AC to DC Converter

1. Rectifiers: Converting AC to DC
1

2. 2.1 Single-Phase Half-Wave
Rectifiers
2.2 Performance Parameters
2.3 Single-Phase Full-Wave
Rectifiers
2.4 Single-Phase Full-Wave
Rectifier With RL Load
2.5 Three-Phase Bridge
Rectifiers
2.6 Three-Phase Bridge Rectifier
With RL Load
2

3. 2.1 To understand the operation and
characteristics of single and three
phases uncontrolled rectifiers
2.2 To understand the performance
parameters of single and three phases
uncontrolled rectifiers
3

4. 4

5. 5
t
i ,vL
vL
vs /R
io
t
io
i=vs/R
t
i ,vC
vs /R
io
vC
vC

6.  Converts AC power signal into DC
power signal.
 Converts signal (voltage/current) in
sinusoid into a useful and reliable
constant (dc) voltage for the successful
operation of electronic circuits and
direct current machines.
 The conversion process is called the
rectification.
6

7.  The rectifiers are classified into two types,
single-phase and three-phase.
 The typical applications of the rectifier
circuits such as dc welder, dc motor drive,
Battery charger, dc power supply, High
Voltage Direct Current (HVDC).
7

8. 8

9. 9
Figure 2.1: Single-phase half-wave rectifier
(b) Waveforms
(a) Circuit diagram
!
!
!
?,
!
!
!
?,
prove
please
I
prove
please
I
dc
rms


Principle of rectifier operation .

10.  A single-phase half-wave rectifier is the
simplest type.
 
 
 
 
m
m
m
m
m
o
dc
V
V
V
t
V
t
d
t
V
V
318
.
0
0
cos
cos
2
cos
2
sin
2
1
0
0
,







 










10

11. m
m
m
m
m
T
m
rms
o
V
t
t
V
t
d
t
V
t
d
t
V
t
d
t
V
t
d
t
V
T
V
5
.
0
2
)
(
2
sin
)
(
4
)
(
2
)
(
2
cos
1
2
)
(
)
(
sin
2
1
)
(
)]
sin(
[
2
1
)
(
)]
sin(
[
1
0
2
0
2
0
2
2
0
2
0
2
,














 


























11

12. 12
Single-Phase Half-Wave
Rectifiers – RL loads
•Due to inductive load, the conduction period of diode D1 will
extend beyond 180o until the current becomes zero at t =  + .
(a) Circuit diagram (b) Waveforms
Figure 2.3: Half-wave rectifier with RL load:

13. 13
(c) Waveforms

14. 14
•Vs, Vo & Io of the single phase half-wave
rectifier using the load is R and RL

15.  The waveforms for the current and
voltage are shown in Figure 2.3(b). The
average, VL of the inductor is zero.
 The average output voltage is
 The average load current is Idc = Vdc/R.
15
     
 

















 

 cos
1
2
cos
2
sin
2
0
0
m
m
m
dc
V
t
V
t
d
t
V
V

16. 16
   





 















 
















 










2
)
(
2
sin
)
(
4
0
2
)
(
2
sin
)
(
4
2
)
(
2
sin
)
(
4
)
(
2
)
(
2
cos
1
2
)
(
)
(
sin
2
1
sin
2
1
2
2
`
0
2
0
2
0
2
2
0
2






























m
m
m
m
m
m
rms
V
V
t
t
V
t
d
t
V
t
d
t
V
t
d
t
V
V

17.  The effect of diode Dm is to prevent a
negative voltage appearing across the
load; and as a result, the magnetic
stored energy is increased.
 At t = t1 = /, the current from D1 is
transferred to Dm and this process is
called commutation of diodes and the
waveforms are shown in Figure 2.3(c).
 Load current i0 is discontinuous with a
resistive load and continuous with a
very high inductive load.
 The continuity of the load current
depends on its time constant  = L/R.
17

18. 18
t
io
io (mode 1)
io (mode 2)
mode 2
mode 1
• Function of Dm : to remove dissipated as
a heat a cross the switch as sparks, it
was caused by energy in the inductive

19. 19

20.  Although the output voltage as
shown is dc, it is discontinuous
and contains harmonics.
 A rectifier is a power processor
that should give a dc output
voltage with a minimum amount of
harmonic contents.
20

21.  At the same time, it should maintain the
input current as sinusoidal as possible and
in phase with the input voltage so that the
power factor is near unity.
 The power-processing quality of a rectifier
requires the determination of harmonic
contents of the input current, the output
voltage and the output current.
21

22. 22
The average value of the output (load) Voltage, Vdc
The average value of the output (load) current, Idc
The output dc power, Pdc = VdcIdc
The root-mean-square (rms) value of the output voltage, Vrms
The rms value of the output current, Irms
The output ac power, Pac = VrmsIrms
efficiency The effective
(rms) value
The form factor
The ripple factor transformer
utilization factor
ac
dc
P
P

 2
2
dc
rms
ac V
V
V 

dc
rms
V
V
FF 
dc
ac
V
V
RF 
1
1 2
2











 FF
V
V
RF
dc
rms
s
s
dc
I
V
P
TUF 

23. 23
The harmonic factor (HF) Crest factor (CF)
The input power factor (PF) The displacement
factor
DF = cos 
2
/
1
2
1
2
/
1
2
1
2
1
2
1

























 

s
s
s
s
s
I
I
I
I
I
HF

 cos
cos 1
1
s
s
s
s
s
s
I
I
I
V
I
V
PF 

s
peak
s
I
I
CF
)
(

 (displacement angle) is the angle between fundament component
of the input current and voltage

24. Finding the performance parameters of a Half-
wave Rectifier – R load
The rectifier in Figure 2.l(a) has a purely resistive
load of R. Determine,
(a) the efficiency,
(b) the FF,
(c) the RF,
(d) the TUF,
(e) the PIV of diode D1,
(f) the CF of the input current
(g) input PF.
24

25. 25
Solution 2.1
 
 
%
5
.
40
5
.
0
5
.
0
318
.
0
318
.
0
)
















R
V
V
R
V
V
I
V
I
V
P
P
a
m
m
m
m
ac
ac
dc
dc
ac
dc


26. 26
%
157
57
.
1
318
.
0
5
.
0
FF
)




m
m
dc
rms
V
V
V
V
b
%
121
21
.
1
1
F
) 2



 FF
R
c

27. 27
Solution 2.1
d) The rms voltage of the transformer secondary is
  m
m
T
m
s V
V
dt
t
V
T
V 707
.
0
2
sin
1
2
/
1
0
2








  
The rms value of the transformer secondary current is the same as
that of the load:
R
V
I m
s
5
.
0


28. 28
The volt-ampere rating (VA) of the transformer,
VA = VsIs = 0.707Vm × 0.5Vm/R.
%
6
.
28
286
.
0
5
.
0
)
707
.
0
(
/
)
318
.
0
( 2










R
V
V
R
V
I
V
P
TUF
m
m
m
s
s
dc

29. 29
Solution 2.1
e) The peak reverse (or inverse) blocking voltage PIV = Vm
f) Is(peak) = Vm/R and Is = 0.5Vm/R
The CF of the input current is CF = Is(peak)/Is = 1/0.5 =2
g) The input PF for a resistive load can be found from
707
.
0
5
.
0
707
.
0
5
.
0 2




VA
P
PF ac

30.  If the output is
connected to a
battery, the rectifier
can be used as a
battery charger. This
is shown in Figure
2.4(a). For vs > E,
diode D1 conducts.
30

31. The angle () when the diode starts
conducting can be found from the
condition
31
m
V
E
1
sin 


E
Vm 

sin

32.  Diode D1 is turned off when Vs  E
at
  =  – 
 The charging current iL, which is
shown in Figure 2.4(b), can be
found from
32
R
E
t
V
R
E
v
i
i m
s
L






sin
0 for  < t < 

33.  
 
 
 
B
A
B
A
B
A
note
E
E
V
E
V
E
V
t
E
t
V
t
d
E
t
V
V
m
m
m
m
m
o
dc
sin
sin
cos
cos
)
cos(
:
2
cos
2
2
1
)
cos
(
))
(
)
cos(
(
2
1
cos
2
1
)
sin
(
2
1
,

















































33

34. 34
   
   
   
 





































 
















 
















































cos
4
2
sin
2
2
2
2
1
cos
2
2
)
(
2
sin
2
)
(
sin
2
2
)
(
2
cos
1
2
)
(
)
sin
2
)
(
sin
(
2
1
sin
2
1
2
2
2
2
2
2
2
2
2
2
2
E
V
V
E
V
t
E
t
EV
t
t
V
t
d
E
t
EV
t
V
t
d
E
t
EV
t
V
t
d
E
t
V
V
m
m
m
m
m
m
m
m
m
rms

35. 35
E
V
PIV
P
P
P
hours
time
h
h
P
W
EI
P
P
R
I
P
R
V
I
R
V
I
m
diode
R
dc
dc
o
o
dc
battery
dc
dc
battery
rms
resistor
R
o
rms
rms
o
dc
dc












)
(
;
;
2
)
(
,
,

36.  Question for Battery Charger:
 E=20volt, Pbattery=200Wh, Idc=10A,
Vmax(source)=120volt, f=60hz,
transformer ratio 2:1, determine:
 1) Conduction angle (-)
 2) The current limiting resistance R
 3) The power rating of R (PR)
 4) the charging time ho
 5) Efficiency of the rectifier
 6) PIV of the diode

36

37. Finding the Performance Parameters of a Battery
Charger
The battery voltage in Figure 2.4(a) is E = 12 V
and its capacity is 100 Wh. The average charging
current should be Idc = 5 A. The primary input
voltage is Vp = 120 V, 60 Hz, and the
transformer has a turn ratio of n = 2:1. Calculate
(a) the conduction angle  of the diode.
(b) the current-limiting resistance R.
(c) the power rating PR of R.
(d) the charging time h0 in hours.
(e) the rectifier efficiency .
(f) the PIV of the diode.
37

38.  When thyristor T1 is fired at t = , thyristor T1
conducts and the input voltage appears across the
load.
 When the input voltage starts to be negative at t = ,
the thyristor is negative with respect to its cathode
and thyristor T1 said to be reverse biased and it is
turned off.
 The time after the input voltage starts to go negative
until the thyristor is fired at t =  is called the delay
or firing angle . .
38

39. 39
Figure 2.14: Single-phase thyristor converter with a resistive load

40.  If Vm is the peak input voltage, the average
output voltage Vdc can be found from.
 The maximum output voltage Vdm is (a=0)
 The normalized output voltage
40
     











cos
1
2
cos
2
sin
2
1




 
m
m
m
dc
V
t
V
t
td
V
V

m
dm
V
V 
 

cos
1
5
.
0 


dm
dc
n
V
V
V

41.  The root-mean-square (rms) output
voltage.
 
   
2
/
1
2
/
1
2
2
/
1
2
2
2
2
sin
1
2
2
cos
1
4
sin
2
1














































m
m
m
rms
V
t
d
t
V
t
td
V
V
41

42. Find the Performances of a Single-Phase
Controlled Converter
If the converter of Figure 2.14(a) has a
purely resistive load of R and the delay
angle is  = /2. Determine
(a) the rectification efficiency.
(b) the form factor (FF).
(c) the ripple factor (RF).
(d) the TUF.
(e) the peak inverse voltage (PIV) of
thyristor T1.
42

43. m
m
m
rms
V
V
V
V
3536
.
0
2
)
2
/
(
2
sin
2
1
2
2
2
sin
1
2
2
/
1
2
/
1







































43
Delay angle is  = /2
Vdc = Vm/2π(1 + cos ) = 0.1592Vm.
Idc = 0.1592Vm /R
Irms = 0.3536Vm/R

44. 44
(a) The efficiency
(b)
(c)
(d)
(e) PIV = Vm
%
27
.
20
)
3536
.
0
(
)
1592
.
0
(
/
)
(
/
)
(
2
2
2
2




m
m
ac
dc
ac
dc
V
V
R
V
R
V
P
P

%
1
.
222
221
.
2
1592
.
0
3536
.
0




m
m
dc
rms
V
V
V
V
FF
983
.
1
1
221
.
2
1 2
2




 FF
RF
  1014
.
0
/
)
3536
.
0
(
2
/
/
)
1592
.
0
( 2



R
V
V
R
V
I
V
P
TUF
m
m
m
s
s
dc

45. 45

46.  A full-wave rectifier circuit
with a center-tapped
transformer.
 Because there is no dc current
flowing through the
transformer, there is no dc
saturation problem of
transformer core. The average
output voltage is
46
 


2
/
0
6366
.
0
2
sin
2
T
m
m
m
dc V
V
dt
t
V
T
V



47.  Instead of using a center-tapped transformer,
we could use four diodes .
 This circuit is known as a bridge rectifier, and
it is commonly used in industrial applications.
47

48. 48
Figure 2.5: Full-wave rectifier with center-tapped transformer
During the positive half cycle: D1 conduct and D2 does not
During the negative half cycle: D1 does not and D2 conduct
This enables the load current to be in the same direction for both
half cycles

49. 49
With this circuit, when the AC supply voltage is positive at point A
and negative at point B, current flows from point B, through D2, the
load D1 and to point A.
Ehwn the AC supply voltage is positive at point B and negative at
point A, current flow from point A, through D4, the load, D3 and to
point B.
Figure 2.6: Full-wave bridge rectifier

50. !
!
!
?,
!
!
!
?,
!
!
!
?,
prove
please
I
prove
please
V
prove
please
I
rms
rms
dc



50
(a) Circuit diagram (b) Waveforms
Figure 2.5: Full-wave rectifier with center-tapped transformer

51. !
!
!
?,
!
!
!
?,
!
!
!
?,
prove
please
I
prove
please
V
prove
please
I
rms
rms
dc



51
(a)Circuit diagram (b) Waveforms
Figure 2.6: Full-wave bridge rectifier

52.  
 
 
 
m
m
m
m
m
o
dc
V
V
V
t
V
t
d
t
V
V
636
.
0
2
0
cos
cos
cos
sin
1
0
0
,







 










52

53. m
m
m
m
m
T
m
rms
o
V
t
t
V
t
d
t
V
t
d
t
V
t
d
t
V
t
d
t
V
T
V
707
.
0
2
)
(
2
sin
)
(
2
)
(
2
)
(
2
cos
1
)
(
)
(
sin
1
)
(
)]
sin(
[
1
)
(
)]
sin(
[
1
0
2
0
2
0
2
2
0
2
0
2
,














 


























53

54. Finding the Performance Parameters of a Full-
Wave Rectifier with Center-Tapped Transformer
If the rectifier in Figure 2.5(a) has a purely
resistive load of R, determine
(a) the efficiency.
(b) the FF.
(c) the RF.
(d) the TUF.
(e) the PIV of diode D1.
(f) the CF of the input current.
54

55. 55
Single-Phase Full-wave Controlled Converter – R Load

56.  If Vm is the peak input voltage, the average
output voltage Vdc can be found from.
 The maximum output voltage Vdm is (a=0)
 The normalized output voltage
56
     











cos
1
cos
sin
1




  m
m
m
dc
V
t
V
t
td
V
V

m
dm
V
V
2

 

cos
1
5
.
0 


dm
dc
n
V
V
V

57. 57

58. 58
Figure 2.7:
Full-bridge
with RL-load

59.  In practice, most loads are R&L
 The load current depends on the values
of load resistance R and load inductance
L.
59

60.  If the input voltage,
 vs = Vm sin t = 2 Vs sin t
 The load current i0 can be found
from
   
R
E
e
A
t
Z
V
i t
L
R
s



  /
1
0 sin
2


60

61.  where load impedance:
Z = [R2 + (L)2]1/2
 load impedance angle:  = tan-1(L/R)
 Vs is the rms value of the input voltage.
61

62. Continuous load current
 The constant A1 at t = , i0 = I0.
62
  


 /
/
0
1 sin
2 L
R
s
e
Z
V
R
E
I
A












63.  Under a steady-state condition,
 i0(t = 0) = i0(t = )= I0.
63
  
  
R
E
e
e
Z
V
I L
R
L
R
s



 





 /
/
/
/
0
1
1
sin
2
for I0 ≥ 0
    
 
R
E
e
e
t
Z
V
i
t
L
R
L
R
s










 

/
/
/
0 sin
1
2
sin
2


 

and
for 0 ≤ (t – ) ≤  and i0 ≥ 0

64. Continuous load current
 The rms diode current.
64
 
2
/
1
0
2
0
2
1






 



t
d
i
Ir

65.  The rms output current .
 The average diode current
65
  r
r
r
rms I
I
I
I 2
2
/
1
2
2



 




 0
0
2
1
t
d
i
Id

66. 66

67. Discontinuous load current
 The load current flows only during the
period  ≤ t ≤ .
 Let define x = E/Vm = E/2Vs as the
load battery (emf) constant. The diodes
start to conduct at t =  given by:
67
 
x
V
E
m
1
1
sin
sin 





68.  At t = , i0(t) = 0
    



 /
/
1 sin
2 L
R
s
e
Z
V
R
E
A 








68

69. Discontinuous load current
 The load current
 The rms diode current
69
      
R
E
e
Z
V
R
E
t
Z
V
i t
L
R
s
s













 





 /
/
0 sin
2
sin
2
 
2
/
1
0
2
0
2
1








 



t
d
i
Ir

70.  The average diode current
70
 




 0
0
2
1
t
d
i
Id
Boundary conditions
 
R
E
e
e
Z
V
L
R
L
R
s













































1
1
sin
2
0

71. Finding the Performance Parameters of a Full-
Wave Rectifier with an RL Load
The single-phase full-wave rectifier of Figure
2.7(a) has L = 6.5 mH, R = 2.5 H, and E = 10 V.
The input voltage is Vs = 120 V at 60 Hz.
Determine
(a) the steady-state load current I0 at t = 0,
(b) the average diode current Id.
(c) the rms diode current Ir.
(d) the rms output current Irms.
(e) Use PSpice to plot the instantaneous output
current i0.Assume diode parameters IS = 2.22E -
15, BV = 1800 V.
71

72. 72
Single-Phase Full-wave Controlled Converter – RL Load

73.  During the period from  to , the input
voltage vs and input current is are +ve and the
power flows from the supply to the load.
 The converter is said to be operated in
rectification mode.
 During the period from  to  + , the input
voltage vs is -ve and the input current is is
positive and reverse power flows from the load
to the supply.
 The converter is said to be operated in
inversion mode.
 Depending on the value of , the average
output voltage could be either positive or
negative and it provides two-quadrant
operation.
73

74.  The average output voltage
 The rms value of the output voltage
74
    












cos
2
cos
2
2
sin
2
2 m
m
m
dc
V
t
V
t
td
V
V 






 
    s
m
m
m
rms
V
V
t
d
t
V
t
td
V
V





















2
2
cos
1
2
sin
2
2
2
/
1
2
2
/
1
2
2













75.  The load current iL.
 The steady-state condition iL (t =  + )
= IL1 = IL0.
75
mode 1 : when T1 and T2 conduct [ ≤ t ≤ ( + )]
      
t
L
R
s
L
s
L e
Z
V
R
E
I
R
E
t
Z
V
i 














 




 /
/
0 sin
2
sin
2
      
  
R
E
e
e
Z
V
I
I L
R
L
R
s
L
L 






 









/
/
/
/
1
0
1
sin
sin
2
for
0
0 
L
I

76.  The rms current.
 The rms output current.
76
 
2
/
1
2
2
1






 





t
d
i
I L
R
  R
R
R
rms I
I
I
I 2
2
/
1
2
2




77.  The average current
 The average output current.
 








t
d
i
I L
A
2
1
77
A
A
A
dc I
I
I
I 2




78. Finding the Current Ratings of Single-Phase
Controlled Full Converter with an RL load
The single-phase full converter of Figure 215(a)
has a RL load having L = 6.5 mH, R = 0.5 , and
E = 10 V. The input voltage is Vs = 120 V at (rms)
60 Hz. Determine
(a) the load current IL0 at t =  = 60°.
(b) the average thyristor current IA.
(c) the rms thyristor current IR.
(d) the rms output current Irms.
(e) the average output current Idc.
(f) the critical delay angle c.
78

79. 79

80.  It can operate with or without a
transformer and gives six-pulse ripples
on the output voltage.
 The diodes are numbered in order of
conduction sequences and each one
conducts for 120°.
 .
80

81.  The conduction sequence for diodes is
D1 – D2, D3 – D2, D3 – D4, D5 – D4, D5 – D6
and D1 – D6.
 The pair of diodes which are connected
between that pair of supply lines having the
highest amount of instantaneous line-to-
line voltage will conduct
81

82.  The line-to-line voltage is 3 times the
phase voltage of a three-phase Y-
connected source.
 If Vm is the peak value of the phase
voltage, then the instantaneous phase
voltages can be described by
van = Vm sin(t),
vbn = Vm sin(t – 120o),
vcn = Vm sin(t – 240o)
82
Positive sequence
or abc sequence

83. 83
Figure 2.10: Three-phase bridge rectifier

84. 84
Figure 2.11: Waveforms and conduction times of diodes

85.  Because the line-line voltage leads the phase
voltage by 30o, the instantaneous line-line
voltages can be described by
vab = 3Vm sin(t + 30o),
vbc = 3Vm sin(t – 90o),
vca = 3Vm sin(t – 210o)
85

86.  The average output voltage
86
 
m
m
m
dc
V
V
t
d
t
V
V
654
.
1
3
3
cos
3
6
/
2
2
6
/
0


 





where Vm, is the peak phase voltage.

87.  The rms output voltage
87
 
m
m
m
rms
V
V
t
d
t
V
V
6554
.
1
4
3
9
2
3
cos
3
6
/
2
2
2
/
1
2
/
1
6
/
0
2
2

















 






88.  If the load is purely resistive, the peak
current through a diode is Im = 3Vm/R
and the rms value of the diode current
is
 
m
m
m
r
I
I
t
d
t
I
I
5518
.
0
6
2
sin
2
1
6
1
cos
2
4
2
/
1
2
/
1
6
/
0
2
2





















 







88

89.  the rms value of the transformer secondary
current,
89
 
m
m
m
s
I
I
t
d
t
I
I
7804
.
0
6
2
sin
2
1
6
2
cos
2
8
2
/
1
2
/
1
6
/
0
2
2





















 







where Im is the peak secondary line current.

90. Finding the Performance Parameters of a Three-
Phase Bridge Rectifier
A three-phase bridge rectifier has a purely
resistive load of R. Determine
(a) the efficiency.
(b) the FF.
(c) the RF.
(d) the TUF.
(e) the peak inverse (or reverse) voltage (PIV) of
each
diode.
(f) the peak current through a diode.
The rectifier delivers Idc = 60 A at an output
voltage of Vdc = 280.7 V and the source frequency
90

91. 91

92.  The derivation of this Three-Phase
Bridge Rectifiers With RL Load is similar
to Single-Phase Bridge Rectifiers With
RL Load.
92
      
  
R
E
e
e
Z
V
I L
R
L
R
ab





 









3
/
/
3
/
/
0
1
3
/
sin
3
/
2
sin
2
for I0 ≥ 0

93. 93
     
  
  
R
E
e
e
t
Z
V
i t
L
R
t
L
R
ab













 











 3
/
/
3
/
/
0
1
3
/
sin
3
/
2
sin
sin
2
for /3 ≤ t ≤ 2/3 and i0 ≥ 0

94.  The rms diode current.
 the rms output current
94
 
2
/
1
3
/
2
3
/
2
0
2
2






 




t
d
i
Ir
  r
r
r
r
rms I
I
I
I
I 3
2
/
1
2
2
2





95.  The average diode current
 Boundary conditions
95
 


3
/
2
3
/
0
2
2




t
d
i
Id
0
1
3
sin
3
2
sin
2
3
3


























































R
E
e
e
Z
V
L
R
L
R
AB









96. 96

97. 97

98. 98

99. 99