PowerElectronicsConvertersandRegulatorsThirdEdition-1

382 6 DC/AC Converters–Inverters

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Table 6.8 Normalized Fourier coefficients Vh3/VDC of the line voltages of a three-phase PWM inverter [2]

hma Harmonic 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
h =1 0.866 0.799 0.693 0.606 0.520 0.433 0.346 0.260 0.173 0.087
0.030 0.013 0.003
h = mf ± 2 0.275 0.232 0.190 0.150 0.114 0.081 0.053 0.232 0.165 0.086
h = 2mf ± 1 0.157 0.221 0.272 0.307 0.321 0.313 0.282

6.3 Three-Phase Inverters 383

Table 6.9 The calculated values of Vh, L–N and Ih Ih,rms(A)
4.17
n Vh,L-N (V) Zn(W) Ih(A) 0.27
6.39 0.14
1 63.6 11.8 0.38 0.06
0.20 0.04
5 12.73 33.0 0.08
0.06
7 9.09 45.1

11 5.79 69.8

13 4.90 82.3

For ma = 1: V1;LLrms ¼ 0:612VDC:

6.3.1 Overmodulation (ma > 1)

In the case of overmodulation (ma > 1) the amplitude of the control signal is larger
than the amplitude of the triangular signal and the amplitude of the first harmonic is

not a linear function of the amplitude modulation factor. With an increasing ma the
output voltage has a waveform similar to the waveform of the voltage of the six-

step inverter. The dependence of the effective value of the line-to-line voltage

fundamental harmonic on the amplitude modulation factor is shown in Fig. 6.17.

With the transition to the overmodulation region the amplitude of the output

voltage fundamental harmonic increases, but this ÀdependeÁnce ispnffiffiot linear. For
ma = 3.24 the amplitude has the maximal value V1;LLrms max¼ 6 VDC and this
p

value is equal to that of a six-step inverter. Also, unlike the linear modulation, in

this case there are harmonics at lower frequencies besides the harmonics at the

frequency of the carrier signal and its integer multiple.

V1LL,rms
VDC

6 =0,78

0,612 (over-modulation)
ma>1
(square) (six-step)
linear

01 3,24 ma

Fig. 6.17 Effective value of the fundamental harmonic of line voltage versus amplitude
modulation factor [2]

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384 6 DC/AC Converters–Inverters

6.3.2 Asynchronous PWM

For synchronous PWM the frequency modulation factor is an integer. In asyn-

chronous PWM the switching frequency is constant, while the frequency of fun-
damental harmonic is changed, and mf may not be an integer. As a consequence
subharmonics may occur with a frequency lower than the frequency of the first
harmonic. Although small in amplitude, these harmonics can cause significant

currents in predominantly inductive loads, which should be taken into account.

6.4 Space Vector Modulation

Space vector modulation (SVM) is a very important contemporary control tech-
nique of power converters. Its main application is control of three-phase voltage
controlled DC/AC converters, but it is also used for control of other types of
converters such as AC/DC power converters, resonant three-phase inverters, mul-
tilevel converters, matrix converters, etc.

Today there are a number of different variants of SVM, which differ with respect
to the complexity of realization and the quality of the results.

6.4.1 Space Vector Modulation—Basic Principles

A three-phase symmetrical power supply system is observed. In this system, only

two currents and two voltages are linearly independent, while the third current or

voltage can be expressed as a linear combination of the other two. Introducing an
algorithm that includes coupling of the third phase with the first two better results
can be achieved.

By applying the Clark transformation three dependent variables a, b, c are trans-
formed into two independent variables α, β (Fig. 6.18). For simplicity it is assumed
that the a-axis coincides with the α- axis in the orthogonal α, β coordinate system. The
projection of the current vector is ¼ iaa0 þ ibb0 þ icc0, where a0; b0ic0 are the unit
vectors, which lie on the a, b and c axis, respectively, to the α and β axis gives

ia ¼ ia À 1 ib À1 ic;
pffiffi 2 pffiffi2
ð6:34Þ
3 3
ib ¼ 2 ib À 2 ic;

or presented in the matrix form

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6.4 Space Vector Modulation 385

Fig. 6.18 Transformation of bi
current vector from three- 120 is i
phase a, b, c system to
stationary α, β coordinate
system

a

120
c

Fig. 6.19 Three-phase
inverter bridge

Sa1 Sb1 Sc1

+ va
vb
VDC vc

Sa2 Sb2 Sc2

ia !" 1 Àpffiffi21 À 1 #2 ia 3
ib 0 À 4 ib 5:
¼ 3 2pffiffi ic ð6:35Þ
2
3
2

Analogously, by the same transformation of voltages it is obtained that

va !" 1 Àpffiffi21 À 1 #2 va 3
vb 0 À 4 vb 5:
¼ 3 2pffiffi vc ð6:36Þ
2
3
2

A three-phase inverter bridge controlled by the space vector modulation tech-

nique is shown in Fig. 6.19. The input voltage is DC and the output voltages are the
phase voltages va, vb and vc.

It is known that the switches in the same branch of the inverter bridge must not

simultaneously lead, because in that case a DC voltage source is short-circuited.
This means that if one knows the states of the upper switches in the bridge, the

states of the lower switches of the inverter bridge are also known and vice versa.
One binary variable is assigned to each switch of the inverter bridge, A (for Sa1),

B (for Sb1) and C (for SC1) and the state of the binary variable is 1 if the

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386 6 DC/AC Converters–Inverters

Table 6.10 Applied switching sequence and phase voltages associated with the corresponding
switching states

Switching sequence Vector Phase voltages vα vβ

ABC V0 va vb vc 0 0
ABC V1 00 0
ABC V2 0
A BC V3 VDC 0 0 VDC 0
A BC V4 VDC VDC pffiffi
ABC V5 0 1 VDC 3 VDC
ABC V6 2
ABC V7 VDC p2ffiffi
0 VDC VDC À 1 VDC 3 VDC
VDC 2 2
VDC
0 VDC −VDC 0 pffiffi
00
À 1 VDC À 3 VDC
2
p2ffiffi
VDC 0 1 VDC À 3 VDC
VDC VDC 2 2

0 0

Fig. 6.20 Vectors of stator V3(ABC) V2(ABC)
voltages for applied control
sequence 000-111 in α, β II
coordinate system III

V4(ABC) I V1(ABC)

V0,V7
IV VI

V

V5(ABC) V6(ABC)

corresponding switch is closed, and 0 if the switch is opened. With these three
binary variables eight different states can be presented, i.e., any situation in which
the converter can be found (Table 6.10). Vectors of the stator voltage applied to the
control sequence of binary variables 000-111 in the α, β coordinate system are
presented in Fig. 6.20.

The vectorsV1 À V6 end up in the points, which represent vertexes of a regular
hexagon and they divide the regular hexagon into six segments, regular triangles
(Fig. 6.20). Vectors of the stator voltage applied to the control sequence of binary
variables 000-111 in the α, β coordinate system are presented in Fig. 6.20. The
amplitudes of these vectors are equal among themselves and their value is

Vab ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ VDC: ð6:37Þ
v2a þ vb2

The vectors V0 and V7 are zero vectors and upon the application of these vectors
the line to line voltages are equal to 0.

Using two adjacent vectors and possibly a zero vector any voltage vector whose
tip lies inside a regular hexagon defined by the vectors V1 À V6 can be obtained. In

other words, we can get a voltage vector of a desired amplitude and phase. The

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6.4 Space Vector Modulation 387

obtained voltages at the inverter output vab, vbc, and vca must make a symmetrical
three-phase system, which means that the output voltage vector describes a circle

with the corresponding frequency. This determines the voltage amplitude of the

three-phase inverter controlled by the space vector modulation technique, and this
amplitude is equal to the radius of a circle inscribed in the hexogen defined by the
vectors V1 À V6:

Vab ¼ pffiffi VDC: ð6:38Þ
3
max 2

The maximum value of the three-phase inverter phase voltage controlled by the
space vector modulation technique is

ðVLNmaxÞSVM ¼ 2 Vab ¼ pffiffi VDC; ð6:39Þ
3 3
max 3

and the maximum rms value of the line to line voltage is equal to

pffiffi pffiffi p1ffiffi
pffi3ffi 3 2
ðVLLrmsÞSVM ¼ 2 3 VDC ¼ VDC: ð6:40Þ

The maximum rms value of the line to line voltage that can be obtained for the
same input voltage by the sine pulse width modulation technique (for ma = 1) (6.33) is

pffiffi
p3ffiffi
ðVLLrmsÞSM ¼ 22 VDC: ð6:41Þ

Comparing the obtained expressions for the rms values of the line to line
voltages, one obtains from (6.40) and (6.41)

ðVLLrmsÞSVM ¼ p1ffiffi VDC ¼ p2ffiffi : ð6:42Þ
ðVLLrmsÞSM pffi2ffi 3
p3ffiffi VDC

22

From Eq. (6.42) one can conclude that for the space vector modulation technique
and the same input voltage, a higher effective value of the line to line voltage can be
obtained, than if the sine PWM technique is applied.

6.4.2 Application of Space Vector Modulation Technique

Suppose that the voltage vector V whose amplitude is V ¼ jVjand the phase is uðVÞ ¼
# should be realized (Fig. 6.21). The vector V is in first sector, so it can be realized by
applying the vectors V1, V2 and one zero vector V0 or V7 (Fig. 6.22). The period

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388 6 DC/AC Converters–Inverters

V2 V V

V1

Fig. 6.21 Reference vector and vectors defining the boundaries of the sector in which is the
reference vector

t1 t2 t0 t1 t2 t0

0T 2T t
V1(ABC) V2(ABC) V0(ABC) V1
or V2 V0
or
V7(ABC) V7

Fig. 6.22 Control sequence generating the reference voltage vector V

T divided into three time intervals, the interval t1 in which the vector V1 is applied, t2 in
which the vector V2 is applied and the interval t0 when a zero vector is applied.

The times t1, t2, and t0 are determined by the following expressions

t1 ¼ VabV T sin p À
3 #

max

t2 ¼ VabV Tsinð#Þ ð6:43Þ

max

t0 ¼ T À t1 À t2:

In the next period, it was assumed that the amplitude of the vector V is constant

and that it remains in the first quadrant, the angle θ is changed, and thus the time

intervals t1 and t2 are changed. The implementation of the space vector modulation
algorithm is presented by the block d iagram in Fig. 6.23. Based on the reference
value obtained from the controller vaÃ; vbà , the amplitude jVÃj and the phase
hà ¼ uðVÃÞ the reference vector Và is determined. The angle hà determines the
segment containing the vector VÃ from the condition

0 #Ã À ðN À 1Þ p \ p : ð6:44Þ
33

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6.4 Space Vector Modulation 389

Fig. 6.23 Implementation of Regulator reference values
space vector modulation v*, v*
algorithm
=arctg (v *⁄v*)
V=|v * |= v*2+v*2

Determination of segment
number N

Determination of applied
vector: VN, VN+1, V0(7)

Determination of time
interval: tN, tN+1, t0

Generation of vector
V=VN (tN/T) + VN+1 (tN+1/T) + V0(7) (t0/T)

When the segment N is determined, the vectors VN and VNþ1 are determined as
well as the time intervals in the which vectors VN, VNþ1 and the zero vector V0ð7Þ
should be applied.

The control of switches in the inverter is such that the following voltage vector is

realized:

VÃ ¼ VN tN þ V Nþ1 tNþ1 þ V 0ð7Þ T À ðtN þ tNþ1Þ : ð6:45Þ
T T T

In order to obtain the vector V, the vectors V1, V2and V0ð7Þ can be applied in a
different order, direct sequencing V1, V2 and V0ð7Þ and inverse sequencing V2, V1

and V0ð7Þ. The waveforms of phaseÀvoltaÁges when direct sequencing and one zero
vector are applied, for example V0 ABC , are shown in Fig. 6.24.

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390 6 DC/AC Converters–Inverters

V1 V2 V0 V1 V2 V0 V1 V2 V0

t11 t12 t10 t21 t22 t20 t31 t32 t30
t
ABC ABC ABC
3T
T 2T

va

VDC
t

0
vb

VDC
t

0
vc

VDC

t
0

ÀÁ
Fig. 6.24 Waveforms of phase voltages when direct sequencing and zero vector V0 ABC are
applied

6.4.3 Direct and Inverse Sequencing

To prevent greater heating, the zero vector can be alternatively applied first to one
group of switches and then to another group (one group of switches remains longer
in the on state), so that the vector V0 is used in the even segments and the vector V7
in the odd segments. If direct and inverse sequencing are alternatively applied we
get d–i or i–d sequencing. Figure 6.24 shows d–i sequencing with changing of zero
vector. By such sequencing, there is only one switch commutation during the period
T, i.e., a total 3-switch commutation in the inverter during one period is obtained.
Choosing the appropriate sequencing techniques and the zero vector, optimization
with respect to the number of switch commutation and switch stresses can be
achieved. Inverse sequencing with a change of zero vector is shown in Fig. 6.25.

As previously mentioned, there are several different variants of SVM. A very
important application of this technique is for the multilevel converter, which will be
more discussed in Chap. 10.

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6.5 Real Drive Influence 391

V1 V2 V7 V2 V1 V0 V1 V2 V7

ABC ABC ABC ABC ABC ABC ABC ABC ABC t

T 2T 3T

va

VDC
t

0
vb

VDC
t

0
vc

VDC
t

0

Fig. 6.25 Waveforms of phase voltages when inverse sequencing and changing of zero vector are
applied

6.5 Real Drive Influence

In the previous considerations of inverters the switches have been assumed ideal.
Therefore, simultaneous closing of one switch and opening of the other in the same
arm of the bridge was possible. In reality, however, this cannot be done because
there exist finite closing and opening times of the switches. During the transient
regimes of ideally driven switches (Fig. 6.26a) the primary source could be short-
circuited by the switches S1 and S4 (Fig. 6.26a). Namely, while one of the switches
is closing, the other is opening and within certain interval they are both conducting.
To avoid this, the closed switch has to open before the open switch starts closing.
Therefore, in reality there has to be a time shift Δt (Fig. 6.26c) between the change
of states of the switches in the same arm. The interval Δt ranges from several to
several tens of microseconds for fast switches, while for the slow switches it is
several hundreds of microseconds. Several names are used for the interval Δt: idling
time, dead time, non-overlap time, etc. This interval introduces a certain offset in
the output voltage. For PWM inverters this offset is approximately determined by

( 2Dt VDC ; i0 [ 0
TS
DV0 ¼ 2Dt ð6:46Þ
À TS VDC ; i0\0;

where Ts is the cycle of the carrier signal.

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392 6 DC/AC Converters–Inverters

(a) S1 (b)
(c)
S1 D1
VDC + t
S4
S4
D4 t

Fig. 6.26 One arm of the inverter bridge (a) and the states of ideal (b), and real switches (c)

(a) V0 (b) Ideal

i0<0 Δt=0 V0(t)
ΔV0 i0>0
Real
ΔV0 i0(t)
VC

Fig. 6.27 The real (dashed) and ideal (full line) characteristics V0 (Vc) of PWM inverter (a) and
the influence of idling time Δt on output voltage (b)

The transfer characteristic V0(Vc) for the case with ideal switches crosses the
origin (Fig. 6.27a), whereas with real switches it is shifted by +ΔV0, where the sign
depends on the direction of the output current. The influence of Δt to the output
voltage is illustrated in Fig. 6.27b.

In the design of an inverter, or more precisely of its control module, it is
necessary to introduce the non-overlap time. It is determined on the basis of the
dynamic pulse characteristic of the semiconductor switches at given loads.

Problems

6:1 The single-phase full-bridge inverter from Fig. 6.2a has a square-wave voltage
across a series R-L load. The switching frequency is 50 Hz, the DC input
voltage is 250 V, and the load is a serial connection of R = 10 Ω and
L = 50 mH.

(a) Determine expression for the load current and
(b) The power absorbed by the load.

6:2 The square-wave inverter from Fig. 6.2 has a DC source VDC = 250 V and the
load is a serial R-L connection with R = 10 Ω and L = 20 mH. The switching
frequency is 50 Hz.

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6.5 Real Drive Influence 393

(a) Determine the peak value of the load current in a steady state.
(b) Draw the waveform of the load current and determine the time intervals

when each switch component is conducting.

6:3 A single-phase full-bridge inverter has a DC source 150 V, and the load is a
serial R-L connection with R = 10 Ω and L = 20 mH. The switching frequency
is 50 Hz, and the switching scheme is shown in Fig. 6.7a.

(a) Determine the angle α, so the rms value of the output voltage at the
fundamental frequency is 117 V.

(b) Determine the THD of the load current.

6:4 Calculate the rms values of the fundamental harmonic and some of the higher
harmonics of a single-phase inverter based on PWM if VDC = 300 V, ma = 0.8,
mf = 39, and f1 = 47 Hz.

6:5 A DC source supplies an inverter with bipolar PWM control. The effective
value of the output voltage on the fundamental frequency of 50 Hz is 220 V.
The load is a serial R-L connection with a resistance R = 10 Ω and an
inductance L = 50 mH.

(a) Specify the amplitude modulation ratio if the DC voltage is 444 V.
(b) If the frequency modulation ratio is 21, determine the THD of the load

current.

6:6 A full-bridge single-phase PWM inverter has a DC source 150 V, and the load
is a serial R-L connection with R = 10 Ω and L = 20 mH. The fundamental
frequency is 50 Hz and the amplitude modulation factor is 0.8. Determine the
THD of the load current if mf = 15 when

(a) bipolar PWM is applied,
(b) unipolar PWM is applied.

6:7 The switches of the six-step inverter from Fig. 6.14 are controlled according to
the time diagram shown in Fig. 6.15b.

(a) Draw the waveforms of currents through the switches and the diodes.
(b) Draw the waveforms of the phase voltages va, vb, and vC.
(c) For the voltage va calculate the rms value and the amplitude of the first

harmonic.

6:8 The six-step three-phase inverter from Fig. 6.14 has a 511 V DC voltage
source, and its output frequency varies in the range 25–50 Hz. The load is Y-
connected with a serial connection of R = 10 Ω and L = 20 mH in each phase.
Determine the range of the rms values of the fundamental component of the
load current when the frequency is varied.

6:9 To control the three-phase inverter the space vector modulation technique is
used. The reference voltage vector V has an amplitude of 0.6 VDC, a frequency
50 Hz, and a phase θ = 75°.

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394 6 DC/AC Converters–Inverters

(a) Determine the reference phase voltages va, vb and vc.
(b) Define the switching frequency and the corresponding vectors, which

should be applied to realize the reference vector.
(c) Define the time intervals in which the corresponding vectors are applied.

References

1. Mohan N. et al.: Power Electronic-Converters, Applications and Design. Willey, New York
(1995)

2. Hart D.W.: Introduction to Power Electronics, Prentice-Hall (1997)

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Chapter 7

AC/DC Converters–Rectifiers

AnAC/DC converter usually interconnects the primary source and a DC/DC con-
troller (Fig. 7.1). In the supply systems where a high stability of the DC voltage is
not necessary, the application of DC/DC controllers is not required. An AC/DC
converter usually consists of:

• transformer,
• rectifier,
• filter, and
• control block of the rectifier.

The transformer adjusts the primary AC source to the input of the rectifier and
separates galvanically the primary source (usually the mains network) from the
load. The rectifier converts (rectifies) the AC energy to DC. In that sense, it is the
basic subassembly of an AC/DC converter and because of that many people imply
that the rectifier is a complete AC/DC converter. A rectifier comprises diodes and/or
thyristors. The voltage and the current at the output of a rectifier are subject to
considerable pulsations and for that reason their filtering is necessary. The filters are
usually very simple and comprise of a capacitor and/or achoke.

The control block of the rectifier controls the angle of the thyristor conduction
and in this way controls the output voltage and current. In such a situation, one talks
of controlled rectifiers. The controlled rectifier consists of thyristors or transistors. A
diode rectifier is not controlled. In that case, the rectifier does not have a control
block. In some cases, a transformer is not required. When the primary AC voltage is
directly fed to the input of the rectifier it is then, as a rule, a controlled rectifier.

This chapter is dedicated to rectifiers. They can be divided on the basis of the
mode of their connection to the AC network, or on the basis of the mode of using
the energy from the AC source, or on the basis of the character of the output voltage
and the degree of its control. With respect to the mode of using the energy from the
AC source rectifiers can be one-sided (half-wave) or two-sided (full-wave).

© Springer International Publishing Switzerland 2015 395
B.L. Dokić and B. Blanuša, Power Electronics,
DOI 10.1007/978-3-319-09402-1_7

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396 7 AC/DC Converters–Rectifiers

AC/DC CONVERTER

Primary Transformer Rectifier Filter
AC source

V +

V_ DC/DC
controller

Control block
Fig. 7.1 The block diagram of an AC/DC converter

7.1 Half-Wave Single-Phase Rectifiers

The simplest rectifier includes only one diode (Fig. 7.2a). The diode conducts only
during the positive half-cycle of the secondary voltage. If the voltage drop across
the diode is neglected, the voltage across the load will be equal to the secondary
voltage, i.e.,

vo ¼ vs ¼ VSM sinðxtÞ; 2kp xt ð2k þ 1Þp; k ¼ 0; 1; 2; . . . ð7:1Þ

where VSM is the amplitude of vs. During the negative half-cycle the diode is
blocked and vo = 0. Therefore, one obtains a sequence of pulses (positive half-
cycles) across the load (Fig. 7.2b). These pulses ought to be averaged and for that
purpose filters are used at the output of the rectifier. The DC component of the
output voltage is determined by

1 Zp VSM
2p p
Vo ¼ VSM sinðxtÞdðxtÞ ¼ ; ð7:2Þ

0

(b)

(a) V0
VSM =√2 Vrms
D

+ RL V0 V0 V0av =0.45 Vrms
V VS
t
VS

Fig. 7.2 Half-wave rectifier (a) and its output voltage (b)

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7.1 Half-Wave Single-Phase Rectifiers 397

where as the root mean square value of the voltage VS is

Vrms ¼ utuvu2ffiffi1ffipffiffiffiZffiffiffipffiffiffi½ffiVffiffiffiSffiffiMffiffiffiffisffiffiiffinffiffiffiðffiffixffiffiffitffiÞffiffiŠffiffi2ffiffidffiffiðffiffixffiffiffitffiffiÞffi ¼ VpSMffiffi : ð7:3Þ
2
0

Thus the DC component is

pffiffi
Vo ¼ ð 2=pÞVrms % 0:45Vrms:

The relatively small value of the DC component and the connection of the trans-
former to the load only during one half-cycle point are a poor use of the energy
source. The average value of the current is not zero, which has an adverse influence
on the transformer characteristics. In addition, the output voltage ripple is quite
high. Due to all these reasons, a half-wave rectifier is very seldom used.

7.2 Full-Wave Rectifiers

Two-sided rectifiers rectify both half-cycles. For this reason, they are often called
the full-wave rectifiers. Two types can be distinguished: rectifiers having a center-
tapped secondary winding and two-sided rectifiers having a diode (thyristor) bridge
(Graetz bridge).

A single-phase full-wave rectifier having a center-tapped secondary winding has
two diodes (Fig. 7.3a). When the input voltage is positive D1 is conducting and D2
is blocked. During the negative half-cycle D2 is conducting and D1 is blocked. The
direction of the current in both cases is the same so that the negative half-cycle

(a) iL positive (b)
half-cycle
iL negative
D1 half-cycle V0 Vs -VD
V0M

++

VS RL V0 V0av ≈ 0.45VSrms
V+ iL - t

VS VS δ

D2

V0 = Vs -VD t

Fig. 7.3 Single-phase full-wave rectifier having center-tapped secondary (a) and its output
voltage (b)

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398 7 AC/DC Converters–Rectifiers

appears on the positive side at the output (Fig. 7.3b). During each of the half-cycles,
the current flows only through one half of the secondary winding, consequently the
secondary winding has to contain a double number of turns.

The maximum reverse voltage across the diode is twice the amplitude of the
secondary voltage. Therefore, the breakdown voltage of the diode has to be

BVD [ 2VSM: ð7:4Þ

The maximum permitted current through the diode has to be higher than the

maximum expected load current.

In reality, the voltage across a conducting diode is greater than zero and,
depending upon the load current, it amounts 0.7 to 1 V. The voltage VD of the diode
can justifiably be neglected only if Voav ≫ VD. At small voltages, the influence of
VD should be taken into account. The output voltage is zero during δ/ω, where δ is
the angle when both diodes are blocked (Fig. 7.3b).

The second type of the full-wave rectifier is a Graetz bridge rectifier (Fig. 7.4a).

It consists of four diodes in a bridge circuit. During each half-cycle, one pair of the

diodes is conducting (during the positive one D1 and D2, during the negative D3 and
D4). Here too the load current in both half-cycles flows in the same direction so the
output voltage is of the same polarity. The DC component of the full-wave rectifiers
is twice the current of the half-wave rectifiers, i.e.,

(a) i+ (b)

i- D1 D3
RL V0
D4 D1
D4 D2
V VS i + VS

D2 D3

i- RL

(c) V0

V0M =√2 V0rms
V0av≈0.9 V0rms

D1 , D2 on D3 , D4 on t
D3 , D4 off D1 ,D2 off

Fig. 7.4 A Graetz (bridge) full-wave rectifier (a), an alternative representation (b), and the
waveform (c)

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7.2 Full-Wave Rectifiers 399

V0 ¼ 2 VSM ¼ 0:9Vsrms: ð7:5Þ
p

In addition, the ripple is considerably smaller. The diodes of the Graetz bridge
should have the maximum current higher than the maximum load current and the
breakdown voltage BVD higher than the amplitude of the secondary voltage. i.e.,

BVD [ VSM: ð7:6Þ

By comparing full-wave rectifiers it may be concluded that a Graetz bridge,
although having two more diodes, has an advantage because it uses a transformer
having one half of the number of turns in the secondary and diodes having one half
of the breakdown voltage. Of course the need for four diodes is a disadvantage of
the Graetz bridge, not from the point of view of the used-up material but because of
the doubled voltage across the diodes and the doubled dissipation in them. This is
particularly inconvenient at small DC voltages, when the voltage drop across the
diodes cannot be neglected. Because of that the efficiency factor is considerably
reduced.

It should be emphasized that the manufacturers offer the Graetz bridge as an
element in a single case. Also the diode pairs for the other type of the full-wave
rectifier are manufactured as a single element.

By altering the directions of the diodes, a negative voltage rectifier is obtained.
Some electronic systems (e.g., the systems containing linear electronic circuits)
require positive and negative voltages with respect to the ground. The rectifiers of
such power supplies are designed using one Graetz bridge and a transformer having
the center-tapped secondary winding (Fig. 7.5a). Now, both halves of the secondary
are active throughout the cycle. For instance, during the positive half-cycle by the
upper part of the secondary and the diode D1− the load RL is supplied by the current

(a) (b)

V i- i+ V01 V01
D3 +V0
D1
VS RL
VS
V02 t
D4 t
i- i+

RL
D2

V02
-V0

Fig. 7.5 The double full-wave single-phase rectifier (a) and the waveforms of the output voltages
V01 and V02 (b)

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400 7 AC/DC Converters–Rectifiers

i+. At the same time, by the lower part of the secondary and the diode D2 the load
RL is supplied by the current i+. During the negative half-cycle, the diodes D3 and
D4 provide the flow of the current i− through the load RL in the same direction as the
current i+. The waveforms of the output voltages are shown in Fig. 7.5b.

7.2.1 Commutation of Current

The commutation of the currents (taking over the current by one group of diodes
from the other group of diodes) in ideal rectifiers occurs instantly. In reality,

however, this is not possible owing to the inductances in the commutation loop.

These inductances in the commutation loop are mainly the sums of the inductances

of the commutation network referred to the secondary side of the network and the

stray inductances of the windings. The commutation time is usually expressed in

terms of an angle called the angle of commutation or the angle of coverage.

A real secondary circuit of a transformer can be represented by a series of
connection to a generator vs and an inductance LS (Fig. 7.6a). Let a Graetz bridge
operate in the constant output current mode. Then for an ideal rectifier (LS = 0) the
current of the primary source would be is = +I0 during the positive and is = −I0
during the negative half-cycle (Fig. 7.6e). In reality, however, the inductance LS
resists any change and at the beginning of conduction of one pair of the diodes the

other pair will stay turned on for a while. This happens at the beginning of each
half-cycle. Then all four diodes are conducting, the output voltage is v0 = 0
(Fig. 7.6c), and the inductance LS takes over the voltage (Fig. 7.6d).

The relationships at the beginning of the positive half-cycle will be considered.
For ωt > 0 the diodes D1 and D2 are conducting. Due to the action of LS the diodes
D3 and D4 do not turn off immediately. During their conduction, there are three
current loops (Fig. 7.6b). If diodes were ideal (VD = 0) the following would hold

vL ¼ Ls dis ¼ vs ¼ VSM sinðxtÞ: ð7:7Þ
dt

The current of the turned on diodes D1 and D2 is iD1 = iD2 = ik, whereas the current
of the turned off diodes D3 and D4− is iD3 = iD4 = I0 − ik. The line current is

determined by

is ¼ ÀIo þ 2ik ð7:8Þ

The commutation process will go on until the current through the diodes D1 and D2
becomes I0, when the diodes D3 and D4 are turned off.

Therefore, the process of taking over of the current by the diodes, as a conse-

quence of the real drive of the rectifier circuit, has an important influence on the

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7.2 Full-Wave Rectifiers 401
+
(a) D1 D3 + (b) D3 V0
LS A V0 D1 ik I0
B
VS iS D2 LS I0

I0 iS +- -
VS
B

D4 D4 ik D2

-

(c) V0

V0av = V0av

- 2ωLS I0
π
LS =0
ωt

γ
(d) VL

ωt

(e) IS LS=0
L S=0
ωt

Fig. 7.6 Real drive of a Graetz bridge in the constant output current mode (a and b), constant
current at the beginning of the positive half-cycle and the waveforms of the output voltage (c),
voltage across inductance (d), and source current (e)

rectifier characteristics. The waveforms of the voltage and the current change and
higher harmonics appear in the rectified voltage and the load current. The average
value of the rectified voltage decreases and is determined by

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402 7 AC/DC Converters–Rectifiers

1 Zp
p
Voav ¼ VSM sinðxtÞdðxtÞ

1 2c Zp Zc 3
p
¼ 4 VSM sinðxtÞdðxtÞ À VSM sinðxtÞdðxtÞ5: ð7:9Þ

00

The solution of (7.9) gives

Voav ¼ 2VSM À VSM ð1 À cos cÞ; ð7:10Þ
p p

where γ is the angle of commutation. Taking (7.7) into account it follows that

Zc Zc dis ZI0
dt
VSM sinðxtÞdðxtÞ ¼ LS dðxtÞ ¼xLS dis ¼ 2xLSIo: ð7:11Þ

0 0 ÀI0

By combining (7.11) with the second member of (7.10) one obtains

cos c ¼ 1 À 2xLs Io; ð7:12Þ
VSM

and the average value of the output voltage is

Voav ¼ 0:9Vsrms À 2xLs Io ¼ VoavjLS¼0 À 2xLs Io: ð7:13Þ
p p

Now let the Graetz bridge operate in the constant output voltage mode (Fig. 7.7a).

This situation is very close to reality when the rectifier is loaded by a large capacitor.
The equivalent circuit of the rectifier is represented in Fig. 7.7b. The diodes D1 and
D2 start conducting at an angle θb when |vs| reaches Vo (Fig. 7.7c), i.e.

Vo ¼ VSM sin #b: ð7:14Þ

The current i0 reaches its maximum at ωt = θp when the voltage across the
inductance becomes negative, or when the condition |vs| = Vo is met again. Owing

to the symmetry of the half-cycle

hp ¼ p À hb: ð7:15Þ

Because of the action of LS the diodes D1 and D2 will stay turned off for a while
until the areas A and B (Fig. 7.7d) become equal. While the current flows through

the diodes

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7.2 Full-Wave Rectifiers (b) 403

(a) D3 D
LS +
D1 - V0
LS + VS
- V0
A B
VS i s
D2
D4

(c) VS V0

i0 I0av
ωt

(d) 0 θb θp θf π π+θ b

VL ωt
AB

Fig. 7.7 Graetz bridge in the constant output voltage mode (a), the equivalent circuit (b), and the
corresponding waveforms (c and d)

vL ¼ Ls di ¼ VSM sinðxtÞ À V0: ð7:16Þ
dt

By integrating (7.16) one obtains

1 Zxt
xLs
i0ðxtÞ ¼ ½VSM sinðxtÞ À V0ŠdðxtÞ; ð7:17Þ

Hb

where θb ≤ ωt ≤ θf. The angle θf when i0 is zero again is determined from the
condition

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404 7 AC/DC Converters–Rectifiers

ZHf ZHf ð7:18Þ
vLðxtÞdðxtÞ ¼ ½VSM sinðxtÞ À V0ŠdðxtÞ ¼ 0;

Hb Hb

which corresponds to equating the surfaces A and B (Fig. 7.7d). The average value
of the output current is

1 ZHf
p
Io ¼ i0ðxtÞdðxtÞ: ð7:19Þ

Hb

7.3 Output Filters

The output voltage of a rectifier varies over a very wide range, from practically zero
to the maximum of the secondary voltage. For this reason at the output of a rectifier
use is made of filters which reduce the harmonic currents through the load and the
ripple of the output voltage. In practice, two types of filters are mostly used: the
capacitive filter for small currents and the L filter for large load currents.

7.3.1 Capacitive Filter

The load to be supplied is usually resistive. Thecapacitive filter consists of only one
capacitor of high capacitance, or appropriate connection of capacitors connected in

parallel with the load (Fig. 7.8a). It supports the output voltage across the load and
supplies the current at small values of the rectified secondary voltage. The peak-to-
peak variation of the output voltage is considerably smaller than its average value.

The diode pairs are conducting only at small intervals within a half-cycle when the
rectified secondary voltage is higher than the voltage across the capacitor. Then
vo = vs. During the remaining part of the half-cycle, the diodes are off and the
capacitor discharges exponentially through the load RL. By using the notation of
Fig. 7.8b, the output voltage can be written in the following form

8
< jVSMsinðxtÞj; a xt b and p þ a xt p þ b
ðD3 and D4 conductingÞ
V ðxtÞ ¼ : ðD1 and D2 conductingÞ otherwise ðall diodes are offÞ: ð7:20Þ
ðVSMsinbÞeÀðxtÀbÞ=ðxRLCÞ

The angle β at which the diodes stop conducting can be determined by equating the
slopes of functions (7.20)

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7.3 Output Filters 405
id
Fig. 7.8 Single-phase Graetz (a) D3 I 0
rectifier with a capacitive filter
(a) and the voltage and D1 C RL V0
current waveforms (b) A
VS B
V D2
D4

(b) DV0

V0 V0
VS
VM
V0av

Vm

α π β π+ α 3π π+β ωt
2 I0 2
i

id

ωt

d ½VSM sinðxtފ ¼ VSM cosðxtÞ ð7:21Þ
dðxtÞ

and


d 1
dðxtÞ ½ðVSM sin bÞeÀðxtÀbÞ=ðxRL CÞ Š ¼ ðVSM sin bÞ À xRLC eÀðxtÀbÞ=ðxRLCÞ:

ð7:22Þ

At ωt = β these slopes are equal, thus it follows

VSM cos b ¼ VSM sin b : ð7:23Þ
ÀxRLC

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406 7 AC/DC Converters–Rectifiers

From (7.23) β is

b ¼ tanÀ1ðÀxRLCÞ ¼ ÀtanÀ1ðxRLCÞ þ p: ð7:24Þ

In practice the time constant RLC is quite large, so

b % p=2; VSMsinb ¼ VSM: ð7:25Þ

In other words, the capacitor provides the load current during π/2 + kπ ≤ ωt ≤
(k + 1)π + α; k = 1, 2, … The angle α can be determined from the condition

jVSM sinðp þ aÞj ¼ ðVSM sin bÞeÀðxtÀbÞ=ðxRLCÞ: ð7:26Þ
Since β ≈ π/2, then

jsinðp þ aÞj ¼ eÀðp=2þaÞ=ðxRLCÞ: ð7:27Þ

The minimum value of the output voltage is determined by

Vom % VSMeÀðp=2þaÞ=ðxRLCÞ: ð7:28Þ
Usually ωRLC ≫ π so α ≈ π/2 and ð7:29Þ

Vom % VSMeÀp=ðxRLCÞ ¼ VSM sin a:
The peak-to-peak variation of the output voltage is determined by

DVo ¼ VSM À Vom % VSMð1 À eÀp=ðxRLCÞÞ: ð7:30Þ

Taking into account that ωRLC ≫π, ð7:31Þ
eÀp=ðxRLCÞ % 1 À p ;
xRLC

the variation of the output voltage is

DVo % VSM p ¼ VSM : ð7:32Þ
xRLC 2fRLC

It should be emphasized that (7.30) is a rough estimate of the variation of Vo
basically because of the rough estimate of the angle α at which the diodes start
conducting. Usually α < π/2.

Example 7.1 A fullwave rectifier (Fig. 7.8a) with acapacitive filter has
Vseff = 220 V, f = 50 Hz, R = 220 Ω and C = 100 μF.

(a) Determine an expression for the load current.

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7.3 Output Filters 407

(b) Determine the average value of the load current.
(c) Calculate the maximum change of the load voltage.
(d) Draw the waveforms of the load voltage and the current through the capacitor

using PSPICE software.

(a) Voltage on the load can be calculated following the expression (7.20). The
angle β at which the diodes stop conducting can be determined by equating the
slopes of functions (7.20). Solving Eq. (7.21) one obtains

b ¼ p À arctgðxRCÞ ¼ 1:71 rad:

The angle α is determined from the condition that the capacitor begins to recharge
(7.26). Solving Eq. (7.26) one obtains α = 0.76. The load current is equal to vp/R, so
the expression for the load current can be written as

& 1:41 sinðxtÞ ½AŠ; 0:8 xt 1:71
ipðxtÞ ¼ 1:4eÀÀx6tÀ:911:71 ½AŠ; 1:71 xt p þ 0:8

(b) The average value of the load current is

2 Z1:71 pZþ0:8 3

Ipsr ¼ 1 4 1:41 sinðxtÞdðxtÞþ 1:4eÀÀx6tÀ:911:71 dðxtÞ5 ¼ 1:22 A:
p

0:8 1:71

(c) The maximum change in the load voltage is determined as the difference between
the maximum and the minimum voltage on the load (Figs. 7.9 and 7.10)

DVp ¼ Vpmax À Vpmin ¼ VSM À VSM sin a ¼ 87:9V :

Fig. 7.9 Voltage on the capacitor for the circuit shown in Fig. 7.8a

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408 7 AC/DC Converters–Rectifiers

Fig. 7.10 Current through the capacitor for the circuit shown in Fig. 7.8a

7.3.2 L Filter

For supplying large load currents use is made ofL filters consisting of a series
inductance L and a parallel capacitance C (Fig. 7.11a). The action of these filters
distinctly depends on the load current. When this current is small, the filter behaves
as if it were capacitive since the influence of thechoke can be neglected because the
accumulated energy (LI2/2) is small. Then the current through the diodes flows only
in a fraction of the half-cycle, in the vicinity of the maximum value of the mains
voltage (Fig. 7.11b), when the input voltage of the filter is higher than the output
voltage. By increasing the load current, the energy of thechoke increases and also

increases the angle of conduction of the diodes owing to the counter-electromotive

force supporting the existing current through the choke.

In this process, the output voltage decreases with increasing the load current.

This operating mode is called the mode of the discontinuous choke current. The
current through the choke will flow throughout the whole cycle (the continuous
mode) if the load current is higher than certain critical value I0k. Then the angle of
diode conduction is 180, i.e., the diodes are conducting alternately throughout the
whole half-cycle. In this mode (I0 > I0k), the average value of the output voltage
does not depend on the load current (Fig. 7.12). Then theL filter can be analyzed in
the steady state harmonic mode driven by a full-wave rectified sine function whose
Fourier series expansion is

Vi ¼ 2VSM À 4VSM cosð2xtÞ À 4VSM cosð4xtÞ þ Á Á Á ð7:33Þ
p 3p 15p

To make higher harmonics negligible, it is necessary to have the impedance of the
capacitor at these frequencies much lower compared to the impedance of thechoke,
i.e.,

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7.3 Output Filters 409

(a) L + (b) V0
I0
D1 iL Vi ωt
A D3 V0av
iL
VS
Vi C RL V0 a b
B Vi

V0

D4 D2

ωt

Fig. 7.11 Single-phase Graetz bridge with an L filter (a) and the characteristic waveforms in the
modes of discontinuous (b) and continuous choke current (c)

Fig. 7.12 Regulation Vav
characteristics of the L filter

I0K I0

XC ¼ 1=2xC ( XL ¼ 2xL; ð7:34Þ

This condition is usually satisfied since the capacitance C and the inductance L are
quite large. By neglecting the harmonics higher than the second, it is obtained that
the output voltage is the sum of the average value and the second harmonic
(Fig. 7.11c). The average value is

Vosr ¼ 2 VSM ; ð7:35Þ
p

and the amplitude of the second harmonic across the load, since I/(2ωC) ≪ RL, is

Vo2 % 1 4VSM ¼ VSM : ð7:36Þ
2xC 3p 3px2LC

2xL

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410 7 AC/DC Converters–Rectifiers

The ripple factor is

FT ¼ Vo2 ¼ 1 ð7:37Þ
Vosr 6x2LC

and in this mode is not dependent on the load current.The critical load current is

determined from the condition of continuity of the current through the choke. In this

case the DC load current, Io = Voav/RL, should be higher or equal to the amplitude of
the second harmonic of the choke current IL2ω. If the impedance of the capacitor is
much lower than the load resistance, i.e., I/(2ωC) ≪ RL then

IL2 ¼ 2VSM ¼ I0K: ð7:38Þ
3pxL

From the condition of continuity I0 > I0k it will be met if

RL 3xL: ð7:39Þ

In the design of an L filter, one starts from the maximum load resistance RLmax.
For this value, on the basis of (7.39), the critical inductance is determined as:

LK ¼ RLmax=ð3xÞ: ð7:40Þ

The capacitance is determined from the relation 1/(ωC) ≪ Lkω. If the load
resistance varies over a wide range, and if the output is a fixed load the critical

inductance is calculated according to this value.

7.4 Voltage Doublers

The circuit for doubling a DC voltageconsists of two half-wave rectifiers connected
to the same secondary but with reverse connected diodes and with two series
capacitors (Fig. 7.13a). During the positive half-cycle the capacitor C1 is charged
through the diode D1 and during the negative half-cycle the capacitor C2 is charged
through the diode D2. The capacitances are equal and quite large. Figure 7.13b
shows the voltage waveforms across the capacitor and across the output. Even
though the mains voltage is rectified as if by a full-wave rectifier, each of the
capacitor discharges as in a half-wave rectifier. For this reason the ripple is quite
high and it increases with the reduction of the load resistance. Therefore this circuit
is used only at small load currents.

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7.5 Three-Phase Rectifiers 411
ωt
(a) D1 (b) V0 = VC1 + VC2
+ VC1 VC2
VS +
C1 - V VS

RL V0 =2VSrms√2

+
C2 -

D2

Fig. 7.13 The voltage doubler (a) and voltage waveforms (b)

7.5 Three-Phase Rectifiers

In industrial applications, where three-phase AC voltages are available, it is rec-
ommendable to use three-phase rectifiers. At the output of this type of rectifier
compared to the single-phase rectifiers, the DC component is higher, the ripple of
the output voltage is lower, and the output power is higher. Three-phase rectifiers
have favorable features for equipment and installations requiring high power, where
very high DC currents and relatively high voltages are necessary. Most frequently
these are full-wave bridge rectifiers (Fig. 7.14a). As in the single-phase rectifiers, a
simple capacitive filter is used at the output.

In order to facilitate understanding, a real rectifier circuit (Fig. 7.14a) is sim-
plified by assuming an idealized drive and by replacing a real load by a current
generator (Fig. 7.14b). Due to neglecting the series inductances of the driving
generators (Fig. 7.14b), the output current I0 flows through one diode of the upper

(a) (b) D1

A ia D1 D3 D5 D3
B a
C +- +
ib +- D5 +
ic b C RL V0 n +- a VPn

c b n I0
D4 D6 D2 D2 V0

c D6 VNn

--

D4

Fig. 7.14 Three-phase bridge rectifier (a) and its ideal equivalent circuit (b)

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412 7 AC/DC Converters–Rectifiers

group and one diode of the lower group of diodes. In the upper group the diode
having the anode at the highest potential will conduct. The other two are reverse
biased, thus they are off. In the lower group only the diode having the cathode at the
lowest potential will conduct. The output voltage is equal to the difference of the
voltages at the points P and N compared to the neutral point n, i.e.,

Vo ¼ VPn À VNn: ð7:41Þ

Since one diode of the upper group and one diode of the lower group are always
conducting, the voltages VPn and VNn are equal to one of the AC voltages Van,Vbn or
Vcn (Fig. 7.15a). So, the output voltage consists of the six segments of the line
voltage during one cycle (Fig. 7.15b). For this reason, these rectifiers are often
called the six-pulse rectifiers. The output voltage during one cycle practically

consists of six sinusoidal peaks (Fig. 7.15b) thus its ripple is small and the average

value is close to the amplitude of the line voltages. Each diode conducts during
120°. They conduct in pairs (6, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), and (6, 1). The

phase currents are determined by

ia ¼ id1 À id4; ib ¼ id3 À id6;ic ¼ id5 À id2 ð7:42Þ

and are represented in Fig. 7.15c. Thus the current of, e.g., the phase a is

8 if D1 conducting
>< I0 if D4 conducting
if D1 and D4 are turned off
ia ¼ >: À I0 ð7:43Þ
0

The maximum reverse voltage of a diode is equal to the amplitude of the line
voltage. The output voltage is cyclic with the cycle π/3. The corresponding Fourier
series is

voðtÞ ¼ V0 þ X1 Vn cosðnx0t þ pÞ; ð7:44Þ

n¼6;12;18

where the average value is

V0 ¼ 1 Z2p=3 ¼ 3VLLM ¼ 0:955VLLM: ð7:45Þ
p=3 VLLM sinðxtÞdðxtÞ p

p=3

Therefore, the average value of the output voltage is approximately equal to the
amplitude of the line voltage, VLLM. The amplitudes of the harmonics are

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7.5 Three-Phase Rectifiers vcn 413
ωt
(a) vPn

van vbn

0

ωt=0 vNn

(b) v0

A √ 2VLL =VLLM V0av

- π 0 π D4 ωt
6 6 120 ° ωt
(c) ωt
ia D3
0 120 ° 60 ° D4 120 °
D1 120 ° D1

D6 ib D6
0 D3

ic

D2 D2
0 ωt

D5 D5 D5

Fig. 7.15 Waveforms of voltages and currents of a three-phase rectifier operating in the constant
load current mode

Vn ¼ 6VLLM ; n ¼ 6; 12; 18; . . . ð7:46Þ
pðn2 À 1Þ

The harmonics of the output voltage are at frequencies 6 kω, k = 1, 2, 3, …, because
the cycle of the output voltage is 1/6 of the cycle of the input voltage. This is also an
advantage of a three-phase rectifier compared to a single-phase rectifier. The AC
components are at higher frequencies and are of smaller amplitudes, so they can be
filtered out more readily. Owing to their bipolar shape the currents have odd

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414 7 AC/DC Converters–Rectifiers

harmonics 5, 7, 11, 13, 17, 23, …, because those involving integer factor 3 are zero.
The amplitude of the first harmonic is

Is1 ¼ 2 pffiffi ð7:47Þ
p 3I0;

whereas the amplitudes of the higher harmonics are

In ¼ Is1=n; n ¼ 5; 7; 11; 13; . . . ð7:48Þ

As already stressed at the beginning, the preceding analysis dealt with a sim-
plified circuit of the rectifier (Fig. 7.14b). The analysis of a real rectifier circuit
should be taken into account the series inductance of the line drive and the influence
of the filter capacitor. Owing to the series inductance, like in a single-phase rectifier,
the current commutation is not instantaneous. This leads to a certain reduction of
the output voltage. The filter capacitor at the output supports the output voltage so
that the rectifier operates closer to the constant output voltage mode than to the
constant current mode. Because of that, the diodes will not conduct 120° each but
only while the line voltage is higher than the voltage across the capacitor. The effect
is similar to that in single-phase rectifiers.

Example 7.2 For the battery charger with a three-phase rectifier (Fig. 7.16)

(a) Draw waveforms of the line to line input voltages, the load voltage and the
frequency characteristics of the load current using PSPICE software package.

(b) Using PSPICE software package and Fourier analysis of the circuits from
Fig. 7.16 determine the average value and sixth harmonic of the current
through the load.

(a) The waveforms of the line to line input voltages, the load voltage and the
frequency characteristics of the load current are presented in Figs. 7.17 and
7.18 respectively.

Fig. 7.16 Three-phase io
battery charger D1 D5 D3

3x380V vo Lp
50Hz D4 D2 D6 220μH

RB
5Ω

VB
110V

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7.5 Three-Phase Rectifiers 415

Fig. 7.17 Waveforms of line to line input voltages and load voltage for the rectifier from Fig. 7.16

Fig. 7.18 Frequency characteristics of load current for the rectifier from Fig. 7.16

(b) Fourier components of load current I(r).
DC Component = 8.026711E+01

Harmonic Frequency Fourier Normalized Phase phase Normalized
No (Hz) component component (deg) (deg)
1 5.000E+01 3.962E-01 1.000E+00 0.000E+00
2 1.000E+02 7.468E-01 1.885E+00 9.910E+01 −2.675E+02
3 1.500E+02 1.925E-01 4.860E-01 −6.934E+01 −2.357E+02
4 2.000E+02 7.949E-01 2.006E+00 6.163E+01 −2.932E+02
5 2.500E+02 1.850E-01 4.669E-01 1.032E+02 −4.832E+02
6 3.000E+02 4.913E+00 1.240E+01 1.232E+01 −5.063E+02
7 3.500E+02 3.585E-01 9.049E-01 8.827E+01 −6.808E+02
8 4.000E+02 3.179E-01 8.025E-01 1.293E+01 −6.522E+02
9 4.500E+02 1.411E-01 3.561E-01 1.406E+02 −8.414E+02
10 5.000E+02 2.639E-01 6.662E-01 5.046E+01 −1.042E+03
11 5.500E+02 6.559E-02 1.656E-01 5.142E+01 −1.144E+03
12 6.000E+02 8.898E-01 2.246E+00 −5.347E+01 −1.304E+03
13 6.500E+02 1.317E-01 3.324E-01 −1.151E+02 −1.365E+03
14 7.000E+02 3.288E-01 8.299E-01 −7.653E+01 −1.493E+03
−1.054E+02

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416 7 AC/DC Converters–Rectifiers

On the basis of Fourier analysis it is obtained that the average value and sixth
harmonic are I0 = 80.26 A and I6 = 12.4 A.

7.6 Phase Controlled Rectifiers

In phase controlled rectifiers thyristors are used instead of diodes. The average
value of the output voltage is controlled by controlling the start of thyristor con-
duction. This will be considered in the example of ahalf-wave thyristor rectifier
(Fig. 7.19).

The waveforms of this circuit, assuming that the thyristor was ideal, are shown
in Fig. 7.19b. The thyristor starts conducting when a short pulse is fed to its gate.
The input voltage is then transferred to the load. The thyristor turns off when the
input voltage becomes negative. Therefore, the output voltage during one cycle is

( VSM sinðxtÞ; 0 xt p
0; p xt 2p þ a
vo ¼ ð7:49Þ

Since the load is resistive, the current i0 follows the variation of vo.
If the load is inductive-resistive (Fig. 7.20a), the thyristor will not turn off at the

zero voltage crossing, but will continue conducting until the current i0 becomes zero
(Fig. 7.20a). Owing to that the output contains a negative component. The delayed
turning off of the thyristor is the consequence of the phase shift of the current i0
introduced by the inductance L. The angle is calculated from the condition i0(β) = 0
(Fig. 7.20).

(a) Th i0 (b) π 2π 2π + α ωt
RL
VS V0
Vg
+
V0 α

i0

ωt
Vg

ωt

Fig. 7.19 A half-wave thyristor rectifier (a) and the voltage and current waveforms (b)

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7.6 Phase Controlled Rectifiers 417
ωt
(b)

(a) Vs V0

VS Th i0 V0
i0 R
+ VL
Ig R VL
β
V0 α
L Vg
Vs

ωt

Fig. 7.20 Ahalf-wave thyristor rectifier loaded by inductance and capacitance (a) and the
corresponding voltage waveforms (b)

By varying the angle α the average value of the output voltage can be controlled
from zero to the maximum value which is obtained for α = 0. In order to accomplish

that, there must be a generator of synchronization pulses for triggering of the
thyristor in such a way that the angle α can be varied from 0 to 180° (0 ≤ α ≤ π). A

block diagram of such generator is shown in Fig. 7.21a. It consists of a source of the
mains synchronization voltage vsin, a generator of a saw tooth voltage vts, a com-
parator, and a differentiator. The voltage vsin synchronizes the saw tooth voltage
with the zero crossing of the sin function. When the saw tooth voltage is equal to
the control voltage, i.e., when vts = vsin, the output of the comparator becomes
positive. At the output of the buffer B a short pulse is generated. Its duration
depends on the time constant RC. Since the buffer B is synchronized with vsin, the
pulse vg will appear only during the positive half-cycle.

Since vts = VtsM(ωt/π), from the condition vts(α/ω) = Vcon., it follows that

a ¼ pðVcon:=VtsMÞ ¼ 180 ðVcon:=VtsMÞ; ð7:50Þ

where Vtsm is the amplitude of the saw tooth voltage. Therefore, by varying the
control voltage within the limits 0 ≤ Vcon. ≤ VtsM the angle α can be controlled
within the limits 0 ≤ α ≤ π.

There are several types of integrated circuits, such as TCA780, developed for the

triggering of thyristors and control of the angle α according to (7.50).

7.6.1 Full-Wave Thyristor Rectifiers

Two types of the full-wave thyristor rectifiers are used. A full-wave thyristor rec-
tifier with a purely resistive load and a center-tapped secondary is shown in
Fig. 7.22. The triggering pulses Vg1 and Vg2 are phase shifted by π so that the
thyristor Th1 is triggered at angles 2 kπ + α, and the thyristor Th2 at (2k + 1)π + α,
k = 1, 2, 3, … Such a generator can be made if the circuit in Fig. 7.21a is added to

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418 7 AC/DC Converters–Rectifiers

(a)

Gener. Vts C Vg
sawth. B
voltage VREF +
Comp.

- VK

Vsin. Vcon. R Vt
Diferenc.
(b) Vsin.

α VtsM t
ω t
Vts t

Vcon.
Vts

Vt
Vg

t

Fig. 7.21 A generator of triggering pulses for driving the thyristor (a) and the corresponding
voltage waveforms (b)

(a) Th1 (b)

+

VS Vg1 V0

V - - V0 + V0av

+ RL

VS α π+α 2π+α ωt
- Th2

Vg2

Fig. 7.22 A full-wave single-phase thyristor rectifier with a center-tapped secondary (a) and the
waveform of the output voltage (b)

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7.6 Phase Controlled Rectifiers 419

another differentiator synchronized with the negative half-cycle of Vsin. At its
output, one will obtain pulses Vg2 for triggering Th2.

Before any thyristor is turned on the output voltage is zero, and when one of the

thyristors is on, then vo = |vs| (Fig. 7.22b). The average value of the output voltage is
determined by

Voav ¼ Zp VSM sinðxtÞdðxtÞ ¼ VSM ð1 þ cos aÞ: ð7:51Þ
1 p
p

a

Since α can be controlled within the limits 0 ≤ α ≤ π and in these limits −1 ≤ cosα ≤
1, it follows that

0 Voav VSM=p: ð7:52Þ

The average value of the load current is Ioav = Voav/RL, and its root mean square
value is

Iorms ¼ tvuuuffi1ffiffiffiZffiffiffipffiffiffiffiffiVffiffiffiSffiffiMffiffiffiffisffiffiiffinffiffiffiðffixffiffiffiffitffiÞffiffi!ffiffiffidffiffiðffiffixffiffiffitffiffiÞffi ¼ VSM rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð7:53Þ
pp R 1 À a þ sinð2aÞ:
2 2p 4p

a

The allowed load of the thyristor bridge rectifier is the R-L type (Fig. 7.23b).
With this load, two operating modes are possible:

• discontinuous and
• continuous.

In the first mode, the load current does not flow during the whole of the cycle
(Fig. 7.23b), and in the second it is continuous (Fig. 7.23c). Owing to the reactive
component of the load the conducting pair of thyristors remains on after the polarity
of the input voltage has changed. They conduct until the current through them drops
to zero. If this happens before the second thyristor pair is triggered, the operating
mode is discontinuous. If the conducting pair of thyristors is still conducting at the
moment of triggering of the second pair, the current through the load will be
continuous (Fig. 7.23c).

In the first case, the load current is non-zero during α ≤ ωt ≤ β and it amounts to

i0ðxtÞ ¼ VSM ½sinðxt À HÞ À eÀðxtÀaÞ=ðxsÞ sinða À Hފ; a xt b ð7:54Þ
ZL

where

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420 7 AC/DC Converters–Rectifiers

i0 β π+ α ωt

(a) (b)

Th1 α
V0

Th3 π 2π

L ωt
ωt
VS V0 i0

(c)

R απ π+α

Th4 Th2 V0

π 2π ωt

Fig. 7.23 A thyristor single-phase bridge rectifier with an R-L load (a) and the output current and
voltage in the discontinuous (b) and continuous mode (c)

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ZL ¼ R2 þ ðxRÞ2; H ¼ tanÀ1ðxsÞ and s ¼ L=R:

During the positive half-cycle of the input voltage while α ≤ ωt ≤ β the thyristors

Th1 and Th2 are conducting whereas the other pair Th3 and Th4 are off. The pair Th1
and Th2 is turned off at the instant β/ω. The thyristors Th3 and Th4 are turned on at
the instant (π + α)/ω, so that during the time interval β/ω ≤ t ≤ (π + α)/ω all

thyristors are off and the load current is zero. Thus, the discontinuous mode will

exist if

b\p þ a: ð7:55Þ

The boundary between the continuous and the discontinuous modes is defined by
β = π + α, i.e. by

ioðp þ aÞ [ 0: ð7:56Þ
On the basis of (7.54) and (7.56), it follows that

sinðp þ a À HÞ À eÀðpþaÀaÞ=ðxsÞ sinða À HÞ ! 0: ð7:57Þ

Since sin(π + α − θ) = sin(θ − α), (7.57) reduces to

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7.6 Phase Controlled Rectifiers 421

ð7:58Þ
1 À eÀp=ðxsÞ sinð# À aÞ ! 1;

where from it follows that

a h ¼ tanÀ1ðxL=RÞ: ð7:59Þ

Therefore, the continuous mode will be sustained if condition (7.59) is satisfied.
The DC component (average value) of the output voltage for the continuous mode
is determined by

1 Zpþa 2VSM
p p
Voav ¼ VSM sinðxtÞdðxtÞ ¼ cos a: ð7:60Þ

a

Example 7.3 For the rectifier in Fig. 7.24 determine the average value of the
voltage across the load, the average load current, and the average values of the
thyristor and the diode currents if the load is:

(a) resistance R,
(b) a series connection of the resistance R and the inductance L.

It is known that VSM = 311 V, f = 50 Hz, R = 10 Ω, ωL > R, and α = π/3.
(a) The load is the resistor R.

The average value of the voltage across the load is

1 Zp VSM 311
p p p
Vo ¼ VSM sinðxtÞdðxtÞ ¼ ð1 þ cos aÞ ¼ ð1 þ 0:5Þ ¼ 148:5 V:

a

Fig. 7.24 Full-wave thyristor
rectifier

T1 T3

VS D ZL VO

T4 T2

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422 7 AC/DC Converters–Rectifiers

The average value of the load current is

I0 ¼ V0=R ¼ 14:85V:

For a purely resistive load, there is no current through the diode, so Id0 = 0.
The average value of the thyristor current is one half the load current.

(b) The load is a series R-L circuit.

The average values of the load voltage and the load current are the same as in the

case (a). Since the load is inductive, the current through the load is continuous and

the diode D conducts when all thyristors in the bridge are off.
The average value of the current through the diode is Id0 = α/π · I0 = 4.95 A, and

the average value of the current through the thyristor is also It0 = (π − α)/
π · I0 = 4.95 A.

Example 7.4 Determine the average value of the load current for the rectifier from

Fig. 7.25 and draw the waveforms of the load current and the load voltage using

PSPICE software package.

It is known that vs(ωt) = 311sin(ωt)V, f = 50 Hz, R = 10 Ω, Ls = 10 mH, ωL > R,
and α = π/3.

Let us consider the instant of turning on of the thyristors Th1 and Th2, if the
thyristors Th3 and Th4 were conducting before them. Due to the inductivity of the
source, the source current will lag behind the source voltage, so the thyristors Th3
and Th4 are turned on for some time while the thyristors Th1 and Th2 take over the
load current. If the angle of commutation is denoted by γ, the average value of the

load current is

1 Zpþa VSM
pR pR
Io ¼ VSM sinðxtÞdðxtÞ ¼ ½cosða þ cÞ þ cos aŠ:

aþc

During the commutation process, all four thyristors are conducting so the voltage
across the inductance Ls is equal to the source voltage vs, i.e.,

Fig. 7.25 Full-wave rectifier LS T1 T3
with the effect of source
inductance

VS I O VO

T4 T2

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7.6 Phase Controlled Rectifiers 423

vLs ¼ vs for a xt a þ c:

The current through the inductance Ls can be determined following the
Eq. (7.11)

1 Zxt
xLs
iLs ðxtÞ ¼ VSM sinðxtÞdðxtÞ À Io: ð7:61Þ

a

The commutation will be finished when the current through the inductance Ls
becomes equal to the load current, i.e.

iLs ða þ cÞ ¼ Io ¼ VSM ½cos a À cosða þ cފ À Io: ð7:62Þ
xLs

From (7.62) the angle of commutation is

cos a À cosða þ cÞ ¼ 2xLsIo ) cosða þ cÞ ¼ cos a À 2xLsIo : ð7:63Þ
VSM VSM

From (7.62) and (7.63), the average load current is (Figs. 7.26, 7.27)

2VSM xLsIo ! 2VSM 2xLsIo
pR VSM pR pR
Io ¼ cos a À ¼ cos a À ;

and

Io ¼ 2VSM cos a ¼ 2 Â 311 Â 0:5 ¼ 8:25 A:
pR þ 2xLs 10p þ 4p  50  0:01

Fig. 7.26 Waveform of the load voltage for the rectifier shown in Fig. 7.25

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424 7 AC/DC Converters–Rectifiers

Fig. 7.27 Waveform of the load current for the rectifier shown in Fig. 7.25

7.6.2 Three-Phase Thyristor Bridge Rectifiers

The topology of the thyristor bridge rectifier (Fig. 7.28a) is the same as that of the
corresponding diode rectifier. The difference in the principle of operation is that the
diodes are conducting when forward biased and the thyristors conducts if a trig-
gering pulse is present in the gate circuit together with the forward bias. This is the
way of controlling the angle of conduction and consequently the average value of
the output voltage. If the angle of triggering is α, the output voltage waveform is as
shown in Fig. 7.28b and its average value is determined by

1 2pZ=3þa 3VLLM
p=3 p
Vo ¼ VLLM sinðxtÞdðxtÞ ¼ cos a: ð7:64Þ

p=3þa

Compared to Vo of the three-phase diode rectifier (7.45) the factor cosα is present
here. If the angle of triggering is greater than π/3 (α > π/3), the rectifier operates in
the discontinuous mode of the load current. Then the average value of the output

voltage is determined by

1 Zp 3VLLM
p=3 p
Vo ¼ VLLM sinðxtÞdðxtÞ ¼ ½1 þ cosðp=3 þ aފ: ð7:65Þ

p=3þa

Example 7.5 For the battery charger realized with a three-phase thyristor bridge
rectifier (Fig. 7.29)

(a) Determine the change in the angle of triggering the thyristors so the load
current is held constant during charging of the battery s (for the minimum
battery voltage, the angle of thyristor triggering is α = 45°).

(b) Draw the waveform of the load voltage if the angle of thyristor triggering is
α = 45° using PSPICE software package.

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7.6 Phase Controlled Rectifiers 425

(a) Th1 Th3 Th5 +
a V0
i sa
A b ZL

i sb c
B Th4 Th6 Th2

i sc
C

(b)

V0
α

V0av
ωt

Fig. 7.28 A three-phase thyristor bridge rectifier (a) and the output voltage waveforms for a
resistive load (b)

Fig. 7.29 Battery charger io
with three-phase thyristor
bridge rectifier Lp
220μH
3x380V TH1 TH 5 TH 3
50Hz vo R
20 Ω
TH 4 TH 2 TH6
VB
55*(1.8 2.4)V

(a) The average value of the battery charging current is

Io ¼ Vo À Vbat :
R

If the voltage of the battery cell at the beginning of charging is 1.8 V, and the angle
of thyristor triggering is α = 45°, then the average charging current is equal to

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426 7 AC/DC Converters–Rectifiers

Fig. 7.30 Line to line input voltages and load voltage for the rectifier from Fig. 7.29

Io ¼ ð3VLLM=pÞ Â cosðp=4Þ À ð1:8 Â 55Þ ¼ 13:23 A:
20

When the batteries are full, the cell voltage is 2.4 V. To ensure the same charging
current, the thyristor triggering angle is equal to


R Â Io þ ð55 Â 2:4Þ
a ¼ arccos 3VLLM=p ¼ 0:623 rad % 36 :

The change in the angle of triggering the thyristors during the battery charging is
Δα = 45–36° = 9° (Fig. 7.30).

(b) See Fig. 7.30.

7.7 Twelve-Pulse Rectifiers

In the cases when the primary power source is three-phase and a higher DC output
voltage is required multipulse rectifiers are used. The twelve-pulse rectifiers have
found practical applications. Such a rectifier consists of 2 six-pulse three-phase
rectifier rectifiers connected in series. One six-pulse rectifier is supplied through a
Y-Y transformer, and the other through a Y-δ transformer (Fig. 7.31). So, the two
rectifier output voltages have a phase shift of π/6.

The highest average value of the output voltage is obtained if α = 0, i.e., if the
rectifiers behave as a diode rectifier

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7.7 Twelve-Pulse Rectifiers 427

ABC

Th11 Th31 Th51 iO
iS1

vO1

Y-Y Th41 Th61 Th21
coupling

Load vO

Th12 Th32 Th52
iS2 Th62 vO2

Y-Δ Th42 Th22
coupling

Fig. 7.31 Twelve-pulses thyristor rectifier

Vomax ¼ VLLM p þ p þ VLLM p þ p
¼ VLLM sin sin 4
6 3 6
7p 5p p
sin 12 þ VLLM sin 12 ¼ 2VLLM cos 12 ¼ 1:932VLLM:

ð7:66Þ

In the general case, the average value of the output voltage is (Fig. 7.31):

Vo ¼ Vo1 þ Vo2 ¼ 3VLLM cos a þ 3VLLM cos a ¼ 6VLLM cos a: ð7:67Þ
p p p

Besides the average value (zero harmonic), the spectrum of the six-pulse three-
phase rectifier output voltage containsharmonics with a frequency equal to the
product of the frequency of the source and the number n ¼ 6k; k 2 N.

The output voltage of a twelve-pulse rectifier has harmonic frequencies, which
are multiples of 12 times the frequency of source 12kx1; k 2 N, where ω1 is the
frequency of source.

It is interesting to make a similar analysis of the input currents at six pulses and
twelve-pulses rectifiers. Besides the harmonic at the frequency of source, the input
current of a six-pulses rectifier containsharmonics (Fig. 7.32)

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428 7 AC/DC Converters–Rectifiers

Fig. 7.32 Waveforms of three-phase rectifier output voltages and output voltage of twelve-pulse
rectifier when input voltage is 3 × 380 V, 50 Hz

Isn ¼ Is1 ; Is1 ¼ pffiffi n ¼ 6k þ 1 ðSect: 7:6Þ
n 2 3I0;

The input current of the twelve-pulse rectifier from Fig. 7.31 is

is ¼ is1 þ is2 ð7:68Þ

where

2 pffiffi 1
3 5 1 1 1
is1 ¼ p I0 cos x1t À cos 5x1t þ 7 cos 7x1t À 11 cos 11x1t þ 13 cos 13x1t À ÁÁ Á

pffiffi 1 1 1
2 3 5 1 11 13
is2 ¼ p I0 cos x1t þ cos 5x1t À cos 7x1t À cos 11x1t þ cos 13x1t þ ÁÁ Á :

7

ð7:69Þ

From (7.68) and (7.69) it is obtained that

pffiffi 1
is ¼ 4 p 3 I0 11 1
cos x1t À cos 11x1t þ 13 cos 13x1t À . . . ð7:70Þ

Besides the harmonic at the frequency of source, the input current of the twelve-
pulse rectifier contains harmonics n ¼ 12k Æ 1; k 2 N :

Based on the foregoing it can be concluded that the twelve-pulse rectifier
compared to a six-pulse rectifier has:

• higher average value of the output voltage,
• less ripple of the output voltage,
• less influence of higher harmonics to the output voltage and the supply current.

A power system has the practical limitation of 12 pulses, because of the large
expense of producing high-voltage transformers with an appropriate phase shift. In
industrial applications, such rectifiers are realized up to 48 pulse.

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7.8 Rectifiers with Circuit for Power Factor Correction 429

7.8 Rectifiers with Circuit for Power Factor Correction

Rectifiers are often the input stage of many power electronic devices, such as
indirect AC/AC frequency converters, uninterruptible power supply sources, con-

verters in AC electrical drives, etc.
On the other side, the previously discussed rectifiers have an input current that

significantly deviates from the sinusoidal shape and its spectrum is a rich with
harmonics that are close to the fundamental harmonic. Diode and thyristor rectifiers
are nonlinear loads, and as such they are a source of harmonics in the input current.

These harmonics cause problems in the distribution network as well as an increase

of reactive power, higher losses in the transmission network, voltage distortion,

higher EMI, etc.
In order to better understand the problems, a single-phase diode rectifier is

observed. Seen from the input, the characteristics of the rectifier should be such that
the input current has a sinusoidal shape and in a phase with the input voltage. In the
other words, the rectifier should behave as a pure resistance load (Fig. 7.33). The
resistance Re in Fig. 7.33 is not a real resistance, but shows how the power of an AC
source should be delivered to a DC load. This resistance is known as the Loss Free

Resistor (LFR) [1].
Since the control of the rectifier output variable is required, the resistance Re is

controlled by the signal vc. The waveform of voltage at the output of the Graetz
bridge and its desired shape are shown in Fig. 7.34a, b, respectively.

The output voltage of the Graetz bridge and its desired shape are purely DC, so

the DC/DC converter with the output-to-input voltage ratio as shown in Fig. 7.35
should be put between the rectifier and the load. Therefore, it is necessary that the
DC/DC converter be regulated by a controller. The input variables of the controller
are iG and vG (voltage and current of the Graetz bridge) and its output variable in the
simplest realization is the duty ratio of the switching transistor D(t). The voltage
transfer ratio of the DC/DC converter (Fig. 7.35) is 1:M(D(t)), where

MðDðtÞÞ ¼ V : ð7:71Þ
VSMjsinðxtÞj

In this way, the rectifier topology shown in Fig. 7.36 is obtained. Based on the
obtained relationship between the output and the input voltage of the DC/DC
converter it is concluded that its transmission ratio changes from V/VSM to infinity.

iS Re(vC)
vS~

Fig. 7.33 Loss free resistor

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430 7 AC/DC Converters–Rectifiers

(a) vG(t)=|VSM sin(ωt)|
ωt
vG
VSM vO(t)=V

(b)

vO
V

ωt
Fig. 7.34 Waveform of voltage at output of Graetz bridge rectifier (a) and its desired shape (b)

vO
vG

V
VSM

π 2π 3π ωt

Fig. 7.35 Waveform of DC/DC converter output voltage in rectifier with circuit for power factor
correction

iG iO

iS D1 +
D3

DC /DC R vO
vS vG converter C

D4 D2

D(t)

Controler

Fig. 7.36 DC/DC converter for power factor correction connected to the output of Graetz bridge
rectifier

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7.8 Rectifiers with Circuit for Power Factor Correction 431

One of the converters that can provide such a transmission ratio is the boost
converter, on condition that V/VSM ≥ 1.

If power losses in the rectifier are neglected, the output power is equal to the

input power,

psðtÞ ¼ vsðtÞisðtÞ ¼ vGðtÞiGðtÞ ¼ voðtÞioðtÞ ð7:72Þ

ioðtÞ ¼ VS2M sin2ðxtÞ ¼ VS2M ð1 À cos 2xtÞ : ð7:73Þ
VRe 2VRe

By filtering the current i0, which is a function of the capacitance C connected in
parallel with the load, the average value of the current through the load R is

Io ¼ VS2M ð7:74Þ
2VRe

The topology of the rectifier with a circuit for power factor correctionbased on
the use of a boost DC/DC converter is shown in Fig. 7.37 [1]. The boost converter,
depending on the working conditions, can be in continuous or discontinuous current
mode.

If the boost converter works in continuous mode and the inductance L has
negligible influence to low order harmonics then the duty ratio can be written in the
following form

MðDðtÞÞ ¼ 1 1 : ð7:75Þ
À DðtÞ

From (7.71) and (7.75) it is obtained that

DðtÞ ¼ 1 À vg : ð7:76Þ
V

iG iO
DL
+
iS TC R vO
vS vG

D(t)

Controler

Fig. 7.37 Rectifier with boost converter for power factor correction [1]

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