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UNIT I
Power Semiconductor
Devices
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Introduction
• What are Power Semiconductor Devices (PSD)?
They are devices used as switches or rectifiers in
power electronic circuits
• What is the difference of PSD and low-power
semiconductor device?
Large voltage in the off state
High current capability in the on state
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Classification
Fig. 1. The power semiconductor devices family
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Important Parameters
• Breakdown voltage.
• On-resistance.
Trade-off between breakdown voltage and
on-resistance.
• Rise and fall times for switching between on
and off states.
• Safe-operating area.
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Power MOSFET: Structure
Power MOSFET has much higher current handling capability in
ampere range and drain to source blocking voltage(50-100V)
than other MOSFETs.
Fig.2.Repetitive pattern of the cells
structure in power MOSFET
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Power MOSFET: R-V Characteristics
An important parameter of a power MOSFET is on resistance:
, where on S CH D
R R R R= + +( )
CH
n ox GS T
LR
W C V Vµ=
−
Fig. 3. Typical RDS versus ID characteristics of a MOSFET.
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Thyristor: Structure • Thyristor is a general class of a four-layer pnpn
semiconducting device.
Fig.4 (a) The basic four-layer pnpn structure.
(b) Two two-transistor equivalent circuit.
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Three States:
Reverse Blocking
Forward Blocking
Forward Conducting
Thyristor: I-V Characteristics
Fig.5 The current-voltage
characteristics of the pnpn device.
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Applications
Power semiconductor devices have widespread applications:
Automotive
Alternator, Regulator, Ignition, stereo tape
Entertainment
Power supplies, stereo, radio and television
Appliance
Drill motors, Blenders, Mixers, Air conditioners and Heaters
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Thyristors
• Most important type of power semiconductor device.
• Have the highest power handling capability.they have a rating of 1200V / 1500A with switching frequencies ranging from 1KHz to 20KHz.
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• Is inherently a slow switching device compared to BJT or MOSFET.
• Used as a latching switch that can be turned on by the control terminal but cannot be turned off by the gate.
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Different types of Thyristors
• Silicon Controlled Rectifier (SCR).
• TRIAC.
• DIAC.
• Gate Turn-Off Thyristor (GTO).
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SCR
Symbol of
Silicon Controlled Rectifier
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Structure
µ
µ
µ
µ
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Device Operation
Simplified model of a
thyristor
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V-I
Characteristics
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Effects of gate current
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Two Transistor Model of SCR
⇒
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( ) ( )
1 1
1 1
1 1
1
1
Considering PNP transistor
of the equivalent circuit,
, , ,
,
1 1
E A C C
CBO CBO B B
B A CBO
I I I I
I I I I
I I I
α α
α
= = =
= =
∴ = − − − − −
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( ) ( )
2 2 2
2 2
2 2
2
2
Considering NPN transistor
of the equivalent circuit,
, ,
2
C C B B E K A G
C k CBO
C A G CBO
I I I I I I I I
I I I
I I I I
α
α
= = = = +
= +
= + + − − −
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( )
2 1
2 1 2
1 2
From the equivalent circuit,
we see that
1
C B
g CBO CBO
A
I I
I I II
α
α α
∴ =
+ +⇒ =
− +
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( )1 2
1 2
Case 1: When 0
1
g
CBO CBO
A
I
I II
α α
=
+=
− +
( )2 1 2
1 2
Case 2: When 0
1
G
g CBO CBO
A
I
I I II
α
α α
≠
+ +=
− +
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Turn-on
Characteristics
o n d rt t t= +
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Turn-off
Characteristi
c
!!
"
#$% #
"
& '
(
")$ **
) **
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Methods of Thyristor Turn-on
• Thermal Turn-on.
• Light.
• High Voltage.
• Gate Current.
• dv/dt.
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Thyristor Types
• Phase-control Thyristors (SCR’s).
• Fast-switching Thyristors (SCR’s).
• Gate-turn-off Thyristors (GTOs).
• Bidirectional triode Thyristors (TRIACs).
• Reverse-conducting Thyristors (RCTs).
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• Static induction Thyristors (SITHs).
• Light-activated silicon-controlled rectifiers (LASCRs).
• FET controlled Thyristors (FET-CTHs).
• MOS controlled Thyristors (MCTs).
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Phase Control Thyristor • These are converter thyristors.
• The turn-off time tq is in the order of 50 to
100µsec.
• Used for low switching frequency.
• Commutation is natural commutation
• On state voltage drop is 1.15V for a 600V device.
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• They use amplifying gate thyristor.
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Fast Switching
Thyristors • Also called inverter thyristors.
• Used for high speed switching applications.
• Turn-off time tq in the range of 5 to 50µsec.
• On-state voltage drop of typically 1.7V for
2200A, 1800V thyristor.
• High dv/dt and high di/dt rating.
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Bidirectional Triode
Thyristors (TRIAC)
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Mode-I
Operation
MT2 Positive,
Gate Positive
P1
N1
N2
P2Ig
Ig
MT2 (+)
MT1 ( )−G
V(+)
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Mode-II
Operation
MT2 Positive,
Gate Negative
P1
N1
N2N3
P2
Ig
MT2 (+)
MT1 ( )−G
V
Finalconduction
Initialconduction
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Mode-III Operation
MT2 Negative,
Gate Positive
P1
N1
N4
N2
P2
Ig
MT2 ( )−
MT1 (+)G(+)
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Mode-IV Operation
MT2 Negative,
Gate Negative
P1
N1
N4
P2
Ig
MT2 ( )−
MT1 (+)
N3
G(-)
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Triac Characteristics
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BJT structure
note: this is a current of electrons (npn case) and so the conventional current flows from collector to emitter.
heavily doped ~ 10^15 provides the carriers
lightly doped ~ 10^8 lightly doped ~ 10^6
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BJT characteristics
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BJT characteristics
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BJT modes of operation
Mode
EBJ
CBJ
Cutoff Reverse Reverse
Forward
active
Forward Reverse
Reverse
active
Reverse Forward
Saturation Forward Forward
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Cutoff: In cutoff, both junctions reverse biased. There is very little current flow, which corresponds to a logical "off", or an open switch.
Forward-active (or simply, active): The emitter-base junction is forward biased and the base-collector junction is reverse biased. Most bipolar transistors are designed to afford the greatest common-emitter current gain, βf in forward-active mode. If this is the case, the collector-emitter current is approximately proportional to the base current, but many times larger, for small base current variations. Reverse-active (or inverse-active or inverted): By reversing the biasing conditions of the forward-active region, a bipolar transistor goes into reverse-active mode. In this mode, the emitter and collector regions switch roles. Since most BJTs are designed to maximise current gain in forward-active mode, the βf in inverted mode is several times smaller. This transistor mode is seldom used. The reverse bias breakdown voltage to the base may be an order of magnitude lower in this region. Saturation: With both junctions forward-biased, a BJT is in saturation mode and facilitates current conduction from the emitter to the collector. This mode corresponds to a logical "on", or a closed switch.
BJT modes of operation
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BJT structure (active)
current of electrons for npn transistor – conventional current flows from collector to emitter.
B
C E
IE IC
IB
-
+
VBE VCB
-
+
+ - VCE
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• A GATE electrode is placed above (electrically insulated
from) the silicon surface, and is used to control the
resistance between the SOURCE and DRAIN regions
• NMOS: N-channel Metal Oxide Semiconductor
L
• L = channel length
“Metal” (heavily
doped poly-Si)
W • W = channel width
MOSFET
SOURCE
DRAIN
GATE
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• Without a gate-to-source voltage applied, no current can
flow between the source and drain regions.
• Above a certain gate-to-source voltage (threshold
voltage VT), a conducting layer of mobile electrons is
formed at the Si surface beneath the oxide. These
electrons can carry current between the source and drain.
N-channel MOSFET
n
p
oxide insulator gate
n
Drain Source
Gate
ID
IG
IS
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N-channel vs. P-channel
MOSFETs
• For current to flow, VGS > VT
• Enhancement mode: VT > 0
• Depletion mode: VT < 0
– Transistor is ON when VG=0V
p-type Si
n+ poly-Si
n-type Si
p+ poly-Si
NMOS PMOS
n+ n+ p+ p+
• For current to flow, VGS < VT
• Enhancement mode: VT < 0
• Depletion mode: VT > 0
– Transistor is ON when VG=0V
(“n+” denotes very heavily doped n-type material; “p+” denotes very heavily doped p-type material)
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MOSFET Circuit Symbols
p-type Si
n+ poly-Si
NMOS
n+ n+
n-type Si
p+ poly-Si
PMOS
p+ p+
G G
G G
S
S S
S
Body
Body
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• The voltage applied to the GATE terminal determines whether current can flow between the SOURCE & DRAIN terminals.
– For an n-channel MOSFET, the SOURCE is biased at a lower potential (often 0 V) than the DRAIN
(Electrons flow from SOURCE to DRAIN when VG > VT)
– For a p-channel MOSFET, the SOURCE is biased at a higher potential (often the supply voltage VDD) than the DRAIN
(Holes flow from SOURCE to DRAIN when VG < VT )
• The BODY terminal is usually connected to a fixed potential.
– For an n-channel MOSFET, the BODY is connected to 0 V
– For a p-channel MOSFET, the BODY is connected to VDD
MOSFET Terminals
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VGS
S
semiconductor oxide
G
VDS
+ − + −
D
always zero!
IG
VGS
The gate is insulated from the semiconductor, so there is no significant steady gate current.
IG
NMOSFET IG vs. VGS Characteristic
Consider the current IG (flowing into G) versus VGS :
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VGS
S
semiconductor oxide
G
VDS
ID
+ − + −
D
ID
zero if VGS < VT
VDS
Next consider ID (flowing into D) versus VDS, as VGS is varied:
Below “threshold” (VGS < VT): no charge no conduction
Above threshold (VGS > VT): “inversion layer” of electrons appears, so conduction between S and D is possible
VGS > VT
NMOSFET ID vs. VDS Characteristics
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The MOSFET as a Controlled Resistor
• The MOSFET behaves as a resistor when VDS is low:
– Drain current ID increases linearly with VDS
– Resistance RDS between SOURCE & DRAIN depends on VGS
• RDS is lowered as VGS increases above VT
NMOSFET Example:
ID
IDS = 0 if VGS < VT
VDS
VGS = 1 V > VT
VGS = 2 V
Inversion charge density Qi(x) = -Cox[VGS-VT-V(x)] where Cox ≡≡≡≡ εεεεox / tox
oxide thickness ≡ tox
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ID vs. VDS Characteristics
The MOSFET ID-VDS curve consists of two regions:
1) Resistive or “Triode” Region: 0 < VDS < VGS −−−− VT
2) Saturation Region:
VDS > VGS −−−− VT
( )
oxnn
TGSn
DSAT
Ck
VVL
WkI
µ=′
−′
=
where
2
2
oxnn
DSDS
TGSnD
Ck
VV
VVL
WkI
µ=′
−−′=
where
2
process transconductance parameter
“CUTOFF” region: VG < VT
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Part I: Bipolar Power Transistors
The Evolution Of IGBT
• Bipolar Power Transistor Uses Vertical Structure For Maximizing Cross Sectional Area Rather Than Using Planar Structure
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Part II: Power MOSFET
The Evolution Of IGBT
• Power MOSFET Uses Vertical Channel Structure Versus The Lateral Channel Devices Used In IC Technology
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Lateral MOSFET structure
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The Evolution Of IGBT
• Discrete BJT + Discrete Power MOSFET In Darlington Configuration
Part III: BJT(discrete) + Power MOSFET(discrete)
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Part IV: BJT(physics) + Power MOSFET(physics) = IGBT
The Evolution Of IGBT
• More Powerful And Innovative Approach Is To Combine Physics Of BJT With The Physics Of MOSFET Within Same Semiconductor Region
• This Approach Is Also Termed Functional Integration Of MOS And Bipolar Physics
• Using This Concept, The Insulated Gate Bipolar Transistor (IGBT) Emerged
• Superior On-State Characteristics, Reasonable Switching Speed And Excellent Safe Operating Area
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The Evolution Of IGBT
• IGBT Fabricated Using Vertical Channels (Similar To Both The Power BJT And MOSFET)
Part IV: BJT(physics) + Power MOSFET(physics) = IGBT
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Device Operation • Operation Of IGBT Can Be Considered Like A PNP Transistor With Base Drive Current Supplied By The MOSFET
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DRIVER CIRCUIT (BASE / GATE)
• Interface between control (low power electronics) and (high power) switch.
• Functions:
– amplifies control signal to a level required to drive power switch
– provides electrical isolation between power switch and logic level
• Complexity of driver varies markedly among switches. MOSFET/IGBT drivers are simple but GTO drivers are very complicated and expensive.
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ELECTRICAL ISOLATION FOR DRIVERS
• Isolation is required to prevent damages on the high power switch to propagate back to low power electronics.
• Normally opto-coupler (shown below) or high frequency magnetic materials (as shown in the thyristor case) are used.
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ELECTRICAL ISOLATION FOR DRIVERS
• Power semiconductor devices can be categorized into 3
types based on their control input requirements:
a) Current-driven devices – BJTs, MDs, GTOs
b) Voltage-driven devices – MOSFETs, IGBTs, MCTs
c) Pulse-driven devices – SCRs, TRIACs
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CURRENT DRIVEN DEVICES (BJT)
• Power BJT devices have low current gain due to constructional consideration, leading current than would normally be expected for a given load or collector current.
• The main problem with this circuit is the slow turn-off time. Many standard driver chips have built-in isolation. For example TLP 250 from Toshiba, HP 3150 from Hewlett-Packard uses opto-coupling isolation.
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ELECTRICALLY ISOLATED DRIVE CIRCUITS
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EXAMPLE: SIMPLE MOSFET GATE DRIVER
• Note: MOSFET requires VGS =+15V for turn on and 0V to turn off. LM311 is a simple amp with open collector output Q1.
• When B1 is high, Q1 conducts. VGS is pulled to ground. MOSFET is off.
• When B1 is low, Q1 will be off. VGS is pulled to VGG. If VGG is set to +15V, the MOSFET turns on.
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UNIT II
PHASE CONTROLLED
CONVERTERS
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Phase-Control Converters
Single-Phase
Semiconverter
Three-Phase
Full converter
Dual converter
Semiconverter
Full converter
Dual converter
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Semiconverter
..is a one-quadrant converter and it has one polarity
Full converter
..is a two-quadrant converter and the polarity of its
output can be either positive or negative.
However
the output current of full converter has
one polarity only
Dual converter ..can operate in four quadrants ; both the output
voltage and current can be either positive or negative
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cos12
sin2
1 mmdc
VttdVV
Average Output Voltage
m
dm
VV Maximum
Output Voltage
cos15.0 dm
dcn
V
VVNormalizing
Output Voltage
2
2sin1
2sin
2
1 22
mmrms
VttdVV
RMS Output Voltage
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If the converter has a purely resistive load of R and
the delay angle is , determine (a) the rectification efficiency
(b) the form factor FF (c) the ripple factor RF
and (d) the peak inverse voltage PIV of thyristor T1
2/
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%27.203536.0
1592.0
3536.02
22
sin2
1
2
1592.0
2cos1
2sin
2
1
2
2
2
2
2
m
m
rms
dc
mm
rms
mdc
mmdc
V
V
V
V
VV
V
VV
VttdVV
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If the converter has a purely resistive load of R and
the delay angle is , determine (a) the rectification efficiency
(b) the form factor FF
(c) the ripple factor RF and (d) the peak inverse voltage PIV of thyristor T1
2/
221.21592.0
3536.0
m
m
dc
rms
V
V
V
VFF
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If the converter has a purely resistive load of R and
the delay angle is , determine (a) the rectification efficiency
(b) the form factor FF
(c) the ripple factor RF
and (d) the peak inverse voltage PIV of thyristor T1
2/
983.11221.21 22 FFRF
mVPIV
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www.Vidyarthiplus.com Semiconverter
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Single-Phase Semiconverter
ttdVV
VttdVV
mrms
mmdc
22 sin2
2
cos1sin2
2
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Single-Phase Semiconverter (RL-load)
L
R
L
R
LLL
LL
eR
EeItiI
ERidt
diL
1
0
011
11
tL
R
SL
SL
SLL
eZ
V
R
EI
R
Et
Z
VI
tVERidt
diL
sin2
sin2
sin2
12
22
Mode 1
t0
Mode 2
tR
L 1tan 22LRZ
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Single-Phase Semiconverter (RL-load)
RMS Current
for Thyristor
tdiI LR
2
22
1
RMS Current
for Thyristor
tdiI LA 2
2
1
RMS Output
Current
tditdiI LLrms
2
20
2
12
1
2
1
AVG Output
Current
tditdiIdc 2
01
2
1
2
1
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The single-phase semiconverter has an RL load of
L = 6.5mH, R = 2.5 Ohm, and E = 10 V. The input
voltage is VS = 120 V(rms) at 60 Hz. Determine
(a) the load current IL0 at , and the load
current IL1 at , (b) the average thyristor current IA
(c) the rms thyristor current IR (d) the rms output current Irms
and (e) the average output current Idc
0t 60t
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Single-Phase
Full Converter
Rectification
Mode
Inversion
Mode
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Single-Phase Full Converter
2
sin2
2
cos2
sin2
2
22 mmrms
mmdc
VtdtVV
VtdtVV
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Single-Phase Full Converter (RL-load)
tL
R
SL
SL e
Z
V
R
EI
R
Et
Z
VI
sin2
sin2
0
Mode 1 = Mode 2 R
L 1tan 22LRZ
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Single-Phase Full Converter (RL-load)
RMS Current
for Thyristor
tdiI LR
2
2
1
RMS Current
for Thyristor
tdiI LA
2
1
RMS Output
Current RRRrms IIII 222
AVG Output
Current AAAdc IIII 2
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Dual Converter
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Single-Phase Dual Converter
High-Power Variable-Speed Drives
21
22
11
cos2
cos2
dcdc
mdc
mdc
VV
VV
VV
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Three-Phase Semiconverter
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3 Phase Controlled Rectifiers
• Operate from 3 phase ac supply voltage.
• They provide higher dc output voltage.
• Higher dc output power.
• Higher output voltage ripple frequency.
• Filtering requirements are simplified for
smoothing out load voltage and load current.
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• Extensively used in high power variable speed industrial dc drives.
• Three single phase half-wave converters
can be connected together to form a three phase half-wave converter.
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3-Phase
Half Wave Converter (3-Pulse Converter)
with RL Load
Continuous & Constant Load Current Operation
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Vector Diagram of 3 Phase Supply Voltages
VAN
VCN
VBN
1200
1200
1200 RN AN
YN BN
BN CN
v v
v v
v v
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3 Phase Supply Voltage
Equations
We deifine three line to neutral voltages
(3 phase voltages) as follows
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0
0
0
sin ;
Max. Phase Voltage
2sin
3
sin 120
2sin
3
sin 120
sin 240
RN an m
m
YN bn m
m
BN cn m
m
m
v v V t
V
v v V t
V t
v v V t
V t
V t
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van vbn vcn van
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io=Ia
Constant Load
Current
Ia
Ia
Each thyristor conducts for 2/3 (1200)
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To Derive an
Expression for the
Average Output Voltage of a
3-Phase Half Wave Converter
with RL Load
for Continuous Load Current
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0
1
0
2
0
3
0
306
5 150
6
7 270
6
2Each thytistor conducts for 120 or radians
3
T is triggered at t
T is triggered at t
T is triggered at t
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5
6
6
5
6
6
3sin .
2
3cos
2
3 5cos cos
2 6 6
mdc
mdc
mdc
VV t d t
VV t
VV
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0 0
0
Note from the trigonometric relationship
cos cos .cos sin .sin
5 5cos cos sin sin
6 63
2co
cos 150 cos sin 150 sin3
2 cos 30
s .cos sin sin6 6
.cos
mdc
mdc
A
VV
B A B A B
VV
0sin 30 sin
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0 0
0 0 0 0
0 0
0 0
0
0
0
0
0 0
Note: cos 1
cos 180 30 cos sin 180 30 sin3
2 cos 30 .cos sin 30 sin
cos 30 cos sin 30 sin3
2 cos 30 .cos sin 30 s
80 30 cos 30
sin 180 30 sin 30
in
mdc
mdc
VV
VV
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032cos 30 cos
2
3 32 cos
2 2
3 3 33 cos cos
2 2
3cos
2
Where 3 Max. line to line supply voltage
mdc
mdc
m mdc
Lmdc
Lm m
VV
VV
V VV
VV
V V
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max
The maximum average or dc output voltage is
obtained at a delay angle 0 and is given by
3 3
2
Where is the peak phase voltage.
And the normalized average output voltage is
mdmdc
m
ddcn n
VV V
V
VV V
cosc
dmV
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15 26
2 2
6
1
2
The rms value of output voltage is found by
using the equation
3sin .
2
and we obtain
1 33 cos 2
6 8
mO RMS
mO RMS
V V t d t
V V
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3 Phase Half Wave
Controlled Rectifier Output
Voltage Waveforms For RL
Load
at
Different Trigger Angles
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0
0
300
300
600
600
900
900
1200
1200
1500
1500
1800
1800
2100
2100
2400
2400
2700
2700
3000
3000
3300
3300
3600
3600
3900
3900
4200
4200
Van
V0
V0
Van
=300
=600
Vbn
Vbn
Vcn
Vcn
t
t
=300
=600
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030
060
090
0120
0150
0180
0210
0240
0270
0300
0330
0360
0390
0420
0
V0
Van
=900
Vbn Vcn
t
=900
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3 Phase Half Wave Controlled Rectifier With
R Load and
RL Load with FWD
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a a
b b
c c
R
V0
L
R V0
+
T1
T2
T3
n n
T1
T2
T3
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3 Phase Half Wave
Controlled Rectifier Output
Voltage Waveforms For R Load
or RL Load with FWD
at
Different Trigger Angles
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0
0
300
300
600
600
900
900
1200
1200
1500
1500
1800
1800
2100
2100
2400
2400
2700
2700
3000
3000
3300
3300
3600
3600
3900
3900
4200
4200
Vs
V0
Van
=0
=150
Vbn Vcn
t
VanVbn Vcn
t
=00
=150
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0
0
300
300
600
600
900
900
1200
1200
1500
1500
1800
1800
2100
2100
2400
2400
2700
2700
3000
3000
3300
3300
3600
3600
3900
3900
4200
4200
V0
=300
VanVbn Vcn
t
V0
=600
VanVbn Vcn
t
=300
=600
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To Derive An
Expression For The Average Or Dc Output Voltage Of A
3 Phase Half Wave Converter With Resistive Load
Or RL Load With FWD
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0
1
0 0
1
0
2
0 0
2
0
306
30 180 ;
sin
5 150
6
150 300 ;
sin 120
O an m
O bn m
T is triggered at t
T conducts from to
v v V t
T is triggered at t
T conducts from to
v v V t
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0
3
0 0
3
0
0
7 270
6
270 420 ;
sin 240
sin 120
O cn m
m
T is triggered at t
T conducts from to
v v V t
V t
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0
0
0
0
0
0
180
30
0 0
180
30
180
30
3.
2
sin ; for 30 to 180
3sin .
2
3sin .
2
dc O
O an m
dc m
mdc
V v d t
v v V t t
V V t d t
VV t d t
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0
0
180
30
0 0
0
0
3cos
2
3cos180 cos 30
2
cos180 1, we get
31 cos 30
2
mdc
mdc
mdc
VV t
VV
VV
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Three Phase Semiconverters
• 3 Phase semiconverters are used in Industrial dc drive applications upto 120kW
power output.
• Single quadrant operation is possible.
• Power factor decreases as the delay angle
increases.
• Power factor is better than that of 3 phase
half wave converter.
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3 Phase
Half Controlled Bridge Converter
(Semi Converter)
with Highly Inductive Load &
Continuous Ripple free Load
Current
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Wave forms of 3 Phase
Semiconverter for
> 600
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0 0
1
3 phase semiconverter output ripple frequency of
output voltage is 3
The delay angle can be varied from 0 to
During the period
30 210
7, thyristor T is forward biased
6 6
Sf
t
t
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1
1 1
If thyristor is triggered at ,6
& conduct together and the line to line voltage
appears across the load.
7At , becomes negative & FWD conducts.
6
The load current contin
ac
ac m
T t
T D
v
t v D
1 1
ues to flow through FWD ;
and are turned off.
mD
T D
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1
2
1 2
If FWD is not used the would continue to
conduct until the thyristor is triggered at
5, and Free wheeling action would
6
be accomplished through & .
If the delay angle , e3
mD T
T
t
T D
ach thyristor conducts
2for and the FWD does not conduct.
3mD
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0
0
0
We deifine three line neutral voltages
(3 phase voltages) as follows
sin ; Max. Phase Voltage
2sin sin 120
3
2sin sin 120
3
sin 240
RN an m m
YN bn m m
BN cn m m
m
v v V t V
v v V t V t
v v V t V t
V t
V
is the peak phase voltage of a wye-connected source.m
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3 sin6
53 sin
6
3 sin2
3 sin6
RB ac an cn m
YR ba bn an m
BY cb cn bn m
RY ab an bn m
v v v v V t
v v v v V t
v v v v V t
v v v v V t
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Wave forms of 3 Phase
Semiconverter for
600
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To derive an
Expression for the
Average Output Voltage
of 3 Phase Semiconverter
for > / 3 and Discontinuous Output Voltage
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76
6
76
6
For and discontinuous output voltage:3
the Average output voltage is found from
3.
2
33 sin
2 6
dc ac
dc m
V v d t
V V t d t
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max
3 31 cos
2
31 cos
2
3 Max. value of line-to-line supply voltage
The maximum average output voltage that occurs at
a delay angle of 0 is
3 3
mdc
mLdc
mL m
mdmdc
VV
VV
V V
VV V
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17 2
62
6
The normalized average output voltage is
0.5 1 cos
The rms output voltage is found from
3.
2
dcn
dm
acO rms
VV
V
V v d t
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Three Phase Dual Converters
• For four quadrant operation in many industrial
variable speed dc drives , 3 phase dual
converters are used.
• Used for applications up to 2 mega watt output
power level.
• Dual converter consists of two 3 phase full
converters which are connected in parallel & in
opposite directions across a common load.
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Outputs of Converters 1 & 2
• During the interval (/6 + 1) to (/2 + 1),
the line to line voltage vab appears across
the output of converter 1 and vbc appears
across the output of converter 2
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0
0
0
We deifine three line neutral voltages
(3 phase voltages) as follows
sin ;
Max. Phase Voltage
2sin sin 120
3
2sin sin 120
3
sin 240
RN an m
m
YN bn m m
BN cn m m
m
v v V t
V
v v V t V t
v v V t V t
V t
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0
0
0
We deifine three line neutral voltages
(3 phase voltages) as follows
sin ;
Max. Phase Voltage
2sin sin 120
3
2sin sin 120
3
sin 240
RN an m
m
YN bn m m
BN cn m m
m
v v V t
V
v v V t V t
v v V t V t
V t
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To obtain an Expression for the Circulating
Current
If vO1 and vO2 are the output voltages of
converters 1 and 2 respectively, the
instantaneous voltage across the current
limiting inductor during the interval
(/6 + 1) t (/2 + 1) is given by
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1 2
3 sin sin6 2
3 cos6
The circulating current can be calculated by
using the equation
r O O ab bc
r m
r m
v v v v v
v V t t
v V t
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1
1
6
6
1
max
1.
13 cos .
6
3sin sin
6
3
t
r r
r
t
r m
r
mr
r
m
r
r
i t v d tL
i t V t d tL
Vi t t
L
Vi
L
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Four Quadrant Operation
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• There are two different modes of
operation.
Circulating current free
(non circulating) mode of operation
Circulating current mode of operation
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Non Circulating Current Mode Of Operation
• In this mode of operation only one converter is
switched on at a time
• When the converter 1 is switched on,
For 1 < 900 the converter 1 operates in the
Rectification mode
Vdc is positive, Idc is positive and hence the
average load power Pdc is positive.
• Power flows from ac source to the load
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• When the converter 1 is on,
For 1 > 900 the converter 1 operates in the Inversion mode
Vdc is negative, Idc is positive and the average load power Pdc is negative.
• Power flows from load circuit to ac source.
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• When the converter 2 is switched on,
For 2 < 900 the converter 2 operates in
the Rectification mode
Vdc is negative, Idc is negative and the
average load power Pdc is positive.
• The output load voltage & load current
reverse when converter 2 is on.
• Power flows from ac source to the load
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• When the converter 2 is switched on,
For 2 > 900 the converter 2 operates in the
Inversion mode
Vdc is positive, Idc is negative and the average
load power Pdc is negative.
• Power flows from load to the ac source.
• Energy is supplied from the load circuit to the ac
supply.
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• Both the converters are switched on at the same
time.
• One converter operates in the rectification mode
while the other operates in the inversion mode.
• Trigger angles 1 & 2 are adjusted such that
(1 + 2) = 1800
Circulating Current
Mode Of Operation
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When 1 < 900, converter 1 operates
as a controlled rectifier. 2 is made greater than 900 and converter 2 operates as an Inverter.
• Vdc is positive & Idc is positive and Pdc is positive.
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• When 2 < 900, converter 2 operates
as a controlled rectifier. 1 is made greater than 900 and converter 1 operates as an Inverter.
• Vdc is negative & Idc is negative and Pdc is positive.
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UNIT III
DC Choppers
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Introduction • Chopper is a static device.
• A variable dc voltage is obtained from a constant dc voltage source.
• Also known as dc-to-dc converter.
• Widely used for motor control.
• Also used in regenerative braking.
• Thyristor converter offers greater efficiency, faster response, lower maintenance, smaller size and smooth control.
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Choppers are of Two Types
• Step-down choppers.
• Step-up choppers.
• In step down chopper output voltage is less than input voltage.
• In step up chopper output voltage is more than input voltage.
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Principle Of
Step-down Chopper
V
i0
V0
Chopper
R
+
−
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• A step-down chopper with resistive load.
• The thyristor in the circuit acts as a switch.
• When thyristor is ON, supply voltage appears across the load
• When thyristor is OFF, the voltage across the load will be zero.
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Vdc
v0
V
V/R
i0
Idc
t
t
tON
T
tOFF
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verage value of output or load voltage.
verage value of output or load current.
Time interval for which SCR conducts.
Time interval for which SCR is OFF.
Period of switching
dc
dc
ON
OFF
ON OFF
V A
I A
t
t
T t t
=
=
=
=
= + = or chopping period.
1 Freq. of chopper switching or chopping freq.f
T= =
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Average Output Voltage
.
duty cycle
ONdc
ON OFF
ONdc
ON
tV V
t t
tV V V d
T
tbut d
t
=
+
= =
= =
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2
0
Average Output Current
RMS value of output voltage
1 ON
dcdc
ONdc
t
O o
VI
R
tV VI d
R T R
V v dtT
=
= =
= ∫
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2
0
2
But during ,
Therefore RMS output voltage
1
.
.
ON
ON o
t
O
ONO ON
O
t v V
V V dtT
tVV t V
T T
V d V
=
=
= =
=
∫
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2
2
Output power
But
Output power
O O O
OO
OO
O
P V I
VI
R
VP
R
dVP
R
=
=
∴
=
=
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Effective input resistance of chopper
The output voltage can be varied by
varying the duty cycle.
i
dc
i
VR
I
RR
d
=
=
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Methods Of Control
• The output dc voltage can be varied by the following methods.
– Pulse width modulation control or constant frequency operation.
– Variable frequency control.
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Pulse Width Modulation
• tON is varied keeping chopping frequency ‘f’ & chopping period ‘T’ constant.
• Output voltage is varied by varying the ON time tON
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V0
V
V
V0
t
ttON
tON tOFF
tOFF
T
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Variable Frequency Control
• Chopping frequency ‘f’ is varied keeping either tON or tOFF constant.
• To obtain full output voltage range, frequency has to be varied over a wide range.
• This method produces harmonics in the output and for large tOFF load current may become discontinuous
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v0
V
V
v0
t
t
tON
tON
T
T
tOFF
tOFF
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Step-down Chopper
With R-L Load
V
i0
V0
Chopper
R
LFWD
E
+
−
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• When chopper is ON, supply is connected across load.
• Current flows from supply to load.
• When chopper is OFF, load current continues to flow in the same direction through FWD due to energy stored in inductor ‘L’.
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• Load current can be continuous or discontinuous depending on the values of ‘L’ and duty cycle ‘d’
• For a continuous current operation, load current varies between two limits Imax and Imin
• When current becomes equal to Imax the chopper is turned-off and it is turned-on when current reduces to Imin.
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Outputvoltage
Outputcurrent
v0
V
i0
Imax
Imin
t
t
tON
T
tOFF
Continuouscurrent
Outputcurrent
t
Discontinuouscurrent
i0
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Expressions For
Load Current
iO For Continuous Current
Operation When
Chopper Is ON (0 ≤ t ≤ tON)
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V
i0
V0
R
L
E
+
-
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( ) ( ) ( )
( )
( )
min
min
Taking Laplace Transform
. 0
At 0, initial current 0
OO
O O O
O
O
diV i R L E
dt
V ERI S L S I S i
S S
t i I
IV EI S
RRSLS S
LL
−
−
= + +
= + − +
= =
−= +
++
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( ) min
Taking Inverse Laplace Transform
1
This expression is valid for 0 ,
i.e., during the period chopper is ON.
At the instant the chopper is turned off,
load c
R Rt t
L L
O
ON
V Ei t e I e
R
t t
− −
−= − +
≤ ≤
( ) maxurrent is O ONi t I=
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When Chopper is OFF
i0
R
L
E
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( )
( ) ( ) ( )
( ) max
When Chopper is OFF 0
0
Talking Laplace transform
0 0
Redefining time origin we have at 0,
initial current 0
OFF
OO
O O O
O
t t
diRi L E
dt
ERI S L SI S i
S
t
i I
−
−
≤ ≤
= + +
= + − +
=
=
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( )
( )
max
max
Taking Inverse Laplace Transform
1
O
R Rt t
L LO
I EI S
R RS LS S
L L
Ei t I e e
R
− −
∴ = −
+ +
= − −
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( ) min
The expression is valid for 0 ,
i.e., during the period chopper is OFF
At the instant the chopper is turned ON or at
the end of the off period, the load current is
OFF
O OFF
t t
i t I
≤ ≤
=
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( )
( )
min
max
max
max min
min
From equation
1
At ,
To Find &
1
R Rt t
L L
O
ON O
dRT dRT
L L
V Ei t e I e
R
t t dT i t I
V EI e I e
I I
R
− −
− −
−= − +
= = =
−∴ = − +
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( )
( )
( )
max
min
From equation
1
At ,
1
R Rt t
L LO
OFF ON O
OFF
Ei t I e e
R
t t T t i t I
t t d T
− − = − −
= = − =
= = −
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( ) ( )1 1
min max
min
max min
max
1
Substituting for in equation
1
we get,
1
1
d RT d RT
L L
dRT dRT
L L
dRT
L
RT
L
EI I e e
R
I
V EI e I e
R
V e EI
R Re
− −− −
− −
−
−
∴ = − −
−= − +
− = −
−
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( ) ( )
( )
max
1 1
min max
min
max min
Substituting for in equation
1
we get,
1
1
is known as the steady state ripple.
d RT d RT
L L
dRT
L
RT
L
I
EI I e e
R
V e EI
R Re
I I
− −− −
= − −
− = −
−
−
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( )
max min
max min
Therefore peak-to-peak ripple current
Average output voltage
.
Average output current
2
dc
dc approx
I I I
V d V
I II
∆ = −
=
+=
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( )( )
min max
min
max minmin
Assuming load current varies linearly
from to instantaneous
load current is given by
. 0
O ON
O
I I
I ti I for t t dT
dT
I Ii I t
dT
∆= + ≤ ≤
− = +
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( )
( )
( )
( )
( )
2
0
0
2
max min
min
0
2
min max min2 2max minmin
0
RMS value of load current
1
1
21
dT
O RMS
dT
O RMS
dT
O RMS
I i dtdT
I I tI I dt
dT dT
I I I tI II I t dt
dT dT dT
=
− = +
−− = + +
∫
∫
∫
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( )
( )( )
12 2
max min2
min min max min
2
0
0
2
max minmin
0
RMS value of output current
3
RMS chopper current
1
1
O RMS
dT
CH
dT
CH
I II I I I I
I i dtT
I II I t dt
T dT
−= + + −
=
− = +
∫
∫
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( )( )
( )
12 2
max min2
min min max min3
Effective input resistance is
CH
CH O RMS
i
S
I II d I I I I
I d I
VR
I
−= + + −
=
=
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Where
Average source currentS
S dc
i
dc
I
I dI
VR
dI
=
=
∴ =
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Principle Of Step-up Chopper
+
−
VOV
Chopper
CLOAD
DLI
+ −
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• Step-up chopper is used to obtain a load voltage higher than the input voltage V.
• The values of L and C are chosen depending upon the requirement of output voltage and current.
• When the chopper is ON, the inductor L is connected across the supply.
• The inductor current ‘I’ rises and the inductor stores energy during the ON time of the chopper, tON.
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• When the chopper is off, the inductor current I is forced to flow through the diode D and load for a period, tOFF.
• The current tends to decrease resulting in reversing the polarity of induced EMF in L.
• Therefore voltage across load is given by
. ., O O
dIV V L i e V V
dt= + >
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• A large capacitor ‘C’ connected across the load, will provide a continuous output voltage .
• Diode D prevents any current flow from capacitor to the source.
• Step up choppers are used for regenerative braking of dc motors.
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Expression For Output Voltage Assume the average inductor current to be
during ON and OFF time of Chopper.
Voltage across inductor
Therefore energy stored in inductor
= . .
Where
When Chopper
period of chopper.
is ON
ON
ON
I
L V
V I t
t ON
=
=
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( )
(energy is supplied by inductor to load)
Voltage across
Energy supplied by inductor
where period of Chopper.
Neg
When Chopper
lecting losses, energy stored in inductor
is OFF
O
O OFF
OFF
L V V
L V V It
t OFF
L
= −
= −
=
= energy supplied by inductor L
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( )
[ ]
Where
T = Chopping period or period
of switching.
ON O OFF
ON OFF
O
OFF
O
ON
VIt V V It
V t tV
t
TV V
T t
∴ = −
+=
=
−
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1
1
1
1
Where duty cyle
ON OFF
OON
O
ON
T t t
V Vt
T
V Vd
td
T
= +
= −
∴ =
−
= =
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For variation of duty cycle ' ' in the
range of 0 1 the output voltage
will vary in the range
O
O
d
d V
V V
< <
< < ∞
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Performance Parameters • The thyristor requires a certain minimum time to
turn ON and turn OFF.
• Duty cycle d can be varied only between a min. & max. value, limiting the min. and max. value of the output voltage.
• Ripple in the load current depends inversely on the chopping frequency, f.
• To reduce the load ripple current, frequency should be as high as possible.
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Problem
• A Chopper circuit is operating on TRC at a frequency of 2 kHz on a 460 V supply. If the load voltage is 350 volts, calculate the conduction period of the thyristor in each cycle.
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3
460 V, = 350 V, f = 2 kHz
1Chopping period
10.5 sec
2 10
Output voltage
dc
ONdc
V V
Tf
T m
tV V
T
−
=
=
= =×
=
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3
Conduction period of thyristor
0.5 10 350
460
0.38 msec
dcON
ON
ON
T Vt
V
t
t
−
×=
× ×=
=
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Problem
• Input to the step up chopper is 200 V. The output required is 600 V. If the conducting
time of thyristor is 200 µsec. Compute
– Chopping frequency,
– If the pulse width is halved for constant frequency of operation, find the new output voltage.
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6
200 , 200 , 600
600 200200 10
Solving for
300
ON dc
dc
ON
V V t s V V
TV V
T t
T
T
T
T s
µ
µ
−
= = =
=
−
=
− ×
=
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6
6
Chopping frequency
1
13.33
300 10
Pulse width is halved
200 10100
2ON
fT
f KHz
t sµ
−
−
=
= =×
×∴ = =
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( )
6
6
Frequency is constant
3.33
1300
Output voltage =
300 10200 300 Volts
300 100 10
ON
f KHz
T sf
TV
T t
µ
−
−
∴ =
= =
∴
−
×= = −
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Problem
• A dc chopper has a resistive load of 20Ω and input voltage VS = 220V. When chopper is ON, its voltage drop is 1.5 volts and chopping frequency is 10 kHz. If the duty cycle is 80%, determine the average output voltage and the chopper on time.
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( )
( )
220 , 20 , 10
0.80
= Voltage drop across chopper = 1.5 volts
Average output voltage
0.80 220 1.5 174.8 Volts
S
ON
ch
ONdc S ch
dc
V V R f kHz
td
T
V
tV V V
T
V
= = Ω =
= =
= −
= − =
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3
3
3
3
Chopper ON time,
1Chopping period,
10.1 10 secs 100 secs
10 10
Chopper ON time,
0.80 0.1 10
0.08 10 80 secs
ON
ON
ON
ON
t dT
Tf
T
t dT
t
t
−
−
−
=
=
= = × =×
=
= × ×
= × =
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Problem
• In a dc chopper, the average load current is 30 Amps, chopping frequency is 250 Hz, supply voltage is 110 volts. Calculate the ON and OFF periods of the chopper if the load resistance is 2 ohms.
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3
30 , 250 , 110 , 2
1 1Chopping period, 4 10 4 msecs
250
&
30 20.545
110
dc
dcdc dc
dc
dc
I Amps f Hz V V R
Tf
VI V dV
R
dVI
R
I Rd
V
−
= = = = Ω
= = = × =
= =
∴ =
×= = =
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3
3 3
3
Chopper ON period,
0.545 4 10 2.18 msecs
Chopper OFF period,
4 10 2.18 10
1.82 10 1.82 msec
ON
OFF ON
OFF
OFF
t dT
t T t
t
t
−
− −
−
= = × × =
= −
= × − ×
= × =
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• A dc chopper in figure has a resistive load
of R = 10Ω and input voltage of V = 200 V. When chopper is ON, its voltage drop is 2 V and the chopping frequency is 1 kHz. If the duty cycle is 60%, determine
– Average output voltage
– RMS value of output voltage
– Effective input resistance of chopper
– Chopper efficiency.
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V
i0
Chopper
+
−
R v0
200 , 10 , 2
0.60, 1 .
chV V R Chopper voltage drop V V
d f kHz
= = Ω =
= =
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( )
[ ]
( )
( )
Average output voltage
0.60 200 2 118.8 Volts
RMS value of output voltage
0.6 200 2 153.37 Volts
dc ch
dc
O ch
O
V d V V
V
V d V V
V
= −
= − =
= −
= − =
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( )22
0
0 0
Effective input resistance of chopper is
118.811.88 Amps
10
20016.83
11.88
Output power is
1 1
i
S dc
dcdc
i
S dc
dT dT
ch
O
V VR
I I
VI
R
V VR
I I
V VvP dt dt
T R T R
= =
= = =
= = = = Ω
−= =∫ ∫
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( )
[ ]
( )
2
2
0
0
0.6 200 22352.24 watts
10
Input power,
1
1
ch
O
O
dT
i O
dT
ch
O
d V VP
R
P
P Vi dtT
V V VP dt
T R
−=
−= =
=
−=
∫
∫
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( )
[ ]0.6 200 200 22376 watts
10
Chopper efficiency,
100
2352.24100 99%
2376
ch
O
O
O
i
dV V VP
R
P
P
Pη
η
−=
× −= =
= ×
= × =
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Problem • A chopper is supplying an inductive load with a
free-wheeling diode. The load inductance is 5 H
and resistance is 10Ω.. The input voltage to the chopper is 200 volts and the chopper is operating
at a frequency of 1000 Hz. If the ON/OFF time
ratio is 2:3. Calculate
– Maximum and minimum values of load current
in one cycle of chopper operation.
– Average load current
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5 , 10 , 1000 ,
200 , : 2 : 3
Chopping period,
1 11 msecs
1000
2
3
2
3
ON OFF
ON
OFF
ON OFF
L H R f Hz
V V t t
Tf
t
t
t t
= = Ω =
= =
= = =
=
=
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3
2
3
5
3
3
5
31 10 0.6 msec
5
ON OFF
OFF OFF
OFF
OFF
T t t
T t t
T t
t T
T−
= +
= +
=
=
= × × =
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( ) 3
3
3
max
1 0.6 10 0.4 msec
Duty cycle,
0.4 100.4
1 10
Maximum value of load current is given by
1
1
ON OFF
ON
ON
dRT
L
RT
L
t T t
t
td
T
V e EI
R Re
−
−
−
−
−
= −
= − × =
×= = =
×
− = −
−
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3
3
max
0.4 10 1 10
5
max 10 1 10
5
Since there is no voltage source in
the load circuit, E = 0
1
1
200 1
101
dRT
L
RT
L
V eI
Re
eI
e
−
−
−
−
× × ×−
× ×−
− ∴ =
−
−
= −
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3
3
0.8 10
max 2 10
max
min
120
1
8.0047A
Minimum value of load current with E = 0
is given by
1
1
dRT
L
RT
L
eI
e
I
V eI
Re
−
−
− ×
− ×
−=
−
=
− =
−
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3
3
0.4 10 1 10
5
min 10 1 10
5
max min
200 17.995 A
101
Average load current
2
8.0047 7.9958 A
2
dc
dc
eI
e
I II
I
−
−
× × ×
× ×
−
= = −
+=
+= ≈
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Problem
• A chopper feeding on RL load is shown in figure,
with V = 200 V, R = 5Ω, L = 5 mH, f = 1
kHz, d = 0.5 and E = 0 V. Calculate
– Maximum and minimum values of load
current.
– Average value of load current.
– RMS load current.
– Effective input resistance as seen by source.
– RMS chopper current.
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3
3
V = 200 V, R = 5 , L = 5 mH,
f = 1kHz, d = 0.5, E = 0
Chopping period is
1 11 10 secs
1 10T
f
−
Ω
= = = ××
i0
v0
Chopper
R
LFWD
E
+
−
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3
3
3
3
max
0.5 5 1 10
5 10
max 5 1 10
5 10
0.5
max 1
Maximum value of load current is given by
1
1
200 10
51
140 24.9 A
1
dRT
L
RT
L
V e EI
R Re
eI
e
eI
e
−
−
−
−
−
× × ×−
×
× ×−
×
−
−
− = −
−
−
= − −
−= =
−
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3
3
3
3
min
0.5 5 1 10
5 10
min 5 1 10
5 10
0.5
min 1
Minimum value of load current is given by
1
1
200 10
51
140 15.1 A
1
dRT
L
RT
L
V e EI
R Re
eI
e
eI
e
−
−
−
−
× × ×
×
× ×
×
− = −
−
−
= − −
−= =
−
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( )
( )( )
1 2
12 2
max min2
min min max min
Average value of load current is
2
for linear variation of currents
24.9 15.120 A
2
RMS load current is given by
3
dc
dc
O RMS
I II
I
I II I I I I
+=
+∴ = =
−= + + −
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( )
( )( )
( )
( )
12 2
2
1
2
24.9 15.115.1 15.1 24.9 15.1
3
96.04228.01 147.98 20.2 A
3
RMS chopper current is given by
0.5 20.2 14.28 A
O RMS
O RMS
ch O RMS
I
I
I d I
−= + + −
= + + =
= = × =
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Effective input resistance is
= Average source current
0.5 20 10 A
Therefore effective input resistance is
20020
10
i
S
S
S dc
S
i
S
VR
I
I
I dI
I
VR
I
=
=
= × =
= = = Ω
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Classification Of Choppers
• Choppers are classified as
– Class A Chopper
– Class B Chopper
– Class C Chopper
– Class D Chopper
– Class E Chopper
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Class A Chopper
V
Chopper
FWD
+
−
v0
v0
i0
i0
LOAD
V
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• When chopper is ON, supply voltage V is connected across the load.
• When chopper is OFF, vO = 0 and the load current continues to flow in the same direction through the FWD.
• The average values of output voltage and current are always positive.
• Class A Chopper is a first quadrant chopper .
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• Class A Chopper is a step-down chopper in which power always flows form source to load.
• It is used to control the speed of dc motor.
• The output current equations obtained in step down chopper with R-L load can be used to study the performance of Class A Chopper.
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Output current
Thyristorgate pulse
Output voltage
ig
i0
v0
t
t
ttON
T
CH ON
FWD Conducts
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Class B Chopper
V
Chopper
+
−
v0
v0
−i0
i0
L
E
R
D
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• When chopper is ON, E drives a current through L and R in a direction opposite to that shown in figure.
• During the ON period of the chopper, the inductance L stores energy.
• When Chopper is OFF, diode D conducts, and part of the energy stored in inductor L is returned to the supply.
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• Average output voltage is positive.
• Average output current is negative.
• Therefore Class B Chopper operates in second quadrant.
• In this chopper, power flows from load to source.
• Class B Chopper is used for regenerative braking of dc motor.
• Class B Chopper is a step-up chopper.
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Output current
D conducts Chopper
conducts
Thyristorgate pulse
Output voltage
ig
i0
v0
t
t
t
Imin
Imax
T
tONtOFF
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Expression for Output Current
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( ) min
For the initial condition i.e.,
During the interval diode 'D' conduc
at 0
The solution of the ab
ts
voltage equation
ove equation is obtained
along similar lines as in s
is given by
OO
O
LdiV Ri E
dt
i t I t
= + +
= =
tep-down chopper
with R-L load
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( ) ( )
min
max
max min
During the interval chopper is ON voltage
equation is g
1 0
At
1
0
iven by
OFF OFF
R Rt t
L LO OFF
OFF O
R Rt t
L L
OO
V Ei t e I e t t
R
t t i t I
V EI e I e
R
LdiRi E
dt
− −
− −
−∴ = − + < <
= =
−= − +
= + +
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( )
( )
( )
max
max
min
min max
Redefining the time origin, at 0
The solution for the stated initial condition is
1 0
At
1ON ON
O
R Rt t
L LO ON
ON O
R Rt t
L L
t i t I
Ei t I e e t t
R
t t i t I
EI I e e
R
− −
− −
= =
= − − < <
= =
∴ = − −
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Class C Chopper
V
Chopper
+
−
v0
D1
D2
CH2
CH1
v0i0
i0
L
E
R
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• Class C Chopper is a combination of Class A and Class B Choppers.
• For first quadrant operation, CH1 is ON or D2 conducts.
• For second quadrant operation, CH2 is ON or D1 conducts.
• When CH1 is ON, the load current is positive.
• The output voltage is equal to ‘V’ & the load receives power from the source.
• When CH1 is turned OFF, energy stored in inductance L forces current to flow through the diode D2 and the output voltage is zero.
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• Current continues to flow in positive direction.
• When CH2 is triggered, the voltage E forces current to flow in opposite direction through L and CH2 .
• The output voltage is zero.
• On turning OFF CH2 , the energy stored in the inductance drives current through diode D1 and the supply
• Output voltage is V, the input current becomes negative and power flows from load to source.
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• Average output voltage is positive
• Average output current can take both positive and negative values.
• Choppers CH1 & CH2 should not be turned ON simultaneously as it would result in short circuiting the supply.
• Class C Chopper can be used both for dc motor control and regenerative braking of dc motor.
• Class C Chopper can be used as a step-up or step-down chopper.
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Gate pulseof CH2
Gate pulseof CH1
Output current
Output voltage
ig1
ig2
i0
V0
t
t
t
t
D1 D1D2 D2CH1 CH2 CH1 CH2
ON ON ON ON
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Class D Chopper
V+ −v0
D2
D1 CH2
CH1
v0
i0
L ER i0
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• Class D is a two quadrant chopper.
• When both CH1 and CH2 are triggered simultaneously, the output voltage vO = V and output current flows through the load.
• When CH1 and CH2 are turned OFF, the load current continues to flow in the same direction through load, D1 and D2 , due to the energy stored in the inductor L.
• Output voltage vO = - V .
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• Average load voltage is positive if chopper ON time is more than the OFF time
• Average output voltage becomes negative if tON < tOFF .
• Hence the direction of load current is always positive but load voltage can be positive or negative.
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Gate pulseof CH2
Gate pulseof CH1
Output current
Output voltage
Average v0
ig1
ig2
i0
v0
V
t
t
t
t
CH ,CH
ON1 2 D1,D2 Conducting
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Gate pulseof CH2
Gate pulseof CH1
Output current
Output voltage
Average v0
ig1
ig2
i0
v0
V
t
t
t
t
CH
CH1
2
D , D1 2
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Class E Chopper
V
v0
i0L ER
CH2 CH4D2 D4
D1 D3CH1 CH3
+ −
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Four Quadrant Operation v0
i0
CH - CH ON
CH - D Conducts1 4
4 2
D D2 3 - Conducts
CH - D Conducts4 2
CH - CH ON
CH - D Conducts3 2
2 4
CH - D Conducts
D - D Conducts2 4
1 4
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• Class E is a four quadrant chopper
• When CH1 and CH4 are triggered, output current iO flows in positive direction through CH1 and CH4, and with output voltage vO = V.
• This gives the first quadrant operation.
• When both CH1 and CH4 are OFF, the energy stored in the inductor L drives iO through D2 and D3 in the same direction, but output voltage vO = -V.
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• Therefore the chopper operates in the
fourth quadrant.
• When CH2 and CH3 are triggered, the load current iO flows in opposite direction & output voltage vO = -V.
• Since both iO and vO are negative, the chopper operates in third quadrant.
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• When both CH2 and CH3 are OFF, the load current iO continues to flow in the same direction D1 and D4 and the output voltage vO = V.
• Therefore the chopper operates in second quadrant as vO is positive but iO is negative.
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Problem • For the first quadrant chopper shown in figure,
express the following variables as functions of V,
R and duty cycle ‘d’ in case load is resistive.
– Average output voltage and current
– Output current at the instant of commutation
– Average and RMS free wheeling diode current.
– RMS value of output voltage
– RMS and average thyristor currents.
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V
i0
v0
Chopper
FWD
+
−
LOAD
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Average output voltage,
Average output current,
The thyristor is commutated at the instant
output current at the instant of commutation is
since V is the output v
ONdc
dcdc
ON
tV V dV
T
V dVI
R R
t t
V
R
= =
= =
=
∴
oltage at that instant.
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( )2
0
0
Free wheeling diode (FWD) will never
conduct in a resistive load.
Average & RMS free wheeling diode
currents are zero.
1
But during
ONt
O RMS
O ON
V v dtT
v V t
∴
=
=
∫
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( )
( )
( )
2
0
2
1
Where duty cycle,
ONt
O RMS
ON
O RMS
O RMS
ON
V V dtT
tV V
T
V dV
td
T
=
=
=
=
∫
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( )
RMS value of thyristor current
= RMS value of load current
Average value of thyristor current
= Average value of load current
O RMSV
R
dV
R
dV
R
=
=
=
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Boost Converter or
Step Up converter
Buck-Boost
Converter
Buck Converter or
Step Down Converter
Simple DC-DC Converter Topologies
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SMPS benefits
– Very wide input voltage range.
• For example: most personal computer power supplies are SMPSs - accepting AC input 90V to 250V.
– Lower Quiescent Current than linear regulators
– Less heat than an equivalent linear regulator.
• Much Lower Green House Gas emissions
– Overall Smaller geometry components are used
– Lighter Weight
– Lower running cost - Lower total cost of ownership (TCO).
– Battery operated devices - longer lifetime.
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SMPS disadvantages
– Significant Output Ripple
• May need a post filter to decrease ripple
• May need a secondary linear low drop out
regulator to ensure damaging voltage transients
keep away from voltage sensitive elements -
electronics.
• An SMPS May add too much cost.
– How much is too much?
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UNIT-IV
INVERTERS
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Single-Phase Inverters
Half-Bridge Inverter
One of the simplest types of inverter. Produces a square wave output.
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Single-Phase Inverters
(cont’d) Full Bridge (H-bridge) Inverter
Two half-bridge inverters combined.
Allows for four quadrant operation.
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Single-Phase Inverters
(cont’d) Quadrant 1: Positive step-down converter
(forward motoring)
Q1-On; Q2 - Chopping; D3,Q1 freewheeling
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Single-Phase Inverters
(cont’d)
Quadrant 2: Positive step-up converter
(forward regeneration)
Q4 - Chopping; D2,D1 freewheeling
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Single-Phase Inverters
(cont’d) Quadrant 3: Negative step-down converter
(reverse motoring)
Q3-On; Q4 - Chopping; D1,Q3 freewheeling
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Single-Phase Inverters
(cont’d) Quadrant 4: Negative step-up converter
(reverse regeneration)
Q2 - Chopping; D3,D4 freewheeling
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Single-Phase Inverters
(cont’d)
Phase-Shift Voltage Control - the output of
the H-bridge inverter can be controlled by
phase shifting the control of the
component half-bridges. See waveforms
on next slide.
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Single-Phase Inverters
(cont’d)
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Single-Phase Inverters
(cont’d)
The waveform of the output voltage vab is a quasi-
square wave of pulse width φ. The Fourier series of vab
is given by:
The value of the fundamental, a1=
The harmonic components as a function of phase
angle are shown in the next slide.
( )1,3,5...
4sin cos
2
dab
n
V nv n t
n
φω
π=
=
∑
( )4
sin / 2dVφ
π
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Single-Phase Inverters
(cont’d)
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Three-Phase Bridge
Inverters
Three-phase bridge inverters are widely
used for ac motor drives. Two modes of
operation - square wave and six-step. The
topology is basically three half-bridge
inverters, each phase-shifted by 2π/3, driving each of the phase windings.
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Three-Phase Bridge Inverters
(cont’d)
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Three-Phase Bridge Inverters
(cont’d)
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Three-Phase Bridge Inverters
(cont’d)
The three square-wave phase voltages can
be expressed in terms of the dc supply
voltage, Vd, by Fourier series as:
1
0
1,3,5...
2( 1) cos( )nd
a
n
Vv n tω
π+
=
= −∑
1
0
1,3,5...
2 2( 1) cos( )
3
ndb
n
Vv n t
πω
π+
=
= − −∑
1
0
1,3,5...
2 2( 1) cos( )
3
ndc
n
Vv n t
πω
π+
=
= − +∑
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Three-Phase Bridge Inverters
(cont’d)
The line voltages can then be expressed as:
0 0
1,3,5...
2 3cos( / 2) cos( 2)d
bc b c
n
Vv v v t n tω π ω π
π =
= − = − − −∑
0 0
1,3,5...
2 3cos( 5 / 6) cos( 5 6)d
ca c a
n
Vv v v t n tω π ω π
π =
= − = + − −∑
0 0
1,3,5...
2 3cos( / 6) cos( 6)d
ab a b
n
Vv v v t n tω π ω π
π =
= − = + − +∑
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Three-Phase Bridge Inverters
(cont’d)
The line voltages are six-step waveforms and
have characteristic harmonics of 6n±±±±1, where n is an integer. This type of inverter is
referred to as a six-step inverter.
The three-phase fundamental and harmonics
are balanced with a mutual phase shift of
2π/3.
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Three-Phase Bridge Inverters
(cont’d)
If the three-phase load neutral n is isolated from the the
center tap of the dc voltage supply (as is normally the
case in an ac machine) the equivalent circuit is shown
below.
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Three-Phase Bridge Inverters
(cont’d)
In this case the isolated neutral-phase
voltages are also six-step waveforms with
the fundamental component phase-shifted
by π/6 from that of the respective line voltage. Also, in this case, the triplen
harmonics are suppressed.
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Three-Phase Bridge Inverters
(cont’d)
For a linear and balanced 3Φ load, the line currents
are also balanced. The individual line current
components can be obtained from the Fourier series
of the line voltage. The total current can be obtained
by addition of the individual currents. A typical line
current wave with inductive load is shown below.
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Three-Phase Bridge Inverters
(cont’d)
The inverter can operate in the usual inverting or
motoring mode. If the phase current wave, ia, is
assumed to be perfectly filtered and lags the phase
voltage by π/3 the voltage and current waveforms are
as shown below:
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Three-Phase Bridge Inverters The inverter can also operate in rectification or regeneration
mode in which power is pushed back to the dc side from the ac side. The waveforms corresponding to this mode of operation with phase angle = 2π/3 are shown below:
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Three-Phase Bridge Inverters
(cont’d)
The phase-shift voltage control principle
described earlier for the single-phase
inverter can be extended to control the
output voltage of a three-phase inverter.
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Three-Phase Bridge Inverters
(cont’d)
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Three-Phase Bridge Inverters
(cont’d)
The three waveforms va0,vb0, and vc0 are of
amplitude 0.5Vd and are mutually phase-
shifted by 2π/3.
The three waveforms ve0,vf0, and vg0 are of
similar but phase shifted by φ.
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Three-Phase Bridge Inverters
(cont’d)
The transformer’s secondary phase voltages,
vA0, vB0, and vc0 may be expressed as follows:
where m is the transformer turns ratio
(= Ns/Np). Note that each of these waves is a
function of φ angle.
0 0 0( )A ad a d
v mv m v v= = −
0 0 0( )B be b e
v mv m v v= = −
0 0 0( )C cf c f
v mv m v v= = −
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Three-Phase Bridge Inverters
(cont’d)
The output line voltages are given by:
While the component voltage waves va0, vd0, vA0 … etc. all
contain triplen harmonics, they are eliminated from the
line voltages because they are co-phasal. Thus the line
voltages are six-step waveforms with order of harmonics
= 6n±1 at a phase angle φ.
0 0AB A Bv v v= −
0 0BC B Cv v v= −
0 0CA C Av v v= −
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Three-Phase Bridge Inverters
(cont’d)
The Fourier series for vA0 and vB0 are given
by:
( )0
1,3,5...
4sin cos
2
dA
n
mV nv n t
n
φω
π=
=
∑
( )0
1,3,5...
4sin cos 2 / 3
2
dB
n
mV nv n t
n
φω π
π=
= −
∑
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Three-Phase Bridge Inverters
(cont’d)
The Fourier series for vAB is given by:
Note that the triplen harmonics are removed
in vAB although they are present in vA0 and
vB0.
( )1,5,7,11...
4 2sin cos cos
2 3
d
n
mV nn t n t
n
φ πω ω
π=
= − −
∑
0 0AB A Bv v v= −
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PWM Technique
While the 3Φ 6-step inverter offers simple control and low switching loss, lower order
harmonics are relatively high leading to high
distortion of the current wave (unless
significant filtering is performed).
PWM inverter offers better harmonic control
of the output than 6-step inverter.
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PWM Principle
The dc input to the inverter is “chopped” by
switching devices in the inverter. The
amplitude and harmonic content of the ac
waveform is controlled by the duty cycle of
the switches. The fundamental voltage v1
has max. amplitude = 4Vd/π for a square wave output but by creating notches, the
amplitude of v1 is reduced (see next slide).
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PWM Principle (cont’d)
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PWM Techniques
Various PWM techniques, include:
• Sinusoidal PWM (most common)
• Selected Harmonic Elimination (SHE)
PWM
• Space-Vector PWM
• Instantaneous current control PWM
• Hysteresis band current control PWM
• Sigma-delta modulation
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Sinusoidal PWM
The most common PWM approach is
sinusoidal PWM. In this method a
triangular wave is compared to a
sinusoidal wave of the desired
frequency and the relative levels of the
two waves is used to control the
switching of devices in each phase leg
of the inverter.
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Sinusoidal PWM
(cont’d)
Single-Phase (Half-Bridge) Inverter
Implementation
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Sinusoidal PWM (cont’d)
when va0> vT T+ on; T- off; va0 = ½Vd
va0 < vT T- on; T+ off; va0 = -½Vd
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Sinusoidal PWM
(cont’d)
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Sinusoidal PWM (cont’d)
Definition of terms:
Triangle waveform switching freq. = fc (also called
carrier freq.)
Control signal freq. = f (also called modulation
freq.)
Amplitude modulation ratio, m = Vp
VT
Frequency modulation ratio,
mf (P)= fc / f
Peak amplitude
of control signal
Peak amplitude
of triangle wave
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Multiple Pulse-Width Modulation
• In multiple-pulse modulation, all pulses are the same width
• Vary the pulse width according to the amplitude of a sine wave evaluated at the center of the same pulse
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Generate the gating signal
2 Reference Signals, vr, -vr
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Comparing the carrier and reference signals
• Generate g1 signal by comparison with vr
• Generate g4 signal by comparison with -vr
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Comparing the carrier and reference signals
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Potential problem if Q1 and Q4 try to turn ON at the same time!
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If we prevent the problem
Output voltage is low when g1 and g4 are
both high
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This composite signal is difficult to generate
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Generate the same gate pulses with one sine wave
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Alternate scheme
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rms output voltage
• Depends on the modulation index, M
2
1
pm
o S Sm
pV V V
δ δ
π π=
∑= →
Where δm is the width of the mth pulse
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Fourier coefficients of the output voltage
( ) ( )2
1
4 3sin sin sin
4 4 4
1, 3, 5, ..
pS m m m
n m mm
V nB n n
n
n
δ δ δα π α
π=
∑ = + − + +
=
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Harmonic Profile
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Compare with multiple-pulse case for p=5
Distortion Factor is considerably less
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Series-Resonant Inverter
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Operation
T1 fired, resonant pulse of current flows through the load. The current falls to zero at t = t1m and T1 is “self – commutated”.
T2 fired, reverse resonant current flows through the load and T2 is also “self-commutated”.
The series resonant circuit must be underdamped,
R2 < (4L/C)
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Operation in Mode 1 – Fire T1
11 1
1
1(0)
(0) 0
(0)
C S
C C
diL Ri i dt v V
dt C
i
v V
+ + + =
=
= −
∫
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21 1
12 2
2
11
0
1
( ) sin
1
4
( ) sin
2
Rt
Lr
r
s c
t r
ts cr
r
i t A e t
R
LC L
V VdiA
dt L
V Vi t e t
L
R
L
α
ω
ω
ω
ωω
α
−
=
−
=
= −
+= =
+=
=
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To find the time when the current is
maximum, set the first derivative = 0
( )
1
1
1
0
sin cos 0
.....
tan
tan
1tan
2
t ts cr r r
r
rr m
r mr m
rm
r
di
dt
V Ve t e t
L
t
tt
t
α αα ω ω ωω
ωω
α
ωω
α
ω
ω
− −
−
−
=
+− + =
=
=
=
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To find the capacitor voltage, integrate the
current
( )
( )
1
1
1
1
1
0
0
1
1 1
1( ) ( )
1( ) sin
...
( ) ( ) ( sin cos ) /
0 ( )
( ) r
t
C c
t
ts cC r C
r
t
C s C r r r r s
m
r
C m C s C s
v t i t dt VC
V Vv t e t dt V
C L
v t V V e t t V
t t
v t V V V e V
α
α
απ
ω
ωω
α ω ω ω ω
π
ω
−
−
−
= −
+= −
= − + + +
≤ ≤
= = + +
∫
∫
The current i1 becomes = 0 @ t=t1m
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Operation in Mode 2 – T1, T2 Both OFF
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t2m
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Operation in Mode 3 – Fire T2
3
3 2 1
33 3
3
1(0) 0
(0) 0
(0)
C
C C C
diL Ri i dt v
dt C
i
v V V
+ + + =
=
= − = −
∫
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1
3 1
1
3
3
3
0
3
( ) sin
1( )
( sin cos )( )
0 ( )m
C t
r
r
t
C C
t
C r r r
C
r
r
Vi t e t
L
v t i dt VC
V e t tv t
t t
α
α
ωω
α ω ω ω
ω
π
ω
−
−
=
= −
− +=
≤ ≤
∫
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3 3 1
1 1
1
1
3
1
( )
( ) ( )
.
.
1
1
1
r
m
r
m
C C C C
C C S C S
C S z
z
C S z
C S C
v t V V V e
v t V V V e V
V Ve
eV V
e
V V V
πα
ω
πα
ω
−
−
= = =
= = + +
=−
=−
+ =
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• Space Vector Diagram
α1V
0V
3V
2V
4V
5V
6V
βj
POO
PPOOPO
OPP
OOP POP
refV
θ
OOOPPP
SECTOR ISECTOR III
SECTOR IV SECTOR VI
SECTOR V
SECTOR
II
ω• Active vectors: to (stationary, not rotating)
• Zero vector:
1V
6V
0V
• Six sectors: I to VI
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• Space Vectors
• Three-phase voltages
0)()()( =++ tvtvtvCOBOAO
• Two-phase voltages
=
)(
)(
)(
3
4sin
3
2sin0sin
3
4cos
3
2cos0cos
3
2
)(
)(
tv
tv
tv
tv
tv
CO
BO
AO
ππ
ππ
β
α
• Space vector representation
)()()( tvjtvtV βα +=
(2) →→→→ (3)
[ ]3/43/20)()()(
3
2)(
ππ j
CO
j
BO
j
AOetvetvetvtV ++=
where xjxejx sincos +=
(3)
(1)
(2)
(4)
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• Space Vectors (Example)
Switching state [POO] →→→→ S1, S6 and S2 ON
dBOdAOVtvVtv
3
1)(,
3
2)( −==
dCOVtv
3
1)( −=and
(5) →→→→ (4)
(7)
(5)
(6) 0
13
2 j
deVV =
Similarly,
3)1(
3
2 π−
=kj
dkeVV
.6...,,2,1=k
α1V
0V
3V
2V
4V
5V
6V
βj
POO
PPOOPO
OPP
OOP POP
refV
θ
OOOPPP
SECTOR ISECTOR III
SECTOR IV SECTOR VI
SECTOR V
SECTOR
II
ω
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• Active and Zero Vectors
Space Vector Switching State (Three Phases)
On-state Switch Vector
Definition
[PPP] 531 ,, SSS Zero
Vector 0V
[OOO] 264 ,, SSS
00 =V
1V
[POO] 261 ,, SSS 01
3
2 jd eVV =
2V
[PPO] 231 ,, SSS 32
3
2π
=j
d eVV
3V
[OPO] 234 ,, SSS 3
2
33
2π
=j
d eVV
4V
[OPP] 534 ,, SSS 3
3
43
2π
=j
d eVV
5V
[OOP] 564 ,, SSS 3
4
53
2π
=j
d eVV
Active
Vector
6V
[POP] 561 ,, SSS 3
5
63
2π
=j
d eVV
• Active Vector: 6
• Zero Vector: 1
• Redundant switching
states: [PPP] and [OOO]
1S
2S
3S 5S
4S 6S
B
C
P
N
dV
A
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(8)
• Reference Vector Vref
• Definition
α1V
0V
3V
2V
4V
5V
6V
βj
POO
PPOOPO
OPP
OOP POP
refV
θ
OOOPPP
SECTOR ISECTOR III
SECTOR IV SECTOR VI
SECTOR V
SECTOR
II
ω
• Angular displacement
∫=t
dtt0
)( ωθ (9)
θjrefref eVV =
• Rotating in space at ω
fπω 2=
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• Relationship Between Vref and VAB
• Vref is approximated by two active
and a zero vectors
• Vref rotates one revolution,
VAB completes one cycle
• Length of Vref corresponds to magnitude of VAB
1V
2V
refV
θ
1VT
T
s
a
2VT
T
s
b
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• Dwell Time Calculation
• Volt-Second Balancing
++=
++=
0
0021
TTTT
TVTVTVTV
bas
basref
(10)
• Ta, Tb and T0 – dwell times for and ,21
VV
0V
• Ts – sampling period
• Space vectors
d
j
refrefVVeVV
3
2,
1==
θ 3
23
2 πj
deVV =
0
0=V
, and
(11) →→→→ (10)
=
+=
bdsref
bdadsref
TVTV
TVTVTV
3
1)(sin
3
1
3
2)(cos
θ
θ
:Im
:Re
(11)
(12)
1V
2V
refV
θ
1VT
T
s
a
2VT
T
s
b
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• Dwell Times
Solve (12)
−−=
=
−=
bas
d
refs
b
d
refs
a
TTTT
V
VTT
V
VTT
0
sin3
)3
(sin3
θ
θπ
3/0 πθ <≤ (13)
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• Vref Location versus Dwell Times
refV
Location 0=θ 6
0π
θ << 6
πθ =
36
πθ
π<<
3
πθ =
Dwell Times 0
0
=
>
b
a
T
T
ba TT > ba TT = ba TT < 0
0
>
=
b
a
T
T
1V
2V
refV
θ
1VT
T
s
a
2VT
T
s
b
SECTOR I
Q
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• Modulation Index
−−=
=
−=
cbs
asb
asa
TTTT
mTT
mTT
0
sin
)3
(sin
θ
θπ
(15)
d
ref
aV
Vm
3= (16)
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• Modulation Range
• Vref,max
32
3
3
2max,
d
dref
VVV =×= (17)
α1V
0V
3V
2V
4V
5V
6V
βj
POO
PPOOPO
OPP
OOP POP
refV
θ
OOOPPP
SECTOR ISECTOR III
SECTOR IV SECTOR VI
SECTOR V
SECTOR
II
ω
(17) →→→→ (16)
• ma,max = 1 →→→→
• Modulation range: 0 ≤≤≤≤ ma ≤≤≤≤ 1 (18)
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• Switching Sequence Design
• Basic Requirement:
Minimize the number of switchings per
sampling period Ts
• Implementation:
Transition from one switching state to
the next involves only two switches in
the same inverter leg.
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• Seven-segment Switching Sequence
dV
2
0T
2
aT
2
bT
2
aT
BNv
ANv
CNv
0
1V
1V
2V
0V
2V
POOOOO PPO PPP PPO POO OOO
dV
dV
4
0T
4
0T
2
bT
sT
0
0
0V
0V
• Total number of switchings: 6
• Selected vectors: V0, V1 and V2
• Dwell times: Ts = T0 + Ta + Tb
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• Undesirable Switching Sequence
• Vectors V1 and V2 swapped
dV
2
0T
2
aT
2
bT
2
aT
BNv
ANv
CNv
0
1V
1V
2V
2V
POOOOO PPO PPP PPOPOO OOO
dV
dV
4
0T
4
0T
2
bT
sT
0
0
0V
0V
0V
• Total number of switchings: 10
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• Switching Sequence Summary (7–segments)
Sector Switching Sequence
0V
1V
2V
0V
2V
1V
0V
I
OOO POO PPO PPP PPO POO OOO
0V
3V
2V
0V
2V
3V
0V
II
OOO OPO PPO PPP PPO OPO OOO
0V
3V
4V
0V
4V
3V
0V
III
OOO OPO OPP PPP OPP OPO OOO
0V
5V
4V
0V
4V
5V
0V
IV
OOO OOP OPP PPP OPP OOP OOO
0V
5V
6V
0V
6V
5V
0V
V
OOO OOP POP PPP POP OOP OOO
0V
1V
6V
0V
6V
1V
0V
VI
OOO POO POP PPP POP POO OOO
Note: The switching sequences for the odd and ever sectors are different.
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• Simulated Waveforms
ABv
AOv
0
0
0
Ai
dV
3/2 dV
π π2 π3
π π2 π3
VIVI
Sector
III
IIIIV
V
III
IIIIV
V
f1 = 60Hz, fsw = 900Hz, ma = 0.696, Ts = 1.1ms
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• Waveforms and FFT
n
ABv
AOv
Ai
dV
3/2 dV
π2
dVVAB 566.01 =
dn VVAB /
π2π π3
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• Waveforms and FFT (Measured)
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dnAB VV /
am
1=n
2=n
dnAB VV /
am
1=n
• Waveforms and FFT (Measured)
Hz601 =f sec720/1=sT ( and )
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• Even-Order Harmonic Elimination
BNv
ANv
CNv
5V
4V
0V
0V
0V
ABvdV−
4V
5V
dV
dV
dV
Type-A sequence (starts and ends with [OOO])
BNv
ANv
CNv
ABvdV−
5V
4V
dV
dV
dV
4V
5V
0V
0V
0V
Type-B sequence (starts and ends with [PPP])
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• Even-Order Harmonic Elimination
1V
3V
2V
4V
5V
6V
°30
°30
Space vector Diagram
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• Even-Order Harmonic Elimination
AOv
ABv
d
n
V
VAB
• Measured waveforms and FFT
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• Even-Order Harmonic Elimination
am
dnAB VV /
1=n
Hz601 =f sec720/1=sT ( and )
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• Five-segment SVM
dV
2
aTbT
2
aT
BNv
ANv
CNv
0
1V
1V
2V
0V
POOOOO PPO POO OOO
dV
sT
0
0
0V
2
0T
2
0T
dV
aT
1V
2V
0V
PPP PPO POO PPP
dV
sT
0V
2
0T
2
0T
(a) Sequence A
2V
PPO
dV
2
bT
2
bT
(b) Sequence B
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• Switching Sequence ( 5-segment)
Sector Switching Sequence (A)
0V
1V
2V
1V
0V
I
OOO POO PPO POO OOO 0=
CNv
0V
3V
2V
3V
0V
II
OOO OPO PPO OPO OOO 0=
CNv
0V
3V
4V
3V
0V
III
OOO OPO OPP OPO OOO 0=
ANv
0V
5V
4V
5V
0V
IV
OOO OOP OPP OOP OOO 0=
ANv
0V
5V
6V
5V
0V
V
OOO OOP POP OOP OOO 0=
BNv
0V
1V
6V
1V
0V
VI
OOO POO POP POO OOO 0=
BNv
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• Simulated Waveforms ( 5-segment)
dV
π2 π4
1gv
3gv
5gv
A Bv
0
0
Ai
3/2π
π2 π4
π2 π4
• No switching for a 120° period per cycle.
• Low switching frequency but high harmonic distortion
• f1 = 60Hz, fsw = 600Hz, ma = 0.696, Ts = 1.1ms
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UNIT V
AC voltage controller and
cycloconverter
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4.1.1 Single-phase AC voltage controller
R u 1 u o
i o
VT 1
VT 2
u
O
u 1
u o
i o
VT
ω t
O ω t
O ω t
O ω t
The phase shift range
(operation range of phase
delay angle): 0 ≤≤≤≤αααα≤≤≤≤ ππππ
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• Resistive load, quantitative analysis
RMS value of output voltage
RMS value of output current
RMS value of thyristor current
Power factor of the circuit
( ) ( )π
απα
πωω
π
π
α
−+== ∫ 2sin
2
1dsin2
11
2
1o UttUU (4-1)
(4-2) R
UI o
o =
(4-3) ( ) )2
2sin1(
2
1sin2
2
1 1
2
1
π
α
π
αω
ω
π
π
α+−=
= ∫ R
Utd
R
tUIT
(4-4) π
απα
πλ
−+==== 2sin
2
1
1
o
o1
oo
U
U
IU
IU
S
P
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Inductive (Inductor- resistor) load , operation principle
R
L
u 1 u o
i o
VT 1
VT 2
O
u 1
u o
i o
u VT
O
O
ω t O
u G1
u G2
O
O
ω t
ω t
ω t
ω t
ω t
The phase shift range:
φ ≤≤≤≤α ≤≤≤≤ ππππ
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4.2 Other AC controllers
4.2.1 Integral cycle control—AC power controller
Circuit topologies are the same as AC voltage controllers.
Only the control method is different.
Load voltage and current are both sinusoidal when thyristors are conducting.
R u 1 u o
i o
VT 1
VT 2
π
M
Line period
Control period = M *Line period = 2 π
4 π
M O
Conduction
angle =
2 π N
M
3 π
M
2 π
M
u o
u 1 u o , i o
ω t
U 1 2
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4.3 Thyristor cycloconverters
4.3.1 Single- phase thyristor-cycloconverter
• Circuit configuration and operation principle
P N
Z
ωt ο
uo ap= 2
π Output
voltage
ap=0
Average
output voltage
ap=
2
π
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• Single- phase thyristor-cycloconverter
Modes of operation
t
t
t
t
t
O
O
O
O
O
u o , i o
u o i o
t 1 t
2 t 3
t 4 t 5
u o u P
u N
u o
i P
i N
u P u N u o
i o i N i P
blocking P
N
Rectifi
cation
Inver
sion
blocking Rectifi
cation
Inver
sion
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Typical waveforms
1
O
O
2
3 4
5
6
u o
i o
ω t
ω t
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• Modulation methods for firing delay angle
Calculation method
– For the rectifier circuit
ωt
ωt
αcosd0o Uu =
tUu oomo sinω=
ttU
Uoo
d0
om sinsincos ωγωα ==
)sin(cos o
1tωγα −=
(4-15)
–For the cycloconverter
output
(4-16)
–Equating (4- 15) and (4-16)
–therefore (4-17)
(4-18)
Principle of cosine
wave-crossing method
u2 u3 u4 u5 u6 u1
ap3 ap4
uo
us2 us3 us4 us5 us6 us1
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Output voltage ratio
(Modulation factor)
)10(0
≤≤= γγd
om
U
U
π
2 2 π π ωο t 3 π
2
0
30
60
90
120
150
Output voltage phase angle
α / ( º) γ=0
γ=0.1 0.2
0.3
0.8 0.9 1.0
0.1 0.2 0.3
0.8
0.9
1.0
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4.3.2 Three- phase thyristor-cyclo converter
• The configuration with common input line
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• The configuration with star-connected output
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Typical waveforms
200 t / ms
Output voltage
Input current with
Single-phase output
Input current with
3-phase output
200 t / ms
200 t / ms
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• Input and output characteristics
The maximum output frequency and the harmonics in the output voltage are the same as in single-phase circuit. Input power factor is a little higher than single-phase circuit. Harmonics in the input current is a little lower thanthe single- phase circuit due to the cancellation of some harmonics among the 3 phases.
To improve the input power factor:
–Use DC bias or 3k order component bias on each of the 3 output phase voltages
• Features and applications
Features:
–Direct frequency conversion—high efficiency
–Bidirectional energy flow, easy to realize 4- quadrant operation
–Very complicated—too many power semiconductor devices
–Low output frequency
–Low input power factor and bad input current waveform
Applications:
–High power low speed AC motor drive
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4.4 Matrix converter
• Circuit configuration
input
output
a) b)
a b c
u
v
w
S 1
1
S 1
2
S 1
3
S 2
1
S 2
2
S 2
3
S 3
1
S 3
2
S 3
3
S ij
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• Usable input voltage
a) b) c)
a) Single-phase input
voltage
b) Use 3 phase voltages
to construct output
voltage
c) Use 3 line-line voltages
to construct output
voltage
Um
U1m
Um 1
2
√3
2 U1m
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• Features
Direct frequency conversion—high efficiency can realize good input and output waveforms, low harmonics, and nearly unity displacement factor
Bidirectional energy flow, easy to realize 4- quadrant operation
Output frequency is not limited by input frequency
No need for bulk capacitor (as compared to indirect frequency converter)
Very complicated—too many power semiconductor devices
Output voltage magnitude is a little lower as compared to indirect frequency converter.
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