Power Electronics
Chapter 2
AC to DC Converters
(Rectifiers)
Power Electronics
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 circuits
2
Power Electronics
2.1 Single-phase controlled
(controllable) rectifier
2.1.1 Single-phase half-wave controlled rectifier
2.1.2 Single-phase bridge fully-controlled rectifier
2.1.3 Single-phase full-wave controlled rectifier
2.1.4 Single-phase bridge half-controlled rectifier
3
Power Electronics
2.1.1 Single-phase half-wave
controlled rectifier
u2
Resistive load
b)
VT
T
u1
uVT
u2
0
ug
ωt1
π
2π
c)
ud
R
ωt
0
ud
id
d)
0 α
ωt
θ
ωt
uVT
e)
a)
Ud =
1
2π
π
∫α
2U 2 sin ω td (ω t ) =
0
2U 2
1 + cos α
(1 + cos α ) = 0.45U 2
2π
2
ωt
(2-1)
Half-wave, single-pulse
Triggering delay angle, delay angle, firing angle
4
Power Electronics
2.1.1 Single-phase half-wave
controlled rectifier
Inductive (resistor-inductor) load
u2
VT
T
id
u1
0
π
ωt1
ωt
2π
ug
uVT
a)
b)
u2
c)
ωt
0
ud
ud
+
+
d)
ωt
0 α
id
e)
0
θ
ωt
uVT
f)
0
ωt
5
Power Electronics
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.
VT
VT
L
L
u2
u2
R
R
b)
a)
d id
+ Ri d =
L
dt
2U 2 sin ω t
(2-2)
ωt = α ，id= 0
R
− (ωt −α )
2U2
2U2
sin(α −ϕ)e ωL
+
sin(ωt −ϕ)
id = −
Z
Z
(2-3)
6
Power Electronics
Single-phase half-wave controlled
rectifier with freewheeling diode
u2
Inductive load (L is large enough)
VT
T
i VD
R
u VT
a)
u1
u2
I
dVT
=
O
ud
L
− α
2π
I
I VT =
π
∫α I
2
d
ωt
π −α
Id
2π
Id
d)
R
O
iVT
(2-5)
d
d (ω t ) =
ωt
O
id
ωt
Id
e)
O
iVD
(2-6)
1
2π
ωt1
c)
ud
VDR
π
b)
id
π-α
π+α
ωt
R
f)
ωt
O
uVT
(2-7)
g)
(2-8)
ωt
O
Maximum forward voltage, maximum reverse voltage
Disadvantages:
– Only single pulse in one line cycle
– DC component in the transformer current
7
Resistive load
ud
id
T
i2
VT 3
VT 1
b)
ud(id)
0 α
id
a
π α
ωt
uVT
1,4
ud
u2
R
c)
b
VT 4
u1
VT2
Power Electronics
2.1.2 Single-phase bridge
fully-controlled rectifier
0
i2
ωt
0
ωt
d)
a)
For thyristor: maximum forward voltage, maximum reverse
voltage
Advantages:
– 2 pulses in one line cycle
– No DC component in the transformer current
8
Power Electronics
2.1.2 Single-phase bridge fullycontrolled rectifier
Resistive load
Average output (rectified) voltage
Ud =
π
1
π ∫α
2U 2 sinωtd(ωt ) =
1 + cosα
2 2U 2 1 + cosα
= 0.9U 2
2
2
π
(2-9)
Average output current
Id =
U d 2 2U 2 1 + cos α
U 1 + cos α
=
= 0 .9 2
R
πR
R
2
2
(2-10)
For thyristor
I dVT =
I VT =
U 2 1 + cos α
1
I d = 0 . 45
R
2
2
1
2π
π
∫α
(
U
2U 2
sin ωt ) 2 d (ωt ) = 2
R
2R
(2-11)
π −α
1
sin 2α +
π
2π
(2-12)
For transformer
I = I2 =
1
π
π ∫α
(
U
2U 2
sin ωt ) 2 d (ωt ) = 2
R
R
π −α
1
sin 2α +
π
2π
(2-13)
9
T
u1
i2
u2
ωt
O
id
ud
VT3
VT1
Inductive load
(L is large enough)
ωt
O
Id
id
a
ud
u2
L
iVT O
ωt
Id
1,4
iVT O
Id
2,3
b
O
i2
R
VT4
VT2
Power Electronics
2.1.2 Single-phase bridge
fully-controlled rectifier
uVT
O
Id
a)
1
π +α
π ∫α
2U 2 sinωtd(ωt ) =
ωt
ωt
Id
1,4
ωt
O
Ud =
ωt
b)
2 2
π
U 2 cosα = 0.9U 2 cosα
(2-15)
Commutation
Thyristor voltages and currents
Transformer current
10
Power Electronics
Electro-motive-force (EMF) load
With resistor
id
ud
E
R
ud
E
O α
θ
δ
ωt
id
Id
ωt
O
a)
b)
Discontinuous current id
11
Power Electronics
Electro-motive-force (EMF) load
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
ud
α
δ
0
θ =π
E
π
ωt
id
O
2 2U 2
−3 U 2
L=
= 2 . 87 × 10
πω I dmin
I dmin
ωt
(2-17)
12
Power Electronics
2.1.3 Single-phase full-wave
controlled rectifier
i1
u1
T
ud
VT1
u2
u2 VT
2
ud
O α
R i
1
ωt
ωt
O
a)
b)
Transformer with center tap
Comparison with single-phase bridge fully-controlled rectifier
13
u2
a
u
id
ud
2
VD4
b
VDR
T
VT 2
i2
VT 1
b)
VD3
Power Electronics
2.1.4 Single-phase bridge
half-controlled rectifier
ωt
O
ud
α
ωt
L
O
id
R
iVTO
iVD1
Id
iVTO
iVD
2
π−α
4
Id
ωt
3
iVDO
R
O
i2
O
Half-control
Comparison with fully-controlled rectifier
Additional freewheeling diode
Id
Id
ωt
π−α
ωt
α
Id
ωt
ωt
Id
14
VD3
VT1
T
load
VD4
u2
VT2
Power Electronics
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
15
Power Electronics
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
Power Electronics
2.2 Three-phase controlled
(controllable) rectifier
2.2.1 Three-phase half-wave controlled rectifier
(the basic circuit among three-phase rectifiers)
2.2.2 Three-phase bridge fully-controlled rectifier
(the most widely used circuit among three-phase
rectifiers)
17
Power Electronics
2.2.1 Three-phase half-wave
controlled rectifier
Resistive load, α = 0º
u2
O ωt1
T
VT1
VT2
VT3
ud
R
id
ua
ub
ωt2
uc
ωt3
ωt
uG
O
ud
ωt
O
iVT
ωt
O
uVT
ωt
O
ωt
1
1
Common-cathode connection
Natural commutation point
uab
uac
18
Power Electronics
Resistive load, α = 30º
u2
ua
ub
uc
ωt
O
T
VT1
VT2
VT3
ud
uG
ωt
O
ud
O
iVT
ωt1
ωt
1
R
id
O
uVT
ωt
O
ωt
1 uac
uab
uac
19
Power Electronics
Resistive load, α = 60º
u2
O
T
ua ub
uc
ωt
VT1
ud
R
VT2
uG
VT3
O
ud
id
O
iVT
ωt
ωt
1
O
ωt
20
ƒ When α ≤ 30º, load current id is continuous.
1
Ud =
2π
3
5π
+α
6
∫π
6
+α
2U 2 sin ω td (ω t ) =
3 6
U 2 cos α = 1 .17U 2 cos α
2π
(2-18)
ƒ When α > 30º, load current id is discontinuous.
1
Ud =
2π
3
π
∫π α
6
+
2U2 sinωtd (ωt ) =
3 2 ⎡
π
π
⎡
⎤
⎤
U2 ⎢1 + cos( + α )⎥ = 0.675⎢1 + cos( + α )⎥
2π
6
6
⎣
⎦
⎣
⎦
(2-19)
1.2
1.17
Average load current
Ud
Id =
R
(2-20)
Ud/U2
Power Electronics
Resistive load, quantitative analysis
0.8
1
3
0.4
Thyristor voltages
2
0
30
60
90
α/(°
)
120
150
1- resistor load 2- inductor load
3- resistor-inductor load
21
Power Electronics
Inductive load, L is large enough
ua
ud
O
T
u2
a
b
VT2
c
VT1
uc
ωt
α
ia
L
eL
ud
ub
id
R
VT3
O
ib
ωt
O
ic
ωt
O
id
ωt
O
uVT
ωt
1
Load current id is always continuous.
Ud
1
=
2π
3
5π
+α
6
∫π
6
+α
2 U 2 sin ω td (ω t ) =
O
uac
ωt
uac
uab
3 6
U 2 cos α = 1 . 17 U 2 cos α
2π
(2-18)
Thyristor voltage and currents, transformer current
I VT
1
(2
23)
I
=
= 0.368 I d (2-24)
I 2 = I VT =
I d = 0.577I d
VT(AV)
1.57
3
U
FM
=U
RM
= 2 . 45 U
2
(2-25)
22
Power Electronics
2.2.2 Three-phase bridge
fully-controlled rectifier
Circuit diagram
d1
VT1
T
id
ia
a
n
VT 4
VT
6
VT3
VT 5
b
c
VT
load
ud
2
d2
Common-cathode group and common-anode group of
thyristors
Numbering of the 6 thyristors indicates the trigger sequence.
23
Power Electronics
Resistive load, α = 0º
u2 α = 0°ua
ud1
ub
uc
O ω t1
ωt
ud2
u2L
ud
Ⅰ Ⅱ
uab uac
Ⅲ Ⅳ Ⅴ Ⅵ
ubc uba uca ucb uab uac
ωt
O
iVT
1
O
uVT
uab uac
ubc uba uca ucb uab uac
ωt
1
ωt
O
uab
uac
24
Power Electronics
Resistive load, α = 30º
ud1 α = 30°ua
O
ud2
ud
ub
uc
ωt1
ωt
Ⅰ Ⅱ
uab uac
Ⅲ Ⅳ
ubc uba
Ⅴ Ⅵ
uca ucb uab uac
ωt
O
uVT
1
uab uac
ubc uba
uca ucb uab uac
ωt
O
ia
uab
uac
ωt
O
25
Power Electronics
Resistive load, α = 60º
ud1
α = 60°
ua
ub
uc
ωt 1
ωt
O
ud2
ud
Ⅰ Ⅱ Ⅲ Ⅳ Ⅴ Ⅵ
uab uac ubc uba uca ucb uab uac
ωt
O
uVT
1
uac
uac
ωt
O
uab
26
Power Electronics
Resistive load, α = 90º
ud1
ua
ub
uc
ua
ub
ωt
O
ud2
ud
O
uab uac
ubc uba uca ucb uab uac ubc uba
ωt
id
O
iVT
ωt
O
ia
ωt
O
ωt
1
27
Power Electronics
Inductive load, α = 0º
u2 α = 0° ua
ud1
ub
uc
O ωt1
ud2
u2L
ud
ωt
Ⅰ Ⅱ Ⅲ Ⅳ Ⅴ Ⅵ
uab uac ubc uba uca ucb uab uac
ωt
O
id
O
iVT
ωt
O
ωt
1
28
Power Electronics
Inductive load, α = 30º
ud1
α = 30°u
a
ub
uc
O ωt 1
ud2
ud
O
ωt
Ⅰ Ⅱ Ⅲ Ⅳ Ⅴ Ⅵ
uab uac ubc uba uca ucb uab uac
ωt
id
O
ia
ωt
O
ωt
29
Power Electronics
Inductive load, α = 90º
α = 90°
ud1
ud
uc
ua
ωt 1
O
ud2
ub
ωt
Ⅰ Ⅱ Ⅲ Ⅳ Ⅴ Ⅵ
uab uac ubc uba uca ucb uab uac
ωt
O
uVT
1
uac
uac
ωt
O
uab
30
Quantitative analysis
Power Electronics
Average output voltage
Ud =
1
π
2π
+α
3
∫π
3
6U 2 sin ω td (ω t ) = 2 .34U 2 cos α
+α
(2-26)
3
For resistive load, When a > 60º, load current id is discontinuous.
3 π
π
⎤
⎡
U d = ∫π
6U 2 sin ω td (ω t ) = 2 .34U 2 ⎢1 + cos( + α ) ⎥
(2-27)
π 3 +α
3
⎦
⎣
Average output current (load current)
Id =
Ud
R
(2-20)
Transformer current
I2 =
1 ⎛ 2 2
2 ⎞
2
⎜ I d × π + (− I d ) × π ⎟ =
2π ⎝
3
3 ⎠
2π
I d = 0.816 I d
3
(2-28)
Thyristor voltage and current
– Same as three-phase half-wave rectifier
EMF load, L is large enough
– All the same as inductive load except the calculation of average output
current
Id =
U
d
− E
R
(2-29)
31
Power Electronics
2.3 Effect of transformer leakage
inductance on rectifier circuits
T
a
ik
b
c
R
ud
α
ub
ua
LB
LB
LB
ud
ia
VT1
ib
VT2
ic
VT3
L
uc
ωt
O
id
O
ic
ia
γ
ib
ic
ia
Id
ωt
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.
32
Power Electronics
Commutation process analysis
Circulating current ik during commutation
ub-ua = 2·LB·dia/dt
ik: 0
Id
ia = Id-ik : Id
0
ib = ik
Id
:
0
Commutation angle
Output voltage during commutation
dik
dik ua + ub
ud = ua + LB
= ub − LB
=
dt
dt
2
(2-30)
33
Power Electronics
Quantitative calculation
Reduction of average output voltage due to the
commutation process
dik
1 α +γ + 56π
3 α +γ + 56π
(ub − ud )d(ωt ) = ∫ 5π [ub − (ub − LB )]d(ωt )
∆Ud =
5π
∫
+
α
2π / 3 6
2π α + 6
dt
3 α +γ + 56π dik
3 Id
3
X BId
d(ωt ) = ∫ ωLBdik =
= ∫ 5π LB
0
+
α
2π 6
dt
2π
2π
(2-31)
Calculation of commutation angle
cos α − cos( α + γ ) =
– Id
– XB
,Χ
,Χ
2X BId
(2-36)
6U 2
– For Α ≤ 90o , Α , Χ
34
Power Electronics
Summary of the effect on rectifier circuits
Circuits
Singlephase full
wave
∆U d
cosα − cos(α + γ )
X
π
B
Id
Id X B
2U 2
Threephase halfwave
Singlephase
bridge
2X
B
π
2I d X B
2U 2
Id
3X B
Id
2π
2X B I d
6U 2
Threephase
bridge
3X B
π
2X
B
6U
Id
m-pulse recfifier
mX B
Id
2π
I
I
d
2
2U
X
d
2
①
B
sin
②
π
m
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 voltage
35
Power Electronics
2.4 Capacitor-filtered uncontrolled
(uncontrollable) rectifier
Emphasis of previous sections
– Controlled rectifier, inductive load
Uncontrolled rectifier: diodes instead of thyristors
Wide applications of capacitor-filtered uncontrolled
rectifier
– AC-DC-AC frequency converter
– Uninterruptible power supply
– Switching power supply
2.4.1 Capacitor-filtered single-phase uncontrolled
rectifier
2.4.2 Capacitor-filtered three-phase uncontrolled
rectifier
36
Power Electronics
2.4.1 Capacitor-filtered single-phase
uncontrolled rectifier
Single-phase bridge, RC load
id
VD1
i2
u1
VD3
iC
ud +
u2
VD2
VD4
a)
i,ud
i
iR
C
ud
R
0
θ
π
2π
ωt
δ
b)
37
Power Electronics
2.4.1 Capacitor-filtered single-phase
uncontrolled rectifier
Single-phase bridge, RLC load
id
u2
L
+ uL VD3
iC
+
ud
VD2
VD4
VD1
i2
u1
i2,u2,ud
iR
a)
ud
i2
R
C
u2
δ
0
π
θ
ωt
b)
38
Power Electronics
2.4.2 Capacitor-filtered three-phase
uncontrolled rectifier
Three-phase bridge, RC load
ud u
ab u uac
d
VD1 VD3 VD5
T
ia
a
b
c
VD4 VD6 VD2
a)
ia
id
iC
iR
ud +
C
δ
0 θ
π
3
π
ωt
R
id
ωt
O
b)
39
Power Electronics
2.4.2 Capacitor-filtered three-phase
uncontrolled rectifier
Three-phase bridge, RC load
Waveform when ωRC≤1.732
ia
ia
O
ωt O
id
id
O
ωt O
a）ωRC= 3
ωt
ωt
b）ωRC< 3
40
Power Electronics
2.4.2 Capacitor-filtered three-phase
uncontrolled rectifier
Three-phase bridge, RLC load
ia
VD 1VD 3VD 5
T
O
id
ia
iC iR
u d+ C
a
b
c
R
ia
ωt
b)
O
ωt
VD 4VD 6VD 2
a)
c)
41
Power Electronics
2.5 Harmonics and power factor of
rectifier circuits
Originating of harmonics and power factor issues in
rectifier circuits
– Harmonics: working in switching states—nonlinear
– Power factor: firing delay angle causes phase delay
Harmful effects of harmonics and low power factor
Standards to limit harmonics and power factor
2.5.1 Basic concepts of harmonics and reactive power
2.5.2 AC side harmonics and power factor of
controlled rectifiers with inductive load
2.5.3 AC side harmonics and power factor of
capacitor-filtered uncontrolled rectifiers
2.5.4 Harmonic analysis of output voltage and current
42
Power Electronics
2.5.1 Basic concepts of harmonics
and reactive power
For pure sinusoidal waveform
u (t ) = 2U sin(ωt + ϕ u )
(2-54)
For periodic non-sinusoidal waveform
∞
or
u (ωt ) = ao + ∑ (an cos nωt + bn sin nωt )
(2-55)
u (ωt ) = ao + ∑ cn sin( nωt + ϕn )
(2-56)
n =1
∞
n =1
where
cn = an 2 + bn 2
an = cn sin ϕ
ϕn = arctan( an / bn)
bn = cn cos ϕ
Fundamental component
Harmonic components (harmonics)
43
Power Electronics
Harmonics-related specifications
Take current harmonics as examples
Content of nth harmonics
In
HRI n = × 100%
I1
(2-57)
In is the effective (RMS) value of nth harmonics.
I1 is the effective (RMS) value of fundamental component.
Total harmonic distortion
THDi =
Ih
× 100%
I1
(2-58)
Ih is the total effective (RMS) value of all the harmonic components.
44
Power Electronics
Definition of power and power factor
For sinusoidal circuits
Active power
1
P=
2π
∫
2π
0
uid (ω t ) = UI cos ϕ
(2-59)
Reactive power
Q=U I sinϕ
(2-61)
Apparent power
Power factor
S=UI
(2-60)
S 2 = P2 + Q2
(2-63)
λ
=
P
S
λ =cos ϕ
(2-62)
(2-64)
45
Power Electronics
Definition of power and power factor
For non-sinusoidal circuits
Active power
P=U I1 cosϕ1
(2-65)
Power factor
P UI1 cosϕ1 I1
(2-66)
=
= cosϕ1 = ν cosϕ1
S
UI
I
Distortion factor (fundamental-component factor)
λ=
ν =I1 / I
Displacement factor (power factor of fundamental component)
λ1 = cosϕ1
Definition of reactive power is still in dispute.
46
Power Electronics
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.
47
Single-phase bridge fully-controlled rectifier
u2
u1
ud
a
ud
u2
b
ωt
O
VT3
i2
L
ωt
O
Id
id
R
VT4
T
VT1
id
VT2
Power Electronics
2.5.2 AC side harmonics and power factor of
controlled rectifiers with inductive load
iV T O
ωt
Id
1,4
iV T O
Id
2,3
O
i2
a)
uVT
O
Id
ωt
ωt
ωt
Id
1,4
ωt
O
b)
48
Power Electronics
AC side current harmonics of single-phase bridge
fully-controlled rectifier with inductive load
4
1
1
i2 =
I d (sin ω t + sin 3ω t + sin 5ω t + " )
π
5
3
4
1
=
Id ∑
sin n ω t = ∑ 2 I n sin n ω t
π
n = 1 , 3 , 5 ," n
n = 1 , 3 , 5 ,"
(2-72)
where
In =
2
2Id
nπ
n=1,3,5,…
(2-73)
Conclusions
– Only odd order harmonics exist
– In ∝ 1/n
– In / I1 = 1/n
49
Power Electronics
Power factor of single-phase bridge fullycontrolled rectifier with inductive load
Distortion factor
I1 2 2
ν=
=
≈ 0.9
I
π
(2-75)
Displacement factor
λ1 = cosϕ 1 = cosα
(2-76)
I1
2 2
λ = νλ1 = cosϕ 1 =
cosα ≈ 0.9 cosα
I
π
(2-77)
Power factor
50
Power Electronics
Three-phase bridge fully-controlled rectifier
ud1
α = 30°u
a
ub
uc
O ωt 1
ud2
ud
O
ωt
Ⅰ Ⅱ Ⅲ Ⅳ Ⅴ Ⅵ
uab uac ubc uba uca ucb uab uac
ωt
id
O
ia
ωt
O
ωt
51
Power Electronics
AC side current harmonics of three-phase bridge
fully-controlled rectifier with inductive load
1
1
1
1
I d [sinωt − sin 5ωt − sin 7ωt + sin11ωt + sin13ωt − "]
π
13
11
7
5
2 3
2 3
1
=
I d sinωt +
I d ∑ (−1)k sin nωt = 2I1 sinωt + ∑ (−1)k 2I n sin nωt
π
π
n
n=6 k ±1
n=6 k ±1
ia =
2 3
k =1, 2,3"
where ⎧
k =1, 2,3"
6
=
I
Id
⎪1
⎪
π
⎨
⎪ I = 6 I , n = 6k ± 1, k = 1,2,3,"
⎪⎩ n nπ d
(2-79)
(2-80)
Conclusions
– Only 6k±1 order harmonics exist (k is positive integer)
– In ∝ 1/n
– In / I1 = 1/n
52
Power Electronics
Power factor of three-phase bridge fully-controlled
rectifier with inductive load
Distortion factor
I1 3
ν = = ≈ 0.955
I π
(2-81)
Displacement factor
λ1 = cosϕ 1 = cosα
(2-82)
Power factor
I1
3
λ = νλ1 = cos ϕ 1 = cos α ≈ 0.955 cos α
I
π
(2-83)
53
Power Electronics
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.
54
Power Electronics
2.5.4 Harmonic analysis of output voltage
and current
u d0 = U d0 +
∞
∑b
n = mk
n
cos n ω t
ud
∞
2 cos k π
⎡
⎤
cos n ω t ⎥
= U d0 ⎢1 − ∑
2
⎣ n = mk n − 1
⎦
2 U2
(2-85)
where
m
π
U d0 =
2U 2
bn = −
2 cos k π
U d0 (2-87)
2
n −1
π
sin
m
(2-86)
π
π
mO m
2π
m
ωt
Output voltage of m-pulse
rectifier when α = 0º
55
Power Electronics
Ripple factor in the output voltage
Output voltage ripple factor
U R
γ u =
U d0
(2-88)
where UR is the total RMS value of all the harmonic
components in the output voltage
UR =
∞
∑ U n2 = U 2 − U d02
(2-89)
n = mk
and U is the total RMS value of the output voltage
Ripple factors for rectifiers with different number of pulses
m
2
3
6
12
∞
γu（%）
48.2
18.27
4.18
0.994
0
56
Power Electronics
Harmonics in the output current
id = I d +
∞
∑d
n = mk
n
cos(nωt − ϕ n )
(2-92)
where
U d0 − E
Id =
R
bn
dn =
=
zn
(2-93)
bn
R + ( nω L )
2
nω L
ϕ n = arctan
R
2
(2-94)
(2-95)
57
Power Electronics
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.
58
Quantitative harmonic
analysis of output voltage
and current is very
complicated for α ≠ 0º.
As an example,
for 3-phase bridge
fully-controlled rectifier
ud = U d +
∞
∑c
n=6k
n
n=6
0.3
2 U2L
cn
Power Electronics
For α ≠ 0º
cos( n ω t + θ n )
0.2
n=12
0.1
n=18
0
30
60
90 120 150 180
α/(°)
(2-96)
59
Power Electronics
2.6 High power controlled rectifier
2.6.1 Double-star controlled rectifier
2.6.2 Connection of multiple rectifiers
60
Waveforms When α = 0º
Circuit
ud1
ua
ub
uc
ωt
O
T
a
b
ia
iP
n
n2
LP
n1
c' u
d
VT2
b'
VT6
VT4
VT1
VT3
1I
6 d
id
uc'
ua'
uc'
ub'
ia'
R
ωt
ωt
O
L
a'
1I
2 d
c
O
ud2
VT5
Power Electronics
2.6.1 Double-star controlled rectifier
1I
2 d
1I
6 d
O
Difference from 6-phase half-wave rectifier
ωt
61
Power Electronics
VT1
Effect of interphase reactor
(inductor, transformer)
ud1,ud2 u '
b
ua
uc'
ub
ua'
uc
ub'
1
u
2 P
n2-
iP
ua
ub'
ud2 ud
n LP
+ - +n
1
L
ud1
R
VT6
a)
O ωt1
ωt
。
up
b)
60
ωt
O
。
360
u p = u d2 − u d1
u d = u d2
(2-97)
1
1
1
− u p = u d1 + U p = ( u d1 + u d2 ) (2-98)
2
2
2
62
Power Electronics
Quantitative analysis when α = 0º
ud1 =
2
1
3 6U 2
1
[1 + cos 3ωt − cos 6ωt + cos 9ωt − ⋅ ⋅ ⋅]
4
35
40
2π
(2-99)
3 6U 2
1
2
1
[1 + cos 3(ω t − 60°) − cos 6(ω t − 60°) + cos 9(ω t − 60°) − ⋅⋅⋅]
2π
4
35
40
3 6U 2
1
2
1
[1 − cos 3ω t − cos 6ω t − cos 9ω t − ⋅⋅⋅]
=
(2-100)
2π
4
35
40
ud2 =
up =
3 6U 2 1
1
[ − cos 3ωt − cos 9ωt − ⋅ ⋅ ⋅]
2π
2
20
3 6U 2
2
ud =
[1 − cos 6ωt − ⋅ ⋅ ⋅]
2π
35
(2-101)
(2-102)
63
Power Electronics
Waveforms
when α > 0º
。
ud α = 30 u u '
a
c
O
ud
Ud = 1.17U 2 cos α
O
ud
O
α = 60。 u '
c
α = 90。 u '
c
ub
ua'
uc
ub'
ωt
ub
ua'
uc
ub'
ωt
ub
ua'
uc
ub'
ωt
64
Power Electronics
Comparison with 3-phase half-wave
rectifier and 3-phase bridge rectifier
Voltage output capability
– Same as 3-phase half-wave rectifier
– Half of 3-phase bridge rectifier
Current output capability
– Twice of 3-phase half-wave rectifier
– Twice of 3-phase bridge rectifier
Applications
Low voltage and high current situations
65
Power Electronics
2.6.2 Connection of multiple rectifiers
To increase the
output capacity
Connection
of multiple
rectifiers
Larger output voltage:
series connection
Larger output current:
parallel connection
To improve the AC side current waveform
and DC side voltage waveform
Power Electronics
Phase-shift connection of multiple rectifiers
Parallel connection
1
2
LP
VT
T
c1
b1
a1
L
c2
b2
a2
M
12-pulse rectifier realized by
paralleled 3-phase bridge rectifiers
67
Power Electronics
iA
▲
Phase-shift connection of multiple rectifiers
Series connection
i a1
a1
I
id
Id
a)
1
0
0
u a1b1
A
ia1
▲
b1
c1
I
ia2
L
*
'
iab2
ud
C
II
30 °lagging
B
c2
▲
i ab2
*
b2
Id
2
I
3 d
ωt
3
3 Id
2 3
3 Id
ωt
0
iA
3
(1+
R
u a2b2
II
360° ωt
1
I
3 d
c)
a2
*
iab2
b)
0
1
180°
2 3
3 )Id
d)
0
3
3 Id
(1+
12-pulse rectifier realized by
series 3-phase bridge rectifiers
3
3
ωt
)Id
68
Power Electronics
Quantitative analysis of 12-pulse rectifier
Voltage
– Average output voltage
Parallel connection:
U d = 2.34U 2 cos α
Series connection: U d = 4.68U 2 cos α
– Output voltage harmonics
Only 12m harmonics exist
Input (AC side) current harmonics
– Only 12k±1 harmonics exist
Connection of more 3-phase bridge rectifiers
– Three: 18-pulse rectifier (20º phase difference)
– Four: 24-pulse rectifier (15º phase difference)
69
VT13
VT14
VT23
VT33
VT24
VT11
VT34
u2
VT21
u2
VT22
i
VT31
u2
VT12
Id
VT32
Power Electronics
Sequential control of multiple
series-connected rectifiers
ud
Ⅰ
L
Ⅱ
O α
π+α
b)
ud
i
load
Id
2Id
Ⅲ
c)
a)
Circuit and waveforms of series-connected
three single-phase bridge rectifiers
70
Power Electronics
2.7 Inverter mode operation
of rectifiers
Review of DC generator-motor system
EG − EM
Id =
R∑
EM − EG
Id =
R∑
should be avoided
71
Power Electronics
Inverter mode operation of rectifiers
Rectifier and inverter mode operation of single-phase
full-wave converter
VT
VT 1
1
0
u10
iVT
1
u20 VT 2
2
iVT
2
ud
α
u10
u20
1
1
L
R
id
ud
iVT
1
0
VT 2
Energy +
M EM
u10
L
id
ud
Energy
2
iVT
2
ud
u20
u10
R
M EM
+
u10
Ud>EM
ωt
O
ωt
O
Ud<EM
id
id=iVT +iVT
1
2
iVT
iVT
1
2
iVT
O
a)
Ud − EG
Id =
R∑
1
Id
ωt
id
α
id=iVT +iVT
iVT
iVT
1
2
O
2
iVT
1
2
b)
Id =
Id
ωt
EM − Ud
R∑
72
Power Electronics
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.
EM > Ud
73
Power Electronics
Inverter mode operation
of 3-phase bridge rectifier
u2
ua
ub
uc
ua
ub
uc
ua
ub
uc
ua
ub
ωt
O
β =π
3
ud uab uac ubc uba uca
β = π4
ucb uab uac ubc uba uca
β = π6
ucb uab uac ubc uba uca
ucb uab uac ubc
ω t 1 ω t 2 ω t3
ωt
O
β =π
3
β = π4
β = π6
Inversion angle (extinction angle) β
α + β =180º
74
Power Electronics
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）
a
b
c
iVT
1
iVT
2
LB VT
1
LB
VT2
LB VT
L
id
3
iVT
ud
3
M EM
+
o
ub
ua
ud
uc
ua
ub
p
O
id
iVT
3
O
β
γ β <γ
β
γ β >γ
α
iVT
1
iVT
2
ωt
iVT
3
iVT
1
ωt
75
Power Electronics
2.8 Thyristor-DC motor system
2.8.1 Rectifier mode of operation
2.8.2 Inverter mode of operation
2.8.3 Reversible DC motor drive system
(four-quadrant operation)
76
Power Electronics
2.8.1 Rectifier mode of operation
Waveforms and equations
ua
ud
ub
U d = E M + R ∑ Id + ∆ U
(2-112)
where
3 XB
R ∑ = RB + RM +
2π
(for 3-phase half-wave)
uc
ud
Ud
E
idR
ωt
O
α
id
O
ic
ia
ib
ic
ωt
Waveforms of 3-phase half-wave
rectifier with DC motor load
77
Power Electronics
Speed-torque (mechanic) characteristic
when load current is continuous
E M = C en
(2-113)
For 3-phase half-wave
Ud = 1.17U 2 cos α
3X I
(RB+RM+ 2πB ) Cd
e
n
E M = 1 .17U 2 cos α − R ∑ Id − ∆ U
a1
(2-114)
a2
1 .17U 2 cos α R ∑ Id + ∆ U
n=
−
Ce
Ce
(2-115)
For 3-phase bridge
2.34U 2 cos α R ∑
−
n=
Id (2-116)
Ce
Ce
a3
a1<a2<a3
O
Id
For 3-phase half-wave
78
Power Electronics
Speed-torque (mechanic) characteristic
when load current is discontinuous
EMF at no load (taking 3-phase
half-wave as example)
For α ≤ 60º
E0 =
2U 2
For α > 60º
E0 =
E
E0
( 2U2)
E0'
(0.585U2)
2U 2 cos(α − 60 D )
Idmin
O
discontinuouts
mode
continuous mode
Id
For 3-phase half-wave
79
Power Electronics
Speed-torque (mechanic) characteristic
when load current is discontinuous
For different α
The point of EMF at no
load is raised up.
E
E0
boundary
a1
a2
a3
a4
The droop rate becomes
steer. (softer than the
continuous mode)
a5
discontinuous
mode
continuous mode
O
Id
For 3-phase half-wave
(α1< α2 < α3 ≤ 60º, α5 > α4 > 60º)
80
n
– 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
EM = 1.17U 2 cos α − R ∑ Id − ∆U
rectifier
mode
α1
α2
α3
α4
α =β = π
2
(2-114)
n=
1 .17U 2 cos α R ∑ Id + ∆ U
−
Ce
Ce
– Or in another form (2-115)
E M = − (U d 0 cos β + Id R ∑) (2-122)
n=−
1
Ce
U d 0 cos β + IdR ∑ (2-123)
α increasing
Equations
Id
β4
β3
β2
inverter
mode
β increasing
Power Electronics
2.8.2 Inverter mode of operation
β1
Speed-torque characteristic of
a DC motor fed by a thyristor
rectifier circuit
81
Power Electronics
a
b
c
converter 1
2.8.3 Reversible DC motor drive system
(4-quadrant operation)
+n
converter 2 inverting
converter 1 rectifying
Id
Id
+
L
converter 1
M EM
+
EM M
-
AC
AC
source source
Energy
Udβ
-
converter2 converter1
+
Energy
Udα
forward braking(regenerating)
converter 2
-T
Back-to-back
connection of two 3phase bridge circuits
Id
converter 1
EM M
+
-
Energy
Udα
converter2
forward motoring
O
converter 2 rectifying
+
M E
M
-
converter 1 inverting
AC
AC
source source
+ converter2
converter1
-
Id
Energy
+
+T
Udβ
M EM
+
converter2
reverse braking(regenerating)
reverse motoring
-n
82
co n v erter 2
n
co n v erter 1
α1
β '2
α3
β '4
α '= β '= π2
α '4
α '3
α '2
α '1
α increasing
α2
β '3
α4
α = β = π2
Id
β4
β3
β2
α 1 = β ' 1 ; α '1 = β 1
α 2 = β ' 2 ; α '2 = β 2
β1
β increasing
β 'increasing
β '1
α 'increasing
Power Electronics
4-quadrant speed-torque characteristic
of Reversible DC motor drive system
83
Power Electronics
Power Electronics
Supplement:
Gate Triggering Control Circuit
for Thyristor Rectifiers
84
Power Electronics
2.9 Gate triggering control circuit for
thyristor rectifiers
Object
How to timely generate triggering pulses with
adjustable phase delay angle
Constitution
Synchronous circuit
Saw-tooth ramp generating and phase shifting
Pulse generating
Integrated gate triggering control circuits are very
widely used in practice.
85
Power Electronics
A typical gate triggering control circuit
R15
VD11~VD14
220V
C7 + C 6
VD15
36V
VD7
+15V
RP2
VS
R3
A
V1
R1
TS
R
VD1
VD2
R4
R7
C2
V 2 R5
C1
R2
R6
VD9
C5
R16
V7
V4
R17
R8
VD6
+15V
R18
R14
V5
R10
VD8
TP
R13
I1c
Q
uts
C3
VD4
V3
R12
R11
R9
B
C3
VD10
V8
V6
VD5
up
RP1
uco -15V
Disable
XY
-15V
86
Electronics
Power
Power Electronics
Waveforms of the typical
gate triggering control circuit
87
Power Electronics
How to get synchronous voltage for the gate
triggering control circuit of each thyristor
uA
uB
uC
UAB
Ua
Usa
TS
TR
-Usb
-Usc
Usb
Ub
D,y 5-11
D,y 11
ua
ub
uc - usa - usc - usb
- usb - usa - usc
Uc
Usc -U
sa
For the typical circuit on page 20, the synchronous voltage of the gate
triggering control circuit for each thyristor should be lagging 180º to the
corresponding phase voltage ofyy
that thyristor.
88