CHAPTER TEN
CONTROLLED RECTIFIERS
Learning
Objectives:
v
v
v
v
To understand the operation and characteristics of controlled rectifiers .
To learn the types of controlled rectifiers.
To understand the performance parameters of controlled rectifiers.
To learn the techniques for analyzing and design of controlled rectifier
circuits.
v To study the effects of load inductance on the load current.
10.1
10.2
Introduction:
v We have seen that diode rectifiers provide a fixed output voltage only.
v Controlled output voltage could be obtained by replacing the diodes by
phase-controlled thyristors.
v The output voltage of thyristor rectifiers is varied by controlling the delay or
firing angle of the thyristor.
v A phase-controlled thyristors is turned on by applying a short pulse to its gate
and turned off due to natural or line commutation ( self commutation).
v These phase-controlled rectifiers are simple and less expensive, and the
efficiency of these rectifiers is, in general, above 95%.
v These rectifiers are called ac-dc converters ( because convert the AC signal to
DC signal).
v These rectifies are applied extensively in industrial applications, especially in
variable-speed drives, ranging from fractional horsepower to megawatt power
level.
v These rectifiers are classified into two types, depending on the input supply :
(1) Single-phase converter, and (2) three-phase converters.
v Each type of these rectifiers can be subdivided into (a) half-converter, (b)
semicinverter, (c) full-converter, and (d) dual converter.
v A semiconverter in one-quadrant converter and it has one polarity of output
voltage and current . A full converter is a two -quadrant converter and he
polarity of its voltage can be either positive and negative. A dual converter can
operate in four quadrants, and both the output voltage and current can be
either positive and negative.
Principle of Phase-controlled Converter
Operation ( Single-Phase Half –Wave)
Single phase half-wave controlled rectifier is simplest circuit, this circuit is not used in
practical applications due to high voltage ripples, and low efficiency. Therefore discussing
this circuit aims to compare further electrical circuits with it.
The electrical circuit illustrating this type rectifier is shown in fig.10.1-a, where a
resistive load is energized. During the positive half-cycle of input voltage, the thyristor
anode is positive with respect to its cathode, and the thyristor is said to be forward
biased. When thyristor T1 is fired at ωt=α, T1 is conduct and the input voltage appears
across the load. When the input voltage starts to be negative , the thyristor anode is
negative with respect to its cathode and T1 is said to be reverse biased; and it is turned
off. The time after the input voltage starts to go to positive until the thyristor is fired at
ωt=α is called the delay or firing time. The circuit waveforms are displayed on
Fig.10.1-b.where Vs, Vo, Io , and VT ( VAK) are illustrated for firing angle of 45°.
fig.10.1-a: Electrical circuit
fig.10.1-b: Circuit waveforms( source voltage, output voltage , load current and thyristor
voltage ).
As well shown from the figure, T1 will conduct only for the time of ωt= α to ωt= π .
During the negative half wave the diode is in reverse biasing and there is no output
voltage. During the negative half wave the source voltage is applied across the
thyristor, therefore the diode must carry the peak value of the source voltage.
There are different types of converter circuits and the performances of a rectifier are
normally evaluated in terms of the following parameters:
1. The average value of the output ( load) voltage Vdc, and current Idc:
π
V DC =
1
T
∫ Vm sin( ω t ) d ω t =
α
I DC =
Vm
(1 + cos α )
2π
V DC
R
2. The normalizing voltage :
This is the ratio between the controlled voltage to the maximum rectified voltage:
Vn =
V DC ( α ≠ 0 )
Vm (1 + cos α )
=
= 0 . 5 (1 + cos α )
V DC ( α = 0 )
2 π . Vm / π
At α=0 …. VDC=Max= Vm/π
α=π …. VDC=0 .
∴ The control range is 0≤ α ≤ π , leads to voltage control from Vdc= Vm/π to 0.
3. The rms value of the output ( load) voltage Vrms, and current Irms:
π
Vrms =
1
T
∫ (Vm
sin ω t )2 d ω t =
α
Vm
2
1
sin 2 α
(π − α +
)
π
2
Irms = Vrms
R
At α=0 …. Vrms =Max= Vm/2
α=π …. Vrms =0 .
1. Generate a pulse-signal at the positive zero crossing of the supply voltage Vs.
2. Delay the pulse by the desired angle α and apply it between the gate and cathode
terminals of T1 through a gate –isolating circuit.
Gating sequence:
4. The load average and rms power:
P
DC
=
V
DC
P
AC
=
V
RMS
.I
DC
.I
RMS
5. The Rectifier Efficiency :
P DC
η=
P RMS
6. The Transformer Utilization:
TUF =
P DC
( VA ) rating
7. The Form Factor:
FF =
V RMS
V DC
8. The Ripple Factor: RF = V
V
2
AC
DC
9. The Harmonic Factor: HF =
=
 V RMS 

 −1 =
 V DC 
 IS − I S 1 


IS 

2
=
FF
 IS 


 I S1 
2
−1
2
−1
10. The Displacement angle: DF = cos φ
11. The Power Factor: PF = V S .I S1 . cos φ = I S 1 cos φ
V S .IS
12. The Crest factor: CF =
Example 10.1:
IS
( IS) peak
IS
Single phase half- wave converter has a purely resistive load of 10 Ω , energized by
voltage source of 220V throughout two windings transformer with ratio 2:1.The firing
angle α =π/4.
Determine : 1- the average and rms voltage and current
2- the efficiency, TUF,
3- FF, RF, and the peak voltage across the diode (PIV).
4- the CF, and the input PF.
Solution:
1. The average and RMS voltage and current:
Vm
110 . 2
( 1 + cos α ) =
(1 + cos 45 ) = 42 . 265 V
2π
2π
;
V DC
42 . 265
I DC =
=
= 4 . 226 A
R
10
Vm 1
sin 2 α
110 . 2 1
sin 90
Vrms =
(π − α +
)=
(π − π / 4 +
) = 74 . 164 V
2
π
2
2
π
2
Irms = Vrms = 74 .164 = 7 . 416 A
R
10
V DC =
2. The efficiency and TUF:
P DC
V DC .I DC
42 .265 * 4 . 265
=
=
* 100 % = 32 . 77 %
P AC V RMS .I RMS 74 .165 * 7 .416
P DC
180 .26
245 .124
TUF =
=
=
* 100 = 24 . 31 %
( VA ) rating
Vs .Is
110 * 7 .416
where Is = IRMS = 7 .416 A .
η=
3. The FF, RF, and PIV
FF =
RF =
V RMS 74 .165
=
= 1 .7547
V DC
42 .265
FF 2 − 1 = 1 .754 2 − 1 = 1 .441 = 144 %
PIV = Vm =
2 Vs =
2 * 110 = 155 .56 V .
. The CF, and input PF
( I S ) peak
Vm / R
=
=
IS
IS
Pac
PF =
= 0 . 67 lag .
( VA ) rating
CF =
2 * 110 / 10
= 2 .0976
7 . 416
Summary
Taking into account the obtained rectifier parameters we conclude that this type
rectifier characterized with bad parameters presented by :
1. Low ( poor) transform utilization 24.31%, which means that the
transformer must be 1/0.243=4.11 times larger that when it is used to
deliver power from a pure ac voltage. And the tr-r TUF is less than that
of half- wave uncontrolled rectifier.
2. Low ( poor) rectification efficiency = 32.77%
3. Presence of current dc component in the secondary current causing
additional losses ( winding and core heating).
4. High ripples 144% greater than that when the source is pure dc
5. High ripple factor, which means that a filter with large capacitance is
required for smoothing the output voltage, therefore this yield high
capacitor starting current problem.
Therefore this type rectifier is rarely used due to the weakness in quality of it's
power and signal parameters.
Half-wave Converter with
R-L Load
Figure 10.2 illustrates half- wave converter with R-L load, where the
existing the phase angle ϕ reduced the average output voltage. as well
shown as follows.
fig.10.2-a: Electrical circuit with R-L load.
fig.10.2-b: Circuit waveforms( source voltage, output voltage , and load current).
π +ϕ
V DC =
1
T
∫ Vm sin( ω t ) d ω t =
α
I DC
10.3.
Vm
(cos ϕ + cos α )
2π
V DC
=
R
Single-Phase Full –Wave Rectifier:
Single phase Full-wave rectifier is the popular circuit, applied in most industrial
application due to good rectifier parameters . There are two types rectifiers
- Full- wave center tap rectifier ( Midpoint)
- Full- Wave converter.
Both circuit characterized with identified rectifier parameters, except the secondary
current, efficiency and diode voltage.
10.3.1
Full-Wave
Center tap
rectifier
The electrical circuit is shown on fig.10.5-a, where a resistive load is energized
throughout two thyristors, and the waveforms obtained from this circuit are
illustrated on fig.10.3-b.
fig.10.3-a: Electrical circuit
Each thyristor will conduct for half period (T1 - positive cycle, T2 negative cycle).
The output voltage of full wave converter with less ripples. As well shown from the
illustrated waveforms on fig.10.3-b, the output rectified voltage has a unidirectional
form among the full period, also the voltage applied across the thyristor is twice the
source value.
1. The average value of the output ( load) voltage Vdc, and current Idc:
π
V DC =
2
T
∫ Vm sin( ω t) d ωt =
α
I DC = V DC
R
2 .Vm
cos α = 0.6366 Vm . cos α
π
10.3.2
Full-Wave
Converter with
R load
Fig.10.3-b: Circuit waveforms
With purpose to increase the converter efficiency, increasing the transformer
capability, and reduce the thyristor breakdown voltage , a full-wave converter
is used. Such a circuit is shown on fig. 10.4-a , where 4 thyristor operates in
predetermined sequence, each two thyristors operates together for the time of
(π+ϕ-α). The waveforms obtained from this circuit are illustrated on fig.10.4-b
fig.10.4-a: Electrical circuit ( full wave onverter).
fig.10.4-b: The main waveforms ( Vs,Vo).
fig.10.4-c: The diode waveforms ( Io, VT1, IT) .
With purpose to evaluate the circuit parameters an example should be described
hereinafter:
Example 10.2:
Full-wave converter with 120 source voltage, 60Hz, 10Ω load resistance, and 25%
normalized voltage.
Determine : 1- the firing angle, average, and rms voltage.
2- the average and rms thyristor current.
3- the input power factor.
Solution:
1. The firing angle, average and rms voltage:
π
∫ Vm . sin
2
=
2π
V DC
Vm
(1 + cos α )
π
ω t .d ω t =
α
V DC ( α ≠ 0 )
1
Vn =
=
(1 + cos α )
V DC ( α = 0 )
2
1
(1 + cos α ) ⇒ α = cos − 1 ( 0 . 50 ) = 120 °
2
V DC
26 . 9
= 26 . 9 V ⇒ I DC =
=
= 2 .69 A
R
10
∴ 0 .25 =
∴ V DC
2
2π
Vrms =
π
∫ (Vm .sin ωt ) .d ωt =
2
Vm
2
α
1
sin 2 α
(π − α +
) = 53.05 V
π
2
Irms = Vrms = 53 .05 = 5.305 A
R
10
2. The thyristor average, rms current and PIV :
I TAV
ITR
PIV
I DC
2 . 69
=
= 1 . 345 A
2
2
I RMS
5 . 305
=
=
= 3 . 74 A
2
2
= Vm
= 155 . 56 V
=
3. The input power factor:
=
PF
Is 1
cos
Is
α ⇒
∑
is ( ω t ) =
Is 1 = ?; Is = ?
2 Isn . sin(
ω t + ϕ n );
n = 1 , 2 ,3
Is
n
1
=
2
n = 1 ⇒
PF
10.3.3.
=
a
Is
1
Is 1
cos
Is
Full-Wave
Converter with R-L
load & back e.m.f.
=
n
2
+ b
n
2
Is .
π
α =
2
≅
2
2
Is
;⇒
nπ
2 ;⇒
2
2
π
cos 120
= 0 . 45 lag .
Full wave converter energized RL load with back induced voltage E is
illustrated on fig.10.5a, where the load current flows when the supply
voltage being greater than the back voltage E. The circuit waveform are
illustrated on fig.10.5b. The load current has discontinuous character due
to the value of back emf E as well shown on fig.10.5.c.
Fig.10.5-a: Bridge rectifier energized R-L load with back e.m.f. E
Fig.10.5-b: Circuit waveforms ( Vs, Vo, Io) with E=0;
By increasing the back induced voltage, the conducting time of the diode decreases with
reduced magnitude as well shown on fig.10.5-c.
Fig .10.5-c: Circuit waveforms ( Vs, Vo, Io) for R-L load with E>0.
At the same time keeping E at constant value, increasing the load inductance leads to
further deformation in the output voltage and current as well shown on fig3.8d.
Fig.10.5-d: Circuit waveforms ( Vs, Vo, Io) with high inductive load and back emf..
10.4.
Single - Phase
Dual Converter
Single phase dual converters provides a four quadrant operation of the system.
This could be realized by back to back connection of two full-wave converters
as well shown on fig. 10.6-a. Dual converters are normally used in high-power
variable-speed drives. If α1 and α2 are the delay angles of converter 1 and 2,
respectively, the corresponding Vdc1 and Vdc2 are obtained. These voltages
must be with the same value with purpose to avoid circulating current. The
delay angles are controlled such that one converter operates as a rectifier and
another operates as an inverter. The circuit waveforms are displayed on
fig.10.6-b.
N- Converter
P-Converter
Fig.10.6-a: Dual converter energized R
Fig.10.6-b: Circuit waveforms ( Vs, Vo, Io, and Is) When operating P-converter
Fig.10.6-c: Circuit waveforms ( Vs, Vo, Io, and Is) When operating N-converter.
With purpose to realized automatically switching –on P-converter and Nconverter consequently a full control circuit is applied as shown on fig.10.7-a,
where the obtained waveforms are displayed on fig.10.7-b.
Fig.10.7-a: Dual converter with Alpha controller.
Fig.10.7-b: Dual converter waveforms ( Vs, Vo) at various control reference voltage.
The average voltage Vdc1=-Vdc2 ( because one converter is rectifier, and the
other one is inverter), therefore the firing angles: α 2 = π- α 1 .
Summary
There are several key points :
1. Varying the delay angle α from 0 to π can vary the average output voltage
from 2Vm/π to -2Vm/π , provided the load is highly inductive and its
current is continuous.
2. For a purely resistive load, the delay angle α can varied from 0 to π /2
producing an output voltage ranging from 2Vm/π to 0.
3. The full wave converter can operate in two quadrant for a highly load and
in one quadrant only for a purely resistive load.
4. With respect to the power and signal parameters, this converter is
characterized with good performance, but quality less than that of
uncontrolled rectifier. These parameters being insufficient when the delay
angle increases.
5. The current discontinuity increases by increases the back induced voltage .
Despite the good performances of this converter especially at small value of
delay angle, but there is a need to improve the system performances, and
increasing the power rating of the converter according to the practical industrial
needs.
This could be achieved by using three-phase converters as well descri bed in the
coming sections:
Three-phase converters provide higher average output voltage, and in
addition the frequency of the ripples on the output voltage is higher
Principle of Three- compared with that of single-phase converters. As a results, the filtering
Phase Converters requirements for smoothing the load current and load voltage are simpler.
.
For these reasons, three-phase converters are used extensively in high-power
variable-speed drives.
10.5.
Three-Phase Half-Wave
Converter
Figure 10.8-a illustrates three-phase half wave converter energized R
load, where three thyristors ( T1, T2, and T3) operates in series
sequence, each one for a time of 120°.
Fig.10.8-a: Three-phase half-wave converter- Wye connected
The output voltage and current are displayed on fig10.8-b for the firing angle of 30°.
α
π/6+α
5π/6+α
Fig.10.9-b: Three-phase voltage, and output rectified voltage.
Fig.10.9-c: Thyristor voltage and load current
Fig.10.9-d: Phase current, and instantaneous rectified current at RL load.
The Main Equations:
1. The average, rms voltage and current:
V DC =
3
2π
∴ V DC =
5 π / 6+ α
∫
Vm . sin ω t d ( ω t ) =
π / 6+α
3 3
Vm = 0 . 8269 Vm . cos α
2π
When α = 0 ⇒ V DC
IDC =
3 3 Vm
cos α
2π
V DC
.
R
3 3
= Vdm =
Vm = Max .
2π
;
5 π / 6 +α
∫ (Vm. sin ωt )
3
2π
V RMS =
2
d (ωt ) = 3 Vm
π / 6+ α
1
3
+
cos 2α
6 8π
When α = 0 ⇒ ∴ V RMS = 0 .84068 Vm = Max.
VRMS
.
R
Vdc
Vn =
= cos α
Vdm
IRMS =
For a resistive load and α ≥ π/6
V DC =
Example 10.3:
3
2π
Vn =
1
VRMS =
3
2π
3
π
∫ Vm
. sin ω t d ( ω t ) =
π / 6+α
[1 +
3 Vm
2π
[1 +
cos( π / 6 + α ) ]
cos( π / 6 + α ) ].
π
∫ (Vm. sin ωt )
2
d( ω t) = 3Vm
π / 6+ α
5
α
1
−
+
sin( π / 3 + 2α)
24 4π 8π
Half-wave converter with 120 source voltage, 60Hz, 20Ω
average voltage is 86% of the maximum.
load resistance, and the
Determine : 1- the firing angle, average and rms voltage and current.
2- the average and rms thyristor current.
3- the rectifier efficiency, TUF, and FF.
Solution:
1. The firing angle, average and rms voltage and current
V DC = 86 %. Vdm ⇒∴ Vn = cos α = 0 . 86 ⇒ α = 30 o .
3 3
3 3
Vm . cos α =
120
2π
2π
V DC
120 . 7
=
=
= 6 .035 A .
R
20
∴ V DC =
IDC
*
2 . cos
30 = 120 .7 V
1
3
+
cos 2α = 0.7767Vm = 131.82V
6 8π
VRMS 131.82
IRMS =
=
= 6.59A.
R
20
VRMS = 3Vm
2. The average and rms voltage and current
I DC
6 .035
=
= 2 . 011 A .
3
3
I RMS
6 .59
=
=
= 3 . 80 A .
3
3
T TAV =
T TR
PIV =
3 Vm =
3
2 Vs =
6 Vs =
3. The rectifier efficiecy , TUF, and FF
6 120 = 294 V
η =
P DC
V DC . I DC
120 . 7 * 6 . 035
=
=
* 100 == 83 . 85 %
P RMS
V RMS . I RMS
131 . 82 * 6 . 59
TUF =
P DC
V DC . I DC
120 . 7 * 6 . 035
=
=
* 100 = 53 . 19 %
( VA ) rating
3 . V S .I S
3 * 120 * 3 . 804
1
2π
Is =
5π / 6+ α
∫ (I
ML . sin(
ω t + π / 6 ) ) . d ω t = I RMS
2
2 / 6 = 3 . 804 A
π / 6+ α
3
Vm .
R
V RMS
131 . 82
FF =
=
= 1 . 0921
V DC
120 . 7
I ML =
PF =
Po
I RMS 2 . R
6 . 59
=
=
( VA ) rating
3 . V S . IS
3 * 120
2
*
* 20
= 0 . 634
3 . 804
( lagging ) .
Key points of this section
Taking into account the obtained rectifier parameters we conclude:
1.
2.
3.
4.
5.
6.
The output rectified voltages depends on the firing angle α .
The Form factor increases dramatically at large values of α .
The frequency of the output voltage is three times the supply frequency.
There is a dc current component is the secondary voltage.
For α >30° with a resistive load, the current is discontinuous.
This converter is not normally used in practical circuits due to relatively
high ripples, and low efficiency.
With purpose to enhanced the rectifier parameters ( reducing the ripples, increasing
TUF, increasing the rectifier capability, and elimination the dc component in the secondary current) a
Three-phase full-wave converter is applied as follows:
10.6
Three-Phase
Full-Converters
Three-phase converters are extensively used in industrial applications up to
the 120-kW level where a two-quadrant operation is required. Figure 10.10-a
shows a full-wave converter circuit energized R load, where six thyristors (
T1, T2, T3, T4, T5, & T6) operates in six groups. The thyristors are fired at
an interval of 60°. Each thyristor groups ( two thyristor) conducts for interval
of 60° (π/2+α ≤ ωt ≤ 5π/6+α). The frequency of the output voltage is 6fs.
The operation (firing) sequence is T1T2, T2T3, T3T4, T4T5, T5T6. and
T6T1. The firing sequence is determined by the criteria : " the thyristor with
maximum positive voltage applied across it's terminals can be fired and will
conduct, and the thyristor with maximum negative voltage applied across it's
terminals also can be fired and will conduct ", therefore each phase will pass
the current for 240°, where for the rest of time this phase will be off and
doesn't participate in the rectification process. The simulation results
displayed on fig 10.10-b are for a firing angle of 45°, it clearly shown that
the output voltage can vary in wide range depending on α, and also the
voltage ripples varies in the same manner, therefore as α increases the filter
parameters rises.
Fig.10.10-a: Three-phase Full- wave converter
α
Fig.10.10-b: Three-phase voltage, output rectified voltage, and current…..
Each thyristor is fired at α= 45° from the crossing point between two phase voltages,
therefore, there is a repeated commutation process every 60°.
Fig.10.10-c: Output voltage , phase currents, and thyristor voltage at RL load.
The rectifier mathematical equations are going to be described by discussing the following
example :
Three- phase full-wave converter with the following parameters :
VLL=220V; Resistive load with R=10Ω; 60Hz; Wye connected; the average voltage
is 25% of the maximum permissible.
Example 10.4:
Determine : 1- the delay angle .
2- the rms and average output current .
3- the thyristor parameters.
4- the rectifier efficiency, TUF, FF, and PF
5- the average voltage reduction if the circuit inductance is 3mH.
6- the overlapping angle µ.
Solution:
1. The delay angle:
V DC =
=
3
π
π / 2+ α
∫
π / 6+ α
V LL .( ω t ) d ( ω t ) =
3
π
π / 2+ α
∫
3 Vm sin( ω t + π / 6 ) d ( ω t ) =
π / 6+ α
3 3 Vm
3 3 Vm
cos α = 1 . 654 Vm . cos α ⇒ V DC = Vdm . cos α ; Vdm =
.
π
π
Since V DC = 25 %. Vdm ⇒∴ cos α = 0 . 25 ⇒ α = 75 .5 ο .
2. The average, rms voltage and current:
3 3Vm 3 3. 2 .120 / 3
=
= 297 . 1V ⇒ VDC = 0.25 * 297 .1 = 74.24 V
π
π
VDC 74. 24
I DC =
=
= 7.427 A.
R
10
Vdm =
π / 2 +α
∫
3
π
VRMS =
V L 2 L.( ω t ) d (ω t ) =
π / 6 +α
3
π
π / 2 +α
∫(
3 Vm sin( ω t + π / 6)
)
2
d ( ωt ) =
π / 6+ α
0.5
1 3 3

= 3 Vm  +
cos 2 α 
= 115 .72 V
2

4π


V RMS 115 .72
IRMS =
=
= 11 . 572 A
R
10
3. The diode parameters (average, rms current and PIV): .
π / 2 +α
∫
V LL .( ω t )
1
d( ω t ) =
R
π
ITAV =
2
2π
∴ I TAV
7 . 427
=
= 2 . 475 A .
3
π / 6 +α
2
2π
ITR =
π / 2 +α
∫
π / 6 +α
V LL .( ω t ) 2
d (ω t ) =
R
∴ I TR = 11 . 572 *
2
= 6 .68 A .
6
PIV =
6 Vs =
3 .Vm =
π / 2 +α
∫
3 Vm
3 3 Vm
I DC
sin( ω t + π / 6 ) d ( ω t ) =
cos α =
;
R
3 . R .π
3
π / 6 +α
1
π
π / 2 +α



π / 6 +α 
∫
2

3 Vm
sin( ω t + π / 6 )  d ( ω t ) = I RMS .

R

6 * 127 = 311 V .
4. The rectifier efficiency, TUF, FF, and PF:
η =
P DC
V DC . I DC
551 . 08
=
=
* 100 = 41 . 30 %
P RMS
V RMS . I RMS
1334
TUF =
Is =
P DC
V DC .I DC
551 . 08
=
=
* 100 = 15 . 31 %
( VA ) rating
3 . V S .I S
3 * 127 * 9 . 448
4
2π
π / 2+ α
∫ (I
ML . sin(
ω t + π / 6 )) .d ω t = I RMS
π / 6+ α
2
4
6
= 9 . 448 A
3
Vm .
R
V RMS
115 . 72
FF =
=
= 1 . 558
V DC
74 . 27
I ML =
PF =
Po
I RMS 2 . R
11 . 572 2 * 10
1339 . 11
=
=
=
= 0 . 37
( VA ) rating
3 . V S . IS
3 * 127 * 9 . 448
3600 . 17
5. The average voltage reduction, and net output voltage
( lagging ) .
2
;
6
V DC ( L ≠ 0 ) = V DC − ∆ V = 74 . 27 − 8 . 021 = 66 . 249 V
Where ∆ V is the average
voltage
reduction
:
∆ V = 6 . fs . I DC . Lc = 6 * 60 * 7 . 427 * 0 . 003 = 8 . 021 V
5. The overlapping angle µ:
This is the angle occur every commutation process due to circuit inductance, where by
Lenz law there is an induced voltage which opposite the current change ( rising and
falling), therefore during commutating process three thyristors operates together ( first
thyristor with falling current – in process of switching-off; second thyristor with rising
current- in process of switching-on; and third thyristor with continuing conduction ).
This time during which three thyristors operates together is called overlapping time , and
can be measured by so called " Overlapping angle ". This angle can be determined as
follows:
The net average voltage taking into account both α and µ is presented by the following
equation in the case of current continuous mode:
V
DC
( α + µ ) = Vdm . cos( α + µ ) = V
DC
(α ) − ∆ V .
The overlapping angle can be determined from the previous equation as follows:
 VDC ( α + µ ) 
V
− α = cos − 1 


Vdm



 66 . 249 
ο
∴ µ = cos − 1 
 − 75 . 5 = 1 . 615 .
297
.
1


µ = cos
−1
DC
(α ) − ∆ V 
 − α =
Vdm

Key points of this section
Taking into account the obtained rectifier parameters we conclude:
1. As in three-phase half wave converter the output rectified voltages depends
on the firing angle α , the form factor increases dramatically at large values
of α .
2. The frequency of the output voltage is six times the supply frequency.
3. There is no dc current component is the secondary voltage.
4. The three-phase full wave converter is commonly used in practical
applications.
5. It can operate in two quadrants provided the load is highly inductive and
maintains continuous current.
10.7
In many variable-speed drives, the four-quadrant operation is generally
required and three-phase dual converters are extensively used in applications
up to 2000-kW level.
Figure 10.11 shows three-phase dual converters where two three-phase
Three-Phase
converters are connected back to back. From this figure due to instantaneous
Dual Converter voltage difference between the output voltages of converters, a circulating
current flows through the converter. This current is normally limited
connecting a circulating converter Lr, as shown in this figure. The practical
circuit containing so called " Alpha controllers" is shown on fig.10.12-a, and
the ciruit voltage and current are displayed on fig.10.12-b.
Fig.10.11: Three-phase Dual converter .
Fig.10.12: Practical Circuit of Three-Phase Dual Converter .
By varying the pulse firing instants and pulse numbers per one period (cycle) the output
voltage varies with respect to the magnitude and frequency. The output voltage for frequency
of 4 Hz is shown on fig.10.12-a.
Fig.3.11-b: the circuit waveforms.
The rectifier mathematical equations are going to be described by discussing the
following example :
10.8
Summary
Taking into account the described rectifiers modifications and circuits we an
conclude :
1. Three- phase dual converter is used for high-power applications up to
2000kW.
2. For a highly inductive load, the dual converter can operate in four quadrants.
3. A dc inductor is needed to reduce the circulating current .
4. frequency of 6 times ( fo=6.f1=300 Hz).
5. the existing of load inductance cause further control range limitation of the
rectified voltage
6. the overlapping angle decreases by increasing the firing angle
7. the voltage reduction due to the load inductance depends only on the
inductance, frequency, load current, and independent of the firing angle .
10.9
Questions
Refer to the Textbook " Ch10", and answer the following :
1. Answer the Review Questions .
2. Solve the problems – 10.1, 10.2, 10.3, 10.5, 10.7, 10.8, 10.9,
10.10, 10.11, 10.13, 10.15, 10.16, 10.18, 10.25, 10.27.
10.10
Project
1. Design and implement a single-phase circuit of ;
Full-wave rectifier with 5A dc current, 48 V dc voltage and the
current ripples factor must less than 5%, for various angles .
2. Design and implement a three-phase circuit of ;
Full-wave controlled rectifier with 10A dc current, 480 V dc
voltage and the current ripples factor must less than 5%.
Useful Links:
Advanced Power
Technology
ABB Semiconductors
www.advancedpower.com
Dynex Semi converter
www.dynexsemi.com
Hitachi, Ltg, Power
Devices
Motorola, Inc.
www.hitachi.co.jp/pdynexsemi.jp/pse
National Semiconductors,
Inc.
Semikron International
www.national.com/
Tokin, Inc.
www.tokin.com/
Nihon International
Electronics Corp.
Unitrode Integrated
Circuits Corp.
Westcode semiconductors
Ltd.
www.abbsem.com/english/salesb.htm
www.abbsem.com
www.motorola.com
www.semikron.com/
www.unitrode.com
www.westcode.com/ws -prod.html