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 =
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
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=+=
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.
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
= =
=
= = − ÷
= −
= = + ÷
= +
= −
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
+
+
=
= +
∫
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.
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
= = − = − ÷
= = − = − ÷
= = − = + ÷
= = − = + ÷
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.
