Three-Phase, Full-Wave Controlled Bridge Rectifier ...read.pudn.com/downloads152/ebook/662309/power converter/Controlle… · Circuits with Passive Load I Impedance 219 load. InFig

7

Three-Phase, Full-Wave ControlledBridge Rectifier Circuits withPassive Load Impedance

Two half-wave, three-pulse controlled rectifiers of the type shown in Fig. 7.1can be combined in the formation of Fig. 7.1a. If the switches in the top half ofthe bridge (Th1Th3Th5) are gated and fired in antiphase with the switches in thebottom half (Th4Th6Th2) and the neutral wire N is omitted, the two half-wavebridge effects are added. The resultant load voltage eL (t) is then the sum ofthe two individual half bridges. The average load voltage Eav is then twice theaverage value of either of the two half-wave bridges acting individually.

Figure 7.1b shows the basic form of a three-phase, full-wave bridge rectifiercircuit. This is the same form as the uncontrolled bridge Fig. 6.1, and the number-ing notation is the same. Although the semiconductor switches in Fig. 7.1 areshown as silicon controlled rectifiers, they could equally well be any of the otherthree-terminal, gate-controlled switches of Table 1.1.

With a supply of zero impedance the three supply-phase voltages for thecircuit of Fig. 7.1 retain balanced sinusoidal form for any load condition. Thesevoltages are defined by the equations

eaN Em sin t (7.1)

ebN Em sin(t 120) (7.2)

ecN Em sin(t 240) (7.3)

The corresponding line-to-line voltages at the supply point are

Copyright 2004 by Marcel Dekker, Inc. All Rights Reserved.

Circuits with Passive Load I Impedance 217

FIG. 1 Three-phase, full-wave controlled bridge rectifier circuit: (a) depicted as two half-wave bridges with neutral connection and (b) conventional formation, without neutral.


Chapter 7218

eab eaN eNb eaN ebN 3Em sin(t 30) (7.4)

ebc 3Em sin(t 90) (7.5)

eca 3Em sin(t 210) (7.6)

Waveforms of the supply voltages are given in Figs. 6.2 and 7.2.The device numbering notation shown in the bridge rectifier circuit of Fig.

7.1 is standard for the three-phase, full-wave controlled bridge in both its rectifierand inverter modes of operation. To provide a current path from the supply sideto the load side requires the simultaneous conduction of at least two appropriateswitches. When one element of the upper group of switches and one of the lowergroup conducts, the corresponding line-to-line voltage is applied directly to the

FIG. 2 Voltage waveforms of the three-phase, full-wave controlled bridge rectifier withresistive load and ideal supply: (a) supply line voltages, (b) load voltage ( 0), (c)load voltage ( 30), (d) load voltage ( 60), and (e) load voltage ( 90).



load. In Fig. 7.1 the switches are depicted as silicon controlled rectifier types ofthyristor. For this reason the terminology Th is used in their description. If, forexample, the switches Th1 and Th6 conduct simultaneously, then line voltageeab is applied across the load. There are some switch combinations that are notpermissible. If, for example, the switches in any leg conduct simultaneously fromboth the top half and the bottom half of the bridge, then this would represent ashort circuit on the ac supply. To provide load current of the maximum possiblecontinuity and smoothness appropriate bridge switches must conduct in pairssequentially, for conduction intervals up to 120 or /3 radius of the supplyvoltage. The average load voltage and current are controlled by the firing-angleof the bridge thyristors, each measured from the crossover point of its respectivephase voltages.

7.1 RESISTIVE LOAD AND IDEAL SUPPLY

When the thyristor firing-angle is 0, the bridge operates like a diode rectifiercircuit with the waveforms given in Fig. 4.2. The corresponding conduction se-quence of the circuit devices is given in Fig. 6.2a, in which the upper tier repre-sents the upper half of the bridge. Supply line current ia (t) for the first cycle(Fig. 6.2f) is made up of four separate components contributed by the four separatecircuits shown in Fig. 7.3. An alternative representation of the device conductionpattern is shown in Fig. 6.2b in which the upper and lower tiers do not representthe upper and lower halves of the bridge, but this pattern shows that the thyristorswitches conduct in chronological order. The circuit operation possesses twodifferent modes, depending on the value of the firing angle. In the range 0 /3 the load voltage and current are continuous (Fig. 7.2c and d ), and anoncoming thyristor will instantly commutate an off-going thyristor. In the range/3 2/3, the load current becomes discontinuous because an off-goingthyristor extinguishes before the corresponding on-coming thyristor is fired. Forresistive loads with negligible supply reactance, both the load current and thesupply current are always made up of parts of sinusoids, patterned from the linevoltages. For all firing angles the sequence order of thyristor conduction in thecircuit of Fig. 7.1 is always that shown in Fig. 6.2a. However, the onset ofconduction is delayed, after the phase voltage crossover at t 30, until theappropriate forward biassed thyristors are gated and fired.

Consider operation at 30, for example. In Fig. 7.2 forward-bias voltageoccurs on thyristors Th1 and Th6 at t 30. If the firing-angle is set at 30, conduction via Th1 and Th6 (Fig.7.3a) does not begin until t 30 60 and then continues for 60. At t 90 120, the dominant linevoltage is eac, thyristor Th6 is reverse biassed, and conduction continues via thenewly fired thyristor Th2 (Fig. 7.3b) for a further 60. At t 180, thyristorTh1 is commutated off by the switching in of Th3 and line current ia (Fig. 7.4c)


Chapter 7220

FIG. 3 Equivalent circuits of conduction with resistive load and ideal supply: (a) 30 t 90 (Th1, Th6 on), (b) 90 t 150 (Th1, Th2 on), (c) 210 t 270 (Th3,Th4 on), and (d) 270 t 330 (Th5, Th4 on).

becomes zero so that the load current path is provided by Th2 and Th3 for afurther 60 interval. When t 240, the dominant line voltage is eba. The firingof Th4 transfers the load current from Th2, (Fig. 7.3c), and supply current resumesin phase a in the opposite direction. After a further 60, at t 300, line voltageeca is dominant (Fig. 7.4a), and the switching in of Th5 causes the commutationof Th3. Thyristors Th4Th5 then provide the load current path which is fed fromphase c to phase a, as shown in Fig. 7.3d.



FIG. 4 Waveforms of the three-phase, full-wave controlled bridge rectifier with resistiveload and ideal supply: (a) supply line voltages, (b) supply line current ( 0), (c) supplyline current ( 30), and (d) supply line current ( 60).

7.1.1 Load-Side Quantities

The sequence of thyristor firing creates the load voltage (and current) waveformsshown in Fig. 7.2. In mode I operation, where 0 60, the average voltagecan be obtained by taking any 60 interval of eL (t).

For 30 t 90,


Chapter 7222

e t E tL m( ) sin( )ω ωα

α= +

+

+3 30

30

90o

o

o

(7.7)

The average value of Eq. (7.7) in terms of peak phase voltage Em is

E E t d t

E

E

av m

m

av

= +

=

=

+

+

∫3

3 30

3 3

90

0

πω ω

πα

α

α

α

30

oo

o

sin( )

cos

cos (7.8)

where

E E Eav m m0

3 31 654= =

π. (7.9)

The average current Iav is, therefore, a function of of

IE

R

E

Ravav av= = 0 cosα (7.10)

With resistive load the instantaneous load voltage is always positive. When theanode voltage of a thyristor goes negative extinction occurs. At a firing angle 60, the load voltage and current therefore become discontinuous, as shownin Fig. 7.2. This represents a different mode of operation from 60 and theload voltage is then described by the following equation:

For 60 120,

e t E tL m( ) sin( )ω ωα

= ++

3 3030

150o

o

o

(7.11)

The average value of Eq. (7.11) is given by

E Eav m= + + 3 3

1 60π

αcos( )o

(7.12)

When 60, Eq. (7.8) and (7.12) give identical results. At 120, theaverage load voltage becomes zero.

The waveforms of Fig. 7.2 show that the load voltage waveform has arepetition rate six times that of the phase voltage. This means that the lowestripple frequency is six times fundamental frequency. If a Fourier analysis is



performed on the load voltage waveform the two lowest order harmonics are thedc level (i.e., the average value) followed by the sixth harmonic.

Load power dissipation can be found from the rms load current. The rmsor effective load current IL is defined as

I i t d tL L= ∫1

22

0

2

πω ω

π( ) (7.13)

where iL eL/R from Eq. (7.7) or (7.11).Comparing waveforms of the supply and load currents at a given firing

angle, one would anticipate that IL Ia because iL (t) has a greater area underthe curve than does ia (t) and therefore i2L (t) is likely to be greater than i2a(t). The substitution of Eq. (7.7) or (7.11) respectively into Eq. (7.13) gives

IE

RLm

0 60

3

2

2 3 3 2

≤ ≤= +

απ α

πo

cos(7.14)

IE

RLm

60 120

3

2

4 6 3 2 60o o

o

≤ ≤= − − −

απ α α

πsin( )

(7.15)

Power dissipation in the bridge circuit of Fig. 7.1 is assumed to occur entirelyin the load resistor. This may be obtained from the rms (not the average) loadcurrent.

PL I2LR (7.16)

Combining Eqs. (7.14) and (7.15) with Eq. (7.16)

PE

RLm

0 60

3

42 3 3 2

2

≤ ≤= +

α ππ α

o( cos )

(7.17)

PE

RLm

60 120

3

44 6 3 2 60

2

o oo

≤ ≤= − − − α π

π α αsin( )(7.18)

The load side properties of the bridge are summarised in Table 7.1.

7.1.2 Supply-Side Quantities

Waveforms of the currents on the supply side of the bridge are shown in Fig.7.4 for mode I operation. The instantaneous supply currents for the two modesof operation are defined by the following: For 0 60,

i tE

Rt

E

Rta

m m( ) sin( ),

,sin( )

...

...ω ω ω

α

α

α= + + −

+

+

+

330

330

30

90o

o

o

o

990

150

o

o

,

,

...

...α+

(7.19)


Chapter 7224

TABLE 7.1 Three-Phase, Full-Wave Controlled Bridge Rectifier with Ideal Supply:Load-Side Properties

Resistive load Highly inductive load

Instantaneous loadvoltage

Average loadvoltage

RMS load current

Load power

0

3 3090

30

≤ ≤ °

+ °+ °+ °

α 60αα

E tm sin( )ω

,

60° 120

3 30150

30

≤ ≤ °

+ °°

+ °

α

αE tm sin( )ω

,

0 60

1 60

0

0

≤ ≤ °

° ≤ ≤ °+ + °[ ]

αα

60 α 120α

E

E

av

av

cos

cos( )

,

,

0 60

3

2

2 3 3 2

3

2

≤ ≤ °

+

° ≤ ≤ °

°

α

π απ

60 α 120

4π − 6α − 3 α

E

R

E

R

m

m

cos

sin (2 –60 )

π

,

,

0 60

3

42 3 3 2

3

4

2

2

≤ ≤ °

+

° ≤ ≤ °

α

π α)

60 α 120

4π − 6α − 3 α −

E

R

E

R

m

m

( cos

sin (2 660 )°[ ]

,

,

E

Rav0

22cos α

E

Rav0 cosα

Eav0cosα

Eav0cosα

For 60 120,

i tE

Rt

E

Rta

m m( ) sin( ) sin( )ω ω ωα α

= + + −+ +

330

330

30

150

90

210o

o

o

o

o

o

(7.20)



The rms values of the supply line currents may be obtained via the definingintegral Eq. (7.13). Substituting Eqs. (7.19) and (7.20) into the form of Eq. (7.13)gives

IE

Ra

m

0 60 2

2 3 3 2

≤ ≤= +

απ α

πo

cos(7.21)

IE

Ra

m

60 120 2

4 6 3 2 60o o

o

≤ ≤= − − −

απ α α

πsin( )

(7.22)

At 60, Eq. (7.21) and (7.22) are found to be identical.Comparison of the rms supply and load currents gives, for both modes of

operation,

I IL a= 3

2 (7.23)

The relationship of Eq. (7.23) is found to be identical to that obtained for anuncontrolled, full-wave bridge with resistive load (Table 6.1).

7.1.3 Operating Power Factor

With perfect switches the power dissipated at the load must be equal to the powerat the supply point. This provides a method of calculating the operating powerfactor

PFP

E IL

a a

=3 (7.24)

Substituting for PL, Eq. (7.17) or (7.18), and for Ia, Eq. (7.21) or (7.22), into Eq.(7.24), noting that Ea Em/2

PF0 60

2 3 3 2

4≤ ≤= +

απ α

πo

cos(7.25)

PF60 120

4 6 3 2 60

4o o

o

≤ ≤= − − −

απ α α

πsin( )

(7.26)

When 0, Eq. (7.25) has the value

PFα

ππ=

= + =0

2 3 3

40 956.

(7.27)


Chapter 7226

which agrees with Eq. (6.11) for the uncontrolled (diode) bridge. The powerfactor of the three-phase bridge rectifier circuit, as for any circuit, linear or nonlin-ear, with sinusoidal supply voltages, can be represented as the product of a distor-tion factor and a displacement factor The distortion factor is largely related toload impedance nonlinearity; in this case, the switching action of the thyristors.The displacement factor is the cosine of the phase angle between the fundamentalcomponents of the supply voltage and current. This angle is partly due to theload impedance phase angle but mainly due here to the delay angle of the currentintroduced by the thyristors. Both the current displacement factor and the currentdistortion factor are functions of the fundamental component of the supply current.This is calculated in terms of the Fourier coefficients a1 and b1, quoted from theAppendix for the order n 1.

a i t t d t

b i t t d t

1 0

2

1 0

2

1

1

=

=

∫

∫

πω ω ω

πω ω ω

π

π

( )cos

( )sin

Expressions for the coefficients a1 and b1 are given in Table 7.2.The current displacement factor and current distortion factor are given by

Displacement factor of supply current = = −cos cos tanψ1

1 1

1

a

b

(7.28)

Distortion factor of supply current = =I

I

I

Ia

a

1 1

(7.29)

where

Ic a b

a1

1 12

12

2 2= =

+(7.30)

Expressions for the current displacement factor and current distortion factor, forboth modes of operation, are also given in Table 7.2. The product of these isseen to satisfy the defining relation

PF (displacement factor)(distortion factor) (7.31)

7.1.4 Shunt Capacitor Compensation

Some degree of power factor correction can be obtained by connecting equallossless capacitors C across the supply terminals (Fig. 7.5). The bridge voltagesand currents are unchanged and so is the circuit power dissipation. The capacitor


Circuits

with

Passive

Load

IIm

pedance

22

7

TABLE 7.2 Three-Phase, Full-Wave Controlled Bridge Rectifier with Ideal Supply: Supply-Side Properties

Resistive Load

Mode I (0 60°) Mode II (0 120°) Highly inductive load

Fourier coefficientsof supply current

a1

b1

Fundamental rmssupply current I1

RMS current I

Current displacementfactor (cos 1)

Current distortionfactor (I1/I)

Power factor

− 3 3

22

E

Rm

παsin

E

Rm

22 3 3 2

ππ α( cos )+

E

Rm

2

2 3 3 2cosπ α+π

2 3 3 2

27 4 12 3 22

π απ π α+

+ +cos

cos

27 4 12 3 2

4 2 3 3 2

+ ++

π π απ π α

2 cos

( cos (

2 3 3 2

4

π απ

+ cos

− + − °[ ]3

21 2 60

E

Rm

παcos( )

E

Rm

24 6 3 2 60

ππ α α− − − °[ ]sin( )

1

212

12a b+

E

Rm

2

4 6 2 60π α απ

− − − °sin(3 (

4 6 3 2 60

9 1 2 60 4 6 3 2 602

π α αα π α α

− − − °+ − °[ ] + − − − °[ ]

sin( )

cos( ) sin( )2

4 6 3 2 60

4

π α απ

− − − °sin( )

91 2

2πα)E

Rm ( cos+

1

212

12a b+

− 92

2παE

Rm sin

3 2

παE

Rm cos

3

π

3

παcos

− 3 3

22

E

Rm

παsin

cos

Copyright

2004

byM

arcelD

ekker,Inc.All

Rights

Reserved.

Chapter 7228

current ic (t) is a continuous function unaffected by thyristor switching. In phasea, for example, the instantaneous capacitor current ica (t) is given by

i tE

Xt

E

Xtca

m

c

m

c

( ) sin( ) cosω ω ω= + =90o

(7.32)

where Xc 1/2fC.The corresponding instantaneous supply current isa

(t) in phase a is now

isa(t) ia (t) ica

(t) (7.33)

Therefore,

iE

Xt

E

Rt

E

Rt

sm

c

m

m

a

0 60 30

90330

3

≤ ≤ +

+= + +

+ −

α α

αω ω

ω

o

o

o

o

cos sin( )

sin( 33090

210o

o

o

)α+ (7.34)

FIG. 5 Three phase-bridge circuit with supply side capacitors.



iE

Xt

E

Rt

E

R

sm

c

m

m

a

60 120 30

150330

3

o o

o

o

o

≤ ≤ += + +

+

α αω ω

ω

cos sin( )

sin( tt −+

3090

210o

o

o

)α (7.35)

The substitution of Eqs. (7.34) and (7.35), respectively, into Eq. (7.13) givesmodified expressions for the rms supply current,

IE

R

R

X

R

Xsm

c ca

0 60 2

2 3 3 2 3 32

2

2≤ ≤

= + + −α

π απ π

αo

cossin

(7.36)

IE

R

R

X

R

Xsm

c ca

60 120 2

4 6 3 2 60 32 30

2

2o o

o

≤ ≤= + − − − − +

α

π α απ π

αsin( )sin( oo ) + 1

(7.37)

Since the system voltages and power are unchanged by the presence of the capaci-tor the power factor will be improved if the ms supply current with the capacitoris reduced below the level of the bridge rms current (which is the supply rmscurrent in the absence of the capacitor).

If the compensated power factor is denoted by PFc, the ratio of compensatedto uncompensated power factor is found to be

PF

PF R Xc

c0 60

1 2 3 3 2

1 2 3 3 22 2≤ ≤

= ++ + −α

π π απ π αo

( / )( cos )

/ ( / )( cos ) (33 3 2R Xc/ )sinπ α (7.38)

PF

PF R Xc

c60 120

1 4 6 3 2 60

1 42 2o o

o

≤ ≤=

− − − +α

π π α α

π

( / ) sin( )

/ ( / ) ππ α α π α− − − − + + 6 3 2 60 3 2 30 1sin( ) ( / ) sin( )o oR Xc

(7.39)

The ratio PFc/PF will be greater than unity, indicating that power factor improve-ment has occurred, when the following inequalities are true:

For 0 60,

R

X

R

Xc c

−

<3 3

2 0π

αsin(7.40)


Chapter 7230

For 60 120,

R

X

R

Xc c

− + +

<32 30 1 0

παsin( )o

(7.41)

For the limiting values of firing angle , being zero in Eq. (7.40) and 120 inEq. (7.41) it is found that R/Xc would need to be negative to cause power factorimprovement. In other words, when 0, the use of capacitance does not giveimprovement but actually makes the power factor worse.

The use of supply-point capacitance aims to reduce the displacement angles1 to zero so that displacement factor cos s1 1.0, which is its highest realizablevalue. From Eq. (7.28) it is seen that s1 0 when as1 0. If Eqs. (7.34) and(7.35) are substituted into Eq. (A.9) in the Appendix, it is found that

aE

X

E

Rsm

c

m

1

0 60

3 3

22

≤ ≤= −

α πα

o

sin(7.42)

aE

X

E

Rsm

c

m

1

60 120

3

21 2 60

o o

o

≤ ≤= − + −

α παcos( )

(7.43)

Unity displacement factor and maximum power factor compensation are thereforeobtained by separately setting Eq. (7.42) and (7.43) to zero:

For 0 60,

R

Xc

− =3 3

22 0

παsin

(7.44)

For 60 120,

R

Xc

− + − =3

21 2 60 0

παcos( )o for 60° ≤ α ≤ 120°

(7.45)

When the conditions of (7.44),(7.45) are realized the power factor has attainedits maximum possible value due to capacitor compensation. For 0 60,

PFccos

cos

( cos ) sinψ

π α

π π α α1 1

2 3 3 2

4 2 3 3 2 27 22== +

+ − (7.46)

For 60 120,

PFccos

sin( )

sin( )

o

oψ

π α α

π π α α1 1

4 6 3 2 60

4 4 6 3 2 60 9 1== − − −

− − − − + ccos( )2 602

α − o

(7.47)



The degree of power factor improvement realizable by capacitor compensationis zero at 0 and is small for small firing angles. For firing angles in the midrange, 30 60, significant improvement is possible.

Note that the criteria of Eqs. (7.44) and (7.45) are not the same as thecriteria of Eqs. (7.40) and (7.41) because they do not refer to the same constraint.

7.1.5 Worked ExamplesExample 7.1 A three-phase, full-wave controlled bridge rectifier has a

resistive load, R 100 . The three-phase supply 415 V, 50 Hz may be consid-ered ideal. Calculate the average load voltage and the power dissipation at (1) 45 and (2) 90.

At 45 from Eq. (7.8),

E Eav m= 3 3

παcos

where Em is the peak value of the phase voltage. Assuming that 415 V representsthe rms value of the line voltage, then Em has the value

Em = ×4152

3

Therefore,

Eav0

3 3415

2

3

1

2396 3= × × × =

π. V

The power is given by Eq. (7.17),

PE

RLm= +

3

42 3 3 2

2

ππ α( cos )

At 45 /4,

PL = × × × =3

4

415

100

2

32 1722

2

ππ W

At 90 /2, from Eq. (7.12),

E Eav m= + −

= × × −

=

3 31 2 60

3 3415

2

31

3

275 1

πα

π

cos( )

.

o

V


Chapter 7232

The power is now given by Eq. (7.18)

PE

RLm= − − −

= × × − −

3

44 6 3 2 60

3

4

415

100

2

34 3

3 3

2

2

2

ππ α α

ππ π

osin( )

= − =415

2002 589 149

2

ππ( . ) W

Example 7.2 For a three-phase, full-wave controlled bridge rectifier withresistive load and ideal supply, obtain a value for the load current ripple factor,when 60, compared with uncontrolled operation.

The rms values of the load current in the two modes of operation are givenby Eq. (7.14) and (7.15). The average values are given in Eqs. (7.8), (7.10), and(7.12). Taking the ratio IL/Iav it is found that for 0 60,

I

I

E R

E RL

av

m

m

=+

= +⋅

( / ) ( cos ) /

( / ) cos

cos

3 2 2 3 3 2

3 3

6

2 3 3 2

π α ππ α

π π απ ccos2 α (Ex. 7.2a)

and for 60 120,

I

I

E R

E RL

av

m

m

=− − −

+ +( / ) [ sin( )] /

( / ) cos( )

3 2 4 6 3 2 60

3 3 1 60

π α α ππ α

o

o

= − − −

+ +

π π α α

π α6

4 6 3 2 60

1 602

sin( )

cos( )

o

o(Ex. 7.2b)

From each of the relations (Ex. 7.2a) and (Ex. 7.2b) at 60, it is found that

I

IL

av

= 1 134.

The ripple factor is, from Eq. (2.11),

RFI

IL

av

=

− =

2

1 0 535.



From ratio (Ex. 7.2a) above, at 0,

I

IL

av

= + = =π ππ

π6

2 3 3

61 9115 1 001( . ) .

This latter result confirms the values given in Eqs. (6.4) and (6.7). The ripplefactor at 0 is, therefore, zero, but it becomes significant as the firing angleis retarded.

Example 7.3 Calculate the operating power factor for the three-phase,full-wave, bridge rectifier of Example 7.1 at (1) 45 and (2) 90. Ifthe maximum possible compensation by capacitance correction is realized, calcu-late the new values of power factor and the values of capacitance required.

At 45, from Eq. (7.25),

PF = +2 3 3 90

4

ππcos o

= =1

20 707.

At 90, from Eq. (7.26),

PF = − −4 3 3 120

4

π ππ

sin o

=−

=

3 32

40 21

π

π.

If the maximum realizable compensation is achieved the power factor is thengiven by Eqs. (7.46) and (7.47).

At 45, from Eq. (7.46),

PFc = +

+ −

=−

= =

2 3 3 90

4 2 3 3 90 27 90

2

8 27

2

7 210

2

2

π

π π

π

π

π

cos

( cos ) sin

.

o

o o

..87

At 90, from Eq. (7.47),


Chapter 7234

PFc = − −

− − − +

= −

4 3 3 120

4 4 3 3 120 9 1 120

3 3

2

π π

π π π

π

sin

( sin ) ( cos )

/

o

o o

22

43 3

29 1

12

0 5446 83 2 25

0 254

2

.. .

.

π π −

− −

=−

=

The criteria for zero displacement factor are given in Eqs. (7.44) and (7.45). At 45, from Eq. (7.44),

12

3 3

22

1

2 50

3 3

2 10052 6

XfC

R

C

c

= =

= =

ππ

α

π πµ F

sin

.

At 90, from Eq. (7.45),

1 3

21 2 60

1

2 50

3

2 1001

1

2

15 2

X R

C

c

= + −

= −

=

πα

π π

µ

cos( )

.

o

F

7.2 HIGHLY INDUCTIVE LOAD AND IDEALSUPPLY

7.2.1 Load-Side Quantities

The three-phase, full-wave, controlled bridge rectifier is most commonly used inapplications where the load impedance is highly inductive. Load inductance isoften introduced in the form of a large inductor in series with the load resistor(Fig. 7.6). If the load-side inductance smooths the load current to make it, verynearly, a pure direct current as shown in Fig. 7.7b, then

iL (t) Iav IL Im (7.48)



FIG. 6 Three-phase, full-wave controlled bridge rectifier circuit with series R-L load.

With a smooth load current there is zero average voltage on the smoothinginductor and the average load voltage falls entirely on the load resistor sothat Eq. (7.10) remains true. The patterns of the load current and supplycurrents are shown in Fig. 7.7 for firing angles up to 60. Unlike thecase with resistive load, the load current is continuous for all values of a inthe control range, and only one mode of operation occurs. The averagevoltage, for all firing-angles, is identical to that derived in Eq. (7.7) with thecorresponding average current in Eq. (7.10).

It is seen from Eq. (7.10) that the average load current becomes zeroat 90. The controlled range with highly inductive load is thereforesmaller than with resistive load, as shown in Fig. 7.8. With a smooth loadcurrent there is no ripple component at all, and the current ripple factor hasthe ideal value of zero.

For 60, the instantaneous load voltage, with highly inductive load,is the same as for resistive load. At 75, eL (t) contains a small negativecomponent for part of the cycle. When 90, the instantaneous load voltagehas positive segments identical to those in Fig. 7.3e, but these are balancedby corresponding negative segments to give an average value of zero. Althoughthe load current ripple factor is zero, the load voltage ripple factor is determinedby the ratio EL/Eav. From Eqs. (7.8) and (7.14), with 60,

E

EL

av

= +π π απ α6

2 3 3 22

cos

cos (7.49)

The load power dissipation is proportional to the square of the load rms current,and therefore, substituting Eqs. (7.10) into Eq. (7.16),


Chapter 7236

FIG. 7 Waveforms of the three-phase, full-wave controlled bridge rectifier circuit withhighly inductive load and ideal supply: (a) supply line voltages, (b) load current ( 0), (c) supply line current ia ( 0), (d) supply line current ia ( 30), and (e) supplyline current ia ( 60).

P I R I R

E

R

E

R

L L av

av m

= =

= =

2 2

2

22

220

27cos cosα

πα (7.50)

The load-side properties are summarized in Table 7.1.



FIG. 8 Average load current versus SCR firing angle for the three-phase, full-wavecontrolled bridge rectifier circuit with ideal supply.

7.2.2 Supply-Side QuantitiesThe supply current ia (t) shown in Fig. 7.7 is defined by the equation

i tE

R

E

Ra

av av( ) cos cosω α αα

α

α

α= −

+

+

+

+0 0

30

150

210

330

o

o

o

o

(7.51)

Since the rms values of the negative and positive parts of the wave are identical,the rms supply current Ia is given by

IE

Rd t

E

R

a

av

av

=

=

+

+

∫

+

1

1

0

0

2

30

150

15

πα ω

απ

ω

α

α

α

cos

cos

o

o

00

30

o

oα +

t


Chapter 7238

=

=

=

2

3

3 2

2

3

0E

R

E

R

I

av

m

av

cos

cos

α

απ

(7.52)

The value in Eq. (7.52) is found to be 2 times the corresponding value for ahalf-wave rectifier, given in Eq. (5.39), and is identical to Eq. (7.23) for the caseof resistive load.

The operating power factor of the bridge can be obtained by substitutingEqs. (7.50) and (7.51) into Eq. (7.24), noting that Ea Em/2.

PFP

E IL

a a

=3

=

=

( / ) cos

( / )( cos /)

cos

27

3 2 3 2

3

2 2 2E R

E E Rm

m m

π αα π

πα (7.53)

The power factor also is found to be 2 times the value for a three-phase, half-wave controlled bridge rectifier and has the well-known value of 3/, or 0.955,for 0 (or diode bridge) operation. A Fourier analysis of the supply currentia (t) shows that the coefficients a1 and b1 below are valid for the fundamental(supply frequency) component

a i t t d t

E

Rt

a

av

1 0

21

20

150

30

=

= ( )

∫

+

+

πω ω ω

πα ω

π

α

α

( )cos

cos sino

o

= −

= −

2 3

92

0

2

πα α

πα

E

R

E

R

av

m

sin cos

sin (7.54)



b i t t d t

E

Rt

a

av

1 0

21

20

150

30

=

= −( )

∫

+

+

πω ω ω

πα ω

π

α

α

( )sin

cos coso

o

=

= +

2 3

91 2

0 2

2

πα

πα

E

R

E

R

av

m

cos

( cos ) (7.55)

Equations (7.54) and (7.55) can be used to obtain a very important relationship

a

b1

11

2

1 2= −

+= − =sin

costan tan

αα

α ψ(7.56)

From Eq. (7.56) it can be seen that the displacement angle 1 of the input currentis equal to the firing angle (the negative sign representing delayed firing):

1 (7.57)

The displacement factor cos 1 is therefore equal to the cosine of the delayedfiring angle

cos 1 cos (7.58)

The relationship of Eq. (7.58) is true for both half-wave and full-wave bridgeswith highly inductive load. It is not true for bridges with purely resistive loading.The distortion factor of the input current is obtained by combining Eqs. (7.29),(7.30), (7.52), (7.54), and (7.55).

Distortion factor of

=

I

Ia

a

1

1 2 9 2( / )( / )π (( / ) sin ( cos )

( / )cos

E R

E Rm

m

2 22 1 2

3 2

3

α απ α

π

+ +

=

=the supply currents

(7.59)

The product of the displacement factor Eq. (7.58) and distortion factor Eq. (7.59)is seen to give the power factor Eq. (7.53). Some of the supply-side propertiesof the inductively loaded bridge are included in Table 7.2.

For any balanced three-phase load with sinusoidal supply voltage, the realor active power P is given by


Chapter 7240

P 3EI1 cos 1 (7.60)

where I1 the rms value of the fundamental component of the supply current andcos1 is the displacement factor (not the power factor). Substituting Eqs. (7.52)and (7.58) into Eq. (7.60) gives

PE

Rm= 27

2

22

παcos (7.61)

which is seen to be equal to the power PL dissipated in the load resistor, Eq.(7.50).

7.2.3 Shunt Capacitor Compensation

If equal capacitors C are connected in star at the supply point (Fig. 7.5), theinstantaneous supply current is given by

i tE

Xt

E

R

E

Rsm

c

m m

a( ) cos cos cosω ω

πα

πα

α

α

α

α= + −

+

+

+

+3 3 3 3

30

150

210o

o

o

3330o

(7.62)

The substitution of Eq. (7.62) into Eq. (7.13) gives an expression for the rmssupply current

IE

Xt

E

Rsm

c

m

a=

+

∫

+

+1 3 3

0 30

150

πω

πα

π

α

α

cos coso

o

= − +

2

2

2 2 22

2

182

36

d t

E R

X

R

Xm

c c

ω

πα

παsin cos

(7.63)

When the capacitance is absent, Xc becomes infinitely large and Eq. (7.63) reducesto Eq. (7.52). The power flow and the terminal voltage are unaffected by theconnection of the capacitors. The compensated power factor is given by combin-ing Eqs. (7.61) and (7.63)

PFP

E I

R X R X

ca s

c c

a

=

=− +

3

9

4 9 2 2 9

2 2

2 2 2 2

( / ) cos

/ ( / )( / )sin ( / )co

π α

π α π ss2 α (7.64)

The ratio of the compensated power factor to the uncompensated power factoris given by the ratio of the load current to the supply current



PF

PF

I

I R X R Xc a

s c ca

= =− +

( / ) cos

/ ( / )( / )sin ( / )c

18

2 9 2 18

2 2

2 2 2 2

π απ α π oos2 α (7.65)

The power factor is therefore improved when PFc/PF 1 which occurs when

R

X

R

Xc c2

182 0

2−

<

παsin

(7.66)

Examination of the inequality (7.66) shows that power factor improvement occurswhen 0 C (18sin)/2RFourier coefficient as1 for the fundamental component of the compensated

supply current is given by

a i t t d t

E

Xt

E

R

s s

m

c

m

a1

1

2 3 3

0

2

2

0

=

=

+

∫

∫

πω ω ω

πω

π

π

π

( )cos

cos cosαα ω ω

πα

α

α

cos

sin

t d t

E

X

E

Rm

c

m

= −

+

+

30

150

92

2

o

o

(7.67)

When C 0, Eq. (7.67) reduces to Eq. (7.54).To obtain the maximum value of the displacement factor, coefficient as1

must be zero. The condition for maximum realizable capacitor compensator istherefore, from Eq. (7.67),

CR

= 9 12

2π ωαsin (7.68)

Setting Eq. (7.67) to zero and substituting into Eq. (7.64) gives the maximumpower factor achievable by terminal capacitor compensation.

PFcmax

cos

sin

=−

3

19

22

πα

πα

(7.69)

For any nonzero value of , it is seen that the uncompensated power factor (3/)cos is improved due to optimal capacitor compensation, as illustrated in Fig.7.9. Over most of the firing-angle range, the possible degree of power factorimprovement is substantial. A disadvantage of power factor compensation by the


Chapter 7242

FIG. 9 Power factor versus firing angle for the three-phase, full-wave controlled bridgerectifier with highly inductive load and ideal supply.

use of capacitors is that for fixed-load resistance, the value of the optimal capacitorvaries with firing angle.

7.2.4 Worked Examples

Example 7.4 A three-phase, full-wave, controlled bridge rectifier containssix ideal thyristor switches and is fed from an ideal three-phase voltage sourceof 240 V, 50 Hz. The load resistor R 10 is connected in series with a largesmoothing inductor. Calculate the average load voltage and the power dissipationat (1) 30 and (2) 60.

If 240 V. represents the rms value of the line voltage, then the peak phasevoltage Em is given by



Em = 2

3240

From Eq. (7.8)

Eav = × ×3 3 2

3240

παcos

= 324 αcos

At 30, Eav 280.6VAt 60, Eav 162VThe power dissipation is given by Eq. (7.50),

P I RE

Ravav= =22

At 30, P 7.863 kW. At 60, P 2.625 kW.

Example 7.5 For the three-phase bridge of Example 7.4 calculate thedisplacement factor, the distortion factor, and the power factor at (1) 30and (2) 60.

From Eq. (7.58) it is seen that the displacement factor is given by

Displacement factor cos 1 cos

At 30, cos 0.866 3/2At 60, cos 0.5

Because the wave shape of the supply current is not affected by the firing-angle of the bridge thyristors (although the magnitude is affected) the supplycurrent distortion factor is constant. From Eq. (7.59),

Distortion factor = =30 955

π.

For loads with sinusoidal supply voltage the power factor, seen from the supplypoint, is the product of the displacement factor and the distortion factor:

PF = 3

παcos

At 30, PF 0.827At 60, PF 0.478

Example 7.6 For the three-phase bridge rectifier of Example 7.4 calculatethe required voltage and current ratings of the bridge thyristors

In a three-phase, full-wave, thyristor bridge the maximum voltage on anindividual switch is the peak value of the line voltage


Chapter 7244

Emax 2Eline

where Eline is the rms value of the line voltage. Therefore,

Emax 2 240 339.4 V

(Note that Emax is 3 times the peak value Em of the phase voltage.)From Eq. (7.52) the rms value of the supply current is

I IE

Ra avm= =2

3

3 2

παcos

but each thyristor conducts only one (positive) pulse of current every supplyvoltage cycle. In Fig. 7.6, for example, thyristor Th1 conducts only the positivepulses of current ia (t) shown in Fig. 7.7. Therefore,

I i t d tTh a1

1

22

30

150=

+

+

∫πω ω

α

α( )

o

o

The defining expression for ITh above is seen to have the value 1/2 that of Ia

in Eq. (7.52),

II

E

R

Th

a

m

1

1

2

30

3 2

3240

1

1018 7

=

= =

= × × × =

πα

π

at

A

o

.

Example 7.7 The three-phase, full-wave bridge rectifier of Example 7.4is to have its power factor compensated by the connection of equal, star-connectedcapacitors at the supply point. Calculate the maximum value of capacitance thatwill result in power factor improvement and the optimum capacitance that willgive the maximum realizable power factor improvement at (1) 30 and (2) 60. In each ease compare the compensated power factor with the correspond-ing uncompensated value.

The criterion for power factor improvement is defined by Eq. (7.66), whichshows that

CRmax

sin= 18 22

αωπ

At 30,



Cmax 503 F

At 60,

Cmax 503 F

The optimum value of capacitance that will cause unity displacement factor andmaximum power factor is given in Eq. (7.68)

CRopt = 9

2ωπαsin

Note that Copt Cmax/2.For both firing angles,

Copt 251.5 F

In the presence of optimum capacitance the power factor is obtained from Eq.(7.69)

PFcmax=

−

3

92 2

cos

sin

α

π α

At 30, PFc 0.941, which compares with the uncompensated value PF 0.827 (Example 7.4). At 60, PFc 0.85, which compares with theuncompensated value PF 0.478 (Example 7.4).

7.3 HIGHLY INDUCTIVE LOAD IN THE PRESENCEOF SUPPLY IMPEDANCE

Let the three-phase, full-wave bridge rectifier circuit now contain a series induct-ance Ls in each supply line. Because of the nonsinusoidal currents drawn fromthe supply, the voltages at the bridge input terminals abc (Fig. 7.10) are notsinusoidal but are given by

e e Ldi

dtaN AN sa= − (7.70)

e e Ldi

dtbN BN sb= − (7.71)

e e Ldi

dtcN CN sc= − (7.72)


Chapter 7246

FIG. 10 Three-phase, full-wave controlled bridge rectifier circuit including supply in-ductance.

where eAN, eBN, and eCN are now defined by the form of Eqs. (7.1) to (7.3),respectively.

The onset of ignition through any particular rectifier is delayed due bothto the firing angle , as described in the previous sections, and also due to overlapcreated by the supply inductance. For normal bridge operation at full load theoverlap angle is typically 20 to 25 and is usually less than 60. Operation ofthe bridge can be identified in several different modes.

7.3.1 Mode I Operation (0 60)

A common mode of operation is where two thyristors conduct for most of thecycle, except in the commutation or overlap intervals when a third thyristor alsoconducts. Referring to Fig. 7.10, the sequence of conduction is 12, 123, 23, 234,34, 345, 45, 456, 56, 561, 61, 612. The resulting waveforms are given in Fig.7.11.

Consider operation at the instant t of the cycle. Thyristors Th1 andTh2 have been conducting (Fig. 7.11a and c), so that

At t 150,



FIG. 11 Waveforms of the three-phase, full-wave controlled bridge rectifier with highlyinductive load, in the presence of supply inductance: (a) supply phase voltages, (b) supplyline voltages, (c) thyristor firing pattern, (d) supply voltages ( 30, 10), (e) loadvoltage ( 30, 10), and (f) supply currents ( 30, 10).


Chapter 7248

i1 ia Iav (7.73)

i2 ic ia Iav (7.74)

ib 0 (7.75)

eL eac eaN ecN 3Em cos(t 120) (7.76)

At t , thyristor Th3 is gated. Since eb is positive, the thyristor switches onconnecting its cathode to the positive load terminal (Fig. 7.12b). Since points aand b are now joined, they have the same potential with respect to neutral N,which is the average of the corresponding open circuit voltages, (eaN ebN)/2 (Em/2)sin(t 60). But, simultaneously, the negative point of the load (pointc) has a potential with respect to N of Em sin (t 240). Therefore, during theoverlap period

e tE

t E t

E t

Lm

m

m

( ) sin( ) sin( )

sin( )

ω ω ω

ω

= − − −

= −

260 240

3

260

o o

o

(7.77)

This is seen to be three times the value of (eaN ebN)/2 and is in time phasewith it.

At t 150 , thyristor Th1 is extinguished by naturalcommutation after its current ia has fallen to zero. Current ib, which started toflow at t when Th3 was fired, reaches the value Iav at t andtakes over the load current relinquished by Th1. Current then flows from b to cand voltage ebc is impressed upon the load. During the overlap interval 150 t 150 (Fig. 7.12b), the net voltage, proceeding clockwisearound loop BNAB, is

V e e Ldi

dtL

di

dtBN AN sa

sb

BNAB = − + − = 0 (7.78)

But

e e E tBN AN m− = − −3 60cos( )ω o (7.79)

and

ia ib Iav ic (7.80)

Substituting Eqs. (7.79) and (7.80) into Eq. (7.78), noting that dIav/dt 0, gives

− − = =3 60 2 2E t Ldi

dtL

di

d tm sb

sbcos( )ω ω

ωo

(7.81)



FIG. 12 Equivalent circuits of conduction for the three-phase, full-wave controlled bridgerectifier, including supply inductance: (a) t 150 and (b) t 150 .


Chapter 7250

The time variation of a supply line current or a thyristor switch current duringoverlap can be deduced by the use of a running upper limit in the definite integra-tion of Eq. (7.81).

For 150 t 150 ,

diE

Lt d tb

i tm

s

tb

0 150

3

260

( )cos( )

ω

α

ω

ωω ω∫ ∫=

−−

+

oo (7.82)

from which

ib (t) Isc [cos sin(t 60)] Iav ia (7.83)

where Isc 3Em/2Ls is the peak value of the circulating current in the short-circuited section ANBN of Fig. 7.12b. The term Isc was previously used in theanalysis of the uncontrolled rectifier in Chapter 6.

Waveforms of ib (t) and ia (t) are given in Fig. 7.13. It is clearly illustratedthat the portions of ia (t), ib (t) during overlap are parts of sine waves. Thedefinition of ia (t) for a complete period (not given here) would involve fourterms similar in style to Eq. (7.83) plus a term in Iav.

7.3.1.1 Load-Side Quantities

Integrating both sides of Eq. (7.81),

o

o

di iE

Lt d t

E

Lt K

b bm

s

m

s

= =−

−

=−

− +

∫ ∫3

260

3

260

ωω ω

ωω

cos( )

sin( )(7.84)

where K is a constant of integration. At t 150, ib Iav so that

K IE

L

IE

L

avm

s

avm

s

= + + +

= + +

3

290

3

2

ωα µ

ωα µ

sin( )

cos( )

o

(7.85)

At t 150, ib 0 in Fig. 7.11 so that

KE

L

E

L

m

s

m

s

= +

=

3

290

3

2

ωα

ωα

sin( )

cos

o

(7.86)



FIG. 13 Waveforms of the three-phase, full-wave controlled bridge rectifier with highlyinductive load, in the presence of supply inductance 30, 15: (a) generator linevoltage eBC t and (b) supply line currents iA and iB (t).


Chapter 7252

Eliminating K between Eq. (7.85) and (7.86) gives an expression for the averageload current in the presence of supply inductance Ls per phase

IE

Lavm

s

= − +[ ]3

2ωα α µcos cos( )

(7.87)

Expression (7.87) is identical to the corresponding expression (5.70) for half-waveoperation The average load voltage can be found from the eL (t) characteristic ofFig. 7.11e. Consider the 60 section defined by 150 t 210.

E e d t e e d t eav ac aN bN= + + ++

+

+ +

∫ ∫6

2

3

2150

150

150

150

πω ω

α

α

α µ

o

o

o

o

( ) bbc d t150

210

o

o

+ +∫

α µ

ω(7.88)

Elucidation of Eq. (7.88) is found to give

EE

E

avm

av

= + +[ ]

= + +[ ]

3 3

2

20

πα α µ

α α µ

cos cos( )

cos cos( ) (7.89)

The average value Eav in Eq. (7.89) is seen to be twice the value obtained in Eq.(5.61) for a half-wave bridge. The variation of Eav with is demonstrated in Fig.7.14.

If the term cos ( ) is eliminated between Eqs. (7.85) and (7.87), it isfound that

E EL

Iav avs

av= −0

3cosα

ωπ (7.90)

The first term of Eq. (7.90) is seen to represent the average load voltage withideal supply, consistent with Eq. (7.8). The second term of Eq. (7.90) representsthe reduction of average load voltage due to voltage drop in the supply lineinductances and is seen to be consistent with Eq. (6.27) for the diode bridge.Since, as always, Iav Eav/R, Eq. (7.90) can be rearranged to show that

EE

L Rav

av

s

=+

0

1 3

cos

/

αω π (7.91)

The power dissipated in the load PL is conveniently expressed in terms of averageload quantities (since the current is smooth) and Irms Iav

P E IE

RI RL av av

avav= = =

22

(7.92)



FIG. 14 Effect of supply reactance on the normalized average load voltage versus firingangle of a three-phase, full-wave controlled bridge rectifier with highly inductive load.

Substituting Eqs. (7.89) and (7.90) into Eq. (7.92) gives

PE

LLm

s

= − + 9

4

22 2

πωα α µcos cos ( )

(7.93)

7.3.1.2 Supply-Side Quantities

All of the load power passes into the bridge from the supply. For a balancedthree-phase load with sinusoidal voltage of peak value Em per phase and periodicnonsinusoidal current with a fundamental component of rms value I1, the inputpower maybe given by

PE

Iinm=

3

21 1cos ψ

(7.94)


Chapter 7254

where cos 1 is the current displacement factor. Neglecting any power loss inthe conducting thyristors, PL Pin. Equating Eqs. (7.92) and (7.9) gives

I I

I

av1 1

1

6

2

02

coscos cos( )

( )cos cos( )

ψπ

α α µ

α α µ

= + +

= + +(7.95)

The term (6/) Iav in Eq. (7.95) is the rms fundamental supply current withzero overlap, I1 (0).

It is reasonable to assume that overlap makes only a small difference inthe value of I1. In fact, for 30, there is only 1.1% difference. Therefore, if

I I Iav1 1 06≅ =( )

π (7.96)

then, very nearly,

cos cos cos( )ψ α α µ1

1

2= + +[ ] (7.97)

Combining Eqs. (7.96), (7.97), and (7.98) gives

cos cosψ αωπ1

0 0

3≅ = −

E

E

L I

Eav

av

s av

av (7.98)

With an ideal supply Ls 0 and Eq. (7.98) reduces to Eq. (7.58).The rms value of the input current can be obtained from the defining integral

I i t d ta a= ∫1

22

0

2

πω ω

π( ) (7.99)

Each cycle of ia (t) contains six parts and the necessary mathematics to solveEq. (7.99) is very lengthy (see Ref. 4). It is found, after much manipulation, that

I I fa av= −2

31 ( , )α µ (7.100)

where

f ( , )sin cos( ) cos cos( )

cos cos(α µ

πµ α µ µ α α µ

α α µ=

+ +[ ]− + +[ ]− +

3

2

2 2 1 2

))[ ]2(7.101)



The term 2/3 Iav in Eq. (7.100) is seen, from Eq. (7.52), to be the rms supplycurrent with zero overlap. It should be noted that the two numerator parts of Eq.(7.101) are very similar in value and great care is required in a numerical solution.It is found that for 30 the effect of overlap reduces the rms supply currentby less than 5% for all firing angles. One can therefore make the approximation

I Ia av≅ 2

3 (7.102)

For small values of the power factor is given by combining Eq. (7.95) withEq. (7.102)

PFI

Ia

=

= + +[ ]

11

3

2

cos

cos cos( )

ψ

πα α µ

(7.103)

When 0, Eq. (7.103) reduces to Eq. (7.53). For any finite value of thepower factor is reduced compared with operation from an ideal supply.

Terminal capacitance can be used to obtain some measure of power factorimprovement by reducing the displacement angle to zero. For overlap conditionssuch that 30, the equations of this section provide the basis for approximatecalculations. It is significant to note, however, that an important effect of supplyinductance is to render the bridge terminal voltages nonsinusoidal. Equation(7.24), which defines the power factor, remains universally valid, but the relation-ship of Eq. (7.31) no longer has any validity.

7.3.2 Mode II Operation ( 60)

In mode II operation three rectifier thyristors conduct simultaneously, which onlyoccurs during overload or with-short circuited load terminals. The average currentthrough the load is then found to be

IE

Lavm

s

= − − + + 230 30

ωα α µcos( ) cos( )o o

(7.104)

The corresponding average load voltage is found to be

E E

E

L

av av

m

s

= − − + +

= − −

3

230 30

9

230

0cos( ) cos( )

cos( ) c

α α µ

π ωα

o o

o oos( )α µ+ + 30o

(7.105)


Chapter 7256

7.3.3 Mode III Operation 60 120

With mode III operation there is alternate conduction between three and fourrectifier thyristors. Three thyristor operation is equivalent to a line-to-line shortcircuit on the supply, while four thyristor working constitutes a three-phase shortcircuit. Equations (7.104) and (7.105) still apply. If the term cos( 30)is eliminated between these, it is found that

E

E

I

Iav

av

av

sc0

3

23 30+ = −

ˆcos( )α o

(7.106)

7.3.4 Worked Examples

Example 7.8 A fully-controlled, three-phase bridge rectifier suppliespower to a load via a large, series filter inductor. The supply lines contain seriesinductance so as to cause overlap. Sketch the variation of the average load voltagewith firing angle for a range of overlap angle .

The average voltage is given by Eq. (7.89)

EE

av

av= + +[ ]0

2cos cos( )α α µ

The range of firing angles is 0 90, and the realistic range of is 0 60. Values of the ratio Eav/Eavo are given in Fig. 7.14.

Example 7.9 A three-phase, full-wave controlled bridge rectifier withhighly inductive load operates from a 440-V, 50-Hz supply. The load current ismaintained constant at 25 A. If the load average voltage is 400 V when the firingangle is 30, calculate the load resistance, the source inductance, and the overlapangle.

The peak phase voltage Em is

Em = =4402

3359 3. V

From Eq. (7.90),

EL

avs= × × −

×× =3 3

4402

3

3

2

3 2 5025 400

πππ

V

which gives

Ls = − =514 6 400

750015 3

.. mH/phase



The load resistance is given in terms of the (presumed average values of the)load voltage and current

RE

Iav

av

= = =400

2516 Ω

From Eq. (7.89),

cos( ) cosα µ α+ = −2

0

E

Eav

av

NowE

Eav

m

0

3 3594= =

π V

Therefore,

cos( )( )

.

.

.

α µ

α µ

µ

+ = − =

+ =

=

2 400

594

3

20 48

61 3

31 3

o

o

Example 7.10 For the three-phase bridge rectifier of Example 7.9 calcu-late the reduction of the power transferred and the change of power factor dueto overlap compared with operation from an ideal supply.

The average load current and voltage are both reduced due to overlap. FromEq. (7.92),

P E IE

RL av avav= =2

and from Eq. (7.89)

EE

av

av= + +[ ]0

2cos cos( )α α µ

With 30, it was shown in Example 7.9 that 31.3 when Eav 400V. Therefore,

PL = =400

1610

2

kW

This compares with operation from an ideal supply where, from Eq. (7.50),


Chapter 7258

PE

RL

av

µα

== = × =

0

0

2

22594

16

3

416 54cos . kW

The large overlap angle has therefore reduced the power transferred to 60% ofits ‘‘ideal’’ value.

An accurate calculation of power factor is not possible within the presentwork for an overlap angle 31.3. An approximate value, from Eq. (7.103),is

PF ≅ +

=

3

230 61 3

0 642

π(cos cos . )

.

o o

Example 7.11 A three-phase, full-wave controlled bridge rectifier withhighly inductive load is operated with a thyristor firing angle 30. Thesupply line inductance is such that overlap occurs to an extent where 15.Sketch the waveforms of the voltage across a bridge thyristor and the associatedcurrent.

When a thyristor switch is conducting current the voltage drop across, itcan be presumed to be zero. While a thyristor is in extinction, the line-to-linevoltage occurs across it, except during overlap, so that the peak value of a thyristorvoltage is 3Em. During the overlap intervals the peak value of the load (andthyristor) voltages is (3/2) Em, as developed in Eq. (7.77). At the end of the firstsupply voltage cycle t , thyristor Th1 is conducting current ia and the voltageeTh1 is zero (Figs. 7.11 and 7.15). At t , thyristor Th3 is fired and Th3 thenconducts current ib simultaneously. At t the overlap finishes with ia 0 and Th1 switching off. The voltage eTh1 then jumps to the appropriate valueof eab ( ) and follows eab (t) until t 240 and Th4 is switched in.During the overlap of Th2 and Th4, 240 t 240 , the voltage acrossTh1 is 3ebN/2. When Th2 is commutated off at t 240 , a negativecurrent ia (t) is flowing through Th4, and the voltage across Th1 jumps back toeab (240 ). At t 300, thyristor Th5 is switched in and overlaps withTh3. The anode of Th1 is held at point a, but the overlap of Th3 and Th5 causeseTh1 to jump to 3eaN/2 during overlap. When Th3 switches off at t 300 , current ib falls to zero and voltage eTh1 (t ) follows eac and so on. Theoverall waveform for eTh1 is given in Fig. 7.15 with the associated current iTh1also shown.

7.4 SUMMARY OF THE EFFECTS OF SUPPLYREACTANCE ON FULL-WAVE BRIDGERECTIFIER OPERATION (WITH HIGHLYINDUCTIVE LOAD)

The presence of supply line inductance inhibits the process of current commuta-tion from one thyristor switch to the next. Instead of the instantaneous current



FIG. 15 Waveforms of the three-phase, full-wave controlled bridge rectifier with highlyinductive load in the presence of supply inductance 30, 15; (a) load voltage,(b) thyristor voltage eTh1 (t), and (c) thyristor Th1 current.


Chapter 7260

transfer that occurs between thyristors when the supply is ideal a finite time isrequired to accomplish a complete transfer of current. The duration of the currenttransfer time, usually called the overlap period, depends on the magnitude of thesupply voltage, the load resistor, the supply current level, the thyristor firingangle, and the source inductance.

1. The average load voltage Eav is reduced, Eq. (7.89).2. The waveform of the load voltage eL (t), (Fig. 7.11e) is modified

compared with corresponding ‘‘ideal supply’’ operation (Fig. 7.3c).The additional piece ‘‘missing’’ from the waveform in Fig. 7.11e, forexample, can be used to calculate the reduction of the average loadvoltage.

3. Because the waveform eL (t) is changed, its harmonic properties aswell as its average value are changed. The basic ripple frequency (i.e.,six times supply frequency) is unchanged, and therefore the order ofthe harmonics of eL (t) remains 6n, where n 1, 2, 3, … Therefore,the magnitudes of the harmonics of eL (t) are affected by overlap.

4. The average load current Iav is affected by the reduction of averageload voltage because Iav Eav/R.

5. Load power dissipation is reduced due to overlap, Eq. (7.92), sincePL EavIav.

6. The waveform of the supply line currents are modified. Current ia(t) in Fig. 7.7e, for example, is modified to the waveform of Fig.7.11f. Its conduction angle is extended from 120 to 120 ineach half cycle. Because of the change of waveform, the magnitudesof its harmonic components are modified.

7. The rms value of the supply current is reduced, Eq. (7.100).8. The rms value of the fundamental component of the supply current

is reduced.9. The modified shape of the supply current causes the current displace-

ment angle 1 to increase and therefore the displacement factor cos1

to decrease.10. The power factor of operation is reduced due to overlap, Eq. (7.103),

at small values of .11. The waveform of the bridge terminal voltage is no longer sinusoidal

but contains ‘‘notches’’ during the overlap periods. This reduces therms supply voltage and also the rms value of the fundamental compo-nent of the supply voltage.

12. The power factor relationship

PF (displacement factor)(distortion factor)

is no longer valid because of the nonsinusoidal terminal voltage.13. The notching of the supply voltage can give rise to the spurious firing



of silicon controlled rectifiers by forward breakover and also to inter-ference with electronic circuits connected at the same or adjacentsupply points.

PROBLEMS

Three-Phase, Full-Wave, Controlled Bridge withResistive Load and Ideal Supply7.1 A three-phase, full-wave bridge rectifier of six ideal thyristor switches is

connected to a resistive load. The ideal three-phase supply provides bal-anced sinusoidal voltages at the input terminals. Show that the averageload voltage Eav is given by Eqs. (7.8) and (7.12) in the two respectivemodes of operation. Sketch Eav versus firing angle over the full operatingrange.

7.2 A three-phase, full-wave bridge rectifier containing six ideal thyristorssupplies a resistive load R 100 . The ideal supply 240 V, 50 Hzprovides balanced sinusoidal voltages. Calculate the average load currentand power dissipation at (a) 30, (b) 60, and (c) 90.

7.3 For the three-phase bridge circuit of Problem 7.2 deduce and sketch thevoltage waveform across a thyristor at 30.

7.4 For the three-phase bridge circuit of Problem 7.1 show that the rms valuesof the supply current are given by Eqs. (7.21) and (7.22).

7.5 For a three-phase, full-wave bridge circuit with resistive load, show thatfor both modes of operation, the rms supply current Ia is related to therms load current IL by the relation Eq. (7.23).

7.6 Expressions for the fundamental component of the supply current into athree-phase, full-wave controlled bridge rectifier supplying a resistive loadare given in Table 7.2. Calculate the rms values of this fundamental compo-nent with a supply of 240 V, 50 Hz and a load resistor R 100 at (a) 30, (b) 60, and (c) 90.

7.7 The power input to a three-phase, full-wave, controlled bridge rectifier isgiven by the relation P 3EIa1 cos 1, where E is the rms phase voltage,Ia1 is the rms value of the fundamental component of the supply current,and cos 1 is the current displacement factor (not the power factor!). Calcu-late P for the bridge circuit of Problem 7.6 and check that the valuesobtained agree with the power dissipation calculated on the load side.

7.8 Show that the Fourier coefficients a1 and b1 of the fundamental componentof the supply line current for a full-wave controlled bridge with resistiveload R, at firing angle , are given by


Chapter 7262

aE

Rb

E

Rm m

1 1

3 3

22

22 3 3 2= − = +

πα

ππ αsin ( cos )

7.9 Derive expressions for the current displacement factor cos 1 and thecurrent distortion factor I1/I for a three-phase, full-wave controlled bridgerectifier with resistive load. Show that the respective products of these areconsistent with the expressions (7.25) and (7.26) for the power factor.

7.10 Calculate and sketch the variation of the power of a three-phase, full-wavebridge rectifier with resistive load over the operating range of thyristorfiring angles.

7.11 Use the information of Problem 7.7 to derive expressions for the reactivevoltamperes Q into a three-phase, full-wave bridge rectifier with resistiveload, where Q 3EIa1 sin 1. Does a knowledge of real power P andreactive voltamperes Q account for all the apparent voltamperes S (3EIa) at the bridge terminals?

7.12 Three equal capacitors C are connected in star across the terminals of afull-wave, three-phase bridge rectifier with resistive load. If Xc R, sketchwaveforms of a capacitor current, a bridge input current, and the corre-sponding supply current at 30. Does the waveform of the supplycurrent seem to represent an improvement compared with the uncompen-sated bridge?

7.13 For the three-phase bridge circuit of Problem 7.2 what will be the minimumvalue of supply point capacitance per phase that will cause power factorimprovement at (a) 30, (b) 60 and (c) 90?

7.14 For the three-phase bridge circuit of Problem 7.2. what must be the respec-tive values of the compensating capacitors to give the highest realizablepower factor (by capacitor correction) at the three values of firing angle?

7.15 For the three-phase bridge circuit of Problem 7.2 calculate the operatingpower factor at each value of firing angle. If optimum compensation is nowachieved by the use of the appropriate values of supply point capacitance,calculate the new values of power factor.

Three-Phase, Full-Wave Controlled Bridge Rectifierwith Highly Inductive Load and Ideal Supply

7.16 A three-phase, full-wave controlled bridge rectifier contains six ideal thy-ristors and is fed from an ideal, three-phase supply of balanced sinusoidalvoltages. The load consists of a resistor R in series with a large filterinductor. Show that, for all values of thyristor firing angle , the average



load voltage is given by Eq. (7.8). Sketch Eav versus and compare theresult with that obtained for purely resistive load.

7.17 For the three-phase, inductively loaded bridge of Problem 7.16 calculatethe Fourier coefficients a1 and b1 of the fundamental component of thesupply current. Use these to show that the current displacement angle 1

[tan1(a1/b1)] is equal to the thyristor firing angle .

7.18 A three-phase, full-wave controlled bridge rectifier is supplied from anideal three-phase voltage source of 415 V, 50 Hz. The load consists ofresistor R 100 in series with a very large filter inductor. Calculatethe load power dissipation at (a) 30 and (b) 60, and comparethe values with those that would be obtained in the absence of the loadfilter inductor.

7.19 Show that for the inductively loaded bridge of Problem 7.16 the distortionfactor of the supply current is independent of thyristor firing angle.

7.20 Show that the waveform of the supply current into a controlled bridgerectifier with highly inductive load is given by

i tI

t t tav( ) sin( ) sin ( ) sin ( ) ...ωπ

ω α ω α ω α= − − − − −

−2 3 1

55

1

77

where Iav is the average load current.7.21 For the three-phase bridge rectifier of Problem 7.16 show that the power

input is equal to the load power dissipation.

7.22 Derive an expression for the load voltage ripple factor (RF) for a three-phase inductively loaded bridge rectifier and show that this depends onlyon the thyristor firing angle. Obtain a value for the case 0, and therebyshow that the RF is zero within reasonable bounds of calculation.

7.23 For the inductively loaded bridge rectifier of Problem 7.16 show that therms supply current is given by

IE

Rm=

3 2

παcos

Calculate this value for the cases (a) 30 and (b) 60.7.24 For the inductively loaded bridge of Problem 7.18 calculate the rms current

and peak reverse voltage ratings required of the bridge thyristors.

7.25 Show that the average load voltage of a three-phase, full-wave controlledbridge circuit with highly inductive load can be obtained by evaluatingthe integral


Chapter 7264

E E t d tav m=−

+

∫6

2 30

30

πω ω

α

α

o

o

cos

Sketch the waveform of the instantaneous load voltage eL (t) for 75, and show that it satisfies the above relationship.

7.26 A three-phase, full-wave, thyristor bridge is fed from an ideal three-phasesupply and transfers power to a load resistor R. A series inductor on theload side gives current smoothing that may be considered ideal. Derive anexpression for the rms value of the fundamental component of the supplycurrent. Use this expression to show that the reactive voltamperes Q enter-ing the bridge is given by

QE

Rm=

27

42

2

2παsin

7.27 For the three-phase, bridge rectifier of Problem 7.18 calculate the powerfactor. If equal capacitors C are now connected in star at the supply calcu-late the new power factor when Xc R. What is the minimum value offiring angle at which compensation to the degree Xc R renders a powerfactor improvement?

7.28 For the bridge rectifier circuit of Problem 7.16 derive an expression for theterminal capacitance that will give maximum power factor improvement.

7.29 The bridge rectifier circuit of Problem 7.18 is compensated by the use ofequal capacitors C connected in star at the supply terminals. Calculate thevalues of capacitance that will give unity displacement factor at (a) 30 and (b) 60. In each case calculate the degree of power factorimprovement compared with uncompensated operation.

7.30 For the bridge circuit of Problem 7.28 sketch, on squared paper, consistentwaveforms of the bridge line current, the capacitor current and the supplyline current. Does the waveform of the supply current appear less distortedthan the rectangular pulse waveform of the bridge current?

7.31 A three-phase, full-wave, bridge rectifier circuit, Fig. 7.6, supplies powerto load resistor R in the presence of a large load filter inductor. Equalcapacitors are connected at the supply terminals to give power factor im-provement by reducing the current displacement angle 1 to zero at thefixed thyristor firing-angle . Derive a general expression for the supplycurrent distortion factor in the presence of supply capacitance. For the casewhen C has its optimal value so that the displacement factor is increasedto unity is the distortion factor also increased?



Three-Phase, Full-Wave Controlled Bridge Rectifierwith Highly Inductive Load in the Presence of SupplyInductance

7.32 A full-wave controlled bridge rectifier circuit transfers power to a loadresistor R in series with a large filter inductor. The three-phase supplycontains a series inductance Ls in each supply line and has sinusoidal open-circuit voltages where Em is the peak phase voltage. Show that at thyristorfiring angle the average load current is given by

IE

Lavm

s

= − +[ ]3

2ωα α µcos cos( )

where is the overlap angle.

7.33 For the full-wave bridge of Problem 7.32, use Eq. (7.89), or otherwise, toshow that the average load voltage is given by

EE

avm= + +[ ]3 3

2πα α µcos cos( )

7.34 A three-phase, full-wave controlled bridge rectifier with highly inductiveload operates from a 240-V, 50-Hz supply. The load current is requiredto remain constant at 15 A. At firing angle 15, the load voltage isfound to be 200 V. Calculate the source inductance and the overlap angle.

7.35 For the three-phase rectifier of Problem 7.32 show that overlap angle may be obtained from

cos( )

cos

/

/

α µα

ω πω π

+ =−+

1 3

1 3

L R

L Rs

s

7.36 For the three-phase bridge rectifier of Problem 7.34 calculate the reductionof the power transferred due to overlap compared with operation from anideal supply.

7.37 A three-phase bridge rectifier with highly inductive load operates at afiring angle 30 and results in an overlap angle 15. Calculatethe per-unit reduction of rms supply current compared with operation froman ideal supply.

7.38 A resistor R 20 is supplied from a three-phase controlled bridgerectifier containing a large series filter on the load side. The supply is 240V, 50 Hz, and each supply line contains a series inductance Ls 10 mH.


Chapter 7266

Calculate the approximate power factor of operation for (a) 0, (b) 30, and (c) 60.

7.39 A three-phase, full-wave controlled bridge rectifier supplies a highly induc-tive load. Show that in the overlap intervals caused by supply-line induct-ance, the load voltage is 1.5 times the relevant phase voltage.

7.40 For a three-phase, full-wave bridge rectifier with highly inductive load, itwas shown in Eq. (7.58) that the current displacement cos 1 is related tothe thyristor firing angle by a relationship cos 1 cos when thesupply is ideal. Show that in the presence of significant supply inductancethis relationship is no longer valid but that

cos cos cos( )ψ α α µ1

1

2≅ + +[ ]


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