dc to dc conv

1. Chapter 2 AC to DC Converters
Outline
2.1 Single-phase controlled rectifier
2.2 Three-phase controlled rectifier
2.3 Effect of transformer leakage inductance on rectifier circuits
2.4 Capacitor-filtered uncontrolled rectifier
2.5 Harmonics and power factor of rectifier circuits
2.6 High power controlled rectifier
2.7 Inverter mode operation of rectifier circuit
2.8 Thyristor-DC motor system
2.9 Realization of phase-control in rectifier

2. 2.1 Single- phase controlled (controllable) rectifier
2.1.1 Single-phase half-wave controlled rectifier
Resistive load
T
VT
R
a)
u1
u2
uVT
ud
id
0 ωt1 π 2π
ωt
u2
ug
ud
uVT
α θ
0
b)
c)
d)
e)
0
0
ωt
ωt
ωt
∫
+
=+==
π
α
α
α
π
ωω
π 2
cos1
45.0)cos1(
2
2
)(sin2
2
1
2
2
2d U
U
ttdUU
（2-1）

3. Inductive (resistor-inductor) load
a)
u1
T
VT
R
L
u2
uVT
ud
id
u2
0 ωt1
π 2π
ω t
ug
0
ud
0
id
0
uVT
0
θ
α
b)
c)
d)
e)
f)
+ +
ω t
ω t
ω t
ω t

4. Basic thought process of time-domain analysis for power electronic
circuits
The time- domain behavior of a power electronic circuit is actually the
combination of consecutive transients of the different linear circuits
when the power semiconductor devices are in different states.
a) b)
VT
R
L
VT
R
L
u2 u2
tURi
t
i
L ωsin2
d
d
2d
d
=+
ω t= a ，id=0
)sin(
2
)sin(
2 2
)(
2
d ϕωϕα
αω
ω −+−−=
−−
t
Z
U
e
Z
U
i
t
L
R
（2-2）
（2-3）

5. Single- phase half- wave controlled rectifier with freewheeling diode
load (L is large enough) Inductive
a)
L
T VT
R
u1
u2
uVT
ud
VDR
idiVD
u2
ud
id
uVT
iVT
Id
Id
ωtO
O
O
O
O
O
π-α π+α
b)
iVDR
ωt
ωt
ωt
ωt
ωt
g)
c)
e)
f)
d)
ωT1

6. Maximum forward voltage, maximum reverse voltage
Disadvantages:
–Only single pulse in one line cycle
–DC component in the transformer current
ddVT
2
II
π
απ−
=
d
2
dVT
2
)(
2
1
ItdII
π
απ
ω
π
π
α
−
== ∫
d
2
2
dVD
2
)(
2
1
R
ItdII
π
απ
ω
π
απ
π
+
== ∫
+
ddVD
2
R
II
π
απ+
=
（2-5）
（2-6）
（2-7）
（2-8）

7. 2.1.2 Single- phase bridge fully-controlled rectifier
• Resistive load
π ω t
0
0
0
i2
ud
id
b)
c)
d)
ud
(id
)
α α
R
T
u1 u2
a)
i2
a
b
VT3
ud
id
uVT
1,4
ω t
ω t
VT4
VT1
VT2

8. Average output (rectified) voltage:
Average output current:
For thyristor:
For transformer:
∫
+
=
+
==
π
α
αα
π
ωω
π 2
cos1
9.0
2
cos122
)(dsin2
1
2
2
2d U
U
ttUU （2-9）
（2-10）
2
cos1
9.0
2
cos122 22d
d
αα
π
+
=
+
==
R
U
R
U
R
U
I
（2-11）2
cos1
45.0
2
1 2
ddVT
α+
==
R
U
II
π
απ
α
π
ωω
π
π
α
−
+== ∫ 2sin
2
1
2
)(d)sin
2
(
2
1 222
VT
R
U
tt
R
U
I （2-12）
π
απ
α
π
ωω
π
π
α
−
+=== ∫ 2sin
2
1
)()sin
2
(
1 222
2
R
U
tdt
R
U
II (2-13)

9. • Inductive load (L is large enough)
ω t
ω t
ω t
ω t
ω t
ω t
ω t
ο
ο
ο
ο
ο
ο
ο
u2
ud
id
Id
Id
Id
Id
Id
iVT1,4
iVT2,3
uVT1,4
i2
,
b)
R
T
u1 u2
a)
i2
a
b
VT3
ud
id
VT4
VT1
VT2

10. • Electro- motive-force (EMF) load With resistor
∫
+
===
απ
α
αα
π
ωω
π
cos9.0cos
22
)(dsin2
1
222d UUttUU （2-15）
a) b)
R
E
id
ud id
O
E
ud
ωt
Id
O ωt
α θ δ

11. • With resistor and inductor
When L is large enough, the output voltage and current waveforms are the
same as ordinary inductive load.
When L is at a critical value
O
ud
0
E
id
ω t
ω t
π
δ
α θ =π
dmin
23
dmin
2
1087.2
22
I
U
I
U
L −
×==
πω （2-17）

12. 2.1.3 Single- phase full- wave controlled rectifier
a) b)
u1
T
R
u2
u2
i1 VT1
VT2 ud
ud
i1
O
O
α ω t
ω t

13. 2.1.4 Single- phase bridge half-controlled rectifier
a)
T a
b R
L
O
b)
u2
i2
ud
id
VT1
VT2
VD3
VD4
VDR
u2
O
ud
id Id
O
O
O
O
O
i2
Id
Id
Id
Id
Id
α
ωt
ωt
ωt
ωt
ωt
ωt
ωt
α
π−α
π−α
iVT
1iVD
4
iVT
2iVD
3
iVD
R

14. • Another single- phase bridge half-controlled rectifier
Comparison with previous circuit:
–No need for additional freewheeling diode
–Isolation is necessary between the drive circuits of the two thyristors
load
T
u2
VT2 VT4
VT1 VT3

15. Summary of some important points in analysis
When analyzing a thyristor circuit, start from a diode circuit with the
same topology. The behavior of the diode circuit is exactly the same as
the thyristor circuit when firing angle is 0.
A power electronic circuit can be considered as different linear circuits
when the power semiconductor devices are in different states. The
time- domain behavior of the power electronic circuit is actually the
combination of consecutive transients of the different linear circuits.
Take different principle when dealing with different load
– For resistive load: current waveform of a resistor is the same as the
voltage waveform
–For inductive load with a large inductor: the inductor current can
be considered constant

16. 2.2 Three- phase controlled (controllable) rectifier
2.2.1 Three- phase half- wave controlled rectifier
Resistive load, α= 0º
a
b
c
T
R
ud
id
VT2
VT1
VT3
u2
O
O
O
uab uac
O
iVT1
uVT1
ω t
ω t
ω t
ω t
ω t
u a u b u c
u G
u d
ω t1 ω t2 ω t3
Common-cathode connection
Natural commutation point

17. Resistive load, α= 30º
u2
ua ub uc
O
O
O ωt
O
ωt
O
ωt
uG
ud
uab uac
ωt1
iVT1
uVT1 uac
ωt
ωt
a
b
c
T
R
ud
id
VT2
VT1
VT3

18. Resistive load, α= 60º
ωt
u2 ua ub uc
O
O
O
O
uG
ud
iVT
1
ωt
ωt
ωt
a
b
c
T
R
ud
id
VT2
VT1
VT3

19. Resistive load, quantitative analysis
When α≤ 30º , load current id is continuous.
When α > 30º , load current id is discontinuous.
Average load current
Thyristor voltages
αα
π
ωω
π
α
π
α
π cos17.1cos
2
63
)(sin2
3
2
1
22
6
5
6
2d UUttdUU === ∫
+
+
（2-18）






++=





++== ∫ +
)
6
cos(1675.0)
6
cos(1
2
23
)(sin2
3
2
1
2
6
2d α
π
α
π
π
ωω
π
π
α
π UttdUU (2-19)
R
U
I d
d =
（2-20）
0 30 60 90 120 150
0.4
0.8
1.2
1.17
3
2
1
α/(°)
Ud/U2

20. • Inductive load, L is large enough
T
R
L
ud
eL
id
VT3
ud
ia
ua ub uc
ib
ic
id
uac
uab
uac
O
O
O
O
O
O ω tα
uVT
1
ω t
ω t
ω t
ω t
ω t
a
b
c
VT2

21. Thyristor voltage and currents, transformer current :
αα
π
ωω
π
α
π
α
π cos17.1cos
2
63
)(sin2
3
2
1
22
6
5
6
2d UUttdUU === ∫
+
+
（2-18）
ddVT2 577.0
3
1
IIII === d
VT
VT(AV) 368.0
57.1
I
I
I ==
2RMFM 45.2 UUU ==
（2-23）
（2-25）
（2-24）

22. 2.2.2 Three- phase bridge fully-controlled rectifier
Circuit diagram
Common- cathode group and common- anode group of thyristors
Numbering of the 6 thyristors indicates the trigger sequence.
b
a
c
T
n
load
ia
id
ud
VT1 VT3 VT5
VT4 VT6 VT2 d2
d1

23. Resistive load, α= 0º
b
a
c
T
n
load
ia
id
ud
VT1 VT3 VT5
VT4 VT6 VT2 d2
d1

24. u2
ud1
ud2
u2L
ud
uab uac
uab uac ubc uba uca ucb uab uac
uab uac ubc uba uca ucb uab uac
I II III IV V VI
ua ucub
ωt1O ωt
O ωt
O ωt
O ωt
α= 0°
iVT
1
uVT
1

25. Resistive load, α= 30º
b
a
c
T
n
load
ia
id
ud
VT1 VT3 VT5
VT4 VT6 VT2 d2
d1

26. ud1
ud2
α=30¡
ã
ia
O
O
O
O ωt
ud
uab uac
ua ub
uc
ωt1
uab uac ubc uba uca ucb uab uac
І II III IV V VI
uab u ac ubc uba u ca ucb u ab u acuVT
1
ωt
ωt
ωt

27. Resistive load, α= 60º
b
a
c
T
n
load
ia
id
ud
VT1 VT3 VT5
VT4 VT6 VT2 d2
d1

28. α= 60º
ud1
ud2
ud
uac
uac
uab
uab
uac
ubc
uba
uca
ucb
uab
uac
ua
I II III IV V VI
ub uc
O
ωt1
O ωt
O
uVT
1
ωt
ωt

29. Resistive load, α= 90º
b
a
c
T
n
load
ia
id
ud
VT1 VT3 VT5
VT4 VT6 VT2 d2
d1

30. ud1
ud2
ud
ua ub uc ua ub
ωt
O
O
O
O
O
ia
id
uab uac ubc uba uca ucb uab uac ubc uba
iVT1
ωt
ωt
ωt
ωt

31. Inductive load, α= 0º
b
a
c
T
n
load
ia
id
ud
VT1 VT3 VT5
VT4 VT6 VT2 d2
d1

32. ud1
u2
ud2
u2L
ud
id
O
O
O
ωtO
uaα = 0
ub
uc
ωt1
uab uac ubc uba uca ucb uab uac
I II III IV V VI
iVT
1
ωt
ωt
ωt
º

33. Inductive load, α= 30º
b
a
c
T
n
load
ia
id
ud
VT1 VT3 VT5
VT4 VT6 VT2 d2
d1

34. ud1
α= 30
ud2
ud
uab
uac
ubc
uba
uca
ucb
uab
uac
I II III IV V VI
ωtO
ωtO
ωtO
ωtO
id
ia
ωt1
ua ub uc
°

35. Inductive load, α= 90º
b
a
c
T
n
load
ia
id
ud
VT1 VT3 VT5
VT4 VT6 VT2 d2
d1

36. α = 90°
ud1
ud2
uac ubc uba uca ucb uab uac
uab
I II III IV V VI
ud
uac
uab
uac
ωtO
ωtO
ωtO
ub uc ua
ωt1
uVT
1

37. Quantitative analysis
Average output voltage:
For resistive load, When a > 60º, load current id is discontinuous.
everage output current (load current):
Transformer current:
αωω
π
α
π
α
π cos34.2)(sin6
3
1
2
3
2
3
2d UttdUU == ∫
+
+ （2-26）
（2-27）



++== ∫ +
)
3
cos(134.2)(sin6
3
2
3
2d α
π
ωω
π
π
α
π UttdUU
R
U
I d
d = （2-20）
dddd IIIII 816.0
3
2
3
2
)(
3
2
2
1 22
2 ==





×−+×= ππ
π
（2-28）

38. 2.3 Effect of transformer leakage inductance on rectifier circuits
In practical, the transformer leakage inductance has to be taken into account.
Commutation between thyristors, thus can not happen instantly,but with a
commutation process.
a
b
c
T
Lud
ic
ib
iaLB
LB
LB
ik VT1
VT2
VT3
R
id
O γ
ic ia ib ic ia Id
ud
ωt
ua ub uc
α
ωt
O

39. Commutation process analysis
Circulating current ik during commutation
Output voltage during commutation
ik: 0 Id
ub-ua = 2·LB·dia/dt
ia = Id-ik : Id
ib = ik : 0 Id
0
2d
d
d
d bak
Bb
k
Bad
uu
t
i
Lu
t
i
Luu
+
=−=+= （2-30）

40. Quantitative calculation
Reduction of average output voltage due to the commutation process
Calculation of commutation angle
– Id ↑,γ↑
– XB↑, γ↑
– For α ≤ 90۫, α↓, γ↑
dB
0
B
6
5
6
5 B
6
5
6
5 Bbb
6
5
6
5 dbd
2
3
d
2
3
)(d
d
d
2
3
)(d)]
d
d
([
2
3
)(d)(
3/2
1
IXiLt
t
i
L
t
t
i
LuutuuU
I
π
ω
π
ω
π
ω
π
ω
π
π
γα
π
α
π
γα
π
α
π
γα
π
α
===
−−=−=∆
∫∫
∫∫
++
+
++
+
++
+
d
k
k
k
（2-31）
2
dB
6
2
)cos(cos
U
IX
=+− γαα （2-36）

41. Summary of the effect on rectifier circuits
Circuits
Single-
phase
Full wave
Single-
phase
bridge
Three-
phase
half wave
Three-
phase
bridge
m-pulse recfifier
dU∆ d
B
I
X
π d
B2
I
X
π d
B
2
3
I
X
π
d
B3
I
X
π d
B
2
I
mX
π
)cos(cos γαα +−
2
Bd
2U
XI
2
Bd
2
2
U
XI
2
dB
6
2
U
IX
2
dB
6
2
U
IX
m
U
XI
π
sin2 2
Bd

42. • Conclusions
–Commutation process actually provides additional working states of the
circuit.
–di/dt of the thyristor current is reduced.
–The average output voltage is reduced.
–Positive du/dt
– Notching in the AC side voltag

43. 2.4 Capacitor- filtered uncontrolled (uncontrollable) rectifier
2.4.1 Capacitor- filtered single- phase uncontrolled rectifier
Single-phase bridge, RC load:
a)
+
RCu1 u2
i2
VD1 VD3
VD2 VD4
id
iC
iR
ud
b)
0
i
ud
θ
δ
π 2π ω t
i,ud

44. Single-phase bridge, RLC load
a) b)
-
+
R
C
L
+
u1 u2
i2
ud
uL
id
iC iR
VD2 VD4
VD1
VD3
u2 ud
i2
0
δ θ π ωt
i2,u2,ud

45. 2.4.2 Capacitor- filtered three- phase uncontrolled rectifier
Three-phase bridge, RC load
a)
+
a
b
c
T
ia
RCud
id
iC
iR
VD2
b)
O
ia
ud
id
uduab uac
0δ θ ππ
3
ωtVD6VD4
VD1VD3VD5
ωt

46. Three- phase bridge, RC load Waveform when ωRC≤1.732
a) b)
ωt ωt
ωtωt
ia
id
ia
id
O
O
O
O
a）ωRC=
•
b）ωRC<3 3

47. Three- phase bridge, RLC load
a)
b)
c)
+
a
b
c
T ia
RCud
id
iC
iR
VD4
VD6
VD1
VD3
VD5
VD2
ia
ia
O
O ωt
ωt

48. 2.5 Harmonics and power factor of rectifier circuits
2.5.1 Basic concepts of harmonics and reactive power
For pure sinusoidal waveform
For periodic non-sinusoidal waveform
where

49. • Harmonics-related specifications
Take current harmonics as examples
Content of nth harmonics
In is the effective (RMS) value of nth harmonics.
I1 is the effective (RMS) value of fundamental component.
Total harmonic distortion
Ih is the total effective (RMS) value of all the harmonic components.
%100
1
×=
I
I
HRI n
n （2-57）
%100
1
×=
I
I
THD h
i （2-58）

50. • Definition of power and power factor for sinusoidal circuits
Active power
Reactive power
Apparent power
Power factor
∫ ==
π
ϕω
π
2
0
cos)(
2
1
UItuidP （2-59）
Q=U I sinϕ （2-61）
S=UI （2-60）
222
QPS += （2-63）
S
P
=λ （2-62）
λ =cos ϕ （2-64）

51. • Definition of power and power factor For non- sinusoidal circuit
Active power:
Power factor:
Distortion factor (fundamental- component factor):
Displacement factor (power factor of fundamental component):
Definition of reactive power is still in dispute
P=U I1 cosϕ1 （2-65）
（2-66）11
111
coscos
cos
ϕνϕ
ϕ
λ ====
I
I
UI
UI
S
P
ν =I1 / I
λ =cos ϕ1 1

52. • Review of the reactive power concept
The reactive power Q does not lead to net transmission of energy between
the source and load. When Q ≠ 0, the rms current and apparent power
are greater than the minimum amount necessary to transmit the
average power P.
Inductor: current lags voltage by 90°, hence displacement factor is zero.
The alternate storing and releasing of energy in an inductor leads to
current flow and nonzero apparent power, but P = 0. Just as resistors
consume real (average) power P, inductors can be viewed as
consumers of reactive power Q.
Capacitor: current leads voltage by 90°, hence displacement factor is zero.
Capacitors supply reactive power Q. They are often placed in the
utility power distribution system near inductive loads. If Q supplied by
capacitor is equal to Q consumed by inductor, then the net current
(flowing from the source into the capacitor- inductive- load
combination) is in phase with the voltage, leading to unity power
factor and minimum rms current magnitude.

53. 2.5.2 AC side harmonics and power factor of controlled rectifiers with
inductive load
• Single- phase bridge fully-controlled rectifier
R
T
u1 u2
a)
i2
a
b
VT3
ud
id
VT4
VT1
VT2
ωt
ωt
ωt
ωt
ωt
ωt
ωt
ο
ο
ο
ο
ο
ο
ο
u2
ud
id
Id
Id
Id
Id
Id
iVT1,4
iVT2,3
uVT1,4
i2
,
b)

54. AC side current harmonics of single- phase bridge fully-controlled
rectifier with inductive load
Where
Conclusions
–Only odd order harmonics exist
– In∝1/n
– In / I1 = 1/n
∑∑ ==
==
+++=


,5,3,1,5,3,1
d
d2
sin2sin
14
)5sin
5
1
3sin
3
1
(sin
4
n
n
n
tnItn
n
I
tttIi
ωω
π
ωωω
π
（2-72）
πn
I
In
d22
= n=1,3,5,… （2-73）

55. • A typical gate triggering control circuit
220V 36V
+
B
TP
+15V
A
VS
+15V
- 15V - 15VX Y
Disable
R
Q
uts
VD1
VD2
C1 R2
R4
TS
V2 R5
R8
R6
R7
R3
R9
R10
R11
R12
R13
R14
R16
R15
R18
R17
RP1
uco
up
C2
C3
C3
C5
C6C7
R1
RP2
V1
I1c
V3
V4
V6
V5
V7
V8
VD4
VD10 VD5
VD6
VD7
VD9
VD8
VD15
VD11
~VD14

56. • Three- phase bridge fully-controlled rectifier
b
a
c
T
n
load
ia
id
ud
VT1 VT3 VT5
VT4 VT6 VT2 d2
d1

57. ud1
α= 30
ud2
ud
uab
uac
ubc
uba
uca
ucb
uab
uac
I II III IV V VI
ωtO
ωtO
ωtO
ωtO
id
ia
ωt1
ua ub uc
°

58. • AC side current harmonics of three- phase bridge fully- controlled
rectifier with inductive load
where
∑∑
=
±=
=
±=
−+=−+=
−++−−=


3,2,1
16
1
3,2,1
16
dd
da
sin2)1(sin2sin
1
)1(
32
sin
32
]13sin
13
1
11sin
11
1
7sin
7
1
5sin
5
1
[sin
32
k
kn
n
k
k
kn
k
tnItItn
n
ItI
tttttIi
ωωω
π
ω
π
ωωωωω
π （2-79）
（2-80）







=±==
=
,3,2,1,16,
6
6
d
d1
kknI
n
I
II
n
π
π

59. 2.5.3 AC side harmonics and power factor of capacitor- filtered
uncontrolled rectifiers
Situation is a little complicated than rectifiers with inductive load.
Some conclusions that are easy to remember:
–Only odd order harmonics exist in single- phase circuit, and only 6k±1 (k is
positive integer) order harmonics exist in three- phase circuit.
–Magnitude of harmonics decreases as harmonic order increases.
–Harmonics increases and power factor decreases as capacitor increases.
–Harmonics decreases and power factor increases as inductor increases.

60. 2.5.4 Harmonic analysis of output voltage and current






−
−=+= ∑∑
∞
=
∞
=
tn
n
k
UtnbUu
mknmkn
n ω
π
ω cos
1
cos2
1cos 2d0d0d0 （2-85）
（2-86）m
m
UU
π
π
sin2 2d0 =
d02
1
cos2
U
n
k
bn
−
−=
π
（2-87）
u
d
ωtOπ
m
π
m
2π
m
U
2
2
Output voltage of m- pulse
rectifier when α = 0º

61. Ripple factor in the output voltage
Output voltage ripple factor
where UR is the total RMS value of all the harmonic components in the output
voltage
and U is the total RMS value of the output voltage
d0
R
U
U
u =γ （2-88）
（2-89）
2
d0
22
R UUUU
mkn
n −== ∑
∞
=

62. • Harmonics in the output current
where
)cos(dd n
mkn
n tndIi ϕω −+= ∑
∞
=
（2-92）
（2-93）
R
EU
I
−
= d0
d
22
)( LnR
b
z
b
d n
n
n
n
ω+
==
R
Ln
n
ω
ϕ arctan=
（2-94）
（2-95）

63. Conclusions
for α = 0º
Only mk (k is positive integer) order harmonics exist in the output voltage and
current of m- pulse rectifiers
Magnitude of harmonics decreases as harmonic order increases when m is
constant.
The order number of the lowest harmonics increases as m increases. The
corresponding magnitude of the lowest harmonics decreases accordingly.
For α ≠ 0º
Quantitative harmonic analysis of output voltage and current is very
complicated for α ≠ 0º.
As an example,for 3- phase bridge fully- controlled rectifie

64. 2.6 High power controlled rectifier
2.6.1 Double- star controlled rectifier
Circuit Waveforms When α= 0º
T
a b c
L
R
niP
LP
ud
id
VT5
ca'
b'
n1
n2
'
VT4 VT6 VT2VT3 VT1
ud1
ua ub uc
ia
ud2
ia
'
uc' ua
' ub
' uc
'
O ω t
O ω t
O ω t
O ω t
Id
1
2
Id
1
6
Id
1
2
Id
1
6

65. • Effect of interphase reactor(inductor, transformer)
n
L
R
+- +-
ud1
LP
ub
'
ud2
ud
n2
n1
iP
ua
VT1 VT6
uP
1
2
up
ud1
,ud2
O
O
60
360°
ωt1 ωt
ωt
b)
a)
ua ub ucuc
' ua'ub' ub'
°
d1d2p uuu −=
)(
2
1
2
1
2
1
d2d1pd1pd2d uuUuuuu +=+=−=
（2-97）
（2-98）

66. Quantitative analysis when α = 0º
]9cos
40
1
6cos
35
2
3cos
4
1
1[
2
63 2
d1 ⋅⋅⋅−+−+= ttt
U
u ωωω
π
]9cos
40
1
6cos
35
2
3cos
4
1
1[
2
63
])60(9cos
40
1
)60(6cos
35
2
)60(3cos
4
1
1[
2
63
2
2
d2
⋅⋅⋅−−−=
⋅⋅⋅−°−+°−−°−+=
ttt
U
ttt
U
u
ωωω
π
ωωω
π
]9cos
20
1
3cos
2
1
[
2
63 2
p ⋅⋅⋅−−−= tt
U
u ωω
π
]6cos
35
2
1[
2
63 2
d ⋅⋅⋅−−= t
U
u ω
π
（2-99）
（2-100）
（2-101）
（2-102）

67. Waveforms when α > 0º
Ud=1.17 U2 cos α
°90=α
°60=α
°30=αud
ud
ud
ωtO
ωtO
ωtO
ua ub ucuc
' ua
' ub
'
ub
uc
uc
' ua
' ub
'
ub
uc
uc
' ua
' ub
'

68. 2.6.2 Connection of multiple rectifiers
Connection
of multiple
rectifiers
To increase the
output capacity
To improve the AC side current waveform
and DC side voltage waveform
Larger output voltage:
series connection
Larger output current:
parallel connection

69. • Phase-shift connection of multiple rectifiers
Parallel connection
M
LT
VT
1 2
c1
b1
a1
c2
b2
a2
LP
12- pulse rectifier realized by
paralleled 3- phase bridge rectifiers

70. Series connection
C
?
L
RB
A
1
*
?
?
*
*
0
30°
3
iA
c1
b1
a1
1
c2
b2
a2
iab 2
ua2b2
ua1b1
i
a1 id
ud
I
II
I
III
0
a)
b)
c)
d)
ia1
Id
ia2
iab2'
iA
Id
iab2
ωt
ωt
ωt
ωt
0
0
0
Id
2
3
3
3
Id
3
3
Id
Id3
2 3
(1+ ) Id3
2 3
(1+ )Id3
3
Id
1
3
12- pulse rectifier realized by
series 3- phase bridge rectifiers

71. • Sequential control of multiple series-connected rectifiers
L
i
a)
l oad
Ⅰ
Ⅱ
Ⅲ
u
2
u
2
u
2
Id
VT11
u d
b)
c)
i Id
2 Id
ud
O α π + α
VT12
VT13
VT14
VT21
VT22
VT23
VT24
VT31
VT32
VT33
VT34
Circuit and waveforms of series- connected
three single-phase bridge rectifiers

72. 2.7 Inverter mode operationof rectifiers
• Review of DC generator- motor system
c)b)a)
MG MG MG
E G E M
Id
R ¡Æ
E G E M
Id
R ¡Æ
E G
E M
Id
R ¡Æ
Id =
EG EM
RΣ
-
Id =
EM EG
RΣ
- should be avoided

73. • Inverter mode operation of rectifiers
Rectifier and inverter mode operation of single- phase full- wave
converter
a) b)
R
+
-
engry
M
1
0
2
u10
u20
ud
id
L
VT1
VT2
u10ud
u20 u10
α
O
O ωt
ωt
Id
id
Ud
>EM
EM
engry
M
R
+
-
1
0
2
ud
id
LVT1
VT2
u10ud
u20
u10
O
O ωt
ωt
Id
id
Ud
<EM
α
EM
iVT
1
iVT2
iVT
1
iVT
2
iVT1
iVT2
iVT2
id = +iVT1
iVT2
iVT1
iVT2
iVT1
id = +iVT1
iVT2
Id =
Ud EG
RΣ
-
Id =
UdEM
RΣ
-

74. Necessary conditions for the inverter mode operation of controlled
rectifiers
There must be DC EMF in the load and the direction of the DC EMF must be
enabling current flow in
thyristors. (In other word EM must be negative if taking the ordinary output
voltage direction as positive.)
α > 90º so that the output voltage Ud is also negative.

75. • Inverter mode operation of 3- phase bridge rectifier
uab uac ubc uba uca ucb uab uac ubc uba uca ucb uab uac ubc uba uca ucb uab uac ubc
ua ub uc ua ub uc ua ub uc ua ubu2
ud
ωtO
ωtO
β =
π
4
β = π
3
β =
π
6
β = π
4
β = π
3
β = π
6
ωt1 ωt3ωt2

76. Inversion angle (extinction angle) β
α+ β=180º
Inversion failure and minimum inversion angle
Possible reasons of inversion failures
–Malfunction of triggering circuit
–Failure in thyristors
–Sudden dropout of AC source voltage
–Insufficient margin for commutation of thyristors
Minimum inversion angle (extinction angle)
βmin= δ + γ+ θ′ （ 2-109 ）

77. L
a
b
c
+
-
Mud
id
EM
LB
LB
LB
VT1
VT2
VT3
o
ud
O
O
id
ωt
ωt
ua
ub
uc
ua
ub
p
β
γ
β<γ
α γ
β
β>γ
iVT
1
iVT
2
iVT
3
iVT
1
iVT
2
iVT
3
iVT
1
iVT
3
LB
LB

78. 2.8 Thyristor- DC motor system
2.8.1 Rectifier mode of operation
Waveforms and equations
UIREU dMd ∆++= Σ (2-112)
where
RΣ RM RB
3XB
2π
+ +=
(for 3- phase half-wave)
u
d
O
id
ωt
ua ub uc
α
ud
O
ia ib icic
ωt
E
Ud
idR
(Waveforms of 3- phase half- wave
rectifier with DC motor load

79. • Speed- torque (mechanic) characteristic when load current is
continuous
nEM ϕ= Ce （2-113）
For 3- phase half-wave
UIEM −−= cosα1.17 2U ΣR d ∆
EM = cosα1.17 2U
α
e
d
e C
UIR
C
U
n
∆+
−=
Σcos1.17 2
（2-114）
For 3-phase bridge
α
e
d
e C
IR
C
U
n −=
Σcos2.34 2
（2-115）
（2-116）
O
n
a1<a2<a3
a3
a2
a1
Id
(RB+RM+ )
Id
Ce
3XB
2π
For 3- phase half-wave

80. • Speed- torque (mechanic) characteristic when load current is
discontinuous
EMF at no load (taking 3- phase half-wave as example)
For α≤ 60º
22UEo=
For α>60º
)3cos(2 2 πα−UEo=
discontinuouts
mode
continuous mode
E
E0
E0'
O
Idmin
Id
(0.585U2)
( U2)2
For 3- phase half-wave

81. 2.8.2 Inverter mode of operation
Equations
–are just the same as in the
rectifier mode of operation
except that Ud, EM and n
become negative. E.g., in
3- phase half- wave
UIEM −−= cosα1.17 2U ΣR d ∆
α
e
d
e C
UIR
C
U
n
∆+
−=
Σcos1.17 2
（2-114）
（2-115）
– Or in another form
（2-123）
I)0cos β+( ∑RIUE ddM = - （2-122）
eC
n −= ∑+ RIU dd βcos0
1
rectifier
mode
n
α3
α2
α1
Id
α4
β2
β3
β4
β1
α =β =
π
2
inverter
mode
α
increasingβ
increasing
Speed-torque characteristic of
a DC motor fed by a thyristor
rectifier circuit

82. 2.8.3 Reversible DC motor drive system(4-quadrant operation)
L
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
AC
source
converter 2converter 1
converter 2
+T- T
converter 1 rectifying
converter 2 rectifying
converter 2 inverting
converter 1 inverting
forward motoring
reverse motoring
forward braking(regenerating)
reverse braking(regenerating)
converter 1 converter 2
EM
M
a
b
c
a
b
c
+n
Id Id
Udα
M EM
Id
M EM
MEM
Id
Energ
yM
EM
- n
Udβ
Udα Udβ
O
AC
source
converter 1
Energ
y
Energ
y
Energ
y
converter 2 converter 2converter 1
converter 1
AC
source
AC
source
Back-to-back
connection of two 3-
phase bridge circuits

83. converter 1converter 2
n
α3
α2
α1
Id
α4
β2
β3
β4
β1
α =β =
π
2
α '=β '=
π
2
β'3
β'2
β'1
β'4
α'2
α'3
α'4
α'1
α1
=β '1
;α '1
=β1
α2
=β '2
;α '2
=β2
α
increasingβ
increasing
β
increasingα
increasing

84. 2.9 Gate triggering control circuit for thyristor rectifiers
A typical gate triggering control circuit
220V 36V
+
B
TP
+15V
A
VS
+15V
-15V - 15VX Y
Disable
R
Q
uts
VD1
VD2
C1 R2
R4
TS
V2 R5
R8
R6
R7
R3
R9
R10
R11
R12
R13
R14
R16
R15
R
18
R17
RP1
uco
up
C2
C3
C3
C5
C6C7
R1
RP2
V1
I1c
V3
V4
V6
V5
V7
V8
VD4
VD10 VD5
VD6
VD7
VD9
VD8
VD15
VD11
~VD14