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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
!ncontrolled and controlled single-phase and three- hase full-wave converters used as
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4-2 Single-phase full-wave rectifiers
Fig. 4-1 Bridge rectifier :
The lower peak diode voltage ake it ore
suita!le for high-voltage applications.
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Fig. 4-2 center-tapped
transformer rectifier
"ith electrical isolation#
onl$
one
diode
voltage
drop
!etween the source and load#
suita!le for low-voltage# high-
current applications
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Resistive load:
π≤≤π−π≤≤
="
##
wt ,wt sinVm
wt ,wt sinVm )wt ( v
∫ == π
π π #"$%$sin%
1 Vmwt d wt VmVo
$%"
RVm
RVo Io
π ==
"Im= Irms
power absorbed by the load resistor :
rmsR I P R"=
power factor : %f&'
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R-L load: Fig.4-3
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∑∞
=π++=
‧‧4"# , ,n )nwt cos( VnVo )wt ( v
π= VmVo "
+−−= 11
1
1"
nn
Vm
Vn π
RVo Io =
| jnwL R|
Vn
ZnVn In
+==
If & is relatively large, the load current is
essentially dc ( )
R>> L for
Io Irms
R
Vm
R
Vo
Io )wt ( i ω
≈π==≈ "
'ource harmonics are rich in the odd-numbered
harmonics
(ilters: reducing the harmonics
R>> Lω
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R-L source load: Fig.4-
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or con nuous curren opera on, e on y modification to the analysis that was done for
)-& load is in the dc term of the (ourier
series The dc component of current in this circuit
is
R
VdcVm
R
VdcVo Io
−=
−= π
"
The sinusoidal terms in the (ourier
analysis are unchanged by the dc
source, provided that the current is
continuous*iscontinuous current is analyzed li+e
section -
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!apacitance output filter" Fig. 4-#
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Assuming ideal
diodes
θ=
θ−− off diodes ,e sinVm
on pair diode one ,|wt sinVm| )wt ( v
)wRc /( )wt (
#
θ : the angle where the diodes become
reverse biased, which is the same as for the half-wave rectifier and is
π+ω−=ω−=θ −− ) RC ( Tan ) RC ( Tan 11
α π +=wt
) sin( Vme sinVm ) RC /( )( α+π−=θ ωθ−α+π−
#=α−θ ωθ−α+π− sine )(sin ) RC /( )(
α .? solved numerically for α /ea+-to-pea+
variation%ripple$:
$sin1%0$sin%0 α α π −=+−=∆ VmVmVmVo
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In practical circuits where ω)C
,
" ,
"π α π θ ≈≈
minimal output voltage occurs at
α+π=wt
) RC /( ) RC /( )(
eVmeVm )( v ωπ−ω
π−
π+π−
==α+π ""#
[ ]
fRC Vm
RC Vm
RC Vm
eVmeVmVmVo ) RC /( ) RC /(
"
11
1
‧
‧
=ω π=
ωπ
−−=
−=−≈∆ ωπ−ωπ−
f w
e
π "
"1
1"
=
++++=!!!
is half that of the half-wave rectifier
π >>
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Fig. 4-$ %a& 'oltage dou(ler
ig. 4-$ %(& )ual voltage rectifier
*full-wave rectifier%sw. open&+
voltage dou(ler%sw. closed&
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L-! filtered output : Fig.4-,
C holds the output voltage at a constant level# and
the soothes the current fro rectifier and
reduces the peak current in diodes.
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Continuous
Current:
( & " π VmVoV == .# , full-wave rectified )
#, $%
" ==== Ic R
Vm R
Vo I I R L π
LiThe variation in can be estimate from the first
Ac term %n."$ in the (ourier series
The amplitude of the inductor current for n." is
L
Vm
L
/ Vm
L
V
Z
V I
πω=
ω
π=
ω==
"
"
4
"" "
"
"
where "1
1
1
1"=
+−
−π= n ,
nn
VmVn
(or Continuous current, L I I <
"
R
Vm
L
Vm
π<
πω"
"
ω>
R L 1
>
ω R
L
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*iscontinuous current:
2hen is positive ( at ),
Vowt V v m L −= sin
[ ]
( )[ ]
! i
,wt for
wt Vowt Vm
L
wt d Vowt Vm L
wt i
L
wt
L
==
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/rocedure for determining o:
%1$ 3stimate a alue for o slightly below m, and solve =α
%"$ 'olve numerically,β
$%$cos%cos#$% α β β α β −−−== VoVmi L
%$
'olve
[ ]∫
∫ β
α
β
α
α−−−αωπ=
π=
)wt ( d )wt ( Vo )wt cos(cosVm L
)wt ( d )wt ( i I L L
11
1
%4$ 'love o. R I L
%$ )epeat step %1$5%4$ until the computed o in
step%4$ e6uals the estimated o in step%1$
utput *oltage for discontinuous current is larger than
for continuous current.(see Fig4-,%d&)
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4-3 controlled full-wave rectifiers
Resistive load: Fig.4-1
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)cos( Vm
)wt ( d )wt sin( VmVo
α+
π
=
π= ∫
π
α
1
1
an"#e de#a$=α
$cos1% α π
+== R
Vm
R
Vo Io
π
α
π
α
π
π
α
4
$"sin%
""
1
$%$sin%1 "
+−=
= ∫
R
Vm
wt d wt R
Vm I rms
The power delivered to the load rmsR I P "=
The rms current in source is the same as the rms cur
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, e7cept that the
period for the output current is .π
[ ] ) /( )t ( o e ) sin( )t sin( Z
Vm )wt ( i ωτα−ω−θ−α−θ−ω= for β≤ω≤α t
R L , ) R
L( tan
) L( R Z
=τω=θ
ω+=
−1
""
(or discontinuous current
π α β +<
discontinuous current "
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continuous current
#$% , ≥++= α π α π iwt
[ ]
c%rrent contin%o%s for
R
L
Tan
& )'(
& )' sin( e
e
'
$%
#1$sin%
#$sin%$sin%
$8%
$8%$%
ω
θ α
α θ
α θ α θ
θ α θ α π ωτ π
ωτ α α π
=≤
≥≥
≥+−
≥−−−+−
−+−
,9,4,"
1
$1sin%
1
$1sin%"
1
$1cos%
1
$1cos%"
cos"
$%sin1
$cos%$%
""
1
=
−−
−++
=
−−
−++
=
+=
==
++=
∫
∑
+
∞
=
n
n
n
n
nVm
n
n
n
nVm
a
aVn
Vmwt d wt VmVo
nnwt VnVowt v
n
n
nn
n
&
α α
π
α α
π
α π π
θ
α π
α
$%an
nTann '
−=θ
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Fig 4-12
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RVo Io
) In( Io Irms
| jnwL R|Vn
ZnVn In
*** ,n
=
+=
+==
∑∞
= 4"
""
"
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R-L Source load : Fig.4-14
The 'C)' may be turned on at any time that they are forward biased, which is at an angle
$%sin 1Vm
Vdc−≥α
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(or continuous current case, the average bridge output volta
average load current is
The ac voltage terms are unchanged from the
controlled rectifier with an )-& load The ac current
terms are determined from circuit /ower absorbed by the dc voltage is
e# is Lif R IormsR I P arg "" ≈=
α
π
= cosVm
Vo"
RVdcVo Io −=
Vdc Io Pdc =
/ower absorbed by resistor in the load is
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Controlled 'ingle-phase converter operating as an inverter :
seeing Fig 4-14. 4-15
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## :##
Vo
rectifier
operation
## 1;#:#
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4-4 hree-phase rectifiers
Resistive load " Fig 4-1#
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上、下半部 *iode,每次僅一個
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3ach diode conducts one-third of the time, resulting in
av" oav" + I I ,,
1=
rmsorms + I I ,,
1=
rmsorms I I ,,
"=
Apparent power from the three-phase source is
rms , rms , L L I V −=
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*** , , ,n , )n(
V V
V *
V
)wt ( wtd sinV / V
)t nwcos( V Vo )t ( v
L L ,m
n
L L ,m
L L ,m /
/ L L ,m
** , ,n
n
1;1"91
9
:-#
1
"
"
#
#
1;1"9
#
=−π
=
=π=π=
π++=
−
−
−π
π −
∞
=
∫
∑
,ince the output voltage is periodic with period '+
of the ac suppl$ voltage# the haronics in the output
are of order +k# k&'#/#0#1
Adevantage: output is inherentl$ like a dc voltage# and
the high-fre2uenc$ low-aplitude haronics ena!le filters
to !e effective.
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For a dc load current %constant /& --- Fig4.1$
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****t wcost wcost wcost wcost w(cos I i oa ##### 11
111
11
1?
?
1-
-
1"+−+−
π=
which consists of terms at fundamental fre6uency of
the ac system and harmonics of order 9+ ± 1, +.1,",,@
Filters(Fig.4-1,) are fre2uentl$ necessar$ to prevent
haronic currents to enter the ac s$ste.
3esonant filters for th and 5th haronics.
High-pass
filters
for
higher
order
haronics.
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4- !ontrolled three-phase rectifiers
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α π
π
α π
α π
cos$
%
$%sin
1
,
"
,
L Lm
L Lmo
V
wt wtd V V
−
+
+ −
=
= ∫
armonics for output voltage remain of order 9+, but
amplitude are functions of α
seeing Fig. 4-20
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Twelve-pulse rectifiers: using two si6-pulse !ridges
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p g p g
The purpose of the transformer connection is to ∆Υ
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p p
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
∆− Υ##
##
#
#
The delay angles for the bridge are typically the same
α π α π α π cos
9
cos
cos
,,,,,
L Lm L Lm L Lm
o- oo
V V V
V V V
−−−
∆ =+=+=
The pea+ output of the twelve-pulse converter occurs
midway between alternate pea+s of the si7-pulse converters
Adding the voltages at that point for gives °= #α
°==°= −− #:"1$1-cos%" ,,, α for V V V L Lm L Lm pea. o
'ince a transition between conducting 'C)s every°
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'ince a transition between conducting 'C)s every
, there are a total of 1" such transitions for each
period of the ac source The output has harmonic
fre6uencies which are multiple of 1" times the source
fre %1"+ +.1,",@$
°#
,",111"
cos1
cos1
%cos
$%$%$%
$cos1
1cos
11
1cos
?
1cos
-
1%cos
"$%
$cos1
1cos
11
1cos
?
1cos
-
1%cos
"$%
###
#####
#####
=±=
−+=+=
++−+=
−+−+−=
∆
∆
. ,. order armonic ,i
***)t w00
t w
't w I 1
t it it i
t w0t wt w2 't w3t w I t i
t w0t wt w2 t w3t w I t i
ac
o- ac
o
o-
π
π
π
Cancellation of harmonics 9%"n-1$ 1 , n.1, ", @ has resulted
from this transformer and converter configuration
±
This principle can be e7panded to arrangements of hi h l b b i ti i d b f
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higher pulse number by incorporating increased number of
si7-pulse converters with transformers which have the
appropriate phase shifts
The characteristic ac harmonics of a p-pulse converter
will be p+ ' # k&'#/#01± Bore e7pense for producing high-voltage transformers
with the appropriate phase shifts
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ree-phase converter operating as a inverter :
seeing 4-22.
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The bridge output voltage o must be negative
operation Inverter '' &Vo ,
operation Rectifier '' &Vo ,
>
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4-6 DC power transmission
․ >y using controlled twelve-pulse converter %generally$
․ !sed for very long distances of transmission lines
Advantages: %1$ # voltage drop7 in lines (/) # line loss
#= L
4
∞=C 4
% ↓c%rrent #ine %$ Two conductors re6uired rather than three%4$ Transmission towers are smaller% $ /ower flow in a dc transmission line is
controllable by adustment of delay angles
at the terminals
%9$ /ower flow can be modulated during disturbances on
one of the ac system 'ystem stability
increased
%?$ The two ac systems that are connected by
the dc line do not need to be in synchronization
)
Disadvantages: costl$ ac-dc converter# filter# and control
s$ste re2uired at each end of the
line to interface with the ac s$ste.
Fi 4 23 i i l t
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Fig.4-23 using si0-pulse converter
° tifi:##
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°
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Fig.4-24 using twelve-pulse converter
(a !ipolar schee)
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4-7 commutation : effect of source inductance ( &
Single-phase (ridge rectifier" Fig.4-2 s 4
Assume that the load current is constant Io
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Assume that the load current is constant Io
Commutation interval starts at ωt.π
$changed polarity% o%rce
om
t
o s
I )wt cos( Ls
V
I )wt ( wtd sinVm Ls
)wt ( i
++ω
−=
+ω
=
∫
ω
π
1
1
Commutation is completed at ωt.π Du
[ ] ## 1 I )%cos( Ls
V I )%( i m ++π+
ω
−=−=+π
)Vm
4 I ( cos )
Vm
Ls I ( cos% oo
"1
"1 11 −=
ω−= −−
&8 Coutation angle:
Ls 4 ω=
Average load voltage is
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Average load voltage is
)V
4 I (
5V
)%cos( V
)wt ( d wt sinV V
m
som
mm
%o
−π=
+π
=π
= ∫ π
1
11
,ource inductance lowers the average output voltage
of full-wave rectifier.
hree-phase rectifier: Fig 4-2#
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hree phase rectifier : Fig.4 2#
*uring Commutation from The voltage1 +to+
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*uring Commutation from # The voltage
across a is1 +to +
wt sinV v
v L L ,m 67
La""
−==
Current in starts at 9: and decreases ;ero in the
coutation interval
La
)
V
I 4 ( cos )V
I L( cos%
I )wt ( d wt sinV
La )%( i
L L ,m
s
L L ,m
a
% L L ,m
La
−
−
−
−
+π
π
−
−=ω−=
+ω
==+π ∫ #1#1
#
"1"1
"
1#
*uring the commutation interval from the+to+
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*uring the commutation interval from # the
converter output voltage is
1 + to +
"
6C 7C
o
vvv
+=
""
"
#
7C 6C 7C 6C 6C
67 6C
*
*c La L 6C o
7C 6C 67C6 7C 67
vvvvv
vvvvvv
v'vv , vvv
+=
−−=
−=+−=
==++
Average output oltage: () 'ingle-phase rectifier
)V
I 4 (
V V
L L ,m
s L L ,m
o
−
− −π
= #1
,ource inductance lowers the average output voltage
of three-phase rectifiers.