www.getmyuni.com
UNIT-4
Controlled Rectifiers
4.1 Line Commutated AC to DC converters
+
AC
Input
Voltage
Line
Commutated
Converter
DC Output
V0(dc)
-
•
•
•
Type of input: Fixed voltage, fixed frequency ac power supply.
Type of output: Variable dc output voltage
Type of commutation: Natural / AC line commutation
4.1.1Different types of Line Commutated Converters
• AC to DC Converters (Phase controlled rectifiers)
• AC to AC converters (AC voltage controllers)
• AC to AC converters (Cyclo converters) at low output frequency
4.1.2 Differences Between Diode Rectifiers & Phase Controlled Rectifiers
• The diode rectifiers are referred to as uncontrolled rectifiers .
• The diode rectifiers give a fixed dc output voltage .
• Each diode conducts for one half cycle.
• Diode conduction angle = 1800 or π radians.
• We cannot control the dc output voltage or the average dc load current in a diode
rectifier circuit
Single phase half wave diode rectifier gives an
Average dc output voltage VO( dc ) =
Vm
π
Single phase full wave diode rectifier gives an
Average dc output voltage VO( dc ) =
2Vm
π
4.2 Applications of Phase Controlled Rectifiers
• DC motor control in steel mills, paper and textile mills employing dc motor drives.
• AC fed traction system using dc traction motor.
• Electro-chemical and electro-metallurgical processes.
• Magnet power supplies.
Page 130
www.getmyuni.com
• Portable hand tool drives
4.3 Classification of Phase Controlled Rectifiers
• Single Phase Controlled Rectifiers.
• Three Phase Controlled Rectifiers
4.3.1 Different types of Single Phase Controlled Rectifiers.
• Half wave controlled rectifiers.
• Full wave controlled rectifiers.
• Using a center tapped transformer.
• Full wave bridge circuit.
• Semi converter.
• Full converter.
4.3.2 Different Types of Three Phase Controlled Rectifiers
• Half wave controlled rectifiers.
• Full wave controlled rectifiers.
• Semi converter (half controlled bridge converter).
• Full converter (fully controlled bridge converter).
4.4 Principle of Phase Controlled Rectifier Operation
Single Phase Half-Wave Thyristor Converter with a Resistive Load
Page 131
www.getmyuni.com
Equations:
vs = Vm sin ω t = i/p ac supply voltage
Vm = max. value of i/p ac supply voltage
VS =
ω
α
Vm
= RMS value of i/p ac supply voltage
2
vO = vL = output voltage across the load
When the thyristor is triggered at t =
vO = vL = Vm sin ω t ; ω t = α to π
vO
= Load current; ω t = α to π
R
V sin ω t
iO = iL = m
= I m sin ω t ; ω t = α to π
R
V
Where I m = m = max. value of load current
R
iO = iL =
4.4.1 To Derive an Expression for the Average (DC) Output Voltage across the Load
VO( dc )
1
= Vdc =
2π
2π
∫ v .d (ω t );
O
0
vO = Vm sin ω t for ω t = α to π
π
VO( dc )
1
= Vdc =
Vm sin ω t.d (ω t )
2π α∫
VO( dc )
1
=
Vm sin ω t.d (ω t )
2π α∫
VO( dc )
V
= m ∫ sin ω t.d (ω t )
2π α
VO( dc )
V
= m
2π
π
π

 − cos ω t

π


α
Vm
[ − cos π + cos α ] ; cos π = −1
2π
V
= m [1 + cos α ] ; Vm = 2VS
2π
VO( dc ) =
VO( dc )
Page 132
www.getmyuni.com
Maximum average (dc) o/p
voltage is obtained when α = 0
and the maximum dc output voltage
V
Vdc( max ) = Vdm = m (1 + cos 0 ) ; cos ( 0 ) = 1
2π
V
∴Vdc( max ) = Vdm = m
π
Vm
[1 + cos α ] ; Vm = 2VS
2π
The average dc output voltage can be varied
VO( dc ) =
by varying the trigger angle α from 0 to a
maximum of 1800 (π radians )
We can plot the control characteristic
(V (
O dc )
)
vs α by using the equation for VO( dc )
4.5 Control Characteristic of Single Phase Half Wave Phase Controlled Rectifier with
Resistive Load
The average dc output voltage is given by the
expression
Vm
[1 + cos α ]
2π
We can obtain the control characteristic by
VO( dc ) =
plotting the expression for the dc output
voltage as a function of trigger angleα
4.5.1 Control Characteristic
Page 133
www.getmyuni.com
VO(dc)
Vdm
0.6Vdm
0.2 Vdm
0
60
120
180
Trigger angle α in degrees
Normalizing the dc output
voltage with respect to Vdm , the
Normalized output voltage
Vm
1 + cos α )
Vdc 2π (
Vn =
=
Vm
Vdm
π
Vn =
Vdc 1
= (1 + cos α ) = Vdcn
Vdm 2
4.5.2 To Derive an Expression for the RMS Value of Output Voltage of a Single Phase
Half Wave Controlled Rectifier with Resistive Load
The RMS output voltage is given by
 1 2π 2

VO( RMS ) = 
vO .d (ω t ) 
∫
 2π 0

Output voltage vO = Vm sin ω t ; for ω t = α to π
1
VO( RMS )
 1 π 2 2
2
=  ∫ Vm sin ω t.d (ω t ) 
 2π α

By substituting sin 2 ω t =
1 − cos 2ω t
, we get
2
1
VO ( RMS )
 1 π 2 (1 − cos 2ω t )
2
=
Vm
.d (ω t ) 
∫
2
 2π α

VO ( RMS )
V 2
= m
 4π
VO ( RMS )
π
 V 2 π
 2
=  m  ∫ d (ω t ) − ∫ cos 2ω t.d (ω t )  
α
 
 4π α
1
π
2
∫α (1 − cos 2ω t ) .d (ω t )
1
VO( RMS )
V 1 
= m  (ω t )
2  π 
VO( RMS ) =
Vm
2
1
π
α
 2
 sin 2ω t 
−


 2  α 
π
Page 134
1
2
1 
( sin 2π − sin 2α )  ;sin2π = 0
  (π − α ) −

2
 
 π 
www.getmyuni.com
4.5.3 Performance Parameters of Phase Controlled Rectifiers
Output dc power (avg. or dc o/p
power delivered to the load)
PO( dc ) = VO( dc ) × I O ( dc ) ; i.e., Pdc = Vdc × I dc
Where
VO( dc ) = Vdc = avg./ dc value of o/p voltage.
I O ( dc ) = I dc = avg./dc value of o/p current
Output ac power
PO( ac ) = VO( RMS ) × I O( RMS )
Efficiency of Rectification (Rectification Ratio)
Efficiency η =
PO( dc )
PO( ac )
; % Efficiency η =
PO( dc )
PO( ac )
× 100
The o/p voltage consists of two components
The dc component VO( dc )
The ac /ripple component Vac = Vr ( rms )
Output ac power
PO( ac ) = VO( RMS ) × I O( RMS )
Efficiency of Rectification (Rectification Ratio)
Efficiency η =
PO( dc )
PO( ac )
; % Efficiency η =
PO( dc )
PO( ac )
× 100
The o/p voltage consists of two components
The dc component VO( dc )
The ac /ripple component Vac = Vr ( rms )
4.5.4 The Ripple Factor (RF) w.r.t output voltage waveform
Page 135
www.getmyuni.com
Current Ripple Factor ri =
I r ( rms )
I O( dc )
=
I ac
I dc
Where I r ( rms ) = I ac = I O2 ( RMS ) − I O2 ( dc )
Vr ( pp ) = peak to peak ac ripple output voltage
Vr ( pp ) = VO( max ) − VO( min )
I r ( pp ) = peak to peak ac ripple load current
I r ( pp ) = I O( max ) − I O( min )
Transformer Utilization Factor (TUF)
TUF =
PO( dc )
VS × I S
Where
VS = RMS supply (secondary) voltage
I S = RMS supply (secondary) current
Where
vS = Supply voltage at the transformer secondary side
iS = i/p supply current
(transformer secondary winding current)
iS 1 = Fundamental component of the i/p supply current
Page 136
www.getmyuni.com
= Displacement angle (phase angle)
For an RL load
= Displacement angle = Load impedance angle
ωL 
∴ φ = tan −1 
 for an RL load
 R 
Displacement Factor (DF) or
Fundamental Power Factor
DF = Cosφ
Harmonic Factor (HF) or
Total Harmonic Distortion Factor ; THD
1
1
2
 I  2  2
I −I 
S
HF = 
 =   − 1
2
 I S 1 

 I S1 
Where
2
S
2
S1
I S = RMS value of input supply current.
I S 1 = RMS value of fundamental component of
the i/p supply current.
Input Power Factor (PF)
PF =
VS I S 1
I
cos φ = S 1 cos φ
VS I S
IS
The Crest Factor (CF)
CF =
I S ( peak )
IS
=
Peak input supply current
RMS input supply current
For an Ideal Controlled Rectifier
FF = 1; η = 100% ; Vac = Vr ( rms ) = 0 ; TUF = 1;
RF = rv = 0 ; HF = THD = 0; PF = DPF = 1
4.5.5 Single Phase Half Wave Controlled Rectifier with an RL Load
Page 137
www.getmyuni.com
Input Supply Voltage (Vs) & Thyristor (Output) Current Waveforms
Output (Load) Voltage Waveform
4.5.6 To derive an expression for the output (Load) current, during ωt = α to β when
Page 138
www.getmyuni.com
thyristor T1 conducts
ω
ω
φ
α
τ
ω
Assuming T1 is triggered ω t = α ,
β
we can write the equation,
 di 
L  O  + RiO = Vm sin t ; ≤ t ≤
 dt 
General expression for the output current,
V
iO = m sin ( t −
Z
ω
) + A1e
−t
Vm = 2VS = maximum supply voltage.
Z = R 2 + ( L ) =Load impedance.
2
ωL 
 = Load impedance angle.
 R 
φ = tan −1 
L
= Load circuit time constant.
R
∴ general expression for the output load current
τ=
−R
t
Vm
iO =
sin (ω t − φ ) + A1e L
Z
Constant A1 is calculated from
α 
initial condition iO = 0 at ω t = α ; t=  
ω 
−R
t
V
iO = 0 = m sin (α − φ ) + A1e L
Z
−R
t
−Vm
∴
A1e L =
sin (α − φ )
Z
We get the value of constant A1 as
A1 = e
R (α )
ωL
 −Vm

 Z sin (α − φ ) 


Page 139
www.getmyuni.com
ω
Substituting the value of constant A1 in the
φ
α φ
general expression for iO
−R
(ω t −α )  −Vm
Vm
sin ( t − ) + e ω L
 Z sin ( −
Z

∴ we obtain the final expression for the
iO =
ω
φ
α φ

inductive load current
iO =
Vm
Z
β
ω
) 

sin ( t − ) − sin ( −

Where α ≤ ω t ≤ β
−R
) eω L
(ω t −α ) 
;

β
Extinction angle
can be calculated by using
the condition that iO = 0 at
iO =
Vm
Z
t=
−R
(ω t −α ) 

ωL
−
−
−
ω
φ
α
φ
sin
t
sin
e
(
)
(
)

=0


∴ sin ( β − φ ) = e
−R
ωL
( β −α )
× sin (α − φ )
β can be calculated by solving the above eqn.
4.5.7 To Derive an Expression for Average (DC) Load Voltage of a Single Half
Wave Controlled Rectifier with RL Load
VO( dc )
VO( dc )
1
= VL =
2π
2π
∫ v .d (ωt )
O
0
β
2π
α

1 
= VL =
 ∫ vO .d (ω t ) + ∫ vO .d (ω t ) + ∫ vO .d (ω t ) 
2π  0
α
β

vO = 0 for ω t = 0 to α & for ω t = β to 2π
β

1 
∴VO ( dc ) = VL =
 ∫ vO .d (ω t )  ;
2π α

vO = Vm sin ω t for ω t = α to β
VO( dc )
β

1 
= VL =
 ∫ Vm sin ω t.d (ω t ) 
2π α

VO( dc ) = VL =
Vm
2π

 − cos ω t

β


α
Vm
( cos α − cos β )
2π
V
= VL = m ( cos α − cos β )
2π
VO( dc ) = VL =
∴VO ( dc )
Page 140
www.getmyuni.com
Effect of Load Inductance on the Output
During the period ωt = Π to β the instantaneous output voltage is negative and this reduces
the average or the dc output voltage when compared to a purely resistive load.
π
4.5.8 Average
DCβLoad Current
α
I O ( d c ) = I L ( A vg ) =
VO (dc )
RL
=
Vm
( co s
2 RL
− co s
)
4.5.9 Single Phase Half Wave Controlled Rectifier with RL Load & Free Wheeling
Diode
T
i0
+
V0
R
+
Vs
FWD
−
L
~
−
vS
Supply voltage
0
π
α
2π
3π
ωt
iG
Gate pulses
0
iO
ωt
α
Load current
α
ωt=β
0
α
vO
0
ωt
β
β
2π+α
Load voltage
α
π
2π
3π
ωt
Page 141
www.getmyuni.com
The average output voltage
Vm
[1 + cos α ] which is the same as that
2π
of a purely resistive load.
Vdc =
The following points are to be noted
For low value of inductance, the load current
tends to become discontinuous.
During the period α to π
the load current is carried by the SCR.
During the period π to β load current is
carried by the free wheeling diode.
The value of β depends on the value of
R and L and the forward resistance
of the FWD.
For Large Load Inductance the load current does not reach zero, & we obtain
continuous load current.
i0
t1
0
t2
SCR
α
FWD
π
2π
t3
t4
SCR
FWD
2π+α
3π
ωt
Page 142
www.getmyuni.com
4.6 Single Phase Full Wave Controlled Rectifier Using A Center Tapped Transformer
T1
A
+
vO
AC
Supply
L
R
O
T2
B
4.6.1 Discontinuous Load Current Operation without FWD for π <β< (π+α)
vO
Vm
ωt
0
α
iO
α
β
ωt
0
π
α
3π
β 2π
π+α(
()
)
π+β
(i) To derive an expression for the output (load) current, during ωt = α to β when
thyristor
T1 conducts
Assuming T1 is triggered ω t = α ,
we can write the equation,
ω α ω β
 di 
L  O  + RiO = Vm sin t ; ≤ t ≤
 dt 
General expression for the output current,
iO =
Vm ω φ
sin ( t −
Z
τ
) + A1e
−t
Page 143
www.getmyuni.com
Constant A1 is calculated from
α 
initial condition iO = 0 at ω t = α ; t=  
ω 
−R
t
V
iO = 0 = m sin (α − φ ) + A1e L
Z
−R
t
−V
∴
A1e L = m sin (α − φ )
Z
We get the value of constant A1 as
R (α )
 −V

A1 = e ω L  m sin (α − φ ) 
 Z

ω
Vm = 2VS = maximum supply voltage.
Z = R2 + ( L ) =Load impedance.
2
 ωL 
 = Load impedance angle.
 R 
φ = tan −1 
L
= Load circuit time constant.
R
∴general expression for the output load current
τ=
−R
t
Vm
L
iO = sin (ωt − φ ) + Ae
1
Z
Substituting the value of constant A1 in the
general expression for iO
ω
φ
α φ
−R
(ω t −α )  −Vm
V
iO = m sin ( t − ) + e ω L
 Z sin ( −
Z

∴ we obtain the final expression for the
) 

inductive load current
ω
φV 
m
iO =
β
α φ
sin ( t − ) − sin ( −
Z 
Where α ≤ ω t ≤ β
Extinction angle
ω
Vm
Z
(ω t −α ) 
;

can be calculated by using
β
the condition that iO = 0 at
iO =
)e
−R
ωL
t=
−R

(ω t −α ) 
ωL
ω
φ
α
φ
−
−
−
sin
t
sin
e
)
(
)
 (
=0


−R
∴ sin ( β − φ ) = eω L
( β −α )
× sin (α − φ )
β can be calculated by solving the above eqn.
Page 144
www.getmyuni.com
(ii) To Derive an Expression for the DC Output Voltage of A Single Phase Full Wave
Controlled Rectifier with RL Load (Without FWD)
vO
Vm
ωt
0
α
iO
α
0
β
ωt
π
α
2π
β
π+α
(
) (
VO( dc ) = Vdc =
VO( dc )
VO( dc )
VO( dc )
1
π
3π
)
π+β
β
v .d (ω t )
∫
ω α
O
t=
β

1
= Vdc =  ∫ Vm sin ω t.d (ω t ) 
π α

β

Vm 
= Vdc =
 − cos ω t 
π 
α 
V
= Vdc = m ( cos α − cos β )
π
When the load inductance is negligible ( i.e., L ≈ 0 )
Extinction angle β = π radians
Hence the average or dc output voltage for R load
VO ( dc ) =
Vm
VO ( dc ) =
Vm
VO ( dc ) =
Vm
π
π
π
( cos α − cos π )
( cos α − ( −1) )
(1 + cos α ) ; for R load,
when β = π
Page 145
www.getmyuni.com
(iii) To calculate the RMS output voltage we use the expression
(iv) Discontinuous Load Current Operation with FWD
vO
Vm
ωt
0
α
Thyristor T1 is triggered at ωt = α ;
T1 conducts from ωt = α to π
Thyristor T2 is triggered at ωt = (π + α ) ;
iO
T2 conducts from ωt = (π + α ) to 2π
α
β
FWD conducts from ωt = π to β &
ωt
0
α
πβ
3π
2π
(π+α)
vO ≈ 0 during discontinuous load current.
(π+β)
(v) To Derive an Expression for the DC Output Voltage for a Single Phase Full
Wave Controlled Rectifier with RL Load & FWD
VO( dc) = Vdc =
∴ VO( dc) = Vdc =
VO( dc)
VO( dc)
π
1
π ωt∫=0
1
π
π α∫
vO .d (ωt )
Vm sin ωt.d (ωt )
π

Vm 
= Vdc = − cosω t 
π 
α
V
= Vdc = m [ − cosπ + cosα ] ; cosπ = −1
∴ VO( dc) = Vdc =
π
Vm
π
(1+ cosα )
Page 146
www.getmyuni.com
•
•
•
The load current is discontinuous for low values of load inductance and for large
values of trigger angles.
For large values of load inductance the load current flows continuously without
falling to zero.
Generally the load current is continuous for large load inductance and for low
trigger angles.
4.6.2 Continuous Load Current Operation (Without FWD)
Vm
vO
ωt
0
iO
0
α
α
α
α
ωt
π
α
3π
2π
(2π+α)
(π+α)
(i) To Derive an Expression for Average / DC Output Voltage of Single Phase Full
Wave Controlled Rectifier for Continuous Current Operation without FWD
vO
Vm
ωt
0
iO
α
α
α
α
ωt
0
α
π
3π
2π
(π+α)
(2π+α)
Page 147
www.getmyuni.com
VO ( dc ) = Vdc =
1
π
(π +α )
∫
vO .d (ω t )
ω t =α
VO ( dc )
(π +α )

1
= Vdc =  ∫ Vm sin ω t.d (ω t ) 
π  α

VO ( dc )
V 
= Vdc = m  − cos ω t
π 
(π +α ) 
α


VO( dc ) = Vdc
=
∴
•
Vm
cos α − cos (π + α )  ;
π 
cos (π + α ) = − cos α
VO( dc ) = Vdc =
Vm
VO( dc ) = Vdc =
2Vm
π
π
[cos α + cos α ]
cos α
By plotting VO(dc) versus α, we obtain the control characteristic of a single phase
full wave controlled rectifier with RL load for continuous load current operation
without FWD
V
d c
= V
d m
× co s α
Page 148
www.getmyuni.com
VO(dc)
Vdm
0.6Vdm
0.2 Vdm
0
-0.2Vdm
α
30
60
90
120
150
180
-0.6 Vdm
-Vdm
Trigger angle α in degrees
By varying the trigger angle we can vary the
output dc voltage across the load. Hence we can
control the dc output power flow to the load.
For trigger angle α , 0 to 900 ( i.e., 0 ≤ α ≤ 900 ) ;
cos α is positive and hence Vdc is positive
Vdc & I dc are positive ; Pdc = (Vdc × I dc ) is positive
Converter operates as a Controlled Rectifier.
Power flow is from the ac source to the load.
For trigger angle α , 900 to 1800
( i.e., 90
0
≤ α ≤ 1800 ) ,
cosα is negative and hence
Vdc is negative; I dc is positive ;
Pdc = (Vdc × I dc ) is negative.
In this case the converter operates
as a Line Commutated Inverter.
Power flows from the load ckt. to the i/p ac source.
The inductive load energy is fed back to the
i/p source.
Drawbacks of Full Wave Controlled Rectifier with Centre Tapped Transformer
• We require a centre tapped transformer which is quite heavier and bulky.
• Cost of the transformer is higher for the required dc output voltage & output power.
• Hence full wave bridge converters are preferred.
Page 149
www.getmyuni.com
4.7 Single Phase Full Wave Bridge Controlled Rectifier
2 types of FW Bridge Controlled Rectifiers are
Half Controlled Bridge Converter (Semi-Converter)
Fully Controlled Bridge Converter (Full Converter)
The bridge full wave controlled rectifier does not require a centre tapped transformer
4.7.1 Single Phase Full Wave Half Controlled Bridge Converter (Single Phase Semi
Converter)
Trigger Pattern of Thyristors
Thyristor T1 is triggered at
ω t = α , at ω t = ( 2π + α ) ,...
Thyristor T2 is triggered at
ω t = (π + α ) , at ω t = ( 3π + α ) ,...
The time delay between the gating
signals of T1 & T2 = π radians or 1800
Page 150
www.getmyuni.com
Waveforms of single phase semi-converter with general load & FWD for α > 900
Single Quadrant Operation
Page 151
www.getmyuni.com
Thyristor T1 and D1 conduct from ωt = α to π
Thyristor T2 and D2 conduct from ωt = (π + α) to 2 π
FWD conducts during ωt = 0 to α, π to (π + α) , …..
Load Voltage & Load Current Waveform of Single Phase Semi Converter for α < 900
& Continuous load current operation
vO
Vm
ωt
0
α
iO
α
α
α
ωt
0
α
π
3π
2π
(π+α)
(2π+α)
(i) To Derive an Expression for The DC Output Voltage of A Single Phase Semi
Converter with R, L, & E Load & FWD For Continuous, Ripple Free Load Current
Operation
VO( dc) = Vdc =
∴ VO( dc) = Vdc =
π
1
π ωt∫=0
1
π
π α∫
vO .d (ωt )
Vm sin ωt.d (ωt )
π

Vm 
ω
−
cos
t


π 
α
V
= Vdc = m [ − cos π + cos α ] ; cos π = −1
VO( dc) = Vdc =
VO( dc)
∴ VO( dc) = Vdc =
π
Vm
π
(1 + cosα )
Page 152
www.getmyuni.com
Vdc can be varied from a max.
value of
2Vm
π
to 0 by varying α from 0 to π .
For α = 0, The max. dc o/p voltage obtained is
Vdc( max ) = Vdm =
2Vm
π
Normalized dc o/p voltage is
Vm
Vdcn
1 + cos α )
Vdc π (
1
= Vn =
=
= (1 + cos α )
2
Vdn
 2Vm 
 π 


(ii) RMS O/P Voltage VO(RMS)
π
 2

VO( RMS ) =  ∫ Vm2 sin 2 ωt.d (ω t ) 
 2π α

1
2
2 π
m
V
VO( RMS ) = 
 2π
VO( RMS ) =

∫α (1 − cos 2ωt ) .d (ωt )
Vm  1 
sin 2α  
π π − α + 2 
2 

1
2
1
2
4.7.2 Single Phase Full Wave Full Converter (Fully Controlled Bridge Converter) With
R, L, & E Load
Page 153
www.getmyuni.com
Waveforms of Single Phase Full Converter Assuming Continuous (Constant Load
Current) & Ripple Free Load Current.
Page 154
www.getmyuni.com
Constant Load Current
iO=Ia
iO
Ia
ωt
α
iT1
π+α
Ia
Ia
ωt
& iT2
α
iT3
π+α 2π+α
Ia
ωt
& iT4
π+α 2π+α 3π+α
(i) To Derive An Expression For The Average DC Output Voltage of a Single Phase Full
Converter assuming Continuous & Constant Load Current
The average dc output voltage
can be determined by using the expression
2π

1 
 ∫ vO .d (ω t )  ;
2π  0

The o/p voltage waveform consists of two o/p
pulses during the input supply time period of
0 to 2π radians. Hence the Average or dc
o/p voltage can be calculated as
VO( dc ) = Vdc =
VO( dc)
VO( dc)
VO( dc)
π +α

2 
= Vdc =
 ∫ Vm sin ωt.d (ωt ) 
2π  α

2V
π +α
= Vdc = m [ − cos ωt ]α
2π
2V
= Vdc = m cosα
π
Maximum average dc output voltage is
calculated for a trigger angle α = 00
and is obtained as
Vdc( max ) = Vdm =
2Vm
π
∴Vdc( max ) = Vdm =
× cos ( 0 ) =
2Vm
π
2Vm
π
Page 155
www.getmyuni.com
The normalized average output voltage is given by
Vdcn = Vn =
VO( dc)
Vdc( max)
2Vm
∴Vdcn = Vn = π
=
Vdc
Vdm
cosα
2Vm
= cosα
π
By plotting VO(dc) versus α, we obtain the control characteristic of a single phase full
wave fully controlled bridge converter (single phase full converter) for constant &
continuous load current operation.
To plot the control characteristic of a
Single Phase Full Converter for constant
& continuous load current operation.
We use the equation for the average/ dc
output voltage
VO( dc ) = Vdc =
2Vm
π
cos α
Page 156
www.getmyuni.com
VO(dc)
Vdm
0.6Vdm
0.2 Vdm
0
-0.2Vdm
α
30
60
90
120
150
180
-0.6 Vdm
-Vdm
Trigger angle α in degrees
•
•
•
•
During the period from ωt = α to π the input voltage vS and the input current iS are
both positive and the power flows from the supply to the load.
The converter is said to be operated in the rectification mode Controlled Rectifier
Operation for 0 < α < 900
During the period from ωt = π to (π+α), the input voltage vS is negative and the
input current iS is positive and the output power becomes negative and there will be
reverse power flow from the load circuit to the supply.
The converter is said to be operated in the inversion mode.
Line Commutated Inverter Operation for 900 < α < 1800
Two Quadrant Operation of a Single Phase Full Converter
Page 157
www.getmyuni.com
(ii) To Derive an Expression for the RMS Value of the Output Voltage
The rms value of the output voltage
is calculated as
VO( RMS ) =
2π

1  2
 ∫ vO .d (ω t ) 
2π  0

The single phase full converter gives two
output voltage pulses during the input supply
time period and hence the single phase full
converter is referred to as a two pulse converter.
The rms output voltage can be calculated as
VO( RMS ) =
π +α

2 
2
 ∫ vO .d (ω t ) 
2π  α

The single phase full converter gives two
output voltage pulses during the input supply
time period and hence the single phase full
converter is referred to as a two pulse converter.
The rms output voltage can be calculated as
VO( RMS )
π +α

2 
2
=
 ∫ vO .d (ω t ) 
2π  α

Page 158
www.getmyuni.com
VO( RMS )
Vm2 
=
 (ω t )
2π 
π +α
α
 sin 2ω t 
−

 2 
π +α
α



VO( RMS ) =
 sin 2 (π + α ) − sin 2α  
Vm2 
 (π + α − α ) − 

2π 
2

 
VO( RMS ) =
 sin ( 2π + 2α ) − sin 2α  
Vm2 
 (π ) − 
 ;
2π 
2

 
sin ( 2π + 2α ) = sin 2α
VO( RMS ) =
Vm2 
sin 2α − sin 2α  
(π ) − 


2π 
2


Vm2
Vm2 Vm
π
−
0
=
=
( )
2π
2
2
V
∴VO ( RMS ) = m = VS
2
Hence the rms output voltage is same as the
VO( RMS ) =
rms input supply voltage
Page 159
www.getmyuni.com
4.7.3 Thyristor Current Waveforms
Constant Load Current
iO=Ia
iO
Ia
ωt
α
iT1
π+α
Ia
Ia
ωt
& iT2
α
iT3
π+α 2π+α
Ia
ωt
& iT4
π+α 2π+α 3π+α
The rms thyristor current can be
calculated as
I O( RMS )
IT ( RMS ) =
2
The average thyristor current can be
calculated as
I O( dc )
IT ( Avg ) =
2
Page 160
www.getmyuni.com
4.8 Single Phase Dual Converter
Page 161
www.getmyuni.com
The average dc output voltage of converter 1 is
2V
Vdc1 = m cos α1
π
The average dc output voltage of converter 2 is
2V
Vdc 2 = m cos α 2
π
In the dual converter operation one
converter is operated as a controlled rectifier
withα < 900 & the second converter is
operated as a line commutated inverter
in the inversion mode with α > 900
∴
Vdc1 = −Vdc 2
Page 162
www.getmyuni.com
2Vm
π
cos α1 =
−2Vm
π
cos α 2 =
2Vm
π
( − cos α 2 )
cos α1 = − cos α 2
∴
or
cos α 2 = − cos α1 = cos (π − α1 )
α 2 = (π − α1 ) or
∴
(α1 + α 2 ) = π
radians
Which gives
α 2 = (π − α1 )
(i) To Obtain an Expression for the Instantaneous Circulating Current
• vO1 = Instantaneous o/p voltage of converter 1.
• vO2 = Instantaneous o/p voltage of converter 2.
• The circulating current ir can be determined by integrating the instantaneous voltage
difference (which is the voltage drop across the circulating current reactor Lr),
starting from ωt = (2π - α1).
• As the two average output voltages during the interval ωt = (π+α1) to (2π - α1) are
equal and opposite their contribution to the instantaneous circulating current ir is zero.
 ωt

 ∫ vr .d (ω t )  ; vr = ( vO1 − vO 2 )
 ( 2π −α1 )

As the o/p voltage vO 2 is negative
1
ir =
ω Lr
vr = ( vO1 + vO 2 )
 ωt

 ∫ ( vO1 + vO 2 ) .d (ω t )  ;
 ( 2π −α1 )

vO1 = −Vm sin ω t for ( 2π − α1 ) to ω t
1
ir =
ω Lr
∴
ωt
 ωt

 ∫ − sin ω t.d (ω t ) − ∫ sin ω t.d (ω t ) 
 ( 2π −α1 )

( 2π −α1 )
ir =
Vm
ω Lr
ir =
2Vm
( cos ω t − cos α 1 )
ω Lr
The instantaneous value of the circulating current
depends on the delay angle.
For trigger angle (delay angle) α1 = 0,
the magnitude of circulating current becomes min.
when ω t = nπ , n = 0, 2, 4,.... & magnitude becomes
max. when ω t = nπ , n = 1,3, 5,....
If the peak load current is I p , one of the
converters that controls the power flow
Page 163
www.getmyuni.com
10EC73
where
I p = I L( max) =
Vm
,
RL
&
ir( max) =
4Vm
= max. circulating current
ω Lr
The Dual Converter Can Be Operated In Two Different Modes Of Operation
• Non-circulating current (circulating current free) mode of operation.
• Circulating current mode of operation
Non-Circulating Current Mode of Operation
• In this mode only one converter is operated at a time.
• When converter 1 is ON, 0 < α1 < 900
• Vdc is positive and Idc is positive.
• When converter 2 is ON, 0 < α2 < 900
• Vdc is negative and Idc is negative.
Circulating Current Mode Of Operation
• In this mode, both the converters are switched ON and operated at the same time.
• The trigger angles α1 and α2 are adjusted such that (α1 + α2) = 1800 ; α2 = (1800 α1).
• When 0 <α1 <900, converter 1 operates as a controlled rectifier and converter 2
operates as an inverter with 900 <α2<1800.
• In this case Vdc and Idc, both are positive.
• When 900 <α1 <1800, converter 1 operates as an Inverter and converter 2 operated as
a controlled rectifier by adjusting its trigger angle α2 such that 0 <α2<900.
• In this case Vdc and Idc, both are negative.
Page 164
www.getmyuni.com
4.8.1 Four Quadrant Operation
Advantages of Circulating Current Mode of Operation
• The circulating current maintains continuous conduction of both the converters over
the complete control range, independent of the load.
• One converter always operates as a rectifier and the other converter operates as an
inverter, the power flow in either direction at any time is possible.
• As both the converters are in continuous conduction we obtain faster dynamic
response. i.e., the time response for changing from one quadrant operation to another
is faster.
Disadvantages of Circulating Current Mode of Operation
• There is always a circulating current flowing between the converters.
• When the load current falls to zero, there will be a circulating current flowing
between the converters so we need to connect circulating current reactors in order to
limit the peak circulating current to safe level.
• The converter thyristors should be rated to carry a peak current much greater than the
peak load current.
Recommended questions:
1. Give the classification of converters, based on: a) Quadrant operation b) Number of
current pulse c) supply input. Give examples in each case.
2. With neat circuit diagram and wave forms, explain the working of 1 phase HWR
using SCR for R-load. Derive the expressions for Vdc and Idc.
3. With a neat circuit diagram and waveforms, explain the working of 1-phase HCB for
R-load and R-L-load.
4. Determine the performance factors for 1-phase HCB circuit.
Page 165
www.getmyuni.com
5. With a neat circuit diagram and waveforms, explain the working of 1-phase FCB for
R and R-L-loads.
6. Determine the performance factors for 1-phase FCB circuit.
7. What is dual converter? Explain the working principle of 1-phase dual converter.
What are the modes of operation of dual converters? Explain briefly.
8. With a neat circuit diagram and waveforms explain the working of 3 phase HHCB
using SCRs. Obtain the expressions for Vdc and Idc.
9. With a neat circuit diagram and waveforms, explain the working of 3-phase HWR
using SCRs. Obtain the expressions for Vdc and Idc.
10. With a neat circuit diagram and waveforms, explain the working of 3 phase FCB
using SCRs. Obtain the expressions for Vdc and Idc.
11. Draw the circuit diagram of 3 phase dual converter. Explain its working?
12. List the applications of converters. Explain the effect of battery in the R-L-E load in
converters.
13. A single phase half wave converter is operated from a 120V, 60 Hz supply. If the
load resistive load is R=10Ω and the delay angle is α=π/3, determine a) the efficiency
b) the form factor c) the transformer utilization factor and d) the peak inverse voltage
(PIV) of thyristor T1
14. A single phase half wave converter is operated from a 120 V, 60 Hz supply and the
load resistive load is R=10. If the average output voltage is 25% of the maximum
possible average output voltage, calculate a) the delay angel b) the rms and average
output current c) the average and ram thyristor current and d) the input power factor.
15. A single half wave converter is operated from a 120 V, 60Hz supply and freewheeling
diodes is connected across the load. The load consists of series-connected resistance
R=10Ω, L=mH, and battery voltage E=20V. a) Express the instantaneous output
voltage in a Fourier series, and b) determine the rms value of the lowest order output
harmonic current.
16. A single phase semi-converter is operated from 120V, 60 Hz supply. The load current
with an average value of Ia is continuous with negligible ripple content. The turns
ratio of the transformer is unity. If the delay angle is A= π/3, calculate a) the
harmonic factor of input current b) the displacement factor and c) the input power
factor.
17. A single phase semi converter is operated from 120V, 60Hz supply. The load consists
of series connected resistance R=10Ω, L=5mH and battery voltage E=20V. a)
Express the instantaneous output voltage in a Fourier series, b) Determine the rms
value of the lowest order output harmonic current.
18. The three phase half wave converter is operated from a three phase Y connected
220V, 60Hz supply and freewheeling diodes is connected across the load. The load
consists of series connected resistance R=10Ω, L=5mH and battery voltage E=120V.
a) Express the instantaneous output voltage in a Fourier series and b) Determine the
rms value of the lowest order output harmonic current.
Page 166
www.getmyuni.com
13. without a series inductor? What is the ratio of peak resonant to load current for resonant
pulse commutation that would minimize the commutation losses?
14. Why does the commutation capacitor in a resonant pulse commutation get over
charged?
15. How is the voltage of the commutation capacitor reversed in a commutation circuit?
16. What is the type of a capacitor used in high frequency switching circuits?
Page 167