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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.
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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.
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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
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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
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R-L load : Fig.4-3
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‧‧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
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R-L source load : Fig.4-5
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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.
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Capacitance output filter: Fig. 4-6
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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
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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.
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Fig. 4-7 (a) Voltage doubler
Fig. 4-7 (b) Dual voltage rectifier =full-wave rectifier(sw. open)+ voltage doubler(sw. closed)
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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.
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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
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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
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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))
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4-3 controlled full-wave rectifiers
Resistive load : Fig.4-10
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)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.
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R-L load : Fig.4-11
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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 :
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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-
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Fig 4-12
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RVoIo
)In
(IoIrms
|jnwLR|Vn
ZnVnIn
...,n
42
22
2
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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
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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
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Controlled Single-phase converter operating as an inverter :seeing Fig 4-14. 4-15
.
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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
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4-4 Three-phase rectifiers
Resistive load : Fig 4-16
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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
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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
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...,,,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.
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For a dc load current (constant I0) --- Fig4.17
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....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.
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4-5 Controlled three-phase rectifiers
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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
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Twelve-pulse rectifiers : using two six-pulse bridges
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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
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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.
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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.
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Three-phase converter operating as a inverter : seeing 4-22.
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The bridge output voltage Vo must be negative.
operation Inverter -- 0Vo ,
operation Rectifier -- 0Vo ,
18090
900
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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.
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Fig.4-23 using six-pulse converter
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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
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Fig.4-24 using twelve-pulse converter (a bipolar scheme)
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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
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4-7 commutation : effect of source inductance ( ) Single-phase bridge rectifier: Fig.4-25
sX
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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
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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.
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Three-phase rectifier : Fig.4-26
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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
![Page 55: 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,](https://reader035.vdocument.in/reader035/viewer/2022062417/551a9466550346b52d8b5f9e/html5/thumbnails/55.jpg)
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.