Chapter 3 Controlled Rectifier.pdf

Chapter 3
Controlled Rectifier
1

Single-Phase Half-wave Controlled
Converter – R Load
When thyristor T1 is fired at ωt = α, thyristor T1
conducts and the input voltage appears across
the load.
When the input voltage starts to be negative at
ωt = π, the thyristor is negative with respect to
2
ωt = π, the thyristor is negative with respect to
its cathode and thyristor T1 said to be reverse
biased and it is turned off.
The time after the input voltage starts to go
negative until the thyristor is fired at ωt = α is
called the delay or firing angle α. .

Converters
3
Figure 2.14: Single-phase thyristor converter with a resistive load

Figure 2.14(c) shows the waveforms for input
voltage, output voltage, load current, and
voltage across T1.
This converter is not normally used in industrial
applications because its output has high ripple
4
applications because its output has high ripple
content and low ripple frequency.

Converter – R Load (π
π
π
π in radian )
If Vm is the peak input voltage, the average
output voltage Vdc can be found from.
The maximum output voltage Vdm is (a=0)
( ) [ ] ( )
α
π
ω
π
ω
ω
π
π
α
π
α
cos
1
2
cos
2
sin
2
1
+
=
−
=
= ∫
m
m
m
dc
V
t
V
t
td
V
V
5
The maximum output voltage Vdm is (a=0)
The normalized output voltage
π
m
dm
V
V =
( )
α
cos
1
5
.
0 +
=
=
dm
dc
n
V
V
V

The root-mean-square (rms) output
voltage.
( )
2
/
1
2
2
sin
2
1






= ∫ ω
ω
π
π
α
m
rms t
td
V
V
6
( ) ( )
2
/
1
2
/
1
2
2
2
sin
1
2
2
cos
1
4












+
−
=






−
=


∫
α
α
π
π
ω
ω
π
π
α
α
m
m
V
t
d
t
V

Example 2.6
Find the Performances of a Single-Phase Controlled
Converter
If the converter of Figure 2.14(a) has a purely resistive load of
R and the delay angle is α = π/2. Determine
(a) the rectification efficiency.
(b) the form factor (FF).
7
(b) the form factor (FF).
(c) the ripple factor (RF).
(d) the TUF.
(e) the peak inverse voltage (PIV) of thyristor T1.

Example 2.6 Solution
Delay angle is α = π/2
Vdc = Vm/2π(1 + cos α) = 0.1592Vm.
Idc = 0.1592Vm /R
m
rms
V
V
2
2
sin
1
2
2
/
1












+
−
=
α
α
π
π
8
Irms = 0.3536Vm/R
m
m
V
V
3536
.
0
2
)
2
/
(
2
sin
2
1
2
2
2
2
/
1
=












+
−
=

 

π
π
π
π
π

Example 2.6 Solution
(a) The efficiency
%
27
.
20
)
3536
.
0
(
)
1592
.
0
(
/
)
(
/
)
(
2
2
2
2
=
=
=
=
m
m
ac
dc
ac
dc
V
V
R
V
R
V
P
P
η
9
(b)
(c)
(d)
(e) PIV = Vm
%
1
.
222
221
.
2
1592
.
0
3536
.
0
=
=
=
=
m
m
dc
rms
V
V
V
V
FF
983
.
1
1
221
.
2
1 2
2
=
−
=
−
= FF
RF
( ) 1014
.
0
/
)
3536
.
0
(
2
/
/
)
1592
.
0
( 2
=
=
=
R
V
V
R
V
I
V
P
TUF
m
m
m
s
s
dc

Single-Phase Full-wave Controlled
Converter – RL Load
During the period from α to π, the input voltage
vs and input current is are +ve and the power
flows from the supply to the load.
The converter is said to be operated in
rectification mode.
10
rectification mode.
During the period from π to π + α, the input
voltage vs is -ve and the input current is is
positive and reverse power flows from the load
to the supply.
The converter is said to be operated in
inversion mode.

This converter is extensively used in industrial
applications.
Depending on the value of α, the average
output voltage could be either positive or
negative and it provides two-quadrant operation.
11
negative and it provides two-quadrant operation.

Single-Phase Full-wave
Controlled Converters
.
12
Figure 2.15: Single-phase Full-wave Converter

The average output voltage
The rms value of the output voltage
( ) [ ] α
π
ω
π
ω
ω
π
α
π
α
α
π
α
cos
2
cos
2
2
sin
2
2 m
m
m
dc
V
t
V
t
td
V
V =
−
=
=
+
+
∫
13
The rms value of the output voltage
( )
( ) ( ) s
m
m
m
rms
V
V
t
d
t
V
t
td
V
V
=
=






−
=






=
∫
∫
+
+
2
2
cos
1
2
sin
2
2
2
/
1
2
2
/
1
2
2
α
π
α
α
π
α
ω
ω
π
ω
ω
π

The load current iL.
mode 1 : when T1 and T2 conduct [α ≤ ωt ≤ (α + π)]
( ) ( ) ( )( )
t
L
R
s
L
s
L e
Z
V
R
E
I
R
E
t
Z
V
i −








−
−
+
+
−
−
= ω
α
θ
α
θ
ω /
/
0 sin
2
sin
2
for
14
The steady-state condition iL (ωt = π + α) = IL1 =
IL0.




( ) ( ) ( )( )
( )( )
R
E
e
e
Z
V
I
I L
R
L
R
s
L
L −
−
−
−
−
−
=
= −
−
ω
π
ω
π
θ
α
θ
α
/
/
/
/
1
0
1
sin
sin
2
0
0 ≥
L
I

The rms current.
( )
2
/
1
2
2
1






= ∫
+α
π
α
ω
π
t
d
i
I L
R
15
The rms output current.
The average current


( ) R
R
R
rms I
I
I
I 2
2
/
1
2
2
=
+
=
( )
∫
+
=
α
π
α
ω
π
t
d
i
I L
A
2
1

The average output current.
A
A
A
dc I
I
I
I 2
=
+
=
16

Example 2.7
Finding the Current Ratings of Single-Phase Controlled
Full Converter with an RL load
The single-phase full converter of Figure 215(a) has a RL load
having L = 6.5 mH, R = 0.5 Ω, and E = 10 V. The input voltage
is Vs = 120 V at (rms) 60 Hz. Determine
(a) the load current I at ωt = α = 60°.
17
(a) the load current IL0 at ωt = α = 60°.
(b) the average thyristor current IA.
(c) the rms thyristor current IR.
(d) the rms output current Irms.
(e) the average output current Idc.
(f) the critical delay angle αc.

2.10 Principle of Three-
phase Half-wave
Controlled Converter
18

Principle of Three-phase Half-wave
Three-phase converters provide higher
average output voltage and in addition the
frequency of the ripples on the output voltage
is higher compared with that of single-phase
converters.
As a result, the filtering requirements for
19
As a result, the filtering requirements for
smoothing out the load current and load
voltage are simpler.
For these reasons, three-phase converters
are used extensively in high-power variable-
speed drives.

20
Figure 2.16: Three-phase half-wave converter

21
Figure 2.16: Three-phase half-wave converter

When thyristor T1 is fired at ωt = π/6 + α, the
phase voltage van appears across the load until
thyristor T2 is fired at ωt = 5π/6 + α.
When thyristor T2 is fired, thyristor T1 is reverse
biased, because the line-to-line voltage, vab (=
22
biased, because the line-to-line voltage, vab (=
van – vbn), is negative and T1 is turned off.
The phase voltage vbn appears across the load
until thyristor T3 is fired at ωt = 3π/2 + α.
When thyristor T3, is fired, T2 is turned off and
vcn appears across the load until T1 is fired
again at the beginning of next cycle.

Figure 2.16(c) shows the input voltages, output
voltage, and the current through thyristor T1 for
a highly inductive load.
For a resistive load and α π/6, the load
current would be discontinuous and each
23
current would be discontinuous and each
thyristor is self-commutated when the polarity of
its phase voltage is reversed.
However, this converter explains the principle
of the three-phase thyristor converter.

If the phase voltage is van = Vm sin ωt average
output voltage for a continuous load current is
The rms output voltage is found from
( ) α
π
ω
ω
π
α
π
cos
2
3
3
sin
2
3
6
/
5
m
m
dc
V
t
d
t
V
V =
= ∫
+
24
The rms output voltage is found from
( ) α
π
ω
ω
π α
π
cos
2
sin
2 6
/
m
dc t
d
t
V
V =
= ∫+
( )
2
/
1
2
/
1
6
/
5
6
/
2
2
cos
8
3
6
1
3
sin
2
3








+
=






= ∫
+
+
α
π
ω
ω
π
α
π
α
π
m
m
rms V
t
d
t
V
V

For a resistive load and α ≥ π/6:
( )








+
+
=
= ∫+
α
π
ω
ω
π
π
α
π
cos
1
3
sin
2
3
6
/
m
m
dc
V
t
d
t
V
V
25












+
+
= α
π
π 6
cos
1
2
3 m
V
( )
2
/
1
2
/
1
6
/
2
2
2
3
sin
8
1
4
24
5
3
sin
2
3












+
+
−
=






= ∫+
α
π
π
π
α
ω
ω
π
π
α
π
m
m
rms
V
t
d
t
V
V

Three-Phase Full-wave Controlled
Converters
Figure 2.17(a) shows a full-converter circuit with
a highly inductive load.
This circuit is known as a three-phase bridge.
The thyristors are fired at an interval of π/3.
The frequency of output ripple voltage is 6fs
26
The frequency of output ripple voltage is 6fs
and the filtering requirement is less than that of
half-wave converters.
At ωt = π/6 + α, thyristor T6 is already
conducting and thyristor T1 is turned on.

Converters
During interval (π/6 + α) ≤ ωt ≤ (π/2 + α),
thyristors T1 and T6 conduct and the line-to-line
voltage vab(= van – vbn) appears across the load.
At ωt = π/2 + α, thyristor T2 is fired and thyristor
T6 is reversed biased immediately.
T6 is turned off due to natural commutation.
27
T6 is turned off due to natural commutation.
During interval (π/2 + α) ≤ ωt ≤ (5π/6 + α),
thyristors T1 and T2 conduct and the line-to-line
voltage vac appears across the load.

Converters
If the thyristors are numbered, as shown in
Figure 2.17(a), the firing sequence is 12, 23, 34,
45, 56, and 61.
Figure 2.17(b) shows the waveforms for input
voltage, output voltage, input current, and
28
voltage, output voltage, input current, and
currents through thyristors.

Converters
If the line-to-neutral voltages are defined as








−
=
=
2
3
2
sin
sin
π
π
ω
ω
t
V
v
t
V
v
m
bn
m
an
29






+
=
3
2
sin
π
ωt
V
v m
cn






+
=
−
=






−
=
−
=






+
=
−
=
2
sin
3
2
sin
3
6
sin
3
π
ω
π
ω
π
ω
t
V
v
v
v
t
V
v
v
v
t
V
v
v
v
m
an
cn
ca
m
cn
bn
bc
m
bn
an
ab
the corresponding
line-to-line voltages
are

Converters
The average output voltage is found from
( ) ( )
α
π
ω
π
ω
π
ω
π
α
π
α
π
α
π
α
π
cos
3
3
6
sin
3
3
3
2
/
6
/
2
/
6
/
m
m
ab
dc
V
t
d
t
V
t
d
v
V
=






+
=
= ∫
∫
+
+
+
+
30
The rms value of the output voltage is found
from
α
π
cos
=
( )
2
/
1
2
/
1
2
/
6
/
2
2
2
cos
4
3
3
2
1
3
6
sin
3
3








+
=












+
= ∫
+
+
α
π
ω
π
ω
π
α
π
α
π
m
m
rms
V
t
d
t
V
V

Converters
31
Figure 2.17: Three-phase full converter

32
Figure 2.17: Three-
phase full converter

Converters
Figure 2.17(b) shows the waveforms for α = π/3.
For α π/3, the instantaneous output voltage v0
has a negative part.
Because the current through thyristors cannot
be negative, the load current is always positive.
33
be negative, the load current is always positive.
Thus, with a resistive load, the in-stantaneous
load voltage cannot be negative, and the full
converter behaves as a semiconverter.

Q1
Single phase controlled rectifier is connected to
240V, 50Hz, the turn ratio of transformer is 2:1.
If the delay angle is α=600 calculated:
a. Vdc
34
a. Vdc
b. Vrms
c. THD
d. DF
e. PF
A:54.02V,120V,(Is1=0.9Ia,Ia=Is, THD=48.34%),0.5,0.45

Q2
A thyristor half-wave controlled converter has a
supply voltage of 240V at 50Hz and a load
resistance of 100 ohm when the firing delay angle is
300
a) What are the average values of load voltage and current?
35
a) What are the average values of load voltage and current?
b) What are the rms values of load voltage and current?
c) Power factor?
A: a)108 1
b)167.3 1.673
c) PF 0.697

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Chapter 3 Controlled Rectifier.pdf