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UNIT II
PHASE CONTROLLED
CONVERTERS
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Phase-Control ConvertersPhase-Control Converters
Single-PhaseSingle-Phase
SemiconverterSemiconverter
Three-PhaseThree-Phase
Full converterFull converter
Dual converterDual converter
SemiconverterSemiconverter
Full converterFull converter
Dual converterDual converter
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Semiconverter
..is a one-quadrant converter and it has one polarity
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
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
Dual converter
..can operate in four quadrants ; both the output
voltage and current can be either positive or negative
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( ) ( )∫ +==
π
α
α
π
ωω
π
cos1
2
sin
2
1 m
mdc
V
ttdVV
Average Output VoltageAverage Output Voltage
π
m
dm
V
V =Maximum
Output Voltage
Maximum
Output Voltage
( )αcos15.0 +==
dm
dc
n
V
V
VNormalizing
Output Voltage
Normalizing
Output Voltage
( ) 





+−== ∫ 2
2sin1
2
sin
2
1 22 α
απ
π
ωω
π
π
α
m
mrms
V
ttdVV
RMS Output VoltageRMS 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
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/πα =
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( )
( )
( )
%27.20
3536.0
1592.0
3536.0
2
2
2
sin
2
1
2
1592.0
2
cos1
2
sin
2
1
2
2
2
2
2
===
=












×
+−=
=






+== ∫
m
m
rms
dc
m
m
rms
mdc
m
mdc
V
V
V
V
V
V
V
VV
V
ttdVV
η
π
π
π
π
π
π
ωω
π
π
π
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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
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/πα =
221.2
1592.0
3536.0
===
m
m
dc
rms
V
V
V
V
FF
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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
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/πα =
983.11221.21 22
=−=−= FFRF
mVPIV =
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Single-Phase SemiconverterSingle-Phase Semiconverter
( ) ( )
( )∫
∫
=
+==
π
α
π
α
ωω
π
α
π
ωω
π
ttdVV
V
ttdVV
mrms
m
mdc
22
sin
2
2
cos1sin
2
2
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Single-Phase Semiconverter (RL-load)Single-Phase Semiconverter (RL-load)
( ) 





−−===
=++
−−
ω
α
ω
α
αω L
R
L
R
LLL
L
L
e
R
E
eItiI
ERi
dt
di
L
1
0
011
1
1
( ) ( ) tL
R
S
L
S
L
SL
L
e
Z
V
R
E
I
R
E
t
Z
V
I
tVERi
dt
di
L
−






−−++−−=
=++
ω
α
θαθω
ω
sin
2
sin
2
sin2
12
2
2
Mode 1Mode 1
αω ≤≤ t0
Mode 2Mode 2
πωα ≤≤ t
R
Lω
θ 1
tan−
=( )22
LRZ ω+=
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Single-Phase Semiconverter (RL-load)Single-Phase Semiconverter (RL-load)
RMS Current
for Thyristor
RMS Current
for Thyristor
( )∫=
π
α
ω
π
tdiI LR
2
2
2
1
RMS Current
for Thyristor
RMS Current
for Thyristor
( )∫=
π
α
ω
π
tdiI LA 2
2
1
RMS Output
Current
RMS Output
Current
( ) ( )∫∫ +=
π
α
α
ω
π
ω
π
tditdiI LLrms
2
2
0
2
1
2
1
2
1
AVG Output
Current
AVG Output
Current
( ) ( )∫∫ +=
π
α
α
ω
π
ω
π
tditdiIdc 2
0
1
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
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
0=tω
°== 60αωt
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Single-Phase
Full Converter
Rectification
Mode
Rectification
Mode
Inversion
Mode
Inversion
Mode
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Single-Phase Full ConverterSingle-Phase Full Converter
( )
( )
2
sin
2
2
cos
2
sin
2
2
22 m
mrms
m
mdc
V
tdtVV
V
tdtVV
==
==
∫
∫
+
+
ωω
π
α
π
ωω
π
απ
α
απ
α
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Single-Phase Full Converter (RL-load)Single-Phase Full Converter (RL-load)
( ) ( ) tL
R
S
L
S
L e
Z
V
R
E
I
R
E
t
Z
V
I −






−−++−−= ω
α
θαθω sin
2
sin
2
0
Mode 1 = Mode 2Mode 1 = Mode 2
R
Lω
θ 1
tan−
= ( )22
LRZ ω+=
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Single-Phase Full Converter (RL-load)Single-Phase Full Converter (RL-load)
RMS Current
for Thyristor
RMS Current
for Thyristor
( )∫
+
=
απ
α
ω
π
tdiI LR
2
2
1
RMS Current
for Thyristor
RMS Current
for Thyristor
( )∫
+
=
απ
α
ω
π
tdiI LA
2
1
RMS Output
Current
RMS Output
Current RRRrms IIII 222
=+=
AVG Output
Current
AVG Output
Current AAAdc IIII 2=+=
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Dual
Converter
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Single-Phase Dual ConverterSingle-Phase Dual Converter
High-Power Variable-Speed DrivesHigh-Power Variable-Speed Drives
21
22
11
cos
2
cos
2
dcdc
m
dc
m
dc
VV
V
V
V
V
−=
=
=
α
π
α
π
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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
V A N
V C N
V B N
1 2 0
0
1 2 0
0
1 2 0
0 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
2
sin
3
sin 120
2
sin
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
ω
π
ω
ω
π
ω
ω
ω
= =
=
 
= = − ÷
 
= −
 
= = + ÷
 
= +
= −
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van vbn vcn van
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io=Ia
Constant Load
Current
Ia
Ia
Each thyristor conducts for 2π/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
30
6
5
150
6
7
270
6
2
Each thytistor conducts for 120 or radians
3
T is triggered at t
T is triggered at t
T is triggered at t
π
ω α α
π
ω α α
π
ω α α
π
 
= + = + ÷
 
 
= + = + ÷
 
 
= + = + ÷
 
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( )
( )
5
6
6
5
6
6
3
sin .
2
3
cos
2
3 5
cos cos
2 6 6
m
dc
m
dc
m
dc
V
V t d t
V
V t
V
V
π
α
π
α
π
α
π
α
ω ω
π
ω
π
π π
α α
π
+
+
+
+
 
 
=  
  
 
 
= − 
  
    
= − + + + ÷  ÷ 
    
∫
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( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( )
0 0
0
Note from the trigonometric relationship
cos cos .cos sin .sin
5 5
cos cos sin sin
6 63
2
co
cos 150 cos sin 150 sin3
2 cos 30
s .cos sin sin
6 6
.cos
m
dc
m
dc
A
V
V
B A B A B
V
V
π π
α α
π π π
α
α
α
α
π α
+ = −
    
− + ÷  ÷ 
    =
    
+ −  ÷  ÷
   
−
+

+
=
− ( ) ( )0
sin 30 sin α
 
 
 
 
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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
m
dc
m
dc
V
V
V
V
α α
π α α
α α
π α α
− =
 − − + −
 =
 + − 
 + +
 ∴ =
 + −
=

−
−
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( ) ( )
( )
( ) ( )
( )
03
2cos 30 cos
2
3 3
2 cos
2 2
3 3 3
3 cos cos
2 2
3
cos
2
Where 3 Max. line to line supply voltage
m
dc
m
dc
m m
dc
Lm
dc
Lm m
V
V
V
V
V V
V
V
V
V V
α
π
α
π
α α
π π
α
π
 =  
 
= × 
 
 = = 
=
= =
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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
m
dmdc
m
d
dcn n
V
V V
V
V
V V
α
π
=
= =
= = cosc
dmV
α=
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( ) ( )
( )
1
5 2
6
2 2
6
1
2
The rms value of output voltage is found by
using the equation
3
sin .
2
and we obtain
1 3
3 cos2
6 8
mO RMS
mO RMS
V V t d t
V V
π
α
π
α
ω ω
π
α
π
+
+
 
 
=  
  
 
= + 
 
∫
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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
3 0
0
3 0
0
6 0
0
6 0
0
9 0
0
9 0
0
1 2 0
0
1 2 0
0
1 5 0
0
1 5 0
0
1 8 0
0
1 8 0
0
2 1 0
0
2 1 0
0
2 4 0
0
2 4 0
0
2 7 0
0
2 7 0
0
3 0 0
0
3 0 0
0
3 3 0
0
3 3 0
0
3 6 0
0
3 6 0
0
3 9 0
0
3 9 0
0
4 2 0
0
4 2 0
0
V a n
↑
V 0
↑
V 0
V a n
α
α
α = 3 0
0
α = 6 0
0
V b n
V b n
V c n
V c n
ω t
ω t
α=300
α=600
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0
3 0
0
6 0
0
9 0
0
1 2 0
0
1 5 0
0
1 8 0
0
2 1 0
0
2 4 0
0
2 7 0
0
3 0 0
0
3 3 0
0
3 6 0
0
3 9 0
0
4 2 0
0
↑
V 0
V a n
α
α = 9 0
0
V b n V c n
ω t
α=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
V 0
L
R V 0
+
−
T 1
T 2
T 3
n n
T 1
T 2
T 3
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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
3 0
0
3 0
0
6 0
0
6 0
0
9 0
0
9 0
0
1 2 0
0
1 2 0
0
1 5 0
0
1 5 0
0
1 8 0
0
1 8 0
0
2 1 0
0
2 1 0
0
2 4 0
0
2 4 0
0
2 7 0
0
2 7 0
0
3 0 0
0
3 0 0
0
3 3 0
0
3 3 0
0
3 6 0
0
3 6 0
0
3 9 0
0
3 9 0
0
4 2 0
0
4 2 0
0
V s
V 0
V a n
α
α = 0
α = 1 5 0
V b n V c n
ω t
V a n
V b n V c n
ω t
α=00
α=150
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0
0
3 0
0
3 0
0
6 0
0
6 0
0
9 0
0
9 0
0
1 2 0
0
1 2 0
0
1 5 0
0
1 5 0
0
1 8 0
0
1 8 0
0
2 1 0
0
2 1 0
0
2 4 0
0
2 4 0
0
2 7 0
0
2 7 0
0
3 0 0
0
3 0 0
0
3 3 0
0
3 3 0
0
3 6 0
0
3 6 0
0
3 9 0
0
3 9 0
0
4 2 0
0
4 2 0
0
α
α
V 0
α = 3 0
0
V a n
V b n V c n
ω t
V 0
α = 6 0
0
V a n
V b n V c n
ω t
α=300
α=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
30
6
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
π
ω α α
α
ω
π
ω α α
α
ω
 
= + = + ÷
 
+
= =
 
= + = + ÷
 
+
= = −
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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
π
ω α α
α
ω
ω
 
= + = + ÷
 
+
= = −
= +
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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
3
sin .
2
3
sin .
2
dc O
O an m
dc m
m
dc
V v d t
v v V t t
V V t d t
V
V t d t
α
α
α
ω
π
ω ω α
ω ω
π
ω ω
π
+
+
+
 
=  
  
= = = +
 
=  
  
 
=  
  
∫
∫
∫
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( )
( )
0
0
180
30
0 0
0
0
3
cos
2
3
cos180 cos 30
2
cos180 1, we get
3
1 cos 30
2
m
dc
m
dc
m
dc
V
V t
V
V
V
V
α
ω
π
α
π
α
π
+
 
= − 
  
 = − + + 
= −
 = + + 
Q
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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
www.Vidyarthiplus.com
www.Vidyarthiplus.com
Wave forms of 3 Phase
Semiconverter for
α > 600
www.Vidyarthiplus.com
www.Vidyarthiplus.com
www.Vidyarthiplus.com
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
α π
ω
π π
ω
≤ <
≤ <
www.Vidyarthiplus.com
1
1 1
If thyristor is triggered at ,
6
& conduct together and the line to line voltage
appears across the load.
7
At , becomes negative & FWD conducts.
6
The load current contin
ac
ac m
T t
T D
v
t v D
π
ω α
π
ω
 
= + ÷
 
=
1 1
ues to flow through FWD ;
and are turned off.
mD
T D
www.Vidyarthiplus.com
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 , e
3
mD T
T
t
T D
π
ω α
π
α
 
= + ÷
 
≤ ach thyristor conducts
2
for and the FWD does not conduct.
3
mD
π
www.Vidyarthiplus.com
( )
( )
( )
0
0
0
We deifine three line neutral voltages
(3 phase voltages) as follows
sin ; Max. Phase Voltage
2
sin sin 120
3
2
sin 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
ω
π
ω ω
π
ω ω
ω
= = =
 
= = − = − ÷
 
 
= = + = + ÷
 
= −
is the peak phase voltage of a wye-connected source.m
www.Vidyarthiplus.com
( )
( )
( )
( )
3 sin
6
5
3 sin
6
3 sin
2
3 sin
6
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
π
ω
π
ω
π
ω
π
ω
 
= = − = − ÷
 
 
= = − = − ÷
 
 
= = − = + ÷
 
 
= = − = + ÷
 
www.Vidyarthiplus.com
Wave forms of 3 Phase
Semiconverter for
α ≤ 600
www.Vidyarthiplus.com
www.Vidyarthiplus.com
www.Vidyarthiplus.com
www.Vidyarthiplus.com
To derive an
Expression for the
Average Output Voltage
of 3 Phase Semiconverter
for α > π / 3
and Discontinuous Output Voltage
www.Vidyarthiplus.com
( )
( )
7
6
6
7
6
6
For and discontinuous output voltage:
3
the Average output voltage is found from
3
.
2
3
3 sin
2 6
dc ac
dc m
V v d t
V V t d t
π
π
α
π
π
α
π
α
ω
π
π
ω ω
π
+
+
≥
 
 =
 
 
 
  = − ÷
  
 
∫
∫
www.Vidyarthiplus.com
( )
( )
( )max
3 3
1 cos
2
3
1 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
m
dc
mL
dc
mL m
m
dmdc
V
V
V
V
V V
V
V V
α
π
α
π
α
π
= +
= +
= =
=
= =
www.Vidyarthiplus.com
( )
( ) ( )
1
7 26
2
6
The normalized average output voltage is
0.5 1 cos
The rms output voltage is found from
3
.
2
dc
n
dm
acO rms
V
V
V
V v d t
π
π
α
α
ω
π
+
= = +
 
 =
 
 
∫
www.Vidyarthiplus.com
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.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
www.Vidyarthiplus.com
www.Vidyarthiplus.com
Outputs of Converters 1 & 2
• During the interval (π/6 + α1) to (π/2 + α1),
the line to line voltage vab appears across
the output of converter 1 and vbc appears
across the output of converter 2
www.Vidyarthiplus.com
( )
( )
( )
0
0
0
We deifine three line neutral voltages
(3 phase voltages) as follows
sin ;
Max. Phase Voltage
2
sin sin 120
3
2
sin 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
ω
π
ω ω
π
ω ω
ω
= =
=
 
= = − = − ÷
 
 
= = + = + ÷
 
= −
www.Vidyarthiplus.com
( )
( )
( )
0
0
0
We deifine three line neutral voltages
(3 phase voltages) as follows
sin ;
Max. Phase Voltage
2
sin sin 120
3
2
sin 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
ω
π
ω ω
π
ω ω
ω
= =
=
 
= = − = − ÷
 
 
= = + = + ÷
 
= −
www.Vidyarthiplus.com
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
(π/6 + α1) ≤ ωt ≤ (π/2 + α1) is given by
www.Vidyarthiplus.com
1 2
3 sin sin
6 2
3 cos
6
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
π π
ω ω
π
ω
= + = −
    
= + − − ÷  ÷ 
    
 
= − ÷
 
www.Vidyarthiplus.com
( ) ( )
( ) ( )
( )
( )
1
1
6
6
1
max
1
.
1
3 cos .
6
3
sin sin
6
3
t
r r
r
t
r m
r
m
r
r
m
r
r
i t v d t
L
i t V t d t
L
V
i t t
L
V
i
L
ω
π
α
ω
π
α
ω
ω
π
ω ω
ω
π
ω α
ω
ω
+
+
=
 
= − ÷
 
  
= − − ÷ 
  
=
∫
∫
www.Vidyarthiplus.com
Four Quadrant Operation
www.Vidyarthiplus.com
• There are two different modes of
operation.
 Circulating current free
(non circulating) mode of operation
 Circulating current mode of operation
www.Vidyarthiplus.com
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 α1 < 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
www.Vidyarthiplus.com
• When the converter 1 is on,
For α1 > 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.
www.Vidyarthiplus.com
• When the converter 2 is switched on,
For α2 < 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
www.Vidyarthiplus.com
• When the converter 2 is switched on,
For α2 > 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.
www.Vidyarthiplus.com
• 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 α1 & α2 are adjusted such that (α1
+ α2) = 1800
Circulating Current
Mode Of Operation
www.Vidyarthiplus.com
When α1 < 900
, converter 1 operates
as a controlled rectifier. α2 is made
greater than 900
and converter 2
operates as an Inverter.
• Vdc is positive & Idc is positive and Pdc
is positive.
www.Vidyarthiplus.com
• When α2 < 900
, converter 2 operates
as a controlled rectifier. α1 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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Converters

  • 2. www.Vidyarthiplus.com Phase-Control ConvertersPhase-Control Converters Single-PhaseSingle-Phase SemiconverterSemiconverter Three-PhaseThree-Phase Full converterFull converter Dual converterDual converter SemiconverterSemiconverter Full converterFull converter Dual converterDual converter
  • 3. www.Vidyarthiplus.com Semiconverter ..is a one-quadrant converter and it has one polarity 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 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 Dual converter ..can operate in four quadrants ; both the output voltage and current can be either positive or negative
  • 5. www.Vidyarthiplus.com ( ) ( )∫ +== π α α π ωω π cos1 2 sin 2 1 m mdc V ttdVV Average Output VoltageAverage Output Voltage π m dm V V =Maximum Output Voltage Maximum Output Voltage ( )αcos15.0 +== dm dc n V V VNormalizing Output Voltage Normalizing Output Voltage ( )       +−== ∫ 2 2sin1 2 sin 2 1 22 α απ π ωω π π α m mrms V ttdVV RMS Output VoltageRMS Output Voltage
  • 6. www.Vidyarthiplus.com 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 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/πα =
  • 7. www.Vidyarthiplus.com ( ) ( ) ( ) %27.20 3536.0 1592.0 3536.0 2 2 2 sin 2 1 2 1592.0 2 cos1 2 sin 2 1 2 2 2 2 2 === =             × +−= =       +== ∫ m m rms dc m m rms mdc m mdc V V V V V V V VV V ttdVV η π π π π π π ωω π π π
  • 8. www.Vidyarthiplus.com 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 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/πα = 221.2 1592.0 3536.0 === m m dc rms V V V V FF
  • 9. www.Vidyarthiplus.com 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 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/πα = 983.11221.21 22 =−=−= FFRF mVPIV =
  • 11. www.Vidyarthiplus.com Single-Phase SemiconverterSingle-Phase Semiconverter ( ) ( ) ( )∫ ∫ = +== π α π α ωω π α π ωω π ttdVV V ttdVV mrms m mdc 22 sin 2 2 cos1sin 2 2
  • 12. www.Vidyarthiplus.com Single-Phase Semiconverter (RL-load)Single-Phase Semiconverter (RL-load) ( )       −−=== =++ −− ω α ω α αω L R L R LLL L L e R E eItiI ERi dt di L 1 0 011 1 1 ( ) ( ) tL R S L S L SL L e Z V R E I R E t Z V I tVERi dt di L −       −−++−−= =++ ω α θαθω ω sin 2 sin 2 sin2 12 2 2 Mode 1Mode 1 αω ≤≤ t0 Mode 2Mode 2 πωα ≤≤ t R Lω θ 1 tan− =( )22 LRZ ω+=
  • 13. www.Vidyarthiplus.com Single-Phase Semiconverter (RL-load)Single-Phase Semiconverter (RL-load) RMS Current for Thyristor RMS Current for Thyristor ( )∫= π α ω π tdiI LR 2 2 2 1 RMS Current for Thyristor RMS Current for Thyristor ( )∫= π α ω π tdiI LA 2 2 1 RMS Output Current RMS Output Current ( ) ( )∫∫ += π α α ω π ω π tditdiI LLrms 2 2 0 2 1 2 1 2 1 AVG Output Current AVG Output Current ( ) ( )∫∫ += π α α ω π ω π tditdiIdc 2 0 1 2 1 2 1
  • 14. www.Vidyarthiplus.com 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 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 0=tω °== 60αωt
  • 16. www.Vidyarthiplus.com Single-Phase Full ConverterSingle-Phase Full Converter ( ) ( ) 2 sin 2 2 cos 2 sin 2 2 22 m mrms m mdc V tdtVV V tdtVV == == ∫ ∫ + + ωω π α π ωω π απ α απ α
  • 17. www.Vidyarthiplus.com Single-Phase Full Converter (RL-load)Single-Phase Full Converter (RL-load) ( ) ( ) tL R S L S L e Z V R E I R E t Z V I −       −−++−−= ω α θαθω sin 2 sin 2 0 Mode 1 = Mode 2Mode 1 = Mode 2 R Lω θ 1 tan− = ( )22 LRZ ω+=
  • 18. www.Vidyarthiplus.com Single-Phase Full Converter (RL-load)Single-Phase Full Converter (RL-load) RMS Current for Thyristor RMS Current for Thyristor ( )∫ + = απ α ω π tdiI LR 2 2 1 RMS Current for Thyristor RMS Current for Thyristor ( )∫ + = απ α ω π tdiI LA 2 1 RMS Output Current RMS Output Current RRRrms IIII 222 =+= AVG Output Current AVG Output Current AAAdc IIII 2=+=
  • 20. www.Vidyarthiplus.com Single-Phase Dual ConverterSingle-Phase Dual Converter High-Power Variable-Speed DrivesHigh-Power Variable-Speed Drives 21 22 11 cos 2 cos 2 dcdc m dc m dc VV V V V V −= = = α π α π
  • 22. www.Vidyarthiplus.com 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.
  • 23. www.Vidyarthiplus.com • 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.
  • 24. www.Vidyarthiplus.com 3-Phase Half Wave Converter (3-Pulse Converter) with RL Load Continuous & Constant Load Current Operation
  • 26. www.Vidyarthiplus.com Vector Diagram of 3 Phase Supply Voltages V A N V C N V B N 1 2 0 0 1 2 0 0 1 2 0 0 RN AN YN BN BN CN v v v v v v = = =
  • 27. www.Vidyarthiplus.com 3 Phase Supply Voltage Equations We deifine three line to neutral voltages (3 phase voltages) as follows
  • 28. www.Vidyarthiplus.com ( ) ( ) ( ) 0 0 0 sin ; Max. Phase Voltage 2 sin 3 sin 120 2 sin 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 ω π ω ω π ω ω ω = = =   = = − ÷   = −   = = + ÷   = + = −
  • 31. www.Vidyarthiplus.com To Derive an Expression for the Average Output Voltage of a 3-Phase Half Wave Converter with RL Load for Continuous Load Current
  • 32. www.Vidyarthiplus.com ( ) ( ) ( ) 0 1 0 2 0 3 0 30 6 5 150 6 7 270 6 2 Each thytistor conducts for 120 or radians 3 T is triggered at t T is triggered at t T is triggered at t π ω α α π ω α α π ω α α π   = + = + ÷     = + = + ÷     = + = + ÷  
  • 33. www.Vidyarthiplus.com ( ) ( ) 5 6 6 5 6 6 3 sin . 2 3 cos 2 3 5 cos cos 2 6 6 m dc m dc m dc V V t d t V V t V V π α π α π α π α ω ω π ω π π π α α π + + + +     =          = −          = − + + + ÷  ÷       ∫
  • 34. www.Vidyarthiplus.com ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 Note from the trigonometric relationship cos cos .cos sin .sin 5 5 cos cos sin sin 6 63 2 co cos 150 cos sin 150 sin3 2 cos 30 s .cos sin sin 6 6 .cos m dc m dc A V V B A B A B V V π π α α π π π α α α α π α + = −      − + ÷  ÷      =      + −  ÷  ÷     − +  + = − ( ) ( )0 sin 30 sin α        
  • 35. www.Vidyarthiplus.com ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 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 m dc m dc V V V V α α π α α α α π α α − =  − − + −  =  + −   + +  ∴ =  + − =  − −
  • 36. www.Vidyarthiplus.com ( ) ( ) ( ) ( ) ( ) ( ) 03 2cos 30 cos 2 3 3 2 cos 2 2 3 3 3 3 cos cos 2 2 3 cos 2 Where 3 Max. line to line supply voltage m dc m dc m m dc Lm dc Lm m V V V V V V V V V V V α π α π α α π π α π  =     = ×     = =  = = =
  • 37. www.Vidyarthiplus.com ( )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 m dmdc m d dcn n V V V V V V V α π = = = = = cosc dmV α=
  • 38. www.Vidyarthiplus.com ( ) ( ) ( ) 1 5 2 6 2 2 6 1 2 The rms value of output voltage is found by using the equation 3 sin . 2 and we obtain 1 3 3 cos2 6 8 mO RMS mO RMS V V t d t V V π α π α ω ω π α π + +     =        = +    ∫
  • 39. www.Vidyarthiplus.com 3 Phase Half Wave Controlled Rectifier Output Voltage Waveforms For RL Load at Different Trigger Angles
  • 40. www.Vidyarthiplus.com 0 0 3 0 0 3 0 0 6 0 0 6 0 0 9 0 0 9 0 0 1 2 0 0 1 2 0 0 1 5 0 0 1 5 0 0 1 8 0 0 1 8 0 0 2 1 0 0 2 1 0 0 2 4 0 0 2 4 0 0 2 7 0 0 2 7 0 0 3 0 0 0 3 0 0 0 3 3 0 0 3 3 0 0 3 6 0 0 3 6 0 0 3 9 0 0 3 9 0 0 4 2 0 0 4 2 0 0 V a n ↑ V 0 ↑ V 0 V a n α α α = 3 0 0 α = 6 0 0 V b n V b n V c n V c n ω t ω t α=300 α=600
  • 41. www.Vidyarthiplus.com 0 3 0 0 6 0 0 9 0 0 1 2 0 0 1 5 0 0 1 8 0 0 2 1 0 0 2 4 0 0 2 7 0 0 3 0 0 0 3 3 0 0 3 6 0 0 3 9 0 0 4 2 0 0 ↑ V 0 V a n α α = 9 0 0 V b n V c n ω t α=900
  • 42. www.Vidyarthiplus.com 3 Phase Half Wave Controlled Rectifier With R Load and RL Load with FWD
  • 43. www.Vidyarthiplus.com a a b b c c R V 0 L R V 0 + − T 1 T 2 T 3 n n T 1 T 2 T 3
  • 44. www.Vidyarthiplus.com 3 Phase Half Wave Controlled Rectifier Output Voltage Waveforms For R Load or RL Load with FWD at Different Trigger Angles
  • 45. www.Vidyarthiplus.com 0 0 3 0 0 3 0 0 6 0 0 6 0 0 9 0 0 9 0 0 1 2 0 0 1 2 0 0 1 5 0 0 1 5 0 0 1 8 0 0 1 8 0 0 2 1 0 0 2 1 0 0 2 4 0 0 2 4 0 0 2 7 0 0 2 7 0 0 3 0 0 0 3 0 0 0 3 3 0 0 3 3 0 0 3 6 0 0 3 6 0 0 3 9 0 0 3 9 0 0 4 2 0 0 4 2 0 0 V s V 0 V a n α α = 0 α = 1 5 0 V b n V c n ω t V a n V b n V c n ω t α=00 α=150
  • 46. www.Vidyarthiplus.com 0 0 3 0 0 3 0 0 6 0 0 6 0 0 9 0 0 9 0 0 1 2 0 0 1 2 0 0 1 5 0 0 1 5 0 0 1 8 0 0 1 8 0 0 2 1 0 0 2 1 0 0 2 4 0 0 2 4 0 0 2 7 0 0 2 7 0 0 3 0 0 0 3 0 0 0 3 3 0 0 3 3 0 0 3 6 0 0 3 6 0 0 3 9 0 0 3 9 0 0 4 2 0 0 4 2 0 0 α α V 0 α = 3 0 0 V a n V b n V c n ω t V 0 α = 6 0 0 V a n V b n V c n ω t α=300 α=600
  • 47. www.Vidyarthiplus.com 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
  • 48. www.Vidyarthiplus.com ( ) ( ) ( ) ( ) ( ) 0 1 0 0 1 0 2 0 0 2 0 30 6 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 π ω α α α ω π ω α α α ω   = + = + ÷   + = =   = + = + ÷   + = = −
  • 49. www.Vidyarthiplus.com ( ) ( ) ( ) ( ) 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 π ω α α α ω ω   = + = + ÷   + = = − = +
  • 50. www.Vidyarthiplus.com ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 0 180 30 0 0 180 30 180 30 3 . 2 sin ; for 30 to 180 3 sin . 2 3 sin . 2 dc O O an m dc m m dc V v d t v v V t t V V t d t V V t d t α α α ω π ω ω α ω ω π ω ω π + + +   =      = = = +   =        =      ∫ ∫ ∫
  • 51. www.Vidyarthiplus.com ( ) ( ) 0 0 180 30 0 0 0 0 3 cos 2 3 cos180 cos 30 2 cos180 1, we get 3 1 cos 30 2 m dc m dc m dc V V t V V V V α ω π α π α π +   = −      = − + +  = −  = + +  Q
  • 52. www.Vidyarthiplus.com 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.
  • 53. www.Vidyarthiplus.com 3 Phase Half Controlled Bridge Converter (Semi Converter) with Highly Inductive Load & Continuous Ripple free Load Current
  • 55. www.Vidyarthiplus.com Wave forms of 3 Phase Semiconverter for α > 600
  • 58. www.Vidyarthiplus.com 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 α π ω π π ω ≤ < ≤ <
  • 59. www.Vidyarthiplus.com 1 1 1 If thyristor is triggered at , 6 & conduct together and the line to line voltage appears across the load. 7 At , becomes negative & FWD conducts. 6 The load current contin ac ac m T t T D v t v D π ω α π ω   = + ÷   = 1 1 ues to flow through FWD ; and are turned off. mD T D
  • 60. www.Vidyarthiplus.com 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 , e 3 mD T T t T D π ω α π α   = + ÷   ≤ ach thyristor conducts 2 for and the FWD does not conduct. 3 mD π
  • 61. www.Vidyarthiplus.com ( ) ( ) ( ) 0 0 0 We deifine three line neutral voltages (3 phase voltages) as follows sin ; Max. Phase Voltage 2 sin sin 120 3 2 sin 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 ω π ω ω π ω ω ω = = =   = = − = − ÷     = = + = + ÷   = − is the peak phase voltage of a wye-connected source.m
  • 62. www.Vidyarthiplus.com ( ) ( ) ( ) ( ) 3 sin 6 5 3 sin 6 3 sin 2 3 sin 6 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 π ω π ω π ω π ω   = = − = − ÷     = = − = − ÷     = = − = + ÷     = = − = + ÷  
  • 63. www.Vidyarthiplus.com Wave forms of 3 Phase Semiconverter for α ≤ 600
  • 67. www.Vidyarthiplus.com To derive an Expression for the Average Output Voltage of 3 Phase Semiconverter for α > π / 3 and Discontinuous Output Voltage
  • 68. www.Vidyarthiplus.com ( ) ( ) 7 6 6 7 6 6 For and discontinuous output voltage: 3 the Average output voltage is found from 3 . 2 3 3 sin 2 6 dc ac dc m V v d t V V t d t π π α π π α π α ω π π ω ω π + + ≥    =         = − ÷      ∫ ∫
  • 69. www.Vidyarthiplus.com ( ) ( ) ( )max 3 3 1 cos 2 3 1 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 m dc mL dc mL m m dmdc V V V V V V V V V α π α π α π = + = + = = = = =
  • 70. www.Vidyarthiplus.com ( ) ( ) ( ) 1 7 26 2 6 The normalized average output voltage is 0.5 1 cos The rms output voltage is found from 3 . 2 dc n dm acO rms V V V V v d t π π α α ω π + = = +    =     ∫
  • 71. www.Vidyarthiplus.com 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.
  • 75. www.Vidyarthiplus.com Outputs of Converters 1 & 2 • During the interval (π/6 + α1) to (π/2 + α1), the line to line voltage vab appears across the output of converter 1 and vbc appears across the output of converter 2
  • 76. www.Vidyarthiplus.com ( ) ( ) ( ) 0 0 0 We deifine three line neutral voltages (3 phase voltages) as follows sin ; Max. Phase Voltage 2 sin sin 120 3 2 sin 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 ω π ω ω π ω ω ω = = =   = = − = − ÷     = = + = + ÷   = −
  • 77. www.Vidyarthiplus.com ( ) ( ) ( ) 0 0 0 We deifine three line neutral voltages (3 phase voltages) as follows sin ; Max. Phase Voltage 2 sin sin 120 3 2 sin 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 ω π ω ω π ω ω ω = = =   = = − = − ÷     = = + = + ÷   = −
  • 78. www.Vidyarthiplus.com 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 (π/6 + α1) ≤ ωt ≤ (π/2 + α1) is given by
  • 79. www.Vidyarthiplus.com 1 2 3 sin sin 6 2 3 cos 6 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 π π ω ω π ω = + = −      = + − − ÷  ÷         = − ÷  
  • 80. www.Vidyarthiplus.com ( ) ( ) ( ) ( ) ( ) ( ) 1 1 6 6 1 max 1 . 1 3 cos . 6 3 sin sin 6 3 t r r r t r m r m r r m r r i t v d t L i t V t d t L V i t t L V i L ω π α ω π α ω ω π ω ω ω π ω α ω ω + + =   = − ÷      = − − ÷     = ∫ ∫
  • 82. www.Vidyarthiplus.com • There are two different modes of operation.  Circulating current free (non circulating) mode of operation  Circulating current mode of operation
  • 83. www.Vidyarthiplus.com 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 α1 < 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
  • 84. www.Vidyarthiplus.com • When the converter 1 is on, For α1 > 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.
  • 85. www.Vidyarthiplus.com • When the converter 2 is switched on, For α2 < 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
  • 86. www.Vidyarthiplus.com • When the converter 2 is switched on, For α2 > 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.
  • 87. www.Vidyarthiplus.com • 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 α1 & α2 are adjusted such that (α1 + α2) = 1800 Circulating Current Mode Of Operation
  • 88. www.Vidyarthiplus.com When α1 < 900 , converter 1 operates as a controlled rectifier. α2 is made greater than 900 and converter 2 operates as an Inverter. • Vdc is positive & Idc is positive and Pdc is positive.
  • 89. www.Vidyarthiplus.com • When α2 < 900 , converter 2 operates as a controlled rectifier. α1 is made greater than 900 and converter 1 operates as an Inverter. • Vdc is negative & Idc is negative and Pdc is positive.