motor drives dr. nik rumzi nik idris dept. of energy conversion, utm 2013
TRANSCRIPT
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Motor Drives
Dr. Nik Rumzi Nik IdrisDept. of Energy Conversion, UTM
2013
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Conventional electric drives (variable speed)
• Bulky
• Inefficient
• inflexible
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Modern electric drives (With power electronic converters)
• Small
• Efficient
• Flexible
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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)
Electric Drives
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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
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Overview of AC and DC drives
Extracted from Boldea & Nasar
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Armatureterminals
Fieldterminals
DC Machine:• Can be operated as motor• Can be operated as generator
In basic construction, it consist of 4 terminals:• Armature terminals• Field terminals
DC DRIVES
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DC Motors - 2 pole
Stator
Rotor
Field circuit produces field flux
Field flux
DC DRIVES
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DC Motors - 2 pole
X
X
X
X
X
• Current flowing in armature circuit interact with field winding to produce torque
DC DRIVES
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DC Motors - 2 pole
X
X
X
X
X
Torque
Torque
• Torque produced is proportional to the armature current
• The torque will make the rotor to rotate (clockwise)
• Current flowing in armature circuit interact with field winding to produce 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
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Electric torque
Armature back e.m.f.
Lf Rf
if+
ea
_
LaRa
ia+
Vt
_
+
Vf
_
dtdi
LiRv ffff +=
The flux per pole is proportional to if
DC DRIVES
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• 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
kt is the torque constant ke is the back e.m.f. constant
• Most of the time, kt = ke
DC DRIVES
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aaat EIRV • In steady state current will not change with time (dia/dt =0),
2
et
ea
e
t
kk
TR
k
V
• Therefore steady state speed is given by,
• Armature circuit:
2
t
ea
t
t
k
TR
k
V
• For the case kt = ke,
DC DRIVES
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• 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
2
t
ea
t
t
k
TR
k
V
DC DRIVES
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Controlling Vt and if using Controlled Rectifier
3-phasesupply
+
Vt
−
+Vf
−
αt αf
if
Current if is changed by changing Vf
Controlled Rectifier Controlled Rectifier
DC DRIVES
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Controlling Vt using Controlled RectifierControlling if using resistance
3-phasesupply
+
Vt
−
if
+Vf
−
αt
Current if is changed by changing Rf
Rf
Controlled Rectifier Un-controlled Rectifier
DC DRIVES
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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
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Example:
+
Vt
−
αt
Controlled Rectifier
aaat EIRV
Using
where Ra = 1.2 Ω, Ia = 15 A and Ea = keω .
The given speed is in rotation per minute (RPM) and has to be converted to rad/s.
αt can be calculated as:
Torque developed is Te= ktIa = 0.5 *15 = 0.5*15 = 7.5 Nm
From DC machine equation,
DC DRIVES
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+
Vt
−
αt
Controlled Rectifier
+
-78.5V
−
Ra
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,
DC DRIVES
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3-phasesupply
+
Vt
−
ll
DCV
V 3
Vt is change by changing the duty cyle
DC-DC converterUn-controlled Rectifier
Controlling Vt using a DC-DC converter
+
VDC
−
DC DRIVES
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Controlling Vt using a DC-DC converter
3-phasesupply
+
VDC
−
ll
DCV
V 3
T1
D1
+
Vt
-
iaRa
DC DRIVES
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Controlling Vt using a DC-DC converter
When T1 ON, vt = VDC
When T1 OFF, D1 will ON and vt = 0
0
VDC
T1
D1
+
Vt
-
iaRa
T
ton
T
tonIf = duty cycle, then, DCt VV
The voltage Vt canbe controlled by changing the duty cycle
DC DRIVES
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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
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T1
D1
+
Vt
-
iaRa
VEIRV aaat 639183622421 ..).(
12Nm of torque is due to Te/kt = 12/0.5 = 24 A of armature current. 1200 RPM is equivalent to:
sradN
/.)()(
6712560
21200
60
2
Hence the back e.m.f.,
Ea = 125.67 *0.5 = 62.83V
From the motor equation, we can calculate the terminal voltage
Since DCt VV 460200
6391.
.
DC
t
V
V
DC DRIVES
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0
200V
20 s
9.2s
stT
ton
on 291020460460 6 .)(..
DC DRIVES
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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
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AC DRIVES
(i) Six-step voltage source inverter (assuming available source is 3-phase supply)
3-phasesupply
3-phase VSIControlled Rectifier
Variable DC
AC motor
Frequency control
f
Amplitude control
A
a
b
c
+
Vdc
_
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(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:
line-line voltage
phase voltage
AC DRIVES
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(i) Six-step voltage source inverter (assuming available source is 3-phase supply)
Example of amplitude control
A1 < A2 < A3
A1
A2
A3
AC DRIVES
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(i) Six-step voltage source inverter (assuming available source is 3-phase supply)
Example of frequency control
f1> f2 > f3
1/f1 1/f2 1/f3
AC DRIVES
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3-phasesupply
3-phase VSIControlled Rectifier
Fixed DC
AC motor
Frequency and
amplitude control
A, f
a
b
c
+
Vdc
_
(ii) PWM voltage source inverter (assuming available source is 3-phase supply)
AC DRIVES
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With the correct switching signals to the VSI, the following voltages can be obtained:
line-line voltage
phase voltage
(ii) PWM voltage source inverter (assuming available source is 3-phase supply)
AC DRIVES
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(ii) PWM voltage source inverter (assuming available source is 3-phase supply) Variable amplitude
AC DRIVES
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(ii) PWM voltage source inverter (assuming available source is 3-phase supply) Variable frequency
AC DRIVES
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(ii) PWM voltage source inverter (assuming available source is 3-phase supply) Variable frequency
AC DRIVES
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Construction of induction machine
a
b
b’
c’
c
a’
Stator – 3-phase windingRotor – squirrel cage / wound
120o120o
120o
INDUCTION MOTOR DRIVES
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a
a’
Single N turn coil carrying current iSpans 180o elec
Permeability of iron >> o
→ all MMF drop appear in airgap
/2-/2-
Ni / 2
-Ni / 2
INDUCTION MOTOR DRIVES
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Distributed winding – coils are distributed in several slots
Nc for each slot
/2-/2-
(3Nci)/2
(Nci)/2
INDUCTION MOTOR DRIVES
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Phase a – sinusoidal distributed winding
Air–gap mmf
F()
2
INDUCTION MOTOR DRIVES
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• 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:
f2p2
s
p – number of polesf – supply frequency
INDUCTION MOTOR DRIVES
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INDUCTION MOTOR DRIVES
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• 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
rss
INDUCTION MOTOR DRIVES
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Stator voltage equation:
Vs = Rs Is + j(2f)LlsIs + Eag
Eag – airgap voltage or back emf
Eag = k f ag
Rotor voltage equation: Er = Rr Ir + js(2f)Llr
Er – induced emf in rotor circuit
Er /s = (Rr / s) Ir + j(2f)Llr
INDUCTION MOTOR DRIVES
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Per–phase equivalent circuit
Rr/s
+
Vs
–
RsLls
Llr
+
Eag
–
Is
Ir
Im
Lm
Rs – stator winding resistanceRr – rotor winding resistanceLls – stator leakage inductanceLlr – rotor leakage inductanceLm – mutual inductances – slip
+
Er/s
–
INDUCTION MOTOR DRIVES
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We know Eg and Er related by
\ rotor voltage equation becomes
Eg = (Rr’ / s) Ir’ + j(2f)Llr’ Ir’
as
EE
g
r Where a is the winding turn ratio
The rotor parameters referred to stator are:
2lr
lr2r
rrr a'L
L,a
'RR,)'I(aI
INDUCTION MOTOR DRIVES Per–phase equivalent circuit
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Rr’/s+
Vs
–
RsLls Llr’
+
Eag
–
Is Ir’
Im
Lm
Rs – stator winding resistanceRr’ – rotor winding resistance referred to statorLls – stator leakage inductanceLlr’ – rotor leakage inductance referred to statorLm – mutual inductanceIr’ – rotor current referred to stator
INDUCTION MOTOR DRIVES Per–phase equivalent circuit
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Power and Torque
Power is transferred from stator to rotor via air–gap, known as airgap power
s1s
'RI3'RI3
s'R
I3P r2'rr
2'r
r2'rag
Lost in rotor winding
Converted to mechanical power = (1–s)Pag
INDUCTION MOTOR DRIVES
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Power and Torque
Mechanical power, Pm = Tem r
But, ss = s - r r = (1-s)s
Pag = Tem s
s
r2'r
s
agem s
'RI3PT
Therefore torque is given by:
2lrls
2r
s
2s
s
rem
'XXs
'RR
Vs
'R3T
INDUCTION MOTOR DRIVES
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Power and Torque
1 0
rs
Trated
Pull out Torque(Tmax)
Tem
0 rated s
2lrls2
s
rm
XXR
Rs
2lrls
2ss
2s
smax
XXRR
Vs3
T
sm
INDUCTION MOTOR DRIVES
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Control of induction machine based on steady-state model (per phase steady-state equivalent circuit) is known as scalar control
Rr’/s+
Vs
–
RsLls Llr’
+
Eag
–
Is Ir’
Im
Lm
Speed Control of IMINDUCTION MOTOR DRIVES
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rs
Pull out Torque(Tmax)
Te
ssmrotor
TL
Te
Intersection point (Te=TL) determines the steady –state speed
Speed Control of IMINDUCTION MOTOR DRIVES
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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 polesDiscrete step change in speed
Variable voltage (amplitude), frequency fixedE.g. using transformer or triacSlip becomes high as voltage reduced – low efficiency
Variable voltage (amplitude), variable frequencyUsing power electronics converter Operated at low slip frequency
Speed Control of IMINDUCTION MOTOR DRIVES
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0 20 40 60 80 100 120 140 1600
100
200
300
400
500
600
Tor
que
w (rad/s)
Low speed high slip
Therefore, low efficiency at low speed
Speed Control of IMVariable voltage, fixed frequency
INDUCTION MOTOR DRIVES
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T-ω characteristic of IM when air-gap is kept constant:
0 20 40 60 80 100 120 140 1600
100
200
300
400
500
600
700
800
900
Tor
que
50Hz
30Hz
10Hz
Speed Control of IMVariable voltage, variable frequency
INDUCTION MOTOR DRIVES
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How do we keep air-gap flux constant?
Eag = k f ag
f
V
f
Eag = constant
• 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
Speed Control of IMVariable voltage, variable frequency
At high speed, Eag ≈ Vs
INDUCTION MOTOR DRIVES
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Vrated
frated
Vs
f
Speed Control of IMVariable voltage, variable frequency
Constant V/Hz – open-loop
INDUCTION MOTOR DRIVES
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VSIRectifier
3-phase supply IM
Pulse Width
Modulators* +
Rampf
C
V
rate limiter is needed to ensure the slip change within allowable range (e.g. rated value)
Constant V/Hz – open-loop
Speed Control of IMVariable voltage, variable frequency
INDUCTION MOTOR DRIVES
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Speed Control of IMVariable voltage, variable frequency
INDUCTION MOTOR DRIVES
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 = K2. 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?
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Speed Control of IMVariable voltage, variable frequency
INDUCTION MOTOR DRIVES
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?
ii) 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 ?