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Chapter 2
AC to DC CONVERSION
(RECTIFIER)
• Single-phase, half wave rectifier
– Uncontrolled: R load, R-L load, R-C load
– Controlled
– Free wheeling diode
• Single-phase, full wave rectifier
– Uncontrolled: R load, R-L load,
– Controlled
– Continuous and discontinuous current mode
• Three-phase rectifier
– uncontrolled
– controlled
Power Electronics and
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1
Rectifiers
• DEFINITION: Converting AC (from
mains or other AC source) to DC power by
using power diodes or by controlling the
firing angles of thyristors/controllable
switches.
• Basic block diagram
AC input
DC output
• Input can be single or multi-phase (e.g. 3phase).
• Output can be made fixed or variable
• Applications: DC welder, DC motor drive,
Battery charger,DC power supply, HVDC
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Single-phase, half-wave, R-load
+
vs
_
+
vo
_
vs

vo
t
2
io
Output vol tage (DC or average),
Vo  Vavg 

Vm
Vm sin(t )dt 

2 0

1
 0.318Vm
Output vol tage (rms),
Vo , RMS 

2
Vm
1


V
sin(

t
)
d

t

 0.5Vm
m

2 0
2
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Half-wave with R-L load
i
+
vR
_
+
vs
_
+
vL
_
+
vo
_
KVL : vs  v R  v L
di(t )
d t
First order differenti al eqn. Solution :
Vm sin(t )  i (t ) R  L
i (t )  i f (t )  in (t )
i f : forced response; in natural response,
From diagram, forced response is :
V 
i f (t )   m   sin(t   )
 Z 
where :
Z  R 2  (L) 2
  tan 1 
L 

R
 
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R-L load
Natural response is when source  0,
di(t )
i (t ) R  L
0
d t
which results in :
in (t )  Ae t  ;   L R
Hence
V 
i (t )  i f (t )  in (t )   m   sin(t   )  Ae t 
 Z 
A can be solved by realising inductor current
is zero before the diode starts conducting , i.e :
 Vm 
0 
i ( 0)  
  sin( 0   )  Ae
 Z 
 Vm 
 Vm 
 A
  sin(  )  
  sin( )
 Z 
 Z 
Therefore the current is given as,

V 
i (t )   m   sin(t   )  sin( )e t 
 Z 
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R-L waveform
vs,
io
b
vo
vR
vL
0
2

3
4
t
Note :
v L is negative because the current is decreasing , i.e :
vL  L
di
dt
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Extinction angle
Note that the diode remains in forward biased
longer tha n  radians (although the source is
negative during that duration)T he point when
current reaches zero is whendiode turns OFF.
This point is known as theextinc tion angle, b .


 Vm 
 b 
i(b )  
0
  sin( b   )  sin( )e
 Z 
which reduces to :
sin( b   )  sin( )e  b   0
b can only be solved numericall y.
Therefore, the diode conducts between 0 and b
To summarise the rectfier w ith R - L load,

 Vm 
t 

sin(

t


)

sin(

)
e
 Z 

i (t )  for 0  t  b
0

otherwise
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
7
RMS current, Power
The average (DC) current is :
b
1 2
1
Io 
i (t )dt 
i (t )dt
2 0
2 0
The RMS current is :
b
1 2 2
1 2
I RMS 
i (t )dt 
i (t )dt


2 0
2 0
POWER CALCULATIO N
Power absorbed by the load is :
Po   I RMS 2  R
Power Factor is computed from definition :
P
pf 
S
where P is the real power supplied by the source,
which equal to the power absorbed by the load.
S is the apparent power supplied by the
source, i.e
S  Vs , RMS . I RMS 
 pf 
P
Vs,RMS .I RMS 
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Half wave rectifier, R-C Load
+
vs
_
+
vo
_
vs
Vm
 /2
Vmax
Vmin
iD
2 3 /2

v
4
vo
DVo
iD
a
vo
3

w hen diode is ON
Vm sin(t )

V e t   / RC
when diode is OFF
 Vm sin 
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Operation
• Let C initially uncharged. Circuit is
energised at t=0
• Diode becomes forward biased as the
source become positive
• When diode is ON the output is the same
as source voltage. C charges until Vm
• After t=/2, C discharges into load (R).
• The source becomes less than the output
voltage
• Diode reverse biased; isolating the load
from source.
• The output voltage decays exponentially.
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Estimation of 
The slope of the functions are :
d Vm sin t 
 Vm cos t
d (t )
and
d V sin   e t   / RC
m
d (t )

1  t   / RC

 Vm sin    
e
 RC 
At t   , the slopes are equal,
1     / RC

Vm cos   Vm sin    
e
 RC 
V cos 
1
 m

Vm sin  
RC
1
1

tan   RC
  tan 1  RC    tan 1 RC   
For practical circuits, RC is large, then :


  -tan         
2
2
 is very close to the peak of the sine wave. Therefore
and Vm sin   Vm
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Estimation of a
At t  2  a ,
Vm sin( 2  a )  (Vm sin  )e ( 2 a  ) RC
or
sin(a  (sin  )e ( 2 a  ) RC  0
This equation must be solved numericall y for a
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Ripple Voltage
Max output vol tage is Vmax .
Min output vol tage occurs at t  2  a
DVo  Vmax  Vmin
 Vm  Vm sin( 2  a )  Vm  Vm sin a
If V  Vm and    2, and C is large such that
DC output vol tage is constant, then a   2.
The output vol tage evaluated at t  2  a is :
vo (2  a )  Vm
 2  2 2 



RC


e
 Vm
 2 



RC


e
The ripple voltage is approximat ed as :
DVo  Vm  Vm
 2 



RC

e 
Using Series expansoin :
 2  






RC

 Vm 1  e 





 2 



RC


e
2
1
RC
 2  Vm
 DVo  Vm 

 RC  fRC
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Capacitor Current
The current in the capacitor can be expressed as :
dv (t )
ic t   C o
d (t )
In terms of t , :
dv (t )
ic t   C o
d (t )
But
Vm sin(t )
vo (t )  
Vm sin   e t   / RC
when diode is ON
when diode is OFF
Then, substituting vo (t ),
CVm cos(t )
when diode is ON,

 i.e (2  a )  t  (2   )


ic t   

 Vm sin   e t   / RC

R
when diode is OFF,

i.e ( )  t  (2  a )
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Peak Diode Current
Note that :
is  iD  iR  iC
The peak diode current occurs at (2  a ). Hence.
I c, peak  CVm cos (2  a )  CVm cos a
Resistor current at (2  a ) can be obtained :
.
Vm sin (2  a ) Vm sin a
iR (2  a ) 

R
R
The diode peak current is :
Vm sin a
iD, peak  CVm cos a 
R
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Example
A half-wave rectifier has a 120V rms source at 60Hz. The
load is =500 Ohm, C=100uF. Assume a and  are calculated
as 48 and 93 degrees respectively. Determine (a) Expression
for output voltage (b) peak-to peak ripple (c) capacitor
current (d) peak diode current.
vs
Vm
 /2
Vmax
Vmin
2 3 /2

3
4
vo
DVo
iD
Vm  120 2  169 .7V ;
a

  93o  1.62 rad ;
a  48 o  0.843rad
Vm sin   169 .7 sin(1.62 rad )  169 .5V ;
(a) Output vol tage :
Vm sin(t )  169 .7 sin(t )
vo (t )  
Vm sin   e t   / RC
169 .7 sin(t )

t 1.62  /(18.85)
169 .5e
(ON)
(OFF)
(ON)
(OFF)
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Example (cont’)
(b)Ripple :
Using : DVo  Vmax  Vmin
DVo  Vm  Vm sin( 2  a )  Vm  Vm sin a  43V
Using Approximat ion :
169 .7
 2  Vm
DVo  Vm 

 56.7V

 RC  fRC 60  500  100u
(c) Capacitor current :
CVm cos( t )

ic t    Vm sin( ) t   /(RC )
e

R
6.4 cos(t ) A

t 1.62  /(18.85)
 0.339  e
(ON)
(OFF)
(ON)
A
(OFF)
(d) Peak diode current :
V sin a
iD, peak  CVm cos a  m
R
 (2    60)(100u )169 .7 cos( 0.843rad ) 
169 .7 sin(1.62 rad )
500
 (4.26  0.34)  4.50 A
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Controlled half-wave
ig
vs
ia
+
vs
_
+
vo
_
t
vo
t
v
ig
a
Average voltage :
t
Vm
1 


1  cos a 
Vo 
V
sin

t
d

t

m

2 a
2
RMS voltage

2
1
Vm sin t  dt
Vo, RMS 

2 a
Vm2 
Vm
a sin 2a 

[
1

cos(
2

t
]
d

t

1



4 a
2

2
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Controlled h/w, R-L load
i
+
vR
_
+
vs
_
+
vL
_
+
vo
_
vs

t
2
vo
io
a
b
t
V 
i (t )  i f (t )  in (t )   m   sin t     Ae 
 Z 
Initial condition : i a   0,
a
V 
i a   0   m   sin a     Ae 
 Z 
a
 V 

 A    m   sin a    e 
 Z 

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Controlled R-L load
Substituti ng for A and simplifyin g,
(a t ) 


V
 m   sin t     sin a   e   for a  t  b
i t    Z  




0 otherwise
Extinction angle b must be solved numericall y
(a  b ) 

V 
i b   0   m  sin b     sin b   e  

 Z 


Angle   b    is called the conduction angel .
Average voltage :
b
Vm
1


cos a  cos b 
Vo 
V
sin

t
d

t

m
2 a
2
Average current :
b
1
Io 
i t d
2 a
RMS current :
b
1 2
I RMS 
 i t d
2 a
The power absorbed by the load :
Po  I RMS 2  R
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Examples
1. A half wave rectifier has a source of 120V RMS at 60Hz.
R=20 ohm, L=0.04H, and the delay angle is 45 degrees.
Determine: (a) the expression for i(t), (b) average
current, (c) the power absorbed by the load.
2. Design a circuit to produce an average voltage of 40V
across a 100 ohm load from a 120V RMS, 60Hz supply.
Determine the power factor absorbed by the resistance.
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Freewheeling diode (FWD)
• Note that for single-phase, half wave rectifier
with R-L load, the load (output) current is
NOT continuos.
• A FWD (sometimes known as commutation
diode) can be placed as shown below to make
it continuos
io
+
vR
_
+
vs
_
+
vL
_
+
vo
_
io
io
vo= 0
+
vs
_
vo= vs
+
vo
+
vo
io
_
_
D1 is on, D2 is off
D2 is on, D1 is off
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Operation of FWD
• Note that both D1 and D2 cannot be turned
on at the same time.
• For a positive cycle voltage source,
– D1 is on, D2 is off
– The equivalent circuit is shown in Figure (b)
– The voltage across the R-L load is the same as
the source voltage.
• For a negative cycle voltage source,
–
–
–
–
D1 is off, D2 is on
The equivalent circuit is shown in Figure (c)
The voltage across the R-L load is zero.
However, the inductor contains energy from
positive cycle. The load current still circulates
through the R-L path.
– But in contrast with the normal half wave
rectifier, the circuit in Figure (c) does not
consist of supply voltage in its loop.
– Hence the “negative part” of vo as shown in the
normal half-wave disappear.
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FWD- Continuous load current
• The inclusion of FWD results in continuos
load current, as shown below.
• Note also the output voltage has no
negative part.
output
vo
io
t
iD1
Diode
current
iD2
0

2
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3
4
24
io
D1
iD1
Full wave rectifier
D
is
3
+
vs
_
+
vo
_
D
Full Bridge 4
D
2
is
+
vs
_
iD1
D1
+ vD1 
+
vs1
_
+
vs2
_
 vo
io
+ vD2 
iD2
Center-tapped
+
D
• CT: 2 diodes
• FB: 4 diodes.
Hence, CT
experienced
only one diode
volt-drop per
half-cycle
2
•
For both circuits,
Vm sin t
vo  
 Vm sin t
• Center-tapped
(CT) rectifier
requires
center-tap
transformer.
Full Bridge
(FB) does not.
0  t  
  t  2
Conduction
losses for CT
is half.
Average (DC) voltage :
• Diodes ratings
for CT is twice
2Vm
1
Vo   Vm sin t dt 
 0.637Vm
than FB
0

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io
D1
is
iD1
Bridge waveforms
D3
+
vs
_
+
vo
_
Full Bridge
Vm
D4
D2
v
s
2

Vm
3
4
v
o
vD1 vD2
-Vm
vD3 vD4
Vm
io
iD1 iD2
iD3 iD4
i
s
Power Electronics and
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Dr. Zainal Salam, 2003
26
Center-tapped waveforms
is
iD1
D1
+ vD1 
 vo
+
vs1
_
+
vs
_
+
vs2
_
iD2
Center-tapped
Vm
+ vD2 
+
io
D
2
vs
2

Vm
3
4
vo
vD1
-2Vm
vD2
-2Vm
io
iD1
iD2
is
Power Electronics and
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27
Full wave bridge, R-L load
iD1
io
+
vR
_
+
vL
_
is
+
vs
_
+
vo
_
vs

2
t
iD1 , iD2
iD3 ,iD4
io
vo
is
Power Electronics and
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28
Approximation with large L
Using Fourier Series,
vo (t )  Vo 

Vn cos( nt   )
n  2, 4...
where the DC term
2V
Vo  m

and the harmonics terms
2Vm  1
1 



  n  1 n  1
The DC curent
V
Io  o
R
The harmonic currents :
V
Vn
In  n 
Z n R  jn L
Vn 
As n increases, Vn harmonic decreases.
Thus I n decreases rapidly ve ry increasing n.
If L is large enough, it is possible to drop all
the harmonic terms, i.e. :
V
2V
i t   I o  o  m , for L  R,
R
R
Power Electronics and
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29
R-L load approximation
Approximat e current
V
2V
Io  o  m ,
R
R


I RMS  I o 2   I n, RMS 2  I o
Power delivered to the load :
Po  I RMS 2 R
vs

2
t
iD1 , iD2
iD3 ,iD4
io
vo
is
Power Electronics and
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30
Examples
Given a bridge rectifier has an AC source Vm=100V at
50Hz, and R-L load with R=100ohm, L=10mH
a)
determine the average current in the load
b)
determine the first two higher order harmonics of the
load current
c)
determine the power absorbed by the load
Power Electronics and
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Dr. Zainal Salam, 2003
31
T1
io
iD1
Controlled full wave, R load
T3
is
+
vs
_
+
vo
_
T2
T4
Average (DC) voltage :
Vo 
1
Vm


1  cos a 
V
sin

t
d

t

 m
a

RMS Voltage
Vo, RMS 
1

2
Vm sin t  dt

a
a sin 2a 

1

2 2
4
The power absorbed by the R load is :
 Vm
VRMS 2
Po 
R
Power Electronics and
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32
Controlled, R-L load
iD1
io
+
vR
_
is
+
vs
_
+
vL
_
+
vo
_
io
a
 b
a
2
vo
Discontinuous mode
+a
io
a

b
2
vo
Continuous mode
Power Electronics and
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33
Discontinuous mode
Analysis similar to controlled half wave with
R - L load :

 Vm 
i (t )     sin(t   )  sin(a   )e (t a ) 
 Z 
for a  t  b
Z  R 2  (L) 2
L
1  L 
and   tan   ;  
R
 R 
For discontino us mode, need to ensure :
b  (a   )
Note that b is the extinction angle and
must be solved numericall y with condition :
io ( b )  0
The boundary between continous and
discontino us current mode is when b in
the output current expression is (  a ).
For continous operation current at
t  (  a ) must be greater than zero.
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34

Continuous mode
i (  a )  0
sin(  a   )  sin(  a   )e ( a a )   0
Using Trigonomet ry identity :
sin(  a   )  sin(  a ),

sin(  a ) 1  e (
 )
 0,
Solving for a
a  tan 1 
L 

 R 
Thus for continuous current mode,
L
a  tan 1  
 R 
Average (DC) output vol tage is given as :
2Vm
1 a 
Vo 
cos a
 Vm sin t dt 

a

Power Electronics and
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35
Single-phase diode groups
D1
io
+
vs
_
D3
vp
+
vo
_
D4
D2
vn
vo =vp vn
• In the top group (D1, D3), the cathodes (-) of the two
diodes are at a common potential. Therefore, the
diode with its anode (+) at the highest potential will
conduct (carry) id.
• For example, when vs is ( +), D1 conducts id and D3
reverses (by taking loop around vs, D1 and D3).
When vs is (-), D3 conducts, D1 reverses.
• In the bottom group, the anodes of the two diodes
are at common potential. Therefore the diode with
its cathode at the lowest potential conducts id.
• For example, when vs (+), D2 carry id. D4 reverses.
When vs is (-), D4 carry id. D2 reverses.
Power Electronics and
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36
Three-phase rectifiers
D1
+ van
-
io
D3
n
+ vbn
+ vcn
-
D5
-
vpn
D2
D6
vnn
+
vo
_
vo =vp vn
D4
van
Vm
vbn
vcn
vp
Vm
vn
vo =vp - vn
0

2
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3
4
37
Three-phase waveforms
• Top group: diode with its anode at the
highest potential will conduct. The other
two will be reversed.
• Bottom group: diode with the its cathode at
the lowest potential will conduct. The other
two will be reversed.
• For example, if D1 (of the top group)
conducts, vp is connected to van.. If D6 (of the
bottom group) conducts, vn connects to vbn .
All other diodes are off.
• The resulting output waveform is given as:
vo=vp-vn
• For peak of the output voltage is equal to
the peak of the line to line voltage vab .
Power Electronics and
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38
Three-phase, average voltage
vo
vo
/3
Vm, L-L
0
/3
2/3
Considers only one of the six segments. Obtain
its average over 60 degrees or  3 radians.
Average voltage :
Vo 
1
2 3
 Vm,L L sin(t )dt
 3
3

3Vm, L  L

3Vm, L  L


cos(t )233
 0.955Vm, L  L
Note that the output DC voltage component of
a three - phase rectifier is much higher tha n of a
single - phase.
Power Electronics and
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39
Controlled, three-phase
T1
+ van
-
io
T3
+ vbn
-
T5
n
+ vcn
vpn
+
vo
_
T2
vnn
T6
T4
a
van
vbn
Vm
vcn
vo
Power Electronics and
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40
Output voltage of controlled
three phase rectifier
From the previous Figure, let a be the
delay angle of the SCR.
Average voltage can be computed as :
Vo 
1
2 3a
 3
 Vm,L L sin(t )dt
3a
 3Vm, L  L 

  cos a
 

• EXAMPLE: A three-phase controlled rectifier has
an input voltage of 415V RMS at 50Hz. The load
R=10 ohm. Determine the delay angle required to
produce current of 50A.
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41