EE 435- Electric Drives
Dr. Ali M. Eltamaly
Chapter 3
Thyristor Converters or Controlled Converters
3.1 Introduction
The controlled rectifier circuit is divided into three main circuits:(1) Power Circuit
This is the circuit contains voltage source, load and switches as diodes, thyristors or
IGBTs.
(2) Control Circuit
This circuit is the circuit, which contains the logic of the firing of switches that may,
contains amplifiers, logic gates and sensors.
(3) Triggering circuit
This circuit lies between the control circuit and power thyristors. Sometimes this circuit
called switch drivers circuit. This circuit contains buffers, opt coupler or pulse
transformers. The main purpose of this circuit is to separate between the power circuit and
control circuit.
The method of switching off the thyristor is known as Thyristor commutation. The thyristor
can be turned off by reducing its forward current below its holding current or by applying a
reverse voltage across it. The commutation of thyristor is classified into two types:1- Natural Commutation
If the input voltage is AC, the thyristor current passes through a natural zero, and a reverse
voltage appear across the thyristor, which in turn automatically turned off the device due to the
natural behavior of AC voltage source. This is known as natural commutation or line
commutation. This type of commutation is applied in AC voltage controller rectifiers and
cycloconverters.
2- Forced Commutation
In DC thyristor circuits, if the input voltage is DC, the forward current of the thyristor is forced to
zero by an additional circuit called commutation circuit to turn off the thyristor. This technique is
called forced commutation. Normally this method for turning off the thyristor is applied in
choppers.
There are many thyristor circuits we can not present all of them. In the following items we are
going to present and analyze the most famous thyristor circuits.
3.2 Half Wave Single Phase Controlled Rectifier
3.2.1 Half Wave Single Phase Controlled Rectifier With Resistive Load
The circuit with single SCR is similar to the single diode circuit, the difference being that an SCR
is used in place of the diode. Most of the power electronic applications operate at a relative high
voltage and in such cases; the voltage drop across the SCR tends to be small. It is quite often
justifiable to assume that the conduction drop across
the SCR is zero when the circuit is analyzed. It is also
justifiable to assume that the current through the SCR
is zero when it is not conducting. It is known that the
SCR can block conduction in either direction. The
explanation and the analysis presented below are
based on the ideal SCR model. All simulation carried
out by using PSIM computer simulation program.
A circuit with a single SCR and resistive load is shown
in Fig.3.1. The source vs is an alternating sinusoidal Fig.3.1 Half wave single phase controlled rectifier.
30
Chapter Three
source. If vs = Vm sin (ωt ) , vs is positive when 0 < ω t < π, and vs is negative when π < ω t <2π. When vs
starts become positive, the SCR is forward-biased but remains in the blocking state till it is triggered. If
the SCR is triggered at ω t = α, then α is called the firing angle. When the SCR is triggered in the
forward-bias state, it starts conducting and the positive source keeps the SCR in conduction till ω t
reaches π radians. At that instant, the current through the circuit is zero. After that the current tends to
flow in the reverse direction and the SCR blocks conduction. The entire applied voltage now appears
across the SCR. Various voltages and currents waveforms of the half-wave controlled rectifier with
resistive load are shown in Fig.3.2 for α=40o. FFT components for load voltage and current of half wave
single phase controlled rectifier with resistive load at α=40o are shown in Fig.3.3. It is clear from Fig.3.3
that the supply current containes DC component and all other harmonic components which makes the
supply current highly distorted. For this reason, this converter does not have acceptable practical
applecations.
Fig.3.2 Various voltages and currents waveforms for half wave single-phase controlled rectifier with
resistive load at α=40o.
Fig.3.3 FFT components for load voltage and current of half wave single phase controlled rectifier at α =40o.
The average voltage, Vdc , across the resistive load can be obtained by considering the waveform
π
shown in Fig.3.2.
Vdc
V
V
1
Vm sin(ωt ) dωt = m (− cos π + cos(α )) = m (1 + cos α )
=
2π
2π
2π
∫
α
(3.1)
SCR Rectifier or Controlled Rectifier
31
The maximum output voltage and can be acheaved when α = 0 which is the same as diode
(3.2)
case which obtained before in (2.12). Vdm = Vm / π
The normalized output voltage is the DC voltage devideded by maximum DC voltage, Vdm which
can be obtained as shown in equation (3.3).
Vn = Vdc / Vm = 0.5 (1 + cos α )
(3.3)
The rms value of the output voltage is shown in the following equation:π
Vrms
1
=
(Vm sin(ω t ))2 dω t = Vm 1  π − α + sin(2 α ) 
∫
2π α
2 π
2

The rms value of the transformer secondery current and load is: I s = Vrms / R
(3.3)
(3.4)
Example 1 In the rectifier shown in Fig.3.1 it has a load of R=15 Ω and, Vs=220 sin 314 t and
unity transformer ratio. If it is required to obtain an average output voltage of 70% of the
maximum possible output voltage, calculate:- (a) The firing angle, α, (b) The efficiency, (c)
Ripple factor (d) Peak inverse voltage (PIV) of the thyristor
Solution: (a) Vdm is the maximum output voltage and can be acheaved when α = 0 , The normalized
output voltage is shown in equation (3.3) which is required to be 70%. Then,
Vdc
= 0.5 (1 + cos α ) = 0.7 . Then, α=66.42o =1.15925 rad.
Vdm
V
V
49.02
(b) Vm = 220 V , Vdc = 0.7 * Vdm = 0.7 * m = 49.02 V ,
= 3.268 A
I dc = dc =
π
R
15
Vn =
Vm 1 
sin(2 α ) 
o
π − α +
 at α=66.42 , Vrms=95.1217V. Then, Irms=95.122/15=6.34145A
2 π
2

V
VS = m = 155.56 V , I S = I rms = 6.34145 A
2
P
V *I
49.02 * 3.268
Then, the rectification efficiency is:η = dc = dc dc =
= 26.56%
Pac Vrms * I rms 95.121 * 6.34145
V
95.121
π
=
= 1.94
(b) FF = rms =
49.02 2 2
Vdc
V
(c) RF = ac = FF 2 − 1 = 1.94 2 − 1 = 1.6624 , (d) The PIV is Vm
Vdc
Vrms =
3.2.2 Half Wave Single Phase Controlled Rectifier With RL Load
A circuit with single SCR and RL load is
shown in Fig.3.4. The source vs is an
alternating sinusoidal source. If vs = Vm
sin ( ω t), vs is positive when 0 < ω t < π,
and vs is negative when π < ω t <2π.
When vs starts become positive, the SCR
is forward-biased but remains in the
blocking state till it is triggered. If the
SCR is triggered when ω t = α, then it
starts conducting and the positive source
keeps the SCR in conduction till ω t
Fig.3.4 Half wave single phase controlled rectifier with RL load.
Chapter Three
32
reaches π radians. At that instant, the current through the circuit is not zero and there is some
energy stored in the inductor at ω t = π radians. The voltage across the inductor is positive when
the current through it is increasing and it becomes negative when the current through the inductor
tends to fall. When the voltage across the inductor is negative, it is in such a direction as to
forward bias the SCR. There is current through the load at the instant ω t = π radians and the
SCR continues to conduct till the energy stored in the inductor becomes zero. After that the
current tends to flow in the reverse direction and the SCR blocks conduction. Fig.3.5 shows the
output voltage, resistor, inductor voltages and thyristor voltage drop waveforms.
Fig.3.5 Various voltages and currents waveforms for half wave single phase controlled rectifier with RL
load.
3.3 Single-Phase Full Wave Controlled Rectifier
3.3.1 Single-Phase Center Tap Controlled Rectifier With Resistive Load
Center tap controlled rectifier is shown
in Fig.3.8. When the upper half of the
transformer secondary is positive and
thyristor T1 is triggered, T1 will conduct
and the current flows through the load
from point a to point b. When the lower
half of the transformer secondary is
b
a
positive and thyristor T2 is triggered, T2
will conduct and the current flows
through the load from point a to point b.
So, each half of input wave a
unidirectional voltage (from a to b ) is
applied across the load. Various voltages
and currents waveforms for center tap
controlled rectifier with resistive load are
Fig.3.8 Center tap controlled rectifier with resistive load.
shown in Fig.3.9 and Fig.3.10.
33
SCR Rectifier or Controlled Rectifier
Fig.3.9 The output voltgae and thyristor T1 reverse voltage wavforms along with the supply voltage
wavform.
Fig.3.10 Load current and thyristors currents for Center tap controlled rectifier with resistive load.
The average voltage, Vdc, across the resistive load is given by:
Vdc =
1
π
V sin(ω t ) dω t
π∫ m
α
=
Vm
π
(− cos π − cos(α )) =
Vm
π
(1 + cos α )
(3.27)
Vdm is the maximum output voltage and can be acheaved when α=0 in the above equation. The
V
(3.28)
normalized output voltage is: Vn = dc = 0.5 (1 + cos α )
Vdm
34
Chapter Three
From the wavfrom of the output voltage shown in Fug.3.9 the rms output voltage can be obtained as
following: Vrms =
1
π
(V sin(ω t ) )
π∫ m
2
dω t =
α
Vm
2π
π −α +
sin( 2 α )
2
(3.29)
Example 4 The rectifier shown in Fig.3.8 has load of R=15 Ω and, Vs=220 sin 314 t and unity
transformer ratio. If it is required to obtain an average output voltage of 70 % of the maximum
possible output voltage, calculate:- (a) The delay angle α, (b) The efficiency, (c) The ripple factor
(d) The peak inverse voltage (PIV) of the thyristor.
Solution : (a) Vdm is the maximum output voltage and can be acheaved when α=0, the normalized output
voltage is shown in equation (3.28) which is required to be 70%. Then:
Vdc
= 0.5 (1 + cos α ) = 0.7 , then, α=66.42o
Vdm
2 Vm
(b) Vm = 220 , then, Vdc = 0.7 * Vdm = 0.7 *
= 98.04 V
Vn =
π
Vm
sin( 2 α )
π −α +
2
2π
at α=66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A
VS = Vm / 2 = 155.56 V , I S = I rms / 2 = 6.347 A
P
V *I
98.04 * 6.536
= 53.04%
Then, The rectification efficiency is:η = dc = dc dc =
Pac Vrms * I rms 134.638 * 8.976
V
V
134.638
(c) FF = rms =
= 1.3733 and, RF = ac = FF 2 − 1 = 1.37332 − 1 = 0.9413
Vdc
Vdc
98.04
(d) The PIV is 2 Vm
I dc =
Vdc 98.04
=
= 6.536 A ,
R
15
Vrms =
3.3.2 Single-Phase Fully Controlled Rectifier Bridge With Resistive Load
This section describes the operation of a single-phase fully-controlled bridge rectifier circuit with
resistive load. The operation of this circuit can be understood more easily when the load is pure
resistance. The main purpose of the fully controlled bridge rectifier circuit is to provide a variable
DC voltage from an AC source.
The circuit of a single-phase fully controlled bridge rectifier circuit is shown in Fig.3.11. The
circuit has four SCRs. For this circuit, vs is a sinusoidal voltage source. When the supply voltage
is positive, SCRs T1 and T2 triggered then current flows from vs through SCR T1, load resistor R
(from up to down), SCR T2
and back into the source. In
the next half-cycle, the other
pair of SCRs T3 and T4
conducts when get pulse on
their gates. Then current flows
from vs through SCR T3, load
resistor R (from up to down),
SCR T4 and back into the
source. Even though the
direction of current through
the source alternates from one
Fig.3.11 Single-phase fully controlled rectifier bridge with resistive load.
35
SCR Rectifier or Controlled Rectifier
half-cycle to the other half-cycle, the current through the load remains unidirectional (from up to
down).
Fig.3.12 Various voltages and currents waveforms for converter shown in Fig.3.11 with resistive load.
Fig.3.13 FFT components of the output voltage and supply current for converter shown in Fig.3.11.
The main purpose of this circuit is to provide a controllable DC output voltage, which is brought about
by varying the firing angle, α. Let vs = Vm sin ω t, with 0 < ω t < 360o. If ω t = 30o when T1 and T2 are
triggered, then the firing angle α is said to be 30o. In this instance, the other pair is triggered when ω t =
30+180=210o. When vs changes from positive to negative value, the current through the load becomes
zero at the instant ω t = π radians, since the load is purely resistive. After that there is no current flow till
the other is triggered. The conduction through the load is discontinuous. The average value of the output
voltage is obtained as follows.:-
Let the supply voltage be vs = Vm*Sin ( ω t), where ω t varies from 0 to 2π radians. Since the
output waveform repeats itself every half-cycle, the average output voltage is expressed as a
function of α, as shown in equation (3.27).
Vdc =
1
π
V sin(ω t ) dω t
π∫ m
α
=
Vm
π
[− cos π − (− cos(α ) )] = Vm (1 + cos α )
π
Vdm is the maximum output voltage and can be acheaved when α=0,
The normalized output voltage is: Vn = Vdc / Vdm = 0.5 (1 + cos α )
The rms value of output voltage is obtained as shown in equation (3.29).
(3.27)
(3.28)
36
Chapter Three
Vrms =
1
π
(Vm sin(ω t ) )
π∫
α
2
dω t =
Vm
2π
π −α +
sin(2 α )
2
(3.29)
Example 5 The rectifier shown in Fig.3.11 has load of R=15 Ω and, Vs=220 sin 314 t and unity
transformer ratio. If it is required to obtain an average output voltage of 70% of the maximum
possible output voltage, calculate:- (a) The delay angle α, (b) The efficiency, (c) Ripple factor of
output voltage(d) The peak inverse voltage (PIV) of one thyristor.
Solution: (a) Vdm is the maximum output voltage and can be acheaved when α=0,
The normalized output voltage is shown in equation (3.31) which is required to be 70%.
V
Then, Vn = dc = 0.5 (1 + cos α ) = 0.7 , then, α=66.42o
Vdm
2 Vm
V
98.04
(b) Vm = 220 , then, Vdc = 0.7 * Vdm = 0.7 *
= 98.04 V , I dc = dc =
= 6.536 A
R
15
π
Vm
sin( 2 α )
π −α +
2
2π
At α=66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A
V
VS = m = 155.56 V
2
The rms value of the transformer secondery current is: I S = I rms = 8.976 A
P
V *I
98.04 * 6.536
= 53.04%
Then, The rectification efficiency is η = dc = dc dc = =
Pac Vrms * I rms
134.638 * 8.976
V
V
134.638
(c) FF = rms =
= 1.3733 , RF = ac = FF 2 − 1 = 1.37332 − 1 = 0.9413
Vdc
98.04
Vdc
(d) The PIV is Vm
Vrms =
3.3.3 Full Wave Fully Controlled Rectifier With RL Load In Continuous Conduction Mode
The circuit of a single-phase fully controlled bridge rectifier circuit is shown in Fig.3.14. The
main purpose of this circuit is to provide a variable DC output voltage, which is brought about by
varying the firing angle. The circuit has four SCRs. For this circuit, vs is a sinusoidal voltage
source. When it is positive, the SCRs T1 and T2 triggered then current flows from +ve point of
voltage source, vs through SCR T1, load inductor L, load resistor R (from up to down), SCR T2
and back into the –ve point of voltage source.
In the next half-cycle the current flows from ve point of voltage source, vs through SCR
T3, load resistor R, load inductor L (from up
to down), SCR T4 and back into the +ve
point of voltage source. Even though the
direction of current through the source
alternates from one half-cycle to the other
half-cycle, the current through the load
remains unidirectional (from up to down).
Fig.3.15 shows various voltages and currents
waveforms for the converter shown in
Fig.3.14. Fig.3.16 shows the FFT components
Fig.3.14 Full wave fully controlled rectifier with RL load.
of load voltage and supply current.
SCR Rectifier or Controlled Rectifier
37
Fig.3.15 Various voltages and currents waveforms for the converter shown in Fig.3.14 in continuous
conduction mode.
Fig.3.16 FFT components of load voltage and supply current in continuous conduction mode.
Let vs = Vm sin ω t, with 0 < ω t < 360o. If ω t = 30o when T1 and T2 are triggered, then the
firing angle is said to be 30o. In this instance the other pair is triggered when ω t= 210o. When vs
changes from a positive to a negative value, the current through the load does not fall to zero
value at the instant ω t = π radians, since the load contains an inductor and the SCRs continue to
conduct, with the inductor acting as a source. When the current through an inductor is falling, the
voltage across it changes sign compared with the sign that occurs when its current is rising. When
the current through the inductor is falling, its voltage is such that the inductor delivers power to
the load resistor, feeds back some power to the AC source under certain conditions and keeps the
SCRs in conduction forward-biased. If the firing angle is less than the load angle, the energy
stored in the inductor is sufficient to maintain conduction till the next pair of SCRs is triggered.
When the firing angle is greater than the load angle, the current through the load becomes zero
and the conduction through the load becomes discontinuous. Usually the description of this
circuit is based on the assumption that the load inductance is sufficiently large to keep the load
current continuous and ripple-free.
38
Chapter Three
Since the output waveform repeats itself every half-cycle, the average output voltage is
expressed in equation (3.33) as a function of α, the firing angle. The maximum average output
voltage occurs at a firing angle of 0o as shown in equation (3.34). The rms value of output voltage
is obtained as shown in equation (3.35).
Vdc =
1
π +α
∫Vm sin(ωt ) dωt
π
α
=
2Vm
π
cos α
(3.33)
The normalized output voltage is Vn = Vdc / Vdm = cos α
The rms value of output voltage is obtained as shown in equation (3.35).
Vrms =
1
π
π +α
2
∫ (Vm sin(ω t )) dω t =
α
Vm
2π
π +α
∫ (1 − cos(2 ω
α
t ) dω t =
(3.34)
Vm
2
(3.35)
Example 6 The rectifier shown in Fig.3.14 has pure DC load current of 50 A and, Vs=220 sin 314
t and unity transformer ratio. If it is required to obtain an average output voltage of 70% of the
maximum possible output voltage, calculate:- (a) The delay angle α, (b) The efficiency, (c)
Ripple factor (d) The peak inverse voltage (PIV) of the thyristor and (e) Input displacement
factor.
Solution: (a) Vdm is the maximum output voltage and can be acheaved when α=0. The normalized
V
output voltage is shown in equation (3.30) which is required to be 70%. Then, Vn = dc = cos α = 0.7 ,
o
Vdm
then, α=45.5731 = 0.7954
(b) Vm = 220 , Vdc = 0.7 *Vdm = 0.7 * 2 Vm / π = 98.04 V , Vrms = Vm / 2
At α=45.5731o Vrms=155.563 V. Then, Irms=50 A, VS = Vm / 2 = 155.56 V
The rms value of the transformer secondery current is: I S = I rms = 50 A
P
V *I
98.04 * 50
Then, The rectification efficiency is η = dc = dc dc =
= 63.02%
155.563 * 50
Pac Vrms * I rms
V
V
155.563
(c) FF = rms =
= 1.587 , RF = ac = FF 2 − 1 = 1.37332 − 1 = 1.23195
98.04
Vdc
Vdc
(d) The PIV is Vm
3.3.5 Single Phase Full Wave Fully Controlled Rectifier With Source Inductance:
Full wave fully controlled rectifier with source inductance is shown in Fig.3.19. The presence of
source inductance changes the way the circuit operates during commutation time. Let vs = Vm sin
wt, with 0 < ω t < 360o. Let the load inductance be large enough to maintain a steady current
through the load. Let firing angle α be 30o. Let SCRs T3 and T4 be in conduction before ω t <
30o. When T1 and T2 are triggered at ω t = 30o, there is current through the source inductance,
flowing in the direction opposite to that marked in the circuit diagram and hence commutation of
current from T3 and T4 to T1 and T2 would not occur instantaneously. The source current
changes from − I dc to I dc due to the whole of the source voltage being applied across the source
inductance. When T1 is triggered with T3 in conduction, the current through T1 would rise from
zero to I dc and the current through T3 would fall from I dc to zero. Similar process occurs with
the SCRs T2 and T4. During this period, the current through T2 would rise from zero to I dc and,
the current through T4 would fall from I dc to zero.
SCR Rectifier or Controlled Rectifier
39
Fig.3.19 single phase full wave fully controlled rectifier with source inductance
Various voltages and currents waveforms of converter shown in Fig.3.19 are shown in Fig.3.20
and Fig.3.21. You can observe how the currents through the devices and the line current change
during commutation overlap.
Fig.3.20 Output voltage, thyristors current along with supply voltage waveform of a single phase full
wave fully controlled rectifier with source inductance.
Fig.3.21 Output voltage, supply current along with supply voltage waveform of a single phase full wave
fully controlled rectifier with source inductance.
40
Chapter Three

2ωLs I o 
u = cos −1 cos(α ) −
−α
(3.42)
Vm 

4ω Ls I o
Vrd =
= 4 fLs I o
(3.47)
2π
The DC voltage with source inductance taking into account can be calculated as following:
2V
Vdc actual = Vdc without sourceinduc tan ce − Vrd = m cos α − 4 fLs I o
(3.48)
π
The rms value of supply current is the same as obtained before in single phase full bridge
diode rectifier in (2.64) I s =
2 I o2  π u 
−
π  2 3 
(3.49)
The Fourier transform of supply current is the same as obtained for single phase full bridge
diode rectifier in (2.66) and the fundamental component of supply current I s1 is shown in (2.68)
8I o
u
(3.50)
as following: I S1 =
* sin
2
2 πu
The power factor of this rectifier is shown in the following: p. f =
I s1
u

cos α + 
2
Is

(3.51)
3.3.6 Inverter Mode Of Operation
The thyristor converters can also operate in an inverter mode, where Vd has a negative value, and
hence the power flows from the do side to the ac side.
The easiest way to understandd the inverter mode of
operation is to assume that the DC side of the
converter can be replaced by a current source of a
constant amplitude I d , as shown in Fig.3.25. For a
delay angle a greater than 90° but less than 180°, the
voltage and current waveforms are shown in Fig.3.26.
The average value of vd is negative, given by (3.48),
where 90° < α < 180°. Therefore, the average power
Pd = Vd * I d is negative, that is, it flows from the DC
to the AC side. On the AC side, Pac = Vs I S1 cos φ1 is
also negative because φ > 90 o .
Fig.3.25 Single phase SCR inverter.
Fig.3.26 Waveform output from single phase inverter assuming DC load current.
SCR Rectifier or Controlled Rectifier
41
There are several points worth noting here. This inverter mode of operation is possible since
there is a source of energy on the DC side. On the ac side, the ac voltage source facilitates the
commutation of current from one pair of thyristors to another. The power flows into this AC
source.
Generally, the DC current source is not a realistic DC side representation of systems where
such a mode of operation may be encountered. Fig.3.27 shows a voltage source Ed on the DC
side that may represent a battery, a photovoltaic source, or a DC voltage produced by a
wind-electric system. It may also be encountered in a four-quadrant DC motor supplied by a
back-to-back connected thyristor converter.
An assumption of a very large value of Ld allows us to assume id to be a constant DC, and
hence the waveforms of Fig.3.28 also apply to the circuit of Fig.3.27. Since the average voltage
across Ld is zero,
2
E d = Vd = Vdo cos α − ω Ls I d
(3.55)
π
Fig.3.27 SCR inverter with a DC voltage source.
Fig.3.28 Vd versus I d in SCR inverter with a DC voltage source.
42
Chapter Three
Fig.3.29 Voltage across a thyristor in the inverter mode.
Inverter startup
For startup of the inverter in Fig.3.25, the delay angle α is initially made sufficiently large
(e.g.,165o) so that id is discontinuous as shown in Fig.3.30. Then, α is decreased by the
controller such that the desired I d and Pd are obtained.
Fig.3.30 Waveforms of single phase SCR inverter at startup.
3.5 Three Phase Half Wave Controlled Rectifier
3.5.1 Three Phase Half Wave Controlled Rectifier with Resistive Load
Fig.3.31 shows the circuit of a three-phase half wave controlled rectifier, the control circuit of
this rectifier has to ensure that the three gate
pulses for three thyristor are displaced 120o
relative to each other’s. Each thyristor will
conduct for 120o. A thyristor can be fired to
conduct when its anode voltage is positive with
respect to its cathode voltage. The maximum
output voltage occurred when α=0 which is the
same as diode case. This rectfier has
continuous load current and voltage in case of
α ≤ 30. However, the load voltage and current
will be discontinuous in case of α > 30.
Fig.3.31 Three phase half wave controlled rectifier.
In case of α ≤ 30, various voltages and
currents of the converter shown in Fig.3.31 are shown in Fig.3.32. Fig.3.33 shows FFT
components of load voltage, secondary current and primary current. As we see the load voltage
contains high third harmonics and all other triplex harmonics. Also secondary current contains
DC component, which saturate the transformer core. The saturation of the transformer core is the
main drawback of this system. Also the primary current is highly distorted but without a DC
component. The average output voltage and current are shown in equation (3.57) and (3.58)
respectively. The rms output voltage and current are shown in equation (3.59) and (3.60)
respectively.
3
2π
Vdc =
I dc =
Vrms
5π / 6 +α
3
3 3 Vm
cos α = 0.827Vm cos α =
VLL cos α = 0.675VLL cosα (3.57)
2π
2π
∫ Vm sin ω t dω t =
π / 6 +α
3 3 Vm
0.827 * Vm
cos α =
cos α
2 *π *R
R
3
=
2π
5π / 6 +α
∫ (Vm sin ω t )
π / 6 +α
2
dω t = 3 Vm
(3.58)
1
3
+
cos 2α
6 8π
(3.59)
SCR Rectifier or Controlled Rectifier
43
3 Vm 1
3
+
cos 2α
(3.60)
R
6 8π
Then the thyristor rms current is equal to secondery current and can be obtaiend as follows:
I rms =
Ir = IS =
I rms
3
=
Vm
R
1
3
cos 2α
+
6 8π
The PIV of the diodes is
2 VLL = 3 Vm
(3.61)
(3.62)
Fig.3.32 Voltages and currents waveforms for rectifier shown in Fig.3.31 at α ≤ 30.
Fig.3.33 FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.31
for α ≤ 30.
44
Chapter Three
In case of α > 30, various voltages and currents of the rectifier shown in Fig.3.31 are shown in
Fig.3.34. Fig.3.35 shows FFT components of load voltage, secondary current and primary
current. As we can see the load voltage and current equal zero in some regions (i.e. discontinuous
load current). The average output voltage and current are shown in equation (3.63) and (3.64)
respectively. The rms output voltage and current are shown in equation (3.65) and (3.66)
respectively.
The average output voltage is :3
Vdc =
2π
I dc =
π
∫ Vm sin ω t dωt =
π / 6+α
3 Vm
2π


π

π

1 + cos  6 + α  = 0.4775Vm 1 + cos  6 + α 






3 Vm 
π

1 + cos  + α 

2π R 
6

π
Vrms =
3
(Vm sin ω t )2 dω t = 3 Vm 5 − α + 1 sin(π / 3 + 2α )
∫
24 4π 8 π
2π π / 6 +α
3 Vm 5 α
1
−
+
sin(π / 3 + 2α )
R
24 4π 8 π
Then the diode rms current can be obtaiend as follows:
I
V
5 α
1
−
+
sin(π / 3 + 2α )
I r = I S = rms = m
R 24 4π 8 π
3
I rms =
The PIV of the diodes is
2 VLL = 3 Vm
(3.63)
(3.64)
(3.65)
(3.66)
(3.67)
(3.68)
SCR Rectifier or Controlled Rectifier
45
Fig.3.34 Various voltages and currents waveforms for converter shown in Fig.3.22 for α > 30.
Fig.3.35 FFT components of load voltage, secondary current and supply current for the converter shown
in Fig.3.22 for α > 30.
Example 7 Three-phase half-wave controlled rectfier is connected to 380 V three phase supply
via delta-way 380/460V transformer. The load of the rectfier is pure resistance of 5 Ω . The delay
angle α = 25o . Calculate: The rectfication effeciency (b) PIV of thyristors
Solution:
From (3.57) the DC value of the output voltage can be obtained as following:
V
3
3
281.5
Vdc =
VLL cos α =
460 cos 25 = 281.5V Then; I dc = dc =
= 56.3 A
R
5
2π
2π
From (3.59) we can calculate Vrms as following:
Vrms = 3 Vm
1
3
1
3
+
cos 2α = 2 VLL *
+
cos 2α
6 8π
6 8π
46
Chapter Three
Then, Vrms = 2 * 460 *
Then I rms =
1
3
+
cos (2 * 25) = 298.8 V
6 8π
Vrms 298.8
=
= 59.76 A
R
5
Then, the rectfication effeciency is, η =
Vdc I dc
*100 = 88.75%
Vrms I rms
PIV = 2 VLL = 2 * 460 = 650.54 V
Example 8 Solve the previous example (evample 7) if the firing angle α = 60 o
Slution: From (3.63) the DC value of the output voltage can be obtained as following:
 2
 * 460
3 
3 
3 Vm 

 π π 
π


Vdc =
1 + cos  + α  =
1 + cos  +  = 179.33 V


2π 
2π
 6 3 
6


V
179.33
= 35.87 A
Then; I dc = dc =
5
R
From (3.65) we can calculate Vrms as following:
Vrms = 3 Vm
5 α
1
5 π /3 1
−
+
−
+
sin(π / 3 + 2α ) = 2 * 460 *
sin(π / 3 + 2π / 3 ) = 230V
24 4π 8 π
24 4π 8 π
Vrms 230
=
= 46 A
5
R
Then, the rectfication effeciency can be calculated as following
V I
η = dc dc *100 = 60.79 % and PIV = 2 VLL = 2 * 460 = 650.54 V
Vrms I rms
Then I rms =
3.5 Three Phase Half Wave Controlled Rectifier With DC Load Current
The Three Phase Half Wave Controlled Rectifier With DC Load Current is shown in Fig.3.36,
the load voltage will reverse its direction only if α > 30. However if α < 30 the load voltage will
be positive all the time. Then in case of α > 30 the load voltage will be negative till the next
thyristor in the sequence gets triggering pulse. Also each thyristor will conduct for 120o if the
load current is continuous
as shown in Fig.3.37.
Fig.3.38 shows the FFT
components of load
voltage,
secondary
current
and
supply
current for the converter
shown in Fig.3.36 for α >
30 and pure DC current
load.
Fig.3.36 Three Phase Half Wave Controlled Rectifier With DC Load Current
47
t=0
SCR Rectifier or Controlled Rectifier
Fig.3.37 Various voltages and currents waveforms for the converter shown in Fig.3.36 for α > 30 and
pure DC current load.
Fig.3.38 FFT components of load voltage, secondary current and supply current for the converter shown
in Fig.3.36 for α > 30 and pure DC current load.
As explained before the secondary current of transformer contains DC component. Also the
source current is highly distorted which make this system has less practical significance. The
THD of the supply current can be obtained by the aid of Fourier analysis as shown in the
following:If we move y-axis of supply current to be as shown in Fig.3.33, then the waveform can be
represented as odd function. So, an=0 and bn can be obtained as the following:bn =
2
π
Then,
2π / 3
∫ I dc sin(nω t ) dω t =
0
bn =
2 I dc 3
*
πn 2
2 I dc 
2nπ 
1 − cos
 for n=1,2,3,4,…
3 
πn 
for n=1,2,4,5,7,8,10,…..
(3.69)
Chapter Three
48
And b n = 0
For n=3,6,9,12,…..
Then the source current waveform can be expressed as the following equation
(3.70)
i p (ωt ) =
3I dc 
1
1
1
1

sin ωt + sin 2ωt + sin 4ωt + sin 5ωt + sin 7ωt + ......

2
4
5
7
π 

(3.71)
The resultant waveform shown in equation (3.61) agrees with the result from simulation
(Fig.3.38). The THD of source current can be obtained by two different methods. The first
method is shown below:THD =
I 2p − I 2p1
Where, I p =
I 2p1
2
* I dc
3
(3.72)
The rms of the fundamental component of supply current can be obtained from equation (3.71)
and it will be as shown in equation (3.74)
3I
(3.74)
I p1 = dc
2π
Substitute equations (3.73) and (3.74) into equation (3.72), then,
2 2
9 2
I dc −
I dc
3
2π 2
= 68 %
(3.75)
THD =
9 2
I dc
2π 2
Another method to determine the THD of supply current is shown in the following:Substitute from equation (3.71) into equation (3.72) we get the following equation:2
2
2
2
2
2
2
2
2
1 1 1 1 1  1   1   1   1 
THD =   +   +   +   +   +   +   +   +   + .... ≅ 68 % (3.76)
 2   4   5   7   8   10   11   13   14 
The supply current THD is very high and it is not acceptable by any electric utility system. In
case of full wave three-phase converter, the THD in supply current becomes much better than
half wave (THD=35%) but still this value of THD is not acceptable.
Example 9 Three phase half wave controlled rectfier is connected to 380 V three phase supply
via delta-way 380/460V transformer. The load of the rectfier draws 100 A pure DC current. The
(b) Input power factor.
delay angle, α = 30 o . Calculate: (a) THD of primary current.
Solution: The voltage ratio of delta-way transformer is 380/460V. Then, the peak value of
460
2
= 121.05 A . Then, I P , rms = 121.05 *
= 98.84 A .
primary current is 100 *
3
380
3I
3 *121.05
I P1 can be obtained from equation (372) where I P1 = dc =
= 81.74 A .
2π
2π
2
2
 I P, rms 
98.84 

 − 1 *100 = 
Then, (THD )I P = 
 − 1 *100 = 67.98 %
81
.
74
I


 P1 
The input power factor can be calculated as following:
I
π  81.74

π π 
* cos +  = 0.414 Lagging
P. f = P1 * cos α +  =
6  98.84
I P, rms

6 6
SCR Rectifier or Controlled Rectifier
49
3.7 Three Phase Full Wave Fully Controlled Rectifier Bridge
3.7.1 Three Phase Full Wave Fully Controlled Rectifier With Resistive Load
Three-phase full wave controlled rectifier shown in Fig.3.42. As we can see in this figure the
thyristors has labels T1, T2,……,T6. The label of each thyristor is chosen to be identical to
triggering sequence where thyristors are triggered in the sequence of T1, T2,……,T6 which is
clear from the thyristors currents shown in Fig.3.43.
Fig.3.42 Three-phase full wave controlled rectifier.
Fig.3.43 Thyristors currents of three-phase full wave controlled rectifier.
The operation of the circuit explained here depending on the understanding of the reader the
three phase diode bridge rectifier. The Three-phase voltages vary with time as shown in the
following equations:
va = Vm sin (ω t )
vb = Vm sin (ω t − 120)
vc = Vm sin (ω t + 120)
It can be seen from Fig.3.44 that the voltage va is the highest positive voltage of the three
phase voltage when ωt is in the range 30 < ω t < 150o . So, the thyristor T1 is forward bias during
this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In
Fig.3.44 the firing angle α = 40 as an example. So, T1 takes a pulse at
ω t = 30 + α = 30 + 40 = 70 o as shown in Fig.3.44. Also, it is clear from Fig3.38 that thyristor T1
or any other thyristor remains on for 120o .
Chapter Three
50
Fig.3.44 Phase voltages and thyristors currents of three-phase full wave controlled rectifier at α = 40 o .
It can be seen from Fig.3.44 that the voltage vb is the highest positive voltage of the three
phase voltage when ωt is in the range of 150 < ωt < 270 o . So, the thyristor T3 is forward bias
during this period and it is ready to conduct at any instant in this period if it gets a pulse on its
gate. In Fig.3.44, the firing angle α = 40 as an example. So, T3 takes a pulse at
ω t = 150 + α = 150 + 40 = 190o .
It can be seen from Fig.3.44 that the voltage vc is the highest positive voltage of the three
phase voltage when ωt is in the range 270 < ω t < 390 o . So, the thyristor T5 is forward bias
during this period and it is ready to conduct at any instant in this period if it gets a pulse on its
gate. In Fig.3.44, the firing angle α = 40 as an example. So, T3 takes a pulse at
ωt = 270 + α = 310o .
It can be seen from Fig.3.44 that the voltage va is the highest negative voltage of the three
phase voltage when ω t is in the range 210 < ω t < 330 o . So, the thyristor T4 is forward bias
during this period and it is ready to conduct at any instant in this period if it gets a pulse on its
gate. In Fig.338, the firing angle α = 40 as an example. So, T4 takes a pulse at
ω t = 210 + α = 210 + 40 = 250o .
It can be seen from Fig.3.44 that the voltage vb is the highest negative voltage of the three
phase voltage when ω t is in the range 330 < ω t < 450 o or 330 < ω t < 90 o in the next period of
supply voltage waveform. So, the thyristor T6 is forward bias during this period and it is ready to
conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle
α = 40 as an example. So, T6 takes a pulse at ω t = 330 + α = 370o .
It can be seen from Fig.3.44 that the voltage vc is the highest negative voltage of the three
phase voltage when ω t is in the range 90 < ωt < 210o . So, the thyristor T2 is forward bias during
this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In
Fig.3.44, the firing angle α = 40 as an example. So, T2 takes a pulse at ωt = 90 + α = 130 o .
51
SCR Rectifier or Controlled Rectifier
From the above explanation we can conclude that there is two thyristor in conduction at any
time during the period of supply voltage. It is also clear that the two thyristors in conduction one
in the upper half (T1, T3, or, T5) which become forward bias at highest positive voltage
connected to its anode and another one in the lower half (T2, T4, or, T6) which become forward
bias at highest negative voltage connected to its cathode. So the load is connected at any time
between the highest positive phase voltage and the highest negative phase voltage. So, the load
voltage equal the highest line to line voltage at any time which is clear from Fig.3.45. The
following table summarizes the above explanation.
Period, range of wt
SCR Pair in conduction
α + 30o to α + 90o
T1 and T6
α + 90o to α + 150o
T1 and T2
α + 150o to α + 210o
o
o
o
o
o
o
α + 210 to α + 270
α + 270 to α + 330
T2 and T3
T3 and T4
T4 and T5
o
α + 330 to α + 360 and α + 0 to α + 30
o
T5 and T6
o
Fig.3.45 Output voltage along with three phase line to line voltages of rectifier in Fig.3.42 at α = 40 .
The line current waveform is very easy to obtain it by applying kerchief's current law at the
terminals of any phase. As an example I a = I T 1 − I T 4 which is clear from Fig.3.42. The input
current of this rectifier for α = 40, (α ≤ 60 ) is shown in Fig.3.46. Fast Fourier transform (FFT)
of output voltage and supply current are shown in Fig.3.47.
Analysis of this three-phase controlled rectifier is in many ways similar to the analysis of
single-phase bridge controlled rectifier circuit. The average output voltage, the rms output
voltage, the ripple content in output voltage, the total rms line current, the fundamental rms
current, THD in line current, the displacement power factor and the apparent power factor are to
be determined. In this section, the analysis is carried out assuming that the load is pure resistance.
Vdc =
3
π / 2 +α
π π / 6∫+α
π
3 Vm sin(ω t + ) dω t =
6
3 3 Vm
π
cos α
The maximum average output voltage for delay angle α=0 is
(3.81)
52
Chapter Three
Vdm =
3 3 Vm
π
The normalized average output voltage is as shown in (3.83)
Vn =
Vdc
= cos α
Vdm
(3.82)
(3.83)
The rms value of the output voltage is found from the following equation:
Vrms =
π / 2 +α
2
π 

3 Vm sin(ω t + )  dω t = 3 Vm
π
6 

π / 6 +α
3
∫
1 3 3

 +
cos 2α 
2
4π


(3.84)
Fig.3.46 The input current of this rectifier of rectifier in Fig.3.42 at α = 40, (α ≤ 60) .
Fig.3.46 FFT components of output voltage and supply current of rectifier in Fig.3.42 at
α = 40, (α ≤ 60) .
53
SCR Rectifier or Controlled Rectifier
In the converter shown in Fig.3.42 the output voltage will be continuous only and only if
α ≤ 60 o . If α > 60o the output voltage, and phase current will be as shown in Fig.3.47.
Fig.3.47 Output voltage along with three phase line to line voltages of rectifier in Fig.3.42 at α = 75o .
The average and rms values of output voltage is shown in the following equation:
Vdc =
3
π
5π / 6
∫
3 Vm sin(ω t +
π / 6 +α
π
6
) dω t =
3 3 Vm
π
[1 + cos ( π / 3 + α )]
The maximum average output voltage for delay angle α=0 is: Vdm =
(3.85)
3 3 Vm
π
Vdc
= [1 + cos ( π / 3 + α )]
Vdm
The rms value of the output voltage is found from the following equation:
The normalized average output voltage is
Vrms
5π / 6
Vn =
(3.87)
2
3 
π 
π 


3 Vm sin(ω t + )  dω t = 3Vm 1 −
=
 2α − cos 2α +  
∫
6 
4π 
6 
π π / 6 +α 

3
(3.86)
(3.88)
Example 10 Three-phase full-wave controlled rectifier is connected to 380 V, 50 Hz supply to
feed a load of 10 Ω pure resistance. If it is required to get 400 V DC output voltage, calculate the
following: (a) The firing angle, α (b) The rectfication effeciency (c) PIV of the thyristors.
Solution: From (3.81) the average voltage is :
3 3 Vm
3 3
2
Vdc =
cos α =
*
* 380 cos α = 400V .
π
R
3
V
400
Then α = 38.79 o , I dc = dc =
= 40 A
R
10
From (3.84) the rms value of the output voltage is:
Vrms = 3 Vm

1 3 3
2
 +
cos 2 α  = 3 *
* 380 *
2
4π
3


Vrms 412.412
=
= 41.24 A
R
10
400 * 40
*100 =
*100 = 94.07%
412.4 * 41.24
Then, Vrms = 412.412 V Then, I rms =
Then, η =
Vdc * I dc
Vrms * I rms

1 3 3
 +
cos (2 * 38.79 )
2
4π


The PIV= 3 Vm=537.4V
54
Chapter Three
Example 11 Solve the previous example if the required dc voltage is 150V.
Solution: From (3.81) the average voltage is :
2
3 3*
* 380
3 3 Vm
3
Vdc =
cos α =
cos α = 150V . Then, α = 73o
π
π
It is not acceptable result because the above equation valid only for α ≤ 60 . Then we have to use
the (3.85) to get Vdc as following:
3 3*
Vdc ==
π
2
* 380
3
[1 + cos ( π / 3 + α )] = 150V . Then, α = 75.05o
Vdc 150
=
= 15 A
R
10
From (3.88) the rms value of the output voltage is:
π 
3 
2

Vrms = 3Vm 1 −
* 380 *
 2α − cos 2α +   = 3 *
4π 
6 
3

Then I dc =

π
3 

1 −
− cos (2 * 75.05 + 30 ) 
 2 * 75.05 *
180

 4π 
Vrms 198.075
=
= 19.8075 A
R
10
150 *15
*100 =
*100 = 57.35 %
198.075 *19.81
Then, Vrms = 198.075 V , Then, I rms =
Then, η =
Vdc * I dc
Vrms * I rms
The PIV= 3 Vm=537.4V
3.7.1 Three Phase Full Wave Fully Controlled Rectifier With pure DC Load Current
Three-phase full wave-fully controlled rectifier with pure DC load current is shown in Fig.3.48.
Fig.3.49 shows various currents and voltage of the converter shown in Fig.3.48 when the delay
angle is less than 60o. As we see in Fig.3.49, the load voltage is only positive and there is no
negative period in the output waveform. Fig.3.50 shows FFT components of output voltage of
rectifier shown in Fig.3.48 for α < 60o .
Fig.3.48 Three phase full wave fully controlled rectifier with pure dc load current
In case of the firing angle is greater than 60 o , the output voltage contains negaive portion as
shown in Fig.3.51. Fig.3.52 shows FFT components of output voltage of rectifier shown in
Fig.3.48 for α > 60 o . The average and rms voltage is the same as in equations (3.81) and (3.84)
respectively. The line current of this rectifier is the same as line current of three-phase full-wave
diode bridge rectifier typically except the phase shift between the phase voltage and phase current
is zero in case of diode bridge but it is α in case of three-phase full-wave controlled rectifier
SCR Rectifier or Controlled Rectifier
55
with pure DC load current as shown in Fig.3.53. So, the input power factor of three-phase fullwave diode bridge rectifier with pure DC load current is:
I
(3.89)
PowerFactor = s1 cosα
Is
Fig.3.49 Output voltage and supply current waveforms along with three phase line voltages for the rectifier
shown in Fig.3.48 for α < 60o with pure DC current load.
Fig.3.50 FFT components of SCR, secondary, primary currents respectively of rectifier shown in
Fig.3.48.
Chapter Three
56
Fig.3.51 Output voltage and supply current waveforms along with three phase line voltages for the
rectifier shown in Fig.3.48 for α > 60o with pure DC current load.
Fig.3.52 FFT components of SCR, secondary, primary currents respectively of rectifier shown in Fig.3.48
for α > 60o.
Fig.3.53 Phase a voltage, current and fundamental components of phase a of three phase full bridge fully
controlled rectifier with pure DC current load and α > 60 .
SCR Rectifier or Controlled Rectifier
57
In case of three-phase full-wave controlled rectifier with pure DC load and source inductance
the waveform of output voltage and line current and their FFT components are shown in Fig.3.54
and Fig.3.55 respectively. The output voltage reduction due to the source inductance is the same
as obtained before in Three-phase diode bridge rectifier. But, the commutation time will differ
than the commutation time obtained in case of Three-phase diode bridge rectifier. It is left to the
reader to determine the commutation angle u in case of three-phase full-wave diode bridge
rectifier with pure DC load and source inductance. The Fourier transform of line current and
THD will be the same as obtained before in Three-phase diode bridge rectifier with pure DC load
and source inductance which explained in the previous chapter.
Fig.3.54 Output voltage and supply current of rectifier shown n Fig.3.48 with pure DC load and source
inductance the waveforms.
Fig.3.55 FFT components of output voltage of rectifier shown in Fig.3.48 for α > 60o and there is a
source inductance.

2ω LS I o 
u = cos −1 cos(α ) −
 −α
V
LL


Vrd = 6 fLI o
(3.99)
(3.104)
The DC voltage without source inductance tacking into account can be calculated as following:
58
Chapter Three
Vdc
actual
= Vdc
without sourceinduc tan ce
− Vrd =
3 2
π
VLL cosα − 6 fLs I o
2 I o2  π u 
−
π  3 6 
2 6 Io  u 
sin  
I S1 =
πu
2
The power factor can be calculated from the following equation:
Then I S =
pf =
I S1
u

cos  α +  =
2
IS

2 6 Io  u 
sin  
πu
2
u
2 3 * sin  
u

 2  cos α + u 
cos  α +  =


2
2


π u 
2 I o2  π u 
u π − 
−
 3 6
π  3 6 
(3.105)
(3.106)
(3.110)
(3.111)
Note, if we approximate the source current to be trapezoidal as shown in Fig.3.58n, the
u

displacement power factor will be as shown in (3.111) is cos α +  . Another expression for the
2

displacement power factor, by equating the AC side and DC side powers [ ] as shown in the
following derivation:
From (3.98) and (3.105) we can get the following equation:
3 2
3
Vdc =
VLL cos α −
VLL (cos α − cos(α + u ))
π
2π
3 2
3 2
VLL [2 cos α − (cos α − cos(α + u ))] ∴Vdc =
VLL [cos α + cos(α + u )]
2π
2π
Then the DC power output from the rectifier is Pdc = Vdc I o . Then,
∴Vdc =
(3.112)
3 2
VLL * I o [cos α + cos(α + u )]
(3.113)
2π
(3.114)
On the AC side, the AC power is: Pac = 3 VLL I S1 cos φ1
Substitute from (3.110) into (3.114) we get the following equation:
4 3 Io  u 
6 2 VLL I o  u 
sin   cos φ1 =
sin   cos φ1
(3.115)
Pac = 3 VLL
πu
πu 2
2
2
By equating (3.113) and (3.115) we get the following:
u [cos α + cos (α + u )]
cos φ1 =
(3.116)
u
4 * sin  
2
3.7.2 Inverter Mode of Operation
Once again, to understand the inverter mode of operation, we will assume that the do side of the
converter can be represented by a current source of a constant amplitude I d , as shown in
Fig.3.61. For a delay angle a greater than 90° but less than 180°, the voltage and current
waveforms are shown in Fig.3.62a. The average value of Vd is negative according to (3.81). On
the ac side, the negative power implies that the phase angle φ1 , between vs and is , is greater
than 90°, as shown in Fig.3.62b.
Pdc =
SCR Rectifier or Controlled Rectifier
59
Fig.3.61 Three phase SCR inverter with a DC current.
Fig.3.62 Waveforms in the inverter shown in Fig.3.56.
In a practical circuit shown in Fig.3.63, the operating point for a given Ed and α can be
obtained from the characteristics shown in Fig.3.64.
Similar to the discussion in connection with single-phase converters, the extinction angle
γ = 180o − α − u must be greater than the thyristor turn-off interval ω t q in the waveforms of
(
)
Fig.3.54, where v5 is the voltage across thyristor 5.
Fig.3.63 Three phase SCR inverter with a DC voltage source.
Chapter Three
60
Fig.3.64 Vd versus I d of Three phase SCR inverter with a DC voltage source.
Inverter Startup
As discussed for start up of a single-phase inverter, the delay angle α in the three-phase
inverter of Fig.3.63 is initially made sufficiently large (e.g., 165°) so that id is discontinuous.
Then, α is decreased by the controller such that the desired I d and Pd are obtained.