CH.4 Full-wave and Three-

phase rectifiers

(Converting AC to DC)

4-1 Introduction

� The average current in AC source is zeroin the full-wave rectifier, thus avoiding problemsassociated with nonzero average source currents, particularly in transformers.

� The output of the full-wave rectifier has inherentlyless ripple than the half-wave rectifier.

� Uncontrolled and controlled single-phase and three-phase full-wave converters used as rectifiers areanalyzed.

4-2 Single-phase full-wave rectifiers

Fig. 4-1 Bridge rectifier：：：：

The lower peak diode voltage make it more suitablefor high-voltage applications.

Fig. 4-2 center-tapped transformer

rectifier

With electrical isolation, only

one diode voltage drop betweenthe source and load, suitable for

low-voltage, high-current applications

Resistive load：：：：

π≤≤π−

π≤≤=

2

0　0

wt,wtsinVm

wt,wtsinVm)wt(v

∫ ==π

ππ 02)()sin(

1 VmwtdwtVmVo

)(2

RVm

RVoIo π==

2Im=Irms

power absorbed by the load resistor：

rmsRIPR

2=power factor ：Pf=1

R-L load：：：： Fig.4-3

∑∞

=π++=

‧‧420 ,,n)nwtcos(VnVo)wt(v

π= VmVo 2

+

−−

=1

1

1

12

nn

VmVn

π

RVoIo = |jnwLR|

VnZn

VnIn+

==

If L is relatively large, the load current is essentially

dc. ( )

R>> L for

IoIrms

R

Vm

R

VoIo)wt(i

ω

≈π

==≈　　

　　　　　

2

Source harmonics are rich in the odd-numbered harmonics.

Filters：reducing the harmonics.

R>> L ω

R-L source load：：：： Fig.4-5

For continuous current operation, the only modification to

the analysis that was done for R-L load is in the dc

term of the Fourier series .The dc component of current

in this circuit is.

R

VdcVm

R

VdcVoIo

−=

−= π

2

The sinusoidal terms in the Fourier analysis

are unchanged by the dc source, provided

that the current is continuous.

Discontinuous current is analyzed like

section 3-5.

Capacitance output filter: Fig. 4-6

Assuming ideal diodes

θ=

θ−− off diodes , e sinVm

on pair diode one,|wtsinVm|)wt(v

)wRc/()wt(

　　　0

θ：the angle where the diodes become reverse biased,

which is the same as for the half-wave rectifier

and is

π+ω−=ω−=θ −− )RC(Tan)RC(Tan 11

απ +=wt

)sin(Vme sinVm )RC/()( α+π−=θ ωθ−α+π−

0=α−θ ωθ−α+π−sine)(sin

)RC/()(

α=？ �solved numerically forα

Peak-to-peak variation(ripple)：

)sin1(|)sin(| ααπ −=+−=∆ VmVmVmVo

In practical circuits where ωRC

,2

, 2

παπθ ≈≈

minimal output voltage occurs at α+π=wt

)RC/()RC/()(

e Vme Vm)(v ωπ−ωπ

−π

+π−==α+π 22

0

[ ]

fRC

Vm

RC

Vm

RCVm

eVme VmVmVo )RC/()RC/(

2

11

1

‧

‧

=ω

π=

ωπ

−−=

−=−≈∆ ωπ−ωπ−

fw

xxxe x

π2

...321

132

=

++++=！！！

is half that of the half-wave rectifier.

π>>

Fig. 4-7 (a) Voltage doubler

Fig. 4-7 (b) Dual voltage rectifier

=full-wave rectifier(sw. open)+

voltage doubler(sw. closed)

L-C filtered output：：：： Fig.4-8

C holds the output voltage at a constant level, and the L

smoothes the current from rectifier and reduces the peak

current in diodes.

Continuous

Current：

(LV 2 Qπ

VmVoVx == =0 , full-wave rectified )

0 , )(

2 ==== IcR

VmR

VoIIRL π

LiThe variation in can be estimate from the first

Ac term (n=2) in the Fourier series.

The amplitude of the inductor current for n=2 is

L

Vm

L

/Vm

L

V

Z

VI

πω=

ωπ

=ω

==3

2

2

34

22 2

2

2

where 21

1

1

12=

+

−−π

= n , nn

VmVn

For Continuous current, LII <2

R

Vm

L

Vm

π<

πω2

3

2

ω>

3

RL � 1

3>

ωR

L

Discontinuous current：

When is positive ( at ),

VowtVv mL −= sin

[ ]

( )[ ]

? i

,wt for

wtVowtVmL

wtdVowtVmL

wti

L

wt

L

==

<≤≤

−−−=

−= ∫

ββ

πββα

ααω

ω α

,0)(

,

)cos(cos1

)(sin1

)(

Li VowtVm

=sin α=wt

= −

Vm

Vo1sinα

Procedure for determining Vo：

(1) Estimate a Value for Vo slightly below Vm, and solve ?=α

(2) Solve numerically,β )()cos(cos0)( αββαβ −−−== VoVmiL

(3) Solve

[ ]∫

∫β

α

β

α

α−−−αωπ

=

π=

)wt(d)wt(Vo)wtcos(cosVmL

)wt(d)wt(iILL

11

1

(4) Slove Vo= RI L

(5) Repeat step (1)~(4) until the computed Vo in step(4)

equals the estimated Vo in step(1)

Output Voltage for discontinuous current is larger than

for continuous current.(see Fig4-8(d))

4-3 controlled full-wave rectifiers

Resistive load：：：： Fig.4-10

)cos(Vm

)wt(d)wtsin(VmVo

α+π

=

π= ∫

π

α

1

1angle delay=α

)cos1( απ

+==R

Vm

R

VoIo

πα

πα

π

π

α

4

)2sin(

22

1

)()sin(1 2

+−=

= ∫

R

Vm

wtdwtR

VmI rms

The power delivered to the load rmsRIP 2=

The rms current in source is the same as the rms current in

the load.

R-L load : Fig.4-11

Analysis of the controlled full-wave rectifier operating in the

discontinuous current mode is identical to that of the controlled

half-wave rectifier, except that the period for the output current

is .π

[ ])/()t(

o e)sin()tsin(Z

Vm)wt(i

ωτα−ω−θ−α−θ−ω= for β≤ω≤α t

RL , )

R

L(tan

)L(RZ

=τω

=θ

ω+=

−1

22

For discontinuous current παβ +<

discontinuous current :

continuous current

0)( , ≥++= απαπ iwt

[ ]

current continuous for

R

LTan

0 )-(

0 )- sin(

e

e

1- )(

01)sin(

0)sin()sin(

)/(

)/()(

ωθα

αθ

αθαθ

θαθαπωτπ

ωτααπ

=≤

≥

≥

≥+−

≥−−−+−

−+−

,....6,4,2

1

)1sin(

1

)1sin(2

1

)1cos(

1

)1cos(2

cos2

)(sin1

)cos()(

22

1

=

−−

−++

=

−−

−++

=

+=

==

++=

∫

∑

+

∞

=

n

n

n

n

nVmb

n

n

n

nVma

baVn

Vmwtd wtVmVo

nnwtVnVowtv

n

n

nn

n

0

ααπ

ααπ

αππ

θ

απ

α

)(an

bnTann 1- −

=θ

Fig 4-12

RVoIo

)In

(IoIrms

|jnwLR|Vn

ZnVnIn

...,n

=

+=

+==

∑∞

= 42

22

2

R-L Source load ：：：： Fig.4-14

The SCRS may be turned on at any time that they are

forward biased, which is at an angle

)(sin 1

VmVdc−≥α

For continuous current case, the average bridge output voltage is

average load current is

The ac voltage terms are unchanged from the controlled rectifier

with an R-L load. The ac current terms are determined from

circuit.

Power absorbed by the dc voltage is

elisLifRIormsRIP arg 22 ≈=

απ

= cosVm

Vo2

R

VdcVoIo

−=

VdcIoPdc =

Power absorbed by resistor in the load is

Controlled Single-phase converter operating as an inverter：seeing Fig 4-14. 4-15

.

00900 <α< � 0>Vo rectifier operation

0018090 <<α � 0<Vo inverter operation

IoVoPP acbridge −==

For inverter operation, power is supplied by the dc source,

and power is absorbed by the bridge and is transferred to

the ac system.

� Vdc and Vo must be negative

4-4 Three-phase rectifiers

Resistive load : Fig 4-16

上、下半部Diode，每次僅一個ON；同相上、下Diode不可同時ON；Diode ON由瞬間最大線電壓決定。

A transition of the highest line-to-line voltage must take place

every

.

Because of the six transitions that occur for each period

of the source voltage, the circuit is called a six-pulse

rectifier.

vo(t)之基頻為3 電源頻率之6倍

Diode turn on in the sequence 1,2,3,4,5,6,1,..

00606/360 =

φ

−=

−=

−=

25

63

41

DDc

DDb

DDa

iii

iii

iii

Each diode conducts one-third of the time, resulting in

avgoavgD II ,,3

1=

rmsormsDII

,,3

1=

rmsormsS II ,,3

2=

Apparent power from the three-phase source is

rms,Srms,LL IV S −= 3

...,,,n , )n(

V V

V.

V)wt(wtdsinV

/V

)tnwcos(VVo)t(v

LL,m

n

LL,m

LL,m/

/LL,m

..,,n

n

181261

6

950

3

3

1

2

32

30

0

18126

0

=−π

=

=π

=π

=

π++=

−

−

−π

π −

∞

=

∫

∑

Since the output voltage is periodic with period 1/6 of the ac

supply voltage, the harmonics in the output are of order 6kω,

k=1,2,3,1

Adevantage：output is inherently like a dc voltage, and the high-

frequency low-amplitude harmonics enable filters to be effective.

For a dc load current (constant I0) --- Fig4.17

....twcostwcostwcostwcostw(cosI i oa 00000 1313

111

11

17

7

15

5

132+−+−

π=

which consists of terms at fundamental frequency of the ac

system and harmonics of order 6k ± 1, k=1,2,3,1

Filters(Fig.4-18) are frequently necessary to prevent harmonic

currents to enter the ac system.

Resonant filters for 5th and 7th harmonics.

High-pass filters for higher order harmonics.

4-5 Controlled three-phase rectifiers

απ

πα

π

απ

cos)3

(

)(sin

3

1

,

3

2

3

,

LLm

LLmo

V

wtwtdVV

−

+

+−

=

= ∫

Harmonics for output voltage remain of order 6k, but amplitude are

functions of α

. � seeing Fig. 4-20

Twelve-pulse rectifiers：using two six-pulse bridges

The purpose of the transformer connection is to introduce

phase shift between the source and bridge.

This results in inputs to two bridges which are

apart. The two bridge outputs are similar, but also shifted by

∆−Υ030

030

030

.

The delay angles for the bridge are typically the same.

απ

απ

απ

cos6

cos3

cos3

,,,

,,

LLmLLmLLm

oYoo

VVVVVV

−−−∆ =+=+=

The peak output of the twelve-pulse converter occurs midway

between alternate peaks of the six-pulse converters. Adding the

voltages at that point for gives°= 0α

°==°= −− 0932.1)15cos(2 ,,, α for V VV LLmLLmpeako

Since a transition between conducting SCRs every

, there are a total of 12 such transitions for each period of the

ac source. The output has harmonic frequencies which are multiple

of 12 times the source fre. (12k k=1,2,…)

°30

,...2,1112

cos1

cos1

(cos3

)()()(

....)cos13

1cos

11

1cos

7

1cos

5

1(cos

32)(

....)cos13

1cos

11

1cos

7

1cos

5

1(cos

32)(

000

00000

00000

=±=

−+=+=

++−+=

−+−+−=

∆

∆

k ,k order harmonic ,i

...)tw1313

tw1111

-twI4

tititi

tw13tw11tw7-tw5twIti

tw13tw11tw7tw5twIti

ac

oYac

o

oY

π

π

π

Cancellation of harmonics 6(2n-1) 1 , n=1, 2, … has resulted

from this transformer and converter configuration.

±

This principle can be expanded to arrangements of higher pulse

number by incorporating increased number of six-pulse converters

with transformers which have the appropriate phase shifts.

The characteristic ac harmonics of a p-pulse converter will be

pk 1 , k=1,2,31±

� More expense for producing high-voltage transformers with the

appropriate phase shifts.

Three-phase converter operating as a inverter：：：：seeing 4-22.

The bridge output voltage Vo must be negative.

operation Inverter -- 0Vo ,

operation Rectifier -- 0Vo ,

><°<<°

>>°<<

18090

900

αα

4-6 DC power transmission

․ By using controlled twelve-pulse converter (generally).

․ Used for very long distances of transmission lines.

Advantages：(1) , voltage drop↓ in lines

(2) , line loss

0=LX

∞=C

X ( ↓currentline Q

(3) Two conductors required rather than three

(4) Transmission towers are smaller.

(5 ) Power flow in a dc transmission line is controllable

by adjustment of delay angles at the terminals.

(6) Power flow can be modulated during disturbances on

one of the ac system. � System stability increased.

(7) The two ac systems that are connected by the dc

line do not need to be in synchronization.

)

Disadvantages：costly ac-dc converter, filter, and control system

required at each end of the line to interface

with the ac system.

Fig.4-23 using six-pulse converter

°<<°−

°<<+=

inverter

rectifierVV oo

18090 ,

900 , , 21 α

α

For current being ripple free

2

,2

2

1

,1

1

21

cos3

cos3

απ

απ

LLm

o

LLm

o

oo

o

VV

VV

R

VVI

−

−

=

=

+=

Power supplied by the converter at terminal 1 isoo IVP 11 =

Power supplied by the converter at terminal 2 is oo IVP 22 =

Fig.4-24 using twelve-pulse converter

(a bipolar scheme)

One of the lines is energized at and the other is energized

at - . In emergency situations, one pole of the line can operate

without the other pole, with current returning through the ground path.

dcV+dcV

4-7 commutation ：：：：effect of source inductance ( )

Single-phase bridge rectifier: Fig.4-25sX

Assume that the load current is constant Io.

Commutation interval starts at ωt= π )changedpolarity ( SourceQ

o

m

t

os

I)wtcos(Ls

V

I)wt(wtdsinVmLs

)wt(i

++ω

−=

+ω

= ∫ω

π

1

1

Commutation is completed at ωt= π +u

[ ] 00 1 I)ucos(Ls

VI)u(i m ++π+

ω

−=−=+π

)Vm

XI(cos)

Vm

LsI(cosu Soo

21

21

11 −=ω

−= −−

=> Commutation angle：

LsXS

ω=

Average load voltage is

)V

XI(

2V

)ucos(V

)wt(d wtsinVV

m

som

m

muo

−π

=

+π

=π

= ∫π

1

11

Source inductance lowers the average output voltage of full-

wave rectifier.

Three-phase rectifier：：：： Fig.4-26

During Commutation from , The voltage across La

is31

DtoD

wtsinVv

vLL,mAB

La22

−==

Current in starts at I0 and decreases zero in the

commutation interval

La

)V

IX(cos)

V

IL(cosu

I)wt(d wtsinV

La)u(i

LL,m

s

LL,m

a

uLL,m

La

−

−

−

−

+π

π

−

−=ω

−=

+ω

==+π ∫0101

0

21

21

2

10

During the commutation interval from , the converter

output voltage is31 D to D

2

ACBC

o

vvv

+=

22

2

0

BCACBCACAC

ABAC

.

.cLaLACo

BCACABCABCAB

vvvvv

vvvvvv

v-vv , vvv

+=

−−=

−=+−=

==++

Average output Voltage： 類似 Single-phase rectifier

)V

IX(

VV

LL,m

sLL,m

o

−

− −π

= 013

Source inductance lowers the average output voltage of three-

phase rectifiers.