Transcript CH.4 Full-wave and Three- phase rectifiers (Converting AC to DC)

CH.4
Full-wave and Threephase 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 threephase 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
for high-voltage applications.
make
it
more
suitable
Fig. 4-2 center-tapped
rectifier
transformer
With electrical isolation, only
one diode voltage drop between
the source and load, suitable for
low-voltage, high-current applications
Resistive
load：
 Vm sin wt ,0  wt   
v0 ( wt )  

 Vm sin wt ,   wt  2
Vo 
1


0
Io  Vo
Vm sin( wt )d ( wt )  2 Vm
R
 2Vm
Irms  Im

(R)
2
power absorbed by the load resistor：
PR  I 2 rmsR
power
factor ：Pf=1
R-L load：
Fig.4-3

v ( wt )  Vo 
Vn cos( nwt   )
0
n2,4‧‧
,
Vo  2Vm
Io  Vo
Vn 

2Vm  1
1 



  n 1 n  1
In  Vn
R
If L is relatively large, the
dc.
( L >> R )
i( wt )  Io 
Vo 2Vm 

R
R 

Irms  Io 
Vn

Zn | R  jnwL |
load current is essentially
for L >> R
Source harmonics are rich in the odd-numbered harmonics.
Filters：reducing the harmonics.
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.
2Vm
 Vdc
Vo  Vdc
Io 
 
R
R
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
| Vm sin wt | , one diode pair on
v0 ( wt )  
( wt  ) /( wRc )
Vm
sin

e
, diodes off

 ：the angle where the diodes become reverse biased,
which is the same as for the half-wave rectifier
and is
  Tan 1 ( RC )  Tan 1 ( RC )  
wt    
Vm sin  e (  ) /( RC )  Vm sin(    )
(sin  )e (  ) /( RC )  sin   0

=？
solved numerically for

Peak-to-peak variation(ripple)：
Vo  Vm | Vm sin(    ) | Vm(1  sin  )
In practical circuits where
  2
,
minimal
,  
 
ωRC
2
wt    
output voltage occurs at
v0 (    )  Vm e
 
(    ) /( RC )
2 2

 Vm e   /( RC )
Vo  Vm  Vm e   /( RC )  Vm 1  e   /( RC )
 
 
 Vm1  1 

‧
  RC 
Vm
Vm


RC 2 fRC

‧
x x2 x3
e  1 

 ...
1！ 2！ 3！
w  2f
x
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：
Vx  Vo  2Vm
I L  I R  Vo


R
 2Vm
 VL =0 , full-wave rectified
, Ic  0
(R)
i L can be estimate from the first
The variation in
Ac term (n=2) in the Fourier series.
The amplitude of the inductor current for n=2 is
I2 
where
V2
V
4Vm / 3 2Vm
 2 

Z 2 2L
2L
3L
Vn 
2Vm  1
1 


 , n2
  n 1 n  1
For Continuous current,
2Vm 2Vm

3L
R
R
L
3
I2  IL

3L
1
R

Discontinuous current：
When
iL
,
is positive
(
Vm sin wt  Vo
at
wt  
 Vo 
  sin 

 Vm 
1
vL  Vm sin wt  Vo
1 wt
Vm sin wt  Vod (wt )
iL ( wt ) 


L
1
Vm(cos   cos wt )  Vowt   

L
for   wt   ,    ,
iL (  )  0 ,   ?
)
Procedure for determining Vo：
(1) Estimate a
(2) Solve

Value
for
Vo
slightly below Vm, and solve
 ?
iL ( )  0  Vm(cos   cos  )  Vo(   )
numerically,

(3) Solve
1
I L   i L ( wt )d ( wt )

1  1
Vm(cos   cos wt )  Vo( wt   )d ( wt )
 

 L
(4) Slove
Vo=
ILR
(5) Repeat step (1)~(4) until the computed
equals the estimated Vo in step(1)
Vo
Output Voltage for discontinuous current is
for continuous current.(see Fig4-8(d))
in step(4)
larger
than
4-3
controlled full-wave rectifiers
Resistive
load：
Fig.4-10
1 
Vo   Vm sin( wt )d ( wt )
 
Vm

( 1  cos  )

Io 
I rms
  delay
angle
Vo Vm

(1  cos  )
R R
1

Vm
2

(
sin
wt
)
d ( wt )
  R

Vm 1  sin( 2 )


R 2 2
4
The power delivered to the load
P  I 2 rmsR
The rms current in source is the same as the rms current in
the load.
R-L
load : Fig.4-11
discontinuous current :

Vm
io ( wt ) 
sin( t   )  sin(    )e ( t  ) /(  )
Z

for
  t  
Z  R 2  ( L ) 2
  tan 1 (
L
)
R
, L
R
For discontinuous current
  
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  .
continuous current
wt    
, i (   )  0
sin(      )  sin(    )e  (   ) /( )

sin(    ) 1  e  /( )

0
sin(  -  )  0

v0 ( wt )  Vo   Vn cos( nwt  n)
( - )  0
    Tan (
-1
0
n 1
L
Vo 
)
R
for continuous current
1

 

Vm sin wt d ( wt ) 
Vn  an  bn
2
n  Tan -1 (
 bn
)
an
2Vm

cos 
2
an 
2Vm  cos( n  1) cos( n  1) 


  n 1
n  1 
bn 
2Vm  sin( n  1) sin( n  1) 

  n  1
n  1 
n  2,4,6,....
Fig 4-12
In  Vn
Zn
 Vn
Irms  Io 
2
| R  jnwL |


n  2 ,4...
Io  Vo
R
(
In
2
)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 (Vdc Vm)
For continuous current case, the average bridge output voltage is
Vo 
2 Vm
cos 

average load current is
Io 
Vo  Vdc
R
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
Pdc  Io Vdc
Power absorbed by resistor in the load is
P  I 2 rmsR  Io 2 R if
L is l arg e
Controlled Single-phase converter operating as an inverter：
seeing Fig 4-14. 4-15
.
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
00    90 0
900    1800

Vo  0
rectifier operation

Vo  0
inverter operation
Pbridge  Pac   IoVo
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 360 0 / 6  60 0
.
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,..
ia  i D1  i D 4
i  i  i
D3
D6
b
ic  i D 5  i D 2
Each diode conducts one-third of the time, resulting in
I D ,avg 
1
I o ,avg
3
I D ,rms 
1
I S ,rms 
2
I o,rms
3
3
I o,rms
Apparent power from the three-phase source is
S  3 VLL ,rms I S ,rms
v0 ( t )  Vo 

V
n
n  6 ,12 ,18..
cos( nw0 t   )
3Vm ,L  L
1 2 / 3
V0 
Vm ,L  L sin wtd( wt ) 


/
3
/3

 0.95Vm ,L  L
Vn 
6 Vm ,L  L
( n  1 )
2
, n  6 , 12 , 18, ...
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 highfrequency low-amplitude harmonics enable filters to be effective.
For a dc
load current (constant I0) --- Fig4.17
ia 
2 3
1
1
1
1
I o (cos w0 t  cos 5w0 t  cos 7 w0 t  cos 11w0 t  cos 13w0 t ....

5
7
11
13
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
currents to enter the
Resonant filters for
High-pass filters for
frequently necessary to prevent harmonic
ac system.
5th and 7th harmonics.
higher order harmonics.
4-5 Controlled three-phase rectifiers
Vo 
1

(
2

3


3 3
3Vm , L  L

Vm , L  L sin wtd ( wt )
) cos 
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
phase 30 0 shift
This results in
apart. The two
of the    transformer connection is to introduce
between the source and bridge.
30 0
inputs to two bridges which are
bridge outputs are similar, but also shifted by
30 0
.
The delay angles for the bridge are typically the same.
Vo  Vo,Y  Vo, 
3Vm, L  L

cos  
3Vm, L  L

cos  
6Vm, L  L

cos 
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   0 gives
Vo , peak  2Vm, L  L cos(15)  1.932 Vm, L  L
for   0
Since a transition between conducting SCRs every 30
, 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,…)
1
1
1
1
I o (cos w0t  cos 5 w0t  cos 7 w0t  cos 11w0t  cos 13 w0t  ....)

5
7
11
13
2 3
1
1
1
1
i (t ) 
I o (cos w0t  cos 5 w0t - cos 7 w0t  cos 11w0t  cos 13 w0t  ....)

5
7
11
13
4 3
1
1
iac (t )  iY (t )  i (t ) 
I o (cos w0t - cos 11w0t  cos 13 w0t  ...)

11
13
iac , harmonic order  12k  1 , k  1,2,...
iY (t ) 
2 3
Cancellation of harmonics 6(2n-1)  1 , n=1, 2, … has resulted
from this transformer and converter configuration.
This principle can be expanded
number by incorporating increased
with transformers which have the
The characteristic ac harmonics
pk  1
, k=1,2,3…
to arrangements of higher pulse
number of six-pulse converters
appropriate phase shifts.
of a p-pulse converter will be
 More expense for producing high-voltage transformers with the
appropriate phase shifts.
Three-phase converter operating as
seeing 4-22.
a inverter：
The bridge output voltage Vo must be negative.
0    90
, Vo  0 - -  Rectifier
operation
90    180 , Vo  0 - -  Inverter operation
4-6 DC power transmission
․ By using controlled twelve-pulse converter (generally).
․ Used for very long distances of transmission lines.
Advantages：(1)
(2)
X L  0 , voltage drop↓ in lines
(  line current 
X C   , line loss
)
(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
rectifier 
 , 0    90
Vo1 ,Vo 2  


,
90




180

inverter


For current being ripple free
Vo1  Vo 2
R
3Vm1, L  L
Vo1 
cos  1
Io 

Vo 2 
3Vm 2, L  L

cos  2
Power supplied by the converter at terminal 1 is
P1  Vo1 I o
Power supplied by the converter at terminal 2 is
P2  Vo 2 I o
Fig.4-24
using twelve-pulse converter
(a bipolar scheme)
One of the lines is energized at  Vdc and the other is energized
at - V dc . In emergency situations, one pole of the line can operate
without the other pole, with current returning through the ground path.
4-7 commutation ：effect of source inductance (
Single-phase bridge rectifier: Fig.4-25
Xs)
Assume that the load current is constant Io.
Commutation interval starts at ωt=  ( Source polarity changed )
1 t
i s ( wt ) 
Vm sin wtd( wt )  I o


Ls
Vm

( 1  cos wt )  I o
Ls
Commutation
is completed at ωt=
i(   u )   I 0 
=>

+u
 Vm
1  cos(   u )  I 0
Ls
Commutation angle：
2I o Ls
2I o X S
1
u  cos ( 1 
)  cos ( 1 
)
Vm
Vm
1
X S  Ls
Average load voltage
is
Vm
1 
V
sin
wt
d
(
wt
)

( 1  cos u )
m

u


2V
I X
 m (1  o s )

Vm
Vo 
Source inductance lowers the average output voltage of fullwave rectifier.
Three-phase rectifier：
Fig.4-26
During Commutation
is
v La
from
D1 to D3
, The voltage across La
v AB Vm ,L L


sin wt
2
2
Current in La starts at I0 and decreases zero in
commutation interval
1 u Vm ,L  L
i La (   u )  0 
sin wt d ( wt )  I 0


La
2
2L a I 0
2X s I0
u  cos 1 ( 1 
)  cos 1 ( 1 
)
Vm ,L  L
Vm ,L  L
the
During the commutation
output voltage is
v BC  v AC
vo 
2
interval from
,
D1 to D3
v AB  vBC  vCA  0 ,
v AB  v AC - vBC
.
vo  v AC  vL a  vL c  v AC  v AB
.
 v AC 
Average output Voltage：
Vo 
3Vm ,L  L

(1 
the converter
2
v AC  vBC v AC  vBC

2
2
類似 Single-phase rectifier
X s I0
)
Vm ,L  L
Source inductance lowers the average output voltage of threephase rectifiers.