CH.4 Full-wave and Three- phase rectifiers (Converting AC to DC) 4-1 Introduction The average current in AC source is zero in the full-wave rectifier,

CH.4 　 Full-wave 　 and 　 Three- phase 　 rectifiers 　

(Converting 　 AC 　 to 　 DC)

4-1 　 Introduction

The 　 average 　 current 　 in 　 AC 　 source 　 is 　 zero 　 in 　 the 　 full-wave 　 rectifier, thus 　 avoiding　 problems 　 associated 　 with 　 nonzero 　 average 　

source 　 currents, particularly 　 in 　 transformers. The 　 output 　 of 　 the 　 full-wave 　 rectifier 　 has 　

inherently 　 less 　 ripple 　 than 　 the 　 half-wave 　rectifier.

Uncontrolled 　 and 　 controlled 　 single-phase 　 and 　 three-phase 　 full-wave 　 converters 　 used 　 as 　rectifiers 　 are 　 analyzed.

4-2 　 Single-phase 　 full-wave 　 rectifiers

Fig. 　 4-1 Bridge 　 rectifier ：

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

Fig. 　 4-2 center-tapped 　 transformer　 rectifier 　

　　 With 　 electrical 　 isolation, 　 only　 one 　 diode 　 voltage 　 drop 　 between 　 the 　 source 　 and 　 load, suitable　 for 　 low-voltage, high-current 　 applications

Resistive 　 load ：

2

0 0 wt,wtsinVm

wt,wtsinVm)wt(v

02)()sin(

1 VmwtdwtVmVo

)(2 RVm

RVoIo

2ImIrms

　 power 　 absorbed 　 by 　 the 　 load 　 resistor ：

rmsRIPR2

　 power 　 factor ： Pf=1

R-L 　 load ：　 Fig.4-3

‧‧420 ,,n)nwtcos(VnVo)wt(v

　 VmVo 2

1

1

1

12

nn

VmVn

　RVoIo

|jnwLR|

VnZn

VnIn

If 　 L 　 is 　 relatively large, the 　 load current 　 is 　 essentially 　 dc. ( )

R>> L forIoIrmsR

Vm

R

VoIo)wt(i

2

Source 　 harmonics are 　 rich 　 in 　 the 　 odd-numbered 　 harmonics.Filters ： reducing 　 the 　 harmonics.

R>> L

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

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

off diodes , e sinVm

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

)wRc/()wt(

0

： the 　 angle 　 where 　 the 　 diodes 　 become 　 reverse　 biased, which is 　 the 　 same 　 as 　 for 　 the 　 half-wave 　 rectifier 　 and 　 is

)RC(Tan)RC(Tan 11

wt

)sin(Vme sinVm )RC/()(

0 sine)(sin )RC/()(

　　　= ？　　 solved 　 numerically 　 for 　

　 Peak-to-peak 　 variation(ripple) ：　

)sin1(|)sin(| VmVmVmVo

　 In 　 practical 　 circuits 　 where 　ωRC

,　　　　

2 , 2

minimal 　 output 　 voltage occurs 　 at 　

wt

　 )RC/()RC/()(

e Vme Vm)(v

22

0

fRC

Vm

RC

Vm

RCVm

eVme VmVmVo )RC/()RC/(

2

11

1

‧

‧

fw

xxxe x

2

...321

132

！！！

　 is 　 half 　 that 　 of 　 the 　 half-wave 　 rectifier.

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 VmVoVx =0 , full-wave

rectified

0 , )(2 IcRVm

RVoII RL

　Li　 The 　 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

3

2

2

34

22 2

2

2

　 where 　 21

1

1

12

n ,

nn

VmVn

　 For 　 Continuous 　 current, 　LII 2

R

Vm

L

Vm

2

3

2

3

RL 1

3

R

L

Discontinuous 　 current ：　When 　 is 　 positive ( at )　 ,

VowtVv mL sin　

? i

,wt for

wtVowtVmL

wtdVowtVmL

wti

L

wt

L

,0)(

,

)cos(cos1

)(sin1

)(

Li VowtVm sin wt

Vm

Vo1sin

　 Procedure 　 for 　 determining 　 Vo ：

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

?

(2) Solve 　 numerically,

)()cos(cos0)( VoVmiL

(3) Solve　

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

)wt(d)wt(iI LL

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

1

1

angle delay

)cos1(

R

Vm

R

VoIo

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 　 .

)/()t(o e)sin()tsin(

Z

Vm)wt(i 　　　　　　 for t

　 RL , )

R

L(tan

)L(RZ

1

22

　 For 　 discontinuous 　 current　　　

discontinuous 　 current :

continuous 　 current

0)( , iwt

current continuous for R

LTan

0 )-(

0 )- sin(

e

e

1- )(

01)sin(

0)sin()sin()/(

)/()(

,....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

n0

)(an

bnTann 1-

　 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

　　 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

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

　.

00 900 　　　　 0Vo　　　 rectifier 　 operation　

00 18090 　　　　 0Vo 　　　 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,..

00 606/360

25

63

41

DDc

DDb

DDa

iii

iii

iii

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

avgoavgD II ,, 3

1

rmsormsD II ,,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,mn

LL,m

LL,m/

/ LL,m

..,,nn

181261

6

950

3

3

1

2

32

30

018126

0

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,…

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

which 　 consists 　 of 　 terms 　 at 　 fundamental 　 frequency 　 of 　 the　 ac 　 system 　 and 　 harmonics 　 of 　 order 　 6k 1, k=1,2,3,…

　 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

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 　

. 　 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

030

030

030

.

　 The 　 delay 　 angles 　 for 　 the 　 bridge 　 are 　 typically 　 the 　 same.　

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 　0

0932.1)15cos(2 ,,, 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

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=1,2,3…

　 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

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

0LX

CX

　 ( currentline (3) Two 　 conductors 　 required 　 rather 　 than 　 three(4) Transmission 　 towers 　 are 　 smaller.(5 ) Power 　 flow 　 in 　 a 　 dc 　 transmission line 　 is 　 contro

llable 　 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

For 　 current 　 being 　 ripple 　 free 　

2,2

2

1,1

1

21

cos3

cos3

LLmo

LLmo

ooo

VV

VV

R

VVI

　 Power 　 supplied 　 by 　 the 　 converter 　 at 　 terminal 　 1 　 is 　 oo 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.

dcVdcV

4-7 commutation ： effect 　 of 　 source 　 inductance ( )　 Single-phase 　 bridge 　 rectifier: Fig.4-25

sX

　 Assume 　 that 　 the 　 load 　 current 　 is 　 constant 　 Io.Commutation 　 interval 　 starts 　 at ωt=

)changedpolarity ( Source

om

t

os

I)wtcos(Ls

V

I)wt(wtdsinVmLs

)wt(i

1

1

Commutation 　 is 　 completed 　 at 　ωt= +u

00 1 I)ucos(Ls

VI)u(i m

)Vm

XI(cos)

Vm

LsI(cosu Soo 2

12

1 11

=> Commutation 　 angle ：

LsX S

　 Average 　 load 　 voltage 　 is

)V

XI(

2V

)ucos(V

)wt(d wtsinVV

m

som

mmuo

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 is

31 DtoD

wtsinVv

v LL,mABLa 22

　 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

u LL,mLa

0101

0

21

21

2

10

　 During 　 the 　 commutation 　 interval 　 from 　 , 　 the 　 converter 　 output 　 voltage 　 is　

31 D to D

2ACBC

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,mo

013

　 Source 　 inductance 　 lowers 　 the 　 average 　 output 　 voltage 　 of　 three-phase 　 rectifiers.