2. Introduction
• It used to transfer the ac voltage (input) to
other type of ac voltage (output) with different
in amplitude, frequency, phase.
• Can be categorized in two types:
• Can be categorized in two types:
single phase controllers
three phase controller
• to control the output waveform, the switching
mode of the power electronic must be
controllable
3. • The control signals are angle, pulse width, a voltage
and current signals.
• Two types of conversion
direct conversion – desired output
indirect conversion – depend on
indirect conversion – depend on
application
Converters that change the frequency – cycloconverters
Basically the applications are for steel mill roll, light
control, speed control of the motors
4. • If a thyristor is connected between ac supply and
load, the power flow can be controlled by varying the
rms value of ac voltage applied to the load.
• Two types of control that normally used:
On-Off Control
Phase-angle control
• In on-off control, thyristor switches connect the load
to the ac supply for a few cycles and disconnect for
another few cycles.
• In phase-control, thyristor switches connect the load
to the ac supply for a portion of each cycle of input
voltage.
5. Principle of Half Wave AC Phase Control
The power flow to the load is controlled by delaying the firing
angle of T1
D1 is used to turn on when the input in negative half cycle
Gate pulses are applied at T1 to turn on the device
Gate pulses are generated at external source
6. The output voltage and input current are asymmetric and contain
dc component
This circuit is suitable for low power output such as heating and
lighting.
The rms output voltage is found form
( )
2
/
1
2
/
1
2
)
(
2
sin
2
2
sin
2
2
1
∫ ∫
+
=
π
α
π
π
ω
ω
ω
ω
π
t
td
m
V
t
td
m
V
o
V
( )
2
/
1
)
2
2
sin
2
(
1
2
2
/
1
2
)
2
2
sin
(
)
2
2
sin
(
1
2
2
/
1
2
)
(
)
2
cos
1
(
)
2
cos
1
(
4
2
+
−
=
∫ ∫
−
+
−
=
∫ ∫ −
+
−
=
α
α
π
π
π
α
π
π
π
π
α
π
π
ω
ω
ω
ω
π
m
V
t
t
t
t
m
V
t
d
t
t
d
t
m
V
(5.1)
7. And the average value of output voltage is
(
(
(
( )
)
)
)
)
1
(cos
2
2
2
)
(
sin
2
sin
2
2
1
−
−
−
−
=
=
=
=
∫
∫
∫
∫ ∫
∫
∫
∫
+
+
+
+
=
=
=
=
α
α
α
α
π
π
π
π
π
π
π
π
α
α
α
α
π
π
π
π
π
π
π
π
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
π
π
π
π
s
V
t
td
s
V
t
td
s
V
dc
V
Gating sequence work.
(5.2)
Gating sequence work.
1. Output will occurs when a pulse received by T1
2. T1 on during positive half cycle and turn off when the input
crossing zero point
3. D1 operate due to negative half cycle due to connection of
diode from negative terminal to positive terminal
8. b) RMS load current
The rms value of the output voltage: using (5.1)
Example 5.1
A single phase ac voltage controller in Figure 5.1 has a resistive load of R =
10Ω and the input voltage is Vs = 120 , 60 Hz. The delay angle of thyristor T1
is α = π /2. determine (a) the rms output voltage Vo, (b) the input PF and (c)
the average input current
Solution : R = 10Ω, Vs = 120, α = π /2 and Vm = √2 x 120 = 169.7V
V
Vo
92
.
103
4
3
120
=
=
=
=
=
=
=
=
V 92
.
103
The load power
Because the input current is the same as the load current, the input VA rating is
The input PF
V
R
V
Io
392
.
10
10
92
.
103
0
=
=
=
=
=
=
=
=
=
=
=
=
W
x
R
I
Po 94
.
1079
10
392
.
10 2
2
0 =
=
=
=
=
=
=
=
=
=
=
=
VA
x
I
V
I
V
VA o
s
s
s 04
.
1247
392
.
10
120 =
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
(lagging)
866
.
0
04
.
1247
94
.
1079
4
3
)
2
2
sin
2
(
2
1
2
/
1
=
=
=
=
=
=
=
=
=
=
=
=
+
+
+
+
−
−
−
−
=
=
=
=
=
=
=
=
=
=
=
=
α
α
α
α
α
α
α
α
π
π
π
π
π
π
π
π
s
o
o
V
V
VA
P
PF
9. And the average input current
The average output voltage: using (5.2)
V
x
Vdc 27
2
2
120 −
−
−
−
=
=
=
=
−
−
−
−
=
=
=
=
π
π
π
π
A
R
V
I dc
D 7
.
2
10
27
−
−
−
−
=
=
=
=
−
−
−
−
=
=
=
=
=
=
=
=
10. Single Phase Bidirectional/Full wave Phase Control With Resistive load
Figure: single phase full wave AC-C controller with resistive load
1. During positive half cycle the power flow is control by T1 with the a
gate pulse (g1).
2. For negative half cycle the power flow is control by T2 with second
pulse (g2)
11. The operations AC-AC R load:
• The SCRs cannot conduct simultaneously
• The output Vo is started to generate during pulse at G1 until the input
crossing the zero point.
• The load voltage is the same as the source voltage when either SCR is
on and the load voltage is zero when both SCRs are off.
• The switch voltage Vsw is zero when either SCR is on and is equal to the
source voltage when either is on.
source voltage when either is on.
• The output is discontinuous for every cycle.
• The average current in the source and load is zero if the SCRs are on for
equal time intervals.
• The average current in each SCR is not zero because of unidirectional
SCR current.
• The rms current is each SCR is 1/√2 times the rms current if the SCRs
are on for equal time.
13. If is the input voltage with delay angle of T1 and T2 is
the rms output will be
(
(
(
( )
)
)
)
1
2 α
α
α
α
π
π
π
π
α
α
α
α +
+
+
+
=
=
=
=
t
V
V s
s ω
ω
ω
ω
sin
2
=
=
=
=
( )
2
/
1
2
2
/
1
2
2
)
(
)
2
cos
1
(
2
sin
2
2
−
=
=
∫
∫
ω
ω
ω
ω
π
π
π
α
m
m
rms
t
d
t
V
t
td
V
V
(5.3)
2
/
1
)
2
2
sin
(
2
1
)
(
)
2
cos
1
(
4
+
−
=
−
= ∫
α
α
π
π
ω
ω
π α
m
m
V
t
d
t (5.3)
15. :
Solution : R = 10Ω, Vs = 120 (rms), 60 Hz, α = π/2 and Vm = √2 x 120 = 169.7V
Example 5.2- (Tutorial no 5)
A single phase full wave ac voltage controller in Figure 5.3a has a resistive load of R = 10Ω and the
input voltage Vs = 120 (rms), 60 Hz. The delay angles of thyristors T1 and T2 are equal α1 = α2 = π/2.
Determine a) the rms output voltage Vo, b) the input PF, c) the average current of thyristors Ia and d
) the rms current of thyristors Ir
120
The rms value of the load current is and the load power
.
Because the input current is same the load current, the input VA rating is
From equation, the rms output voltage : using (5.3)
V
Vo
85
.
84
2
120
=
=
=
=
=
=
=
=
A
R
V
Io 485
.
8
10
85
.
84
0
=
=
=
=
=
=
=
=
=
=
=
=
W
x
R
I
Po 95
.
719
10
485
.
8 2
2
0 =
=
=
VA
x
I
V
I
V
VA o
s
s
s 2
.
1018
485
.
8
120 =
=
=
=
16. c) The average thyristor current
The PF is
(lagging)
707
.
0
2
.
1018
95
.
719
2
1
=
=
=
=
=
=
s
o
o
V
V
VA
P
PF
R
V
t
td
V
R
I
s
s
A
)
1
(cos
2
2
)
(
sin
2
2
1
+
+
+
+
=
=
=
=
=
=
=
= ∫
∫
∫
∫
α
α
α
α
π
π
π
π
ω
ω
ω
ω
ω
ω
ω
ω
π
π
π
π
π
π
π
π
α
α
α
α
d) The rms value of the thyristor current
A
R
7
.
2
10
2
120
2
2
=
=
=
=
=
=
=
=
π
π
π
π
π
π
π
π
A
x
R
V
t
d
t
R
V
t
td
V
R
I
s
s
s
R
6
10
2
120
2
2
sin
1
2
)
(
)
2
cos
1
(
4
2
[
]
)
(
sin
2
2
1
[
2
/
1
2
/
1
2
2
2
/
1
2
2
2
=
=
=
=
=
=
=
=
+
+
+
+
−
−
−
−
=
=
=
=
−
−
−
−
=
=
=
=
=
=
=
=
∫
∫
∫
∫
∫
∫
∫
∫
α
α
α
α
α
α
α
α
π
π
π
π
π
π
π
π
ω
ω
ω
ω
ω
ω
ω
ω
π
π
π
π
ω
ω
ω
ω
ω
ω
ω
ω
π
π
π
π
π
π
π
π
α
α
α
α
π
π
π
π
α
α
α
α
17. Single Phase Full wave Phase Control with RL loads
( )
2
/
1
2
sin
2
2
2
∫
=
β
α
ω
ω
π
t
td
m
V
rms
V ( )
2
/
1
2
2
sin
2
2
sin
(
2
1
2
/
1
)
(
)
2
cos
1
(
4
2
2
sin
2
−
+
−
=
∫ −
=
∫
=
β
α
α
β
π
β
α
ω
ω
π
α
ω
ω
π
m
V
t
d
t
m
V
t
td
m
V
rms
V
18. T1 is triggered during positive half cycle at : α
T2 is triggered during negative half cycle at : π+α
The current will falls until it reach ωt = β
β is extinction angle
The conduction angle of T1 is δ=β – α
The control range of the firing angle: θ < α < π
θ is the load angle
The minimum value of firing angle α = θ
(I-output Maximum)
θ= tan-1(ωL/R)
19. Instantaneous Thyristor current
( )
θ
ω −
=
+ t
V
Ri
dt
di
L s sin
2
1
1
( )
V
2
(5.6)
The solution of (5.6) is of the form
( ) ( )t
L
R
s
e
A
t
Z
V
i /
1
0 sin
2 −
+
−
= θ
ω
The constant A1 at ωt = α, i1 = 0.
( )( )
ω
α
θ
α /
/
1 )
sin(
2 L
R
s
e
Z
V
A −
−
=
(5.7)
(5.8)
20. ( ) ( )
−
−
−
=
−
− )
(
/
1 )
sin(
sin
2 t
L
R
s
e
t
Z
V
i ω
α
θ
α
θ
ω
Instantaneous Thyristor current
The RMS Thyristor current
The RMS Thyristor current
( ) ( )
2
/
1
2
)
(
/
2
/
1
2
1
)
(
)
sin(
sin
1
)
(
2
1
−
−
−
=
=
∫
∫
−
−
−
−
β
α
ω
α
β
α
ω
θ
α
θ
ω
π
ω
π
t
d
e
t
Z
V
I
t
d
i
I
t
L
R
s
Thy
rms
Thy
rms
21. The RMS Output current
( ) 2
/
1
2
2
R
R
out I
I
I +
=
The Average Thyristor current
The Average Thyristor current
( ) ( )
−
−
−
=
=
∫
∫
−
−
−
−
β
α
ω
α
β
α
ω
θ
α
θ
ω
π
ω
π
)
(
)
sin(
sin
2
2
)
(
2
1
)
(
/
1
t
d
e
t
Z
V
I
t
d
i
I
t
L
R
s
Thy
dc
Thy
dc
22. Example 5.3
For the single phase voltage controller of Figure 5.8, the source is 120V
rms at 60Hz and the load is a series RL combination with R = 20Ω and L =
50mH. The delay angle is α = 900. Determine:
a) an expression for load current for the first half period,
b) the rms load current,
b) the rms load current,
c) the rms SCR current,
d) the average SCR current,
e) the power deliver to the load and
f) the power factor.
23. Solution
the current is expressed from the parameters
A
m
V
rad
R
L
rad
R
L
L
R
Z
18
.
6
2
120
943
.
0
20
05
.
0
)
377
(
756
.
0
20
)
05
.
0
)(
377
(
1
tan
1
tan
5
.
27
2
)]
05
.
0
)(
377
[(
2
)
20
(
2
)
(
2
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
−
−
−
−
=
=
=
=
−
−
−
−
=
=
=
=
=
=
=
=
+
+
+
+
=
=
=
=
+
+
+
+
=
=
=
=
ω
ω
ω
ω
ωτ
ωτ
ωτ
ωτ
ω
ω
ω
ω
θ
θ
θ
θ
ω
ω
ω
ω
The current is then expressed as
The extinction angle β is determined from the numerical solution of i(β) = 0 in
the preceding equation
The conduction angle γ = β – α = 2.26 rad = 1300, which is less the the limit of
1800
A
e
Z
m
V
rad
A
Z
m
V
8
.
23
/
)
sin(
57
.
1
0
90
18
.
6
5
.
27
2
120
=
=
=
=
−
−
−
−
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
ωτ
ωτ
ωτ
ωτ
α
α
α
α
θ
θ
θ
θ
α
α
α
α
α
α
α
α
β
β
β
β
ω
ω
ω
ω
α
α
α
α
ωτ
ωτ
ωτ
ωτ
ω
ω
ω
ω
ω
ω
ω
ω ≤
≤
≤
≤
≤
≤
≤
≤
−
−
−
−
−
−
−
−
−
−
−
−
=
=
=
= t
for
943
.
0
/
8
.
23
)
756
.
0
sin(
18
.
6
)
(
0 A
e
t
t
i
0
220
83
.
3 =
=
=
=
=
=
=
= rad
β
β
β
β
24. the rms load current is determined by
The rms current in each SCR is determine
[[[[ ]]]]
A
t
d
t
e
t
rms
I
71
.
2
83
.
3
57
.
1
)
(
2
943
.
0
/
8
.
23
)
756
.
0
sin(
18
.
6
1
,
0
=
=
=
=
∫
∫
∫
∫
−
−
−
−
−
−
−
−
−
−
−
−
=
=
=
= ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
π
π
π
π
A
rms
o
I
rms
SCR
I 92
.
1
2
71
.
2
2
,
, =
=
=
=
=
=
=
=
=
=
=
=
the average SCR current is obtained
Power absorbed by the load
Power factor is determined from P/S
A
rms
SCR
I 92
.
1
2
2
, =
=
=
=
=
=
=
=
=
=
=
=
[[[[ ]]]]
A
t
d
t
e
t
avg
SCR
I
04
.
1
)
(
83
.
3
57
.
1
943
.
0
/
8
.
23
)
756
.
0
sin(
18
.
6
2
1
,
=
=
=
=
∫
∫
∫
∫
−
−
−
−
−
−
−
−
−
−
−
−
=
=
=
= ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
π
π
π
π
W
R
rms
o
I
P 147
)
20
(
2
)
71
.
2
(
,
2 =
=
=
=
=
=
=
=
=
=
=
=
45
.
0
)
71
.
2
)(
120
(
147
,
,
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
rms
s
I
rms
s
V
P
s
P
pf
25. Three Phase Voltage Controller
It is commonly used in ac motor drives
Figure below shows a three phase voltage controller with Y connection
The phase voltages are
The line-line voltage are
)
3
4
sin(
)
3
2
sin(
sin
π
π
π
π
θ
θ
θ
θ
π
π
π
π
θ
θ
θ
θ
θ
θ
θ
θ
−
−
−
−
=
=
=
=
−
−
−
−
=
=
=
=
=
=
=
=
p
V
CN
v
p
V
BN
v
p
V
AN
v
)
6
7
sin(
3
)
2
sin(
3
)
6
sin(
3
π
π
π
π
θ
θ
θ
θ
π
π
π
π
θ
θ
θ
θ
π
π
π
π
θ
θ
θ
θ
−
−
−
−
=
=
=
=
−
−
−
−
=
=
=
=
+
+
+
+
=
=
=
=
P
V
CA
v
P
V
BC
v
P
V
AB
v
26. If the gate signal is continuously applied the thyristors would behave as diodes
There are two gate control per pulse
The delay angle in all phases is referenced to 0 conduction angle
Delay angle
Current Communication
T5 to T1
π
π
π
π
α
α
α
α
T6 to T2
T1 to T3
T2 to T4
T3 to T5
T4 to T6
3
5
3
4
3
2
3
π
π
π
π
α
α
α
α
π
π
π
π
α
α
α
α
π
π
π
π
α
α
α
α
π
π
π
π
α
α
α
α
π
π
π
π
α
α
α
α
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
27. Thyristors T1, T3 and T5 turn on in the positive half cycles of waveforms Van, Vbn,
and Vcn respectively
T1 conducts from α to π, T3 conducts from α + 2 π /3 to 5 π /3 and T5 conducts
from α + 4 π /3 to 7 π /3
Thyristors T4, T6 and T2 turn on in the negative half cycles of the waveforms Van,
Vbn, Vcn
Vbn, Vcn
T4 conducts from α + π to 2 π, T6 conducts from α + 5 π /3 to 8 π /3 (= 2 π + 2 π
/3) and T2 conducts from α + 7 π /3 (=α +2 π + π /3) to 10 π /3 (=2 π +4 π /3)
The delay angle α can be selected in the range 0 to π
28. The case α > π /3
first mode, the thyristor T6 was already on while T1 is in forward blocking mode due to
the input voltage Van
T1 is turned at α, referenced to the zero crossing point on the Van waveform
T5 is reversed biased and the output current i0 flows from phase A through T1, load
and T6 to phase B
and T6 to phase B
At angle α + π /3 in mode 2, thyristor T2 is in mode forward blocking mode due to the
input phase voltage Vcn
T2 is turned on at α + π /3, α angle referenced to the zero crossing point on the
positive half cycle of the Vcn waveform (figure below)
29. T6 is reversed biased and the output current i0 flows from phase A through T1, load
and T2 to phase C
Likewise from the angle α + 2 π /3 to α +4 π /3 in mode 3 and mode 4, the current i0
flows from the input phase B to other phases
The phase output voltage Van consists of two parts in the positive half cycle
31. Analysis techniques using SCR with three phase voltage controller
By using SCR the six SCRs are turned on in the sequence 1-2-3-4-5-6 at 600 intervals
At any instant, three SCRs, two SCRs and no SCRs are on
The instantaneous load voltages are either a line to neutral voltage (three on), one half
of a line to line voltage (two on) or zero (none on)
When three SCRs are on (one in each phase), all three phase voltages are connected to
the course corresponding to a balanced three-phase source
the course corresponding to a balanced three-phase source
-S1, S2 and S6 are on, van = VAN, vbn = VBN, vcn = VCN
two SCRs are on, the line to line voltage of those phases is equally divided between the
two load resistors which are connected
32.
33. For 0 < α < 600
At ωt = 0, S5 and S6 are conducting and three is no current in Ra making van= 0
At ωt = π/6 (300), S1 receives a gate signal and begins to conduct, S5 and S6 remain
on and van = VAN
The current in S5 reaches zero at 600 turning S5 off while S1 and S6 remaining on,
van = VAB/2
van = VAB/2
At 900 S2 is turned on, the three SCRs S1, S2 and S6 are then on and van = VAN
At 1200, S6 turns off, leaving S1 and S2 on so van = VAC/2
For internals to exist when three SCRs are on, the delay angle must be less than 600
34.
35. •For 600 < α < 900
Only two SCRs conduct at any one time when the delay angle is between 600 and 900
Just prior to 750, S5 and S6 are conducting, and van = 0
S1 is turned on at 750, S6 continues to conduct but S5 must turn off because VCN is
negative
When S2 is turned on at 1350, S6 is forced off, and van = VAC/2
When S2 is turned on at 1350, S6 is forced off, and van = VAC/2
The next SCRs to turn on in S3 which forces S1 off and Van = 0
Load voltage is one half line to line voltages or zero
•For 900 < α <150
•Only two SCRs can conduct at any one time in this mode
•In the intervals just prior to 1200 no SCRs are on, and van = 0. At α = 1200, S1 is given
by a gate signal, and S6 has a gate signal applied
•Since VAB is positive both S1 and S6 are forward biased and begin to conduct and van
= VAB/2
36. Both S1 and S6 become turn off when VAB becomes negative
When a gate signal is applied to S2, it turns on and S1 turns on again
•For α > 1500
There is no time interval when an SCR is forward biased while a gate signal is
applied. Output voltage is zero at this condition
37. AC Voltage Controlellers with PWM Control
A ac voltage controllers rectifiers can be improved by the pulse width modulation (PWM)
type of control
The performance of ac voltage controllers can be improved by PWM control
Switches S1 and S2 are turned on and off several times during the positive and negative
half cycles of the input voltage,
respectively S1’ and S2’ provide freewheeling path for the load current whereas S1 and
respectively S1’ and S2’ provide freewheeling path for the load current whereas S1 and
S2 respectively in the off state
38. For a resistive load current resembles the output voltage
With a RL load, the load current rises in the positive or negative direction when switch
S1 or S2 is turned on respectively