Power Electronics

Power Electronics and Drives
–Modeling & Simulation
A Problem Based and Project Oriented
Learning

B.Chitti Babu
Member IEEE (USA), Student Member IET (UK)
Department of Electrical Engineering,
National Institute of Technology,Rourkela
bcbabunitrkl@ieee.org
B Chitti Babu,
14 August 2009 1
EE NIT Rourkela

CONTENTS

Pre Requisite of Power Electronics System

Power Electronic Systems

Power Electronic Converters in Electrical Drives
:: DC and AC Drives

Modeling and Control of Electrical Drives
:: Current controlled Converters
:: Modeling of Power Converters
:: Scalar control of IM

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14 August 2009 2
EE NIT Rourkela

Power Electronics-An Enabling Technology
Energy System
REFRIGERATOR

SOLAR CELLS TELEVISION

DC
AC

SOLAR LIGHT
ENERGY TRANSFORMER

3 3 3 1-3 MOTOR
POWER STATION TRANSFORMER PUMP

FACTS
ROBOTICS
COMPEN-
SATOR

INDUSTRY

TRANSFORMER FUEL DC
CELLS AC

☯
3 POWER SUPPLY
a d
WIND TURBINE ~
FUEL =
COMMUNICATION

TRANSPORT
COMBUSTION
ENGINE

B Chitti Babu,
14 August 2009 Courtesy:
EE NIT Rourkela
Aalborg University,Denmark
3

Implementation of problem-oriented and
project-organised education

Literature Lectures Group
studies

Problem Problem Report
analysis solving

Field work/ Experiments/
Tutorials Simulation Prototyping

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14 August 2009 4
EE NIT Rourkela

Prerequisite for Power Electronics
• Study of Second Order System, Control
Concepts and Mathematics
• Role of Passive Elements
• Physics concepts of Devices
• Device Selection
………………………………
• Modeling and Simulation
• Build and Evaluate
• Design & Development
• Research and Innovate
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Modeling & Simulation?
• Modeling here refers to the process of analysis and syntheses to arrive at a suitable
mathematical description that encompasses the relevant dynamic characteristics of the
component, preferably in terms of parameters that can be easily determined in practice

• Model supposely imitates or reproduces certain essential characteristics or conditions of
the actual-This is called SIMULATION.

• Modeling & Simulation-Simulation is a technique that involves setting up a model of a real
situation and performing experiments on the model.

• Simulation to be an experiment with logical and mathematical models, especially
mathematical representations of the dynamic kind that are characterized by a mix of
differential and algebraic equations.

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Simulation Formulation

• Observing the Physical system.
• Formulating the hypotheses or mathematical model to
explain the observation.
• Predicting the behavior of the system from solutions
or properties of the mathematical model.
• Testing the validity of the Hypotheses or
Mathematical Model.

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Mathematical Models

• Linear or Nonlinear
• Lumped or Distributed parameters
• Static & Dynamic
• Continuous or Discrete
• Deterministic or Stochastic

Courtesy: Dynamic Simulation of Electric Machinery-
By Chee Mun Ong

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14 August 2009 8
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Simulation Packages
1)General Purpose:
Equation Oriented in that they require input in the form of
differential or algebraic equations. Eg:IESL, SABER, IMSL,
ODEPAK & DASSL etc.
2)Application-Specific Packages:
Ready to use models of commonly used components for a specific
applications. Eg:SPICE2, EMTP, PSCAD etc.

MATLAB & SIMULINK:They are Registered
Trade mark of the THE MATHWORKS. Inc.,
USA

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14 August 2009 9
EE NIT Rourkela


What is Power Electronics ?

A field of Electrical Engineering that deals
with the application of power semiconductor
devices for the control and conversion of
electric power

sensors
Input Power
Source Electronics Load
- AC
- DC Converters Output
- unregulated - AC
- DC
POWER ELECTRONIC
CONVERTERS – the
Reference
Controlle heart of power a power
r electronics system
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14 August 2009 10
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Why Power Electronics ?

Power semiconductor devices Power switches

isw

ON or OFF
+ vsw −
=0

isw = 0
Ploss = vsw× isw = 0
+ vsw −
Losses ideally ZERO !

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14 August 2009 11
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K K K
− − −
G G
Vak Vak Vak

+ + +
ia ia ia
A A A

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14 August 2009 12
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D
C
iD
+ ic
+
VDS G
G VCE
−
−
S

E

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14 August 2009 13
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Passive elements High frequency
+ VL transformer
−
i
L
+ +
Inductor
V V2
1
+ VC −
− −
i
C

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EE NIT Rourkela

Passive Elements In Power Electronics

• Resistors
• Capacitors
• Inductors
• Transformers
• Filters
• Integrated Magnetics

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Resistors in
Power Electronics
• Resistors are mostly used in Power
Electronics to dissipate the trapped
energy from other components as well to
provide damping.
• Thus, resistors can carry significant
amount of high frequency currents.
• Resistors can carry fundamental ac
component currents in ac circuits and also
carry dc component currents under steady
state.
• No resistor is ideal, so their behavior
depends upon the applied frequency The
peak temperature rise depends on the
energy dissipated in the resistors.

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Capacitors in
Power Electronics
• Capacitors are mostly used in Power
Electronics to by-pass high frequency
components of voltages and currents.
• Thus, capacitors can carry significant
amount of high frequency currents
Capacitors can carry fundamental ac
component.
• currents in ac circuits but cannot carry dc
component currents under steady state.
• No capacitor is ideal, so their behavior
depends upon the applied frequency
• The breakdown voltage depends on the peak
voltage charge

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Inductors in
Power Electronics
• Inductors are mostly used in Power
Electronics to block the flow of high
frequency components of currents.
• Thus, inductors can drop significant
amount of high frequency voltages.
• Inductors can have fundamental ac
component voltage drop in ac circuits but
cannot drop dc component voltages under
steady state.
• No inductor is ideal, so their behavior
depends upon the applied frequency
• The peak flux density depends on the peak
instantaneous current.

Courtesy: Dr.Sujit K. Biswas, Lecture Notes, Jadavpur University
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14 August 2009 18
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sensors
Input Power
Source Electronics Load
IDEALLY
- AC
Converters LOSSLESS !
Output
- DC
- unregulated - AC
- DC

Reference
Controlle
r

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14 August 2009 19
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Other factors:
• Improvements in power semiconductors
• fabrication
• Power Integrated Module (PIM),
Intelligent Power Modules (IPM)
• Decline cost in power semiconductor

• Advancement in semiconductor fabrication
• ASICs • FPGA • DSPs
• Faster and cheaper to implement
complex algorithm
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Some Applications of Power Electronics :
Typically used in systems requiring efficient control and conversion of
electric energy:
Domestic and Commercial Applications
Industrial Applications
Telecommunications
Transportation
Generation, Transmission and Distribution of electrical energy

Power rating of < 1 W (portable equipment)
Tens or hundreds Watts (Power supplies for computers /office equipment)
kW to MW : drives
Hundreds of MW in DC transmission system (HVDC)

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Modern Electrical Drive Systems

• About 50% of electrical energy used for drives

• Can be either used for fixed speed or variable speed
• 75% - constant speed, 25% variable speed (expanding)

• Variable speed drives typically used PEC to supply the motors

DC motors (brushed) AC motors
SRM - IM
BLDC - PMSM

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Classic Electrical Drive for Variable Speed Application :

• Bulky
• Inefficient
• inflexible

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Typical Modern Electric Drive Systems

Power Electronic Converters Electric Motor
Electric Energy Electric Energy Electric Mechanical
- Unregulated - - Regulated - Energy Energy

POWER IN Power
Moto Loa
Electronic d
r
Converters

feedback

Reference
Controller

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Example on VSD application

Constant speed Variable Speed Drives

valve

Supply
motor pump

Power out

Power
In

Power loss
Mainly in valve
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14 August 2009 25
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valve

Supply Supply
motor pump motor
PEC pump

Power out
Power out
Power
Power
In
In

Power loss
Power loss
Mainly in valve
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valve

Supply Supply
motor pump motor
PEC pump

Power out
Power out
Power
Power
In
In

Power loss
Power loss
Mainly in valve
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Electric motor consumes more than half of electrical energy in the US

Fixed speed Variable speed

Improvements in energy utilization in electric motors give large
impact to the overall energy consumption

HOW ?
Replacing fixed speed drives with variable speed drives
Using the high efficiency motors

Improves the existing power converter–based drive systems

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Overview of AC and DC drives

Before semiconductor devices were introduced (<1950)
• AC motors for fixed speed applications
• DC motors for variable speed applications

After semiconductor devices were introduced (1960s)
• Variable frequency sources available – AC motors in variable
speed applications
• Coupling between flux and torque control
• Application limited to medium performance applications –
fans, blowers, compressors – scalar control

• High performance applications dominated by DC motors –
tractions, elevators, servos, etc

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After vector control drives were introduced (1980s)
• AC motors used in high performance applications – elevators,
tractions, servos
• AC motors favorable than DC motors – however control is
complex hence expensive
• Cost of microprocessor/semiconductors decreasing –predicted
30 years ago AC motors would take over DC motors

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14 August 2009 30
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Courtesy: Electrical Drives by Ion Boldea ,CRC Press
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14 August 2009 31
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Power Electronic Converters in ED Systems
Converters for Motor Drives
(some possible configurations)

DC Drives AC Drives

AC Source DC Source AC Source DC Source

DC-AC-
DC-DC
DC
AC-DC- AC-DC- DC-DC-
AC-DC AC-AC DC-AC
DC AC AC

Const. Variable NCC FCC
DC
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B
14 August 2009 32
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DC DRIVES

Available AC source to control DC motor (brushed)

AC-DC-
AC-DC DC

Uncontrolled Rectifier
Single-phase Control
Control
Three-phase
Controlled Rectifier DC-DC Switched mode
Single-phase 1-quadrant, 2-quadrant
Three-phase 4-quadrant

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DC DRIVES
AC-DC
400

200

0

+ 2 Vm
Vo = cos α
-200

π
-400
0.4 0.405 0.41 0.415 0.42 0.425 0.43 0.435 0.44

50Hz Vo 10

1-phase Average voltage
5
over 10ms
−
0
0.4 0.405 0.41 0.415 0.42 0.425 0.43 0.435 0.44

500

0

50Hz
+ -500
3-phase 0.4 0.405 0.41 0.415 0.42 0.425 0.43 0.435 0.44
3VL − L , m
Vo Vo = cos α
π
30

20

− Average voltage
10
over 3.33 ms
0
B Chitti Babu, 0.4 0.405 0.41 0.415 0.42 0.425 0.43 0.435 0.44
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DC DRIVES
AC-DC
2 Vm
π

+ 2 Vm
Vo = cos α
π
50Hz Vo 90o 180o
1-phase Average voltage
over 10ms
− 2 Vm
−
π

3VL − L , m
π
50Hz
+
3-phase
3VL − L , m
Vo Vo = cos α
π 90o 180o

− Average voltage
over 3.33 ms 3VL − L , m
−
π
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DC DRIVES
AC-DC

ia

+
Vt
3-phase
Vt Q2 Q1
supply

− Q3 Q4 Ia

- Operation in quadrant 1 and 4 only

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DC DRIVES
AC-DC

+
3-
phase 3-phase
Vt supply
supply
−

ω

Q2 Q1

Q3 Q4
T

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DC DRIVES
AC-DC

F1 R1

3-phase
supply
+ Va -
R2 F2

ω

Q2 Q1

Q3 Q4
T

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DC DRIVES
AC-DC
Cascade control structure with armature reversal (4-quadrant):

iD

ω

ωref + Speed iD,ref + Current
Firing
control Control Circuit
ler _ ler
_

iD,ref
Armature
iD, reversal Babu,
B Chitti
14 August 2009 39
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DC DRIVES
AC-DC-DC

Uncontrolled control
rectifier
Switch Mode DC-DC
1-Quadrant
2-Quadrant
4-Quadrant

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DC DRIVES
AC-DC-DC

control

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DC DRIVES
AC-DC-DC DC-DC: Two-quadrant Converter

Va
T1 D1
+
ia
Vdc Q2 Q1

+ Ia
− D2
T2
Va

-

T1 conducts → va = Vdc

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DC DRIVES

Va
T1 D1
+
ia
Vdc Q2 Q1

+ Ia
− D2
T2
Va

-

D2 conducts → va = 0 T1 conducts → va = Vdc

Va Eb

Quadrant 1 The average voltage is made larger than the back emf
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DC DRIVES

Va
T1 D1
+
ia
Vdc Q2 Q1

+ Ia
− D2
T2
Va

-

D1 conducts → va = Vdc

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DC DRIVES

Va
T1 D1
+
ia
Vdc Q2 Q1

+ Ia
− D2
T2
Va

-

T2 conducts → va = 0 D1 conducts → va = Vdc

Va Eb

Quadrant 2 The average voltage is made smallerr than the back emf, thus
forcing the current to flow in the reverse direction
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DC DRIVES

vc
2vtri

+
vA Vdc
-

0

+
vc

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DC DRIVES
AC-DC-DC DC-DC: Four-quadrant Converter
leg A leg B

+ D1 D3
Q1 Q3
+ Va −
Vdc

− D4 D2
Q4 Q2

Positive current

va = Vdc when Q1 and Q2 are ON

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DC DRIVES
leg A leg B

+ D1 D3
Q1 Q3
+ Va −
Vdc

− D4 D2
Q4 Q2

Positive current

va = Vdc when Q1 and Q2 are ON
va = -Vdc when D3 and D4 are ON
va = 0 when current freewheels through Q and D
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DC DRIVES
leg A leg B

+ D1 D3
Q1 Q3
+ Va −
Vdc

− D4 D2
Q4 Q2

Positive current Negative current

va = Vdc when Q1 and Q2 are ON va = Vdc when D1 and D2 are ON
va = -Vdc when D3 and D4 are ON
va = 0 when current freewheels through Q and D
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DC DRIVES
leg A leg B

+ D1 D3
Q1 Q3
+ Va −
Vdc

− D4 D2
Q4 Q2

Positive current Negative current

va = Vdc when Q1 and Q2 are ON va = Vdc when D1 and D2 are ON
va = -Vdc when D3 and D4 are ON va = -Vdc when Q3 and Q4 are ON
va = 0 when current freewheels through Q and D va = 0 when current freewheels through Q and D
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DC DRIVES
Bipolar switching scheme – output
AC-DC-DC swings between VDC and -VDC

vc
2vtri

Vdc
Vdc
+ + vA
vA vB 0
- - Vdc
vB
0

vc Vdc

+ vAB

_ -Vdc

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DC DRIVES
Unipolar switching scheme – output
AC-DC-DC swings between Vdc and -Vdc

vc
Vtri
-vc

Vdc
+ + Vdc
vA vB
vA
-
0
-

Vdc
vc vB
0
+
Vdc
_
vAB
0

-vc
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DC DRIVES

Armature
200 current 200

150 150 Armature
Vdc 100 Vdc 100 current
50 50

0 0

-50 -50

Vdc -100 -100

-150 -150

-200 -200

0.04 0.0405 0.041 0.0415 0.042 0.0425 0.043 0.0435 0.044 0.0445 0.045 0.04 0.0405 0.041 0.0415 0.042 0.0425 0.043 0.0435 0.044 0.0445 0.045

Bipolar switching scheme Unipolar switching scheme

• Current ripple in unipolar is smaller
• Output frequency in unipolar is effectively doubled

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AC DRIVES
AC-DC-AC

control

The common PWM technique: CB-SPWM with ZSS
14 August 2009 SVPWM
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• Control the torque, speed or position

• Cascade control structure

Example of current control in cascade control structure

θ* ω* T*
+ + +
− − −
position speed current
controller controller controller converter Motor

kT
ω

θ
1/s

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Current controlled converters in DC Drives - Hysteresis-based

+

ia
Vdc
+
iref
− Va

−
va

iref + ierr q
_ q

• High bandwidth, simple implementation,
insensitive to parameter variations
ierr
• Variable switching frequency – depending on
operating conditions B Chitti Babu,
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Current controlled converters in AC Drives - Hysteresis-based

i*a +

Converter
i*b +

i*c +

• For isolated neutral load, ia + ib + ic = 0
∴control is not totally independent 3-phase
• Instantaneous error for isolated neutral load can
AC Motor
reach double the band
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iq

is

Δh Δh Δh Δh

id

• For isolated neutral load, ia + ib + ic = 0
∴control is not totally independent
• Instantaneous error for isolated neutral load can
reach double the band
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• Δh = 0.3 A • Vdc = 600V
Con u s
tin ou • Sinusoidal reference current, 30Hz load
• 10Ω, 50mH
powergui

Scope

iaref

TW
o orkspace1 g
+ i
A + -
D Voltage Source
C B Series R BranchC
LC 3urrent Measurem 3
ent
c1 p1 -
C
i
c2 p2 + -
U ersal Bridge 1
niv

c3 p3 Series R Branch urrent M
LC C1 easurem 1
ent

ina p4 i
+ -
Sine W e
av
inb p5 Series R Branch urrent M
LC C2 easurem 2
ent
inc p6

Subsystem

Sine W e 1
av

Sine W e 2
av
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Actual and reference currents Current error

0.5
10
0.4

0.3
5
0.2
10
0.1

0 0
9
-0.1

-0.2
-5 8
-0.3

7 -0.4
-10
-0.5

0.005 0.01 6
0.015 0.02 0.025 0.03
-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5

5

4

4 6 8 10 12 14 16
-3
x 10
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Actual current locus Current error
10 0.5

5
0 0.6A

-0.5
0

0.04 0.042 0.044 0.046 0.048 0.05 0.052 0.054 0.056 0.058 0.06
-5

0.5
-10

-10 -5 0 5 10 0 0.6A

-0.5

0.04 0.042 0.044 0.046 0.048 0.05 0.052 0.054 0.056 0.058 0.06

0.5

0 0.6A

-0.5

0.04 0.042 0.044 0.046 0.048 0.05 0.052 0.054 0.056 0.058 0.06
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Current controlled converters in DC Drives - PI-based

Vdc

iref + vc vPulse width
tri

PI vc modulator
q
q
q
−

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i*a +
PI PWM

Converter
i*b +
PI PWM

i*c + PWM
PI

• Sinusoidal PWM

Motor
• Interactions between phases → only require 2 controllers
• Tracking error
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• Perform the 3-phase to 2-phase transformation
- only two controllers (instead of 3) are used

• Perform the control in synchronous frame
- the current will appear as DC

• Interactions between phases → only require 2 controllers
• Tracking error
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Current controlled converters in AC Drives - PI-based

i*a +
PI PWM

Converter
i*b +
PI PWM

i*c + PWM
PI

Motor

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i*a

PI
SVM Converter
i*b
3-2 2-3
PI
i*c

3-2

Motor

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va*
id* + PI
controller
−
vb*
id dq→abc SVM
or SPWM IM
iq* + VSI
PI vc*

− iq controller
ωs

Synch speed
ωs
estimator

abc→dq

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Stationary - ia Stationary - id
4 4

2 3

0 2

-2 1

-4 0
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02

4 Rotating - ia 4 Rotating - id

2 3

0 2

-2 1

-4 0
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02

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Modeling of the Power Converters: DC drives with Controlled rectifier

+
vc firing α controlled
circuit rectifier Va

–

vc(s) va(s)
? DC motor

The relation between vc and va is determined by the firing circuit
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14 August 2009 69
It is desirable to have a linear NIT Rourkela
EE relation between vc and va

Cosine-wave crossing control

Vm
Input voltage
0 π 2π 3π 4π

vc vs
Cosine wave compared with vc

Results of comparison trigger SCRs

Output voltage

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Cosine-wave crossing control
cos(ωt)
Vscos(α) = vc
Vm
⎛v ⎞
0 π 2π 3π 4π α = cos −1 ⎜ c ⎟
⎜v ⎟
⎝ s⎠

vc vs
α

2Vm v c ⎛ −1 ⎛ v c ⎞ ⎞
Va = cos⎜α ) ⎜ ⎟ ⎟
(
π vs ⎝ ⎜ cos ⎜ v ⎟ ⎟
⎝ s ⎠⎠
α

A linear relation between vc and Va
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Va is the average voltage over one period of the waveform
- sampled data system

Delays depending on when the control signal changes – normally taken
as half of sampling period

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Va is the average voltage over one period of the waveform
- sampled data system

Delays depending on when the control signal changes – normally taken
as half of sampling period

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T
− s
G H (s) = Ke 2

Single phase, 50Hz
vc(s) Va(s)
2Vm
K= T=10ms
πVs

Three phase, 50Hz
3VL − L ,m
K= T=3.33ms
πVs

Simplified if control bandwidth is reduced to much lower than the
sampling frequency

B Chitti Babu,
14 August 2009 74
EE NIT Rourkela


+
iref current vc firing α controlled
controller Va
circuit rectifier
–

• To control the current – current-controlled converter
• Torque can be controlled
• Only operates in Q1 and Q4 (single converter topology)

B Chitti Babu,
14 August 2009 75
EE NIT Rourkela


• Input 3-phase, 240V, 50Hz • Closed loop current control
with PI controller

Scope3
+
- v Continuous
Voltage Measurement4
+ i powergui
- Scope2
AC Voltage Source Current Measurement 1 Step

s
AC Voltage Source1 +
g -
+ v
A Controlled Voltage Source
Series RLC Branch
AC Voltage Source2 B To Workspace
+ - i
- v C - + ia
Voltage Measurement2 Universal Bridge Current Measurement To Workspace1
+ +
- v - v
alpha_deg
Voltage Measurement Voltage Measurement3
AB ux Scope
BC pulses
+
- v
CA
Block
Voltage Measurement1
Synchronized Mu
6-Pulse Generator
Scope1 ir

To Workspace2

PID acos -K-
Signal
PID Controller Saturation
1
Generator

7

Constant 1

B Chitti Babu,
14 August 2009 76
EE NIT Rourkela


• Input 3-phase, 240V, 50Hz • Closed loop current control
with PI controller
1000
1000

500
500

0
Voltage
0

-500 -500
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.22 0.23 0.24 0.25 0.26 0.27 0.28

15 15

10
10

5
Current
5

0
0.22 0.23 0.24 0.25 0.26 0.27 0.28
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

B Chitti Babu,
14 August 2009 77
EE NIT Rourkela

Modeling of the Power Converters: DC drives with SM Converters

B Chitti Babu,
14 August 2009 78
EE NIT Rourkela


Vdc
Switching signals obtained by comparing
control signal with triangular wave +

Va

−

vtri

q
vc

We want to establish a relation between vc and Va

AVERAGE voltage

vc(s) Va(s)
? DC motor

B Chitti Babu,
14 August 2009 79
EE NIT Rourkela

Ttri

⎧1 Vc > Vtri
q=⎨
vc
⎩0 Vc < Vtri

1 t + Ttri
d=
Ttri ∫ t
q dt
1
t on
=
0 Ttri
ton
Vdc
1 dTtri
Va = ∫ Vdcdt = dVdc
Ttri 0
B Chitti Babu,
14 August 2009 0
EE NIT Rourkela
80

d

0.5

vc
-Vtri

Vtri

-Vtri vc

For vc = -Vtri → d = 0

B Chitti Babu,
14 August 2009 81
EE NIT Rourkela

d

0.5

vc
-Vtri -Vtri

Vtri

vc

Vtri

For vc = 0 → d = 0.5
14 August 2009
EE NIT
→ Rourkela
For vc = VtriChitti d = 1
B Babu,
82

d

0.5

vc
-Vtri -Vtri

Vtri vc

1
d = 0.5 + vc
2Vtri
Vtri

For vc = 0 → d = 0.5
14 August 2009
EE NIT
→ Rourkela
For vc = VtriChitti d = 1
B Babu,
83

Thus relation between vc and Va is obtained as:

V dc
V a = 0 . 5 V dc + vc
2 V tri

Introducing perturbation in vc and Va and separating DC and AC components:

V dc
DC: V a = 0 . 5 V dc + vc
2 V tri

AC: ~ = V dc ~
va vc
2 V tri
B Chitti Babu,
14 August 2009 84
EE NIT Rourkela


Taking Laplace Transform on the AC, the transfer function is obtained as:

v a (s) V dc
=
v c ( s ) 2 V tri

vc(s) V dc va(s)
DC motor
2 V tri

B Chitti Babu,
14 August 2009 85
EE NIT Rourkela

Bipolar switching scheme
Vdc
vc
2vtri
-Vdc
q
vtri
+
Vdc
Vdc vA
+ VAB −
0
−
vc Vdc
vB
0
q
Vdc
vAB
v v
d A = 0.5 + c dB = 1 − d A = 0.5 − c -Vdc
2Vtri 2Vtri

Vdc Vdc Vdc
VA = 0.5Vdc + vc VB = 0.5Vdc − vc VA − VB = VAB = vc
2Vtri 2Vtri Vtri

B Chitti Babu,
14 August 2009 86
EE NIT Rourkela

Bipolar switching scheme

v a ( s ) V dc
=
v c (s) V tri

vc(s) V dc va(s)
DC motor
V tri

B Chitti Babu,
14 August 2009 87
EE NIT Rourkela

Vdc
Unipolar switching scheme vc
Leg b
Vtri
+ -vc

vtri Vdc

qa
vc −

vA
Leg a

vtri

-vc qb vB

vc − vc vAB
d A = 0.5 + dB = 0.5 +
2Vtri 2Vtri

Vdc Vdc Vdc
VA = 0.5Vdc + vc VB = 0.5Vdc − vc VA − VB = VAB = vc
2Vtri 2Vtri Vtri

The same average value we’ve seen for bipolar !
B Chitti Babu,
14 August 2009 88
EE NIT Rourkela

Unipolar switching scheme

v a ( s ) V dc
=
v c (s) V tri

vc(s) V dc va(s)
DC motor
V tri

B Chitti Babu,
14 August 2009 89
EE NIT Rourkela


DC motor – separately excited or permanent magnet

dia dωm
v t = ia R a + L a + ea Te = Tl + J
dt dt
Te = kt ia ee = kt ω
Extract the dc and ac components by introducing small
perturbations in Vt, ia, ea, Te, TL and ωm
ac components dc components
~
~ = ~ R + L d ia + ~
v t ia a ea Vt = Ia R a + E a
a
dt
~ ~
Te = k E ( ia ) Te = k E Ia
~ = k (ω )
ee ~ Ee = k Eω
E

~
~ ~ ~ + J d(ω )
Te = TL + B ω Te = TL + B(ω)
14 August 2009 dt B Chitti Babu,
90
EE NIT Rourkela


Perform Laplace Transformation on ac components
~
~
~ = i R +L d ia ~ Vt(s) = Ia(s)Ra + LasIa + Ea(s)
vt a a a + ea
dt

~ ~ Te(s) = kEIa(s)
Te = k E ( ia )

~ = k (ω )
ee ~ Ea(s) = kEω(s)
E

~
~ ~ ~ + J d(ω )
Te = TL + B ω Te(s) = TL(s) + Bω(s) + sJω(s)
dt

B Chitti Babu,
14 August 2009 91
EE NIT Rourkela



Tl (s )
-
Va (s ) I a (s ) Te (s ) ω (s )
1 1
kT
+ Ra + sL a +
B + sJ
-

kE

B Chitti Babu,
14 August 2009 92
EE NIT Rourkela

q
vtri

Torque +
controller
Tc +
Vdc
–
−

q kt

DC motor
Tl (s )
Converter
T e (s ) Torque V dc Va (s ) 1 I a (s ) Te (s ) -
1 ω (s )
kT
controller Ra + sL a B + sJ
+ V tri ,peak + +
- -

kE

B Chitti Babu,
14 August 2009 93
EE NIT Rourkela


Closed-loop speed control – an example
Design procedure in cascade control structure

• Inner loop (current or torque loop) the fastest –
largest bandwidth

• The outer most loop (position loop) the slowest –
smallest bandwidth

• Design starts from torque loop proceed towards
outer loops

B Chitti Babu,
14 August 2009 94
EE NIT Rourkela


Closed-loop speed control – an example
OBJECTIVES:
• Fast response – large bandwidth
• Minimum overshoot
good phase margin (>65o) BODE PLOTS
• Zero steady state error – very large DC gain

METHOD
• Obtain linear small signal model

• Design controllers based on linear small signal model

• Perform large signal simulation for controllers verification
B Chitti Babu,
14 August 2009 95
EE NIT Rourkela

Modeling of the Power Converters: IM drives

INDUCTION MOTOR DRIVES

Scalar Control Vector Control

Const. V/Hz is=f(ωr) FOC DTC

Rotor Flux Stator Flux
Circular Hexagon DTC
Flux Flux SVM
B Chitti Babu,
14 August 2009 96
EE NIT Rourkela


Control of induction machine based on steady-state
model (per phase SS equivalent circuit):

Is Lls Llr’
Rs Ir’

+
+
Lm
Vs Rr’/s
Eag
– Im –

B Chitti Babu,
14 August 2009 97
EE NIT Rourkela

Te

Pull out
Torque Intersection point
(Tmax) (Te=TL) determines the
Te
steady –state speed

Trated TL

sm ωratedrotorωs
ω ωr
s

B Chitti Babu,
14 August 2009 98
EE NIT Rourkela

Power Electronics

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