VSI
IM
C
f
Ramp
s*
+
V
Pulse
Width
Modulator
rate limiter is needed to ensure the slip
change within allowable range (e.g. rated
value)
INDUCTION MOTOR
DRIVES
Speed Control of IM
Variable voltage, variable frequency
A 4–pole, 3-phase, 50 Hz IM has a rated torque and speed of 20
Nm and 1450 rpm respectively. The motor is supplied by a 3-phase
inverter using a constant V/f control method. It is used to drive a
load with TL– characteristic given by TL = K2. The load torque
demand is such that it equals the rated torque of the motor at the
rated motor speed and frequency.
i)
Find the constant K in the TL– characteristic of the load.
ii) What are the synchronous and motor speed at a load torque of
15Nm?
INDUCTION MOTOR
DRIVES
Speed Control of IM
Variable voltage, variable frequency
A 4-pole 3-phase induction motor has the following ratings:
330V, 50Hz, 1450 rpm
The motor is fed by a 3-phase VSI with constant V/f control
strategy. The input 3-phase voltage to the VSI is 415V. The load
which is coupled to the induction motor has a T- ω characteristic
given by TL= 0.0015ω2 , such that the motor is operated at its rated
speed, when the torque is at its rated value.
i)
If the motor is operated at 1000 rpm, what should be the
applied phase voltage (fundamental amplitude and frequency)
fed to the IM? If the speed to be increased to 1600 rpm, what
should be the amplitude of the fundamental phase voltage?
i)
If the starting torque of 60 Nm is required, what should be the
amplitude and frequency of the fundamental component initially
applied to the induction motor during start-up ?