Embed Size (px)
Citation preview
1
EDUSAT PROGRAMME
LECTURE NOTES
ON
POWER ELECTRONICS
BY
PROF. M. MADHUSUDHAN RAO
DEPARTMENT OF ELECTRONICS &
COMMUNICATION ENGG.
M.S. RAMAIAH INSTITUTE OF TECHNOLOGY
BANGALORE – 560 054
2
AC VOLTAGE CONTROLLER CIRCUITS
(RMS VOLTAGE CONTROLLERS)
AC voltage controllers (ac line voltage controllers) are employed to vary the RMS value of the alternating voltage applied to a load circuit by introducing Thyristors between the load and a constant voltage ac source. The RMS value of alternating voltage applied to a load circuit is controlled by controlling the triggering angle of the Thyristors in the ac voltage controller circuits. In brief, an ac voltage controller is a type of thyristor power converter which is used to convert a fixed voltage, fixed frequency ac input supply to obtain a variable voltage ac output. The RMS value of the ac output voltage and the ac power flow to the load is controlled by varying (adjusting) the trigger angle ‘α’
ACVoltage
Controller
V0(RMS)
fS
Variable AC RMS O/P Voltage
ACInput
Voltagefs
Vs
fs
There are two different types of thyristor control used in practice to control the ac
power flow
• On-Off control • Phase control
These are the two ac output voltage control techniques. In On-Off control technique Thyristors are used as switches to connect the load circuit
to the ac supply (source) for a few cycles of the input ac supply and then to disconnect it for few input cycles. The Thyristors thus act as a high speed contactor (or high speed ac switch). PHASE CONTROL In phase control the Thyristors are used as switches to connect the load circuit to the input ac supply, for a part of every input cycle. That is the ac supply voltage is chopped using Thyristors during a part of each input cycle. The thyristor switch is turned on for a part of every half cycle, so that input supply voltage appears across the load and then turned off during the remaining part of input half cycle to disconnect the ac supply from the load. By controlling the phase angle or the trigger angle ‘α’ (delay angle), the output RMS voltage across the load can be controlled. The trigger delay angle ‘α’ is defined as the phase angle (the value of ωt) at which the thyristor turns on and the load current begins to flow.
Thyristor ac voltage controllers use ac line commutation or ac phase commutation. Thyristors in ac voltage controllers are line commutated (phase commutated) since the input supply is ac. When the input ac voltage reverses and becomes negative during the negative half cycle the current flowing through the conducting thyristor decreases and
3
falls to zero. Thus the ON thyristor naturally turns off, when the device current falls to zero.
Phase control Thyristors which are relatively inexpensive, converter grade Thyristors which are slower than fast switching inverter grade Thyristors are normally used.
For applications upto 400Hz, if Triacs are available to meet the voltage and current ratings of a particular application, Triacs are more commonly used. Due to ac line commutation or natural commutation, there is no need of extra commutation circuitry or components and the circuits for ac voltage controllers are very simple. Due to the nature of the output waveforms, the analysis, derivations of expressions for performance parameters are not simple, especially for the phase controlled ac voltage controllers with RL load. But however most of the practical loads are of the RL type and hence RL load should be considered in the analysis and design of ac voltage controller circuits. TYPE OF AC VOLTAGE CONTROLLERS The ac voltage controllers are classified into two types based on the type of input ac supply applied to the circuit.
• Single Phase AC Controllers. • Three Phase AC Controllers.
Single phase ac controllers operate with single phase ac supply voltage of 230V RMS at 50Hz in our country. Three phase ac controllers operate with 3 phase ac supply of 400V RMS at 50Hz supply frequency.
Each type of controller may be sub divided into • Uni-directional or half wave ac controller. • Bi-directional or full wave ac controller.
In brief different types of ac voltage controllers are • Single phase half wave ac voltage controller (uni-directional controller). • Single phase full wave ac voltage controller (bi-directional controller). • Three phase half wave ac voltage controller (uni-directional controller). • Three phase full wave ac voltage controller (bi-directional controller).
APPLICATIONS OF AC VOLTAGE CONTROLLERS
• Lighting / Illumination control in ac power circuits. • Induction heating. • Industrial heating & Domestic heating. • Transformer tap changing (on load transformer tap changing). • Speed control of induction motors (single phase and poly phase ac induction
motor control). • AC magnet controls.
PRINCIPLE OF ON-OFF CONTROL TECHNIQUE (INTEGRAL CYC LE CONTROL) The basic principle of on-off control technique is explained with reference to a single phase full wave ac voltage controller circuit shown below. The thyristor switches
1T and 2T are turned on by applying appropriate gate trigger pulses to connect the input
ac supply to the load for ‘n’ number of input cycles during the time interval ONt . The
4
thyristor switches 1T and 2T are turned off by blocking the gate trigger pulses for ‘m’
number of input cycles during the time interval OFFt . The ac controller ON time ONt usually consists of an integral number of input cycles.
LR R= = Load Resistance
Fig.: Single phase full wave AC voltage controller circuit
Vs
Vo
io
ig1
ig2
wt
wt
wt
wt
Gate pulse of T1
Gate pulse of T2
n m
Fig.: Waveforms
Example Referring to the waveforms of ON-OFF control technique in the above diagram,
n = Two input cycles. Thyristors are turned ON during ONt for two input cycles.
5
m= One input cycle. Thyristors are turned OFF during OFFt for one input cycle
Fig.: Power Factor
Thyristors are turned ON precisely at the zero voltage crossings of the input supply. The thyristor 1T is turned on at the beginning of each positive half cycle by
applying the gate trigger pulses to 1T as shown, during the ON time ONt . The load current flows in the positive direction, which is the downward direction as shown in the circuit diagram when 1T conducts. The thyristor 2T is turned on at the beginning of each
negative half cycle, by applying gating signal to the gate of 2T , during ONt . The load
current flows in the reverse direction, which is the upward direction when 2T conducts. Thus we obtain a bi-directional load current flow (alternating load current flow) in a ac voltage controller circuit, by triggering the thyristors alternately.
This type of control is used in applications which have high mechanical inertia and high thermal time constant (Industrial heating and speed control of ac motors). Due to zero voltage and zero current switching of Thyristors, the harmonics generated by switching actions are reduced.
For a sine wave input supply voltage,
sin 2 sins m Sv V t V tω ω= =
SV = RMS value of input ac supply = 2mV
= RMS phase supply voltage.
If the input ac supply is connected to load for ‘n’ number of input cycles and disconnected for ‘m’ number of input cycles, then
,ON OFFt n T t m T= × = ×
Where 1
Tf
= = input cycle time (time period) and
f = input supply frequency.
ONt = controller on time = n T× .
OFFt = controller off time = m T× .
OT = Output time period = ( ) ( )ON OFFt t nT mT+ = + .
6
We can show that,
Output RMS voltage ( ) ( )ON ON
SO RMS i RMSO O
t tV V V
T T= =
Where ( )i RMSV is the RMS input supply voltage = SV .
TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF OUTPUT VOLTAGE, FOR ON-OFF CONTROL METHOD.
Output RMS voltage ( ) ( )2 2
0
1.
ONt
mO RMSO t
V V Sin t d tT
ω
ω
ω ωω =
= ∫
( ) ( )2
2
0
.ONt
mO RMS
O
VV Sin t d t
T
ω
ω ωω
= ∫
Substituting for 2 1 2
2
CosSin
θθ −=
( ) ( )2
0
1 2
2
ONt
mO RMS
O
V Cos tV d t
T
ω ω ωω
− = ∫
( ) ( ) ( )2
0 0
2 .2
ON ONt t
mO RMS
O
VV d t Cos t d t
T
ω ω
ω ω ωω
= −
∫ ∫
( ) ( )2
0 0
222
ON ONt tm
O RMSO
V Sin tV t
T
ω ωωωω
= −
( ) ( )2 sin 2 sin 0
02 2
m ONONO RMS
O
V tV t
T
ωωω
− = − −
Now ONt = An integral number of input cycles; Hence
,2 ,3 ,4 ,5 ,.....ONt T T T T T= & 2 ,4 ,6 ,8 ,10 ,......ONtω π π π π π= Where T is the input supply time period (T = input cycle time period). Thus we note that sin 2 0ONtω =
( )
2
2 2m ON m ON
O RMSO O
V t V tV
T T
ωω
= =
7
( ) ( )ON ON
SO RMS i RMSO O
t tV V V
T T= =
Where ( ) 2m
Si RMS
VV V= = = RMS value of input supply voltage;
( )
ON ON
O ON OFF
t t nT nk
T t t nT mT n m= = = =
+ + += duty cycle (d).
( ) ( )S SO RMS
nV V V k
m n= =
+
PERFORMANCE PARAMETERS OF AC VOLTAGE CONTROLLERS
• RMS Output (Load) Voltage
( ) ( ) ( )122
2 2
0
sin .2 mO RMS
nV V t d t
n m
π
ω ωπ
= +
∫
( ) ( ) ( )2m
SO RMS i RMS
V nV V k V k
m n= = =
+
( ) ( ) SO RMS i RMSV V k V k= =
Where ( )S i RMSV V= = RMS value of input supply voltage.
• Duty Cycle
( ) ( )ON ON
O ON OFF
t t nTk
T t t m n T= = =
+ +
Where, ( )n
km n
=+
= duty cycle (d).
• RMS Load Current
( )( ) ( )O RMS O RMS
O RMSL
V VI
Z R= = ; for a resistive load LZ R= .
• Output AC (Load) Power
( )2
O LO RMSP I R= ×
8
• Input Power Factor
output load power
input supply volt amperesO O
S S
P PPF
VA V I= = =
( )
( ) ( )
2LO RMS
i RMS in RMS
I RPF
V I
×=
×; ( )S in RMSI I= = RMS input supply current.
The input supply current is same as the load current in O LI I I= =
Hence, RMS supply current = RMS load current; ( ) ( )in RMS O RMSI I= .
( )
( ) ( )
( )
( )
( )
( )
2LO RMS O RMS i RMS
i RMS in RMS i RMS i RMS
I R V V kPF k
V I V V
×= = = =
×
nPF k
m n= =
+
• The Average Current of Thyristor ( )T AvgI
( ) ( ) ( )0
sin .2 mT Avg
nI I t d t
m n
π
ω ωπ
=+ ∫
( ) ( ) ( )0
sin .2
mT Avg
nII t d t
m n
π
ω ωπ
=+ ∫
( ) ( ) 0
cos2
mT Avg
nII t
m n
π
ωπ
= − +
( ) ( ) [ ]cos cos02
mT Avg
nII
m nπ
π= − +
+
0 π 2π 3π ωt
Im
nm
iT
Waveform of Thyristor Current
9
( ) ( ) ( )1 12
mT Avg
nII
m nπ= − − + +
( ) ( ) [ ]22 mT Avg
nI I
m nπ=
+
( ) ( ).m m
T Avg
I n k II
m nπ π= =
+
( ) ( )duty cycle ON
ON OFF
t nk
t t n m= = =
+ +
( ) ( ).m m
T Avg
I n k II
m nπ π= =
+,
Where mm
L
VI
R= = maximum or peak thyristor current.
• RMS Current of Thyristor ( )T RMSI
( ) ( ) ( )12
2 2
0
sin .2 mT RMS
nI I t d t
n m
π
ω ωπ
= +
∫
( ) ( ) ( )122
2
0
sin .2
mT RMS
nII t d t
n m
π
ω ωπ
= +
∫
( ) ( )( ) ( )
122
0
1 cos 2
2 2m
T RMS
tnII d t
n m
π ωω
π −
= + ∫
( ) ( ) ( ) ( )122
0 0
cos 2 .4
mT RMS
nII d t t d t
n m
π π
ω ω ωπ
= − +
∫ ∫
( ) ( ) ( )122
0 0
sin 224
mT RMS
nI tI t
n m
π πωωπ
= − +
( ) ( ) ( )122 sin 2 sin 0
04 2
mT RMS
nII
n m
πππ
− = − − +
10
( ) ( ) 122
0 04
mT RMS
nII
n mπ
π
= − − +
( ) ( ) ( )
1 12 22 2
4 4m m
T RMS
nI nII
n m n m
ππ
= = + +
( ) ( )2 2m m
T RMS
I InI k
m n= =
+
( ) 2m
T RMS
II k=
PROBLEM
1. A single phase full wave ac voltage controller working on ON-OFF control technique has supply voltage of 230V, RMS 50Hz, load = 50Ω. The controller is ON for 30 cycles and off for 40 cycles. Calculate
• ON & OFF time intervals. • RMS output voltage. • Input P.F. • Average and RMS thyristor currents.
( ) 230in RMSV V= , 2 230 325.269mV V= × = V, 325.269mV V= ,
1 1
0.02sec50
Tf Hz
= = = , 20T ms= .
n= number of input cycles during which controller is ON; 30n = . m= number of input cycles during which controller is OFF; 40m= .
30 20 600 0.6secONt n T ms ms= × = × = =
0.6secONt n T= × = = controller ON time.
40 20 800 0.8secOFFt m T ms ms= × = × = =
0.8secOFFt m T= × = = controller OFF time.
Duty cycle ( ) ( )
300.4285
40 30
nk
m n= = =
+ +
RMS output voltage
( ) ( ) ( )O RMS i RMS
nV V
m n= ×
+
11
( ) ( )30 3
230 23030 40 7O RMSV V= × =
+
( ) 230 0.42857 230 0.65465O RMSV V= = ×
( ) 150.570O RMSV V=
( )( ) ( ) 150.570
3.011450
O RMS O RMS
O RMSL
V V VI A
Z R= = = =
Ω
( )2 23.0114 50 453.426498O LO RMSP I R W= × = × =
Input Power Factor .P F k=
( )
300.4285
70
nPF
m n= = =
+
0.654653PF = Average Thyristor Current Rating
( )m m
T Avg
I k InI
m nπ π× = × = +
where 2 230 325.269
50 50m
mL
VI
R
×= = =
6.505382mI A= = Peak (maximum) thyristor current.
( )6.505382 3
7T AvgIπ
= ×
( ) 0.88745T AvgI A=
RMS Current Rating of Thyristor
( ) ( )6.505382 3
2 2 2 7m m
T RMS
I InI k
m n= = = ×
+
( ) 2.129386T RMSI A=
12
PRINCIPLE OF AC PHASE CONTROL The basic principle of ac phase control technique is explained with reference to a single phase half wave ac voltage controller (unidirectional controller) circuit shown in the below figure. The half wave ac controller uses one thyristor and one diode connected in parallel across each other in opposite direction that is anode of thyristor 1T is connected to the
cathode of diode 1D and the cathode of 1T is connected to the anode of 1D . The output voltage across the load resistor ‘R’ and hence the ac power flow to the load is controlled by varying the trigger angle ‘α’.
The trigger angle or the delay angle ‘α’ refers to the value of tω or the instant at which the thyristor 1T is triggered to turn it ON, by applying a suitable gate trigger pulse between the gate and cathode lead.
The thyristor 1T is forward biased during the positive half cycle of input ac supply. It can be triggered and made to conduct by applying a suitable gate trigger pulse only during the positive half cycle of input supply. When 1T is triggered it conducts and the
load current flows through the thyristor 1T , the load and through the transformer secondary winding.
By assuming 1T as an ideal thyristor switch it can be considered as a closed switch
when it is ON during the period tω α= to π radians. The output voltage across the load follows the input supply voltage when the thyristor 1T is turned-on and when it conducts
from tω α= to π radians. When the input supply voltage decreases to zero at tω π= , for a resistive load the load current also falls to zero at tω π= and hence the thyristor 1T
turns off at tω π= . Between the time period tω π= to 2π , when the supply voltage reverses and becomes negative the diode 1D becomes forward biased and hence turns ON
and conducts. The load current flows in the opposite direction during tω π= to 2π radians when 1D is ON and the output voltage follows the negative half cycle of input supply.
Fig.: Halfwave AC phase controller (Unidirectional Controller)
13
Equations Input AC Supply Voltage across the Transformer Secondary Winding. sins mv V tω=
( ) 2m
S in RMS
VV V= = = RMS value of secondary supply voltage.
Output Load Voltage 0o Lv v= = ; for 0tω = to α
sino L mv v V tω= = ; for tω α= to 2π .
Output Load Current
sino m
o LL L
v V ti i
R R
ω= = = ; for tω α= to 2π .
0o Li i= = ; for 0tω = to α . TO DERIVE AN EXPRESSION FOR RMS OUTPUT VOLTAGE ( )O RMSV
( ) ( )2
2 21sin .
2 mO RMSV V t d tπ
α
ω ωπ
= ∫
( ) ( )22 1 cos 2
.2 2
mO RMS
V tV d t
π
α
ω ωπ − = ∫
14
( ) ( ) ( )22
1 cos 2 .4
mO RMS
VV t d t
π
α
ω ωπ
= − ∫
( ) ( )2 2
cos 2 .2
mO RMS
VV d t t d t
π π
α α
ω ω ωπ
= −
∫ ∫
( ) ( )2 2sin 2
22m
O RMS
V tV t
π π
α α
ωωπ
= −
( ) ( )2sin 2
222
mO RMS
V tV
π
α
ωπ απ
= − −
( ) ( ) sin 4 sin 22 ;sin 4 0
2 22m
O RMS
VV
π απ α ππ
= − − − =
( ) ( ) sin 22
22m
O RMS
VV
απ απ
= − +
( ) ( ) sin 22
22 2m
O RMS
VV
απ απ
= − +
( ) ( )1 sin 22
2 22m
O RMS
VV
απ απ = − +
( ) ( ) ( )1 sin 22
2 2O RMS i RMSV Vαπ α
π = − +
( ) ( )1 sin 22
2 2SO RMSV Vαπ α
π = − +
Where, ( ) 2m
Si RMS
VV V= = = RMS value of input supply voltage (across the
transformer secondary winding).
Note: Output RMS voltage across the load is controlled by changing ' 'α as indicated by the expression for ( )O RMSV
15
PLOT OF ( )O RMSV VERSUS TRIGGER ANGLE α FOR A SINGLE PHASE HALF-
WAVE AC VOLTAGE CONTROLLER (UNIDIRECTIONAL CONTROLL ER)
( ) ( )1 sin 22
2 22m
O RMS
VV
απ απ = − +
( ) ( )1 sin 22
2 2SO RMSV Vαπ α
π = − +
By using the expression for ( )O RMSV we can obtain the control characteristics,
which is the plot of RMS output voltage ( )O RMSV versus the trigger angle α . A typical
control characteristic of single phase half-wave phase controlled ac voltage controller is as shown below
Trigger angle α in degrees
Trigger angle α in radians
( )O RMSV
0 0 2m
S
VV =
030 6π ( )1; 6
π 0.992765 SV
060 3π ( )2; 6
π 0.949868 SV
090 2π ( )3; 6
π 0.866025 SV
0120 23
π ( )4; 6π 0.77314 SV
0150 56
π ( )5; 6π 0.717228 SV
0180 π ( )6; 6π 0.707106 SV
VO(RMS)
Trigger angle in degreesα
0 60 120 180
100% VS
20% VS
60% VS
70.7% VS
Fig.: Control characteristics of single phase half-wave phase controlled ac voltage controller
16
Note: We can observe from the control characteristics and the table given above that the range of RMS output voltage control is from 100% of SV to 70.7% of SV when we vary
the trigger angle α from zero to 180 degrees. Thus the half wave ac controller has the draw back of limited range RMS output voltage control. TO CALCULATE THE AVERAGE VALUE (DC VALUE) OF OUTPUT VOLTAGE
( ) ( )21
sin .2 mO dcV V t d t
π
α
ω ωπ
= ∫
( ) ( )2
sin .2
mO dc
VV t d t
π
α
ω ωπ
= ∫
( )
2
cos2
mO dc
VV t
π
α
ωπ
= −
( ) [ ]cos 2 cos2
mO dc
VV π α
π= − + ; cos 2 1π =
[ ]cos 12
mdc
VV α
π= − ; 2m SV V=
Hence ( )2cos 1
2S
dc
VV α
π= −
When ' 'α is varied from 0 to π . dcV varies from 0 to mV
π−
DISADVANTAGES OF SINGLE PHASE HALF WAVE AC VOLTAGE CONTROLLER.
• The output load voltage has a DC component because the two halves of the output voltage waveform are not symmetrical with respect to ‘0’ level. The input supply current waveform also has a DC component (average value) which can result in the problem of core saturation of the input supply transformer.
• The half wave ac voltage controller using a single thyristor and a single diode provides control on the thyristor only in one half cycle of the input supply. Hence ac power flow to the load can be controlled only in one half cycle.
• Half wave ac voltage controller gives limited range of RMS output voltage control. Because the RMS value of ac output voltage can be varied from a maximum of 100% of SV at a trigger angle 0α = to a low of 70.7% of SV at
Radiansα π= .
These drawbacks of single phase half wave ac voltage controller can be over come by using a single phase full wave ac voltage controller.
17
APPLICATIONS OF RMS VOLTAGE CONTROLLER • Speed control of induction motor (polyphase ac induction motor). • Heater control circuits (industrial heating). • Welding power control. • Induction heating. • On load transformer tap changing. • Lighting control in ac circuits. • Ac magnet controls.
Problem
1. A single phase half-wave ac voltage controller has a load resistance 50R = Ω , input ac supply voltage is 230V RMS at 50Hz. The input supply transformer has a turns ratio of 1:1. If the thyristor 1T is triggered at 060α = . Calculate
• RMS output voltage. • Output power. • RMS load current and average load current. • Input power factor. • Average and RMS thyristor current.
Given,
0
S
230 , primary supply voltage.
Input supply frequency = 50Hz.
50
60 radians.3
V RMS secondary voltage.
p
L
V V RMS
f
R
πα
=
== Ω
= =
=
1
11
p p
S S
V N
V N= = =
Therefore 230p SV V V= =
Where, pN = Number of turns in the primary winding.
SN = Number of turns in the secondary winding.
18
• RMS Value of Output (Load) Voltage ( )O RMSV
( ) ( )2
2 21sin .
2 mO RMSV V t d tπ
α
ω ωπ
= ∫
We have obtained the expression for ( )O RMSV as
( ) ( )1 sin 22
2 2SO RMSV Vαπ α
π= − +
( )
01 sin120230 2
2 3 2O RMSVππ
π = − +
( ) [ ]1230 5.669 230 0.94986
2O RMSVπ
= = ×
( ) 218.4696 218.47 O RMSV V V= ≈
• RMS Load Current ( )O RMSI
( )( ) 218.46966
4.36939 50
O RMS
O RMSL
VI Amps
R= = =
• Output Load Power OP
( ) ( )22 4.36939 50 954.5799 O LO RMSP I R Watts= × = × =
0.9545799 OP KW=
• Input Power Factor
O
S S
PPF
V I=
×
SV = RMS secondary supply voltage = 230V.
SI = RMS secondary supply current = RMS load current. ( ) 4.36939 S O RMSI I Amps∴ = =
( )
954.5799 W0.9498
230 4.36939 WPF∴ = =
×
19
• Average Output (Load) Voltage
( ) ( )21
sin .2 mO dcV V t d t
π
α
ω ωπ
= ∫
We have obtained the expression for the average / DC output voltage as,
( ) [ ]cos 12
mO dc
VV α
π= −
( ) ( ) [ ]02 230 325.2691193cos 60 1 0.5 1
2 2O dcVπ π
× = − = −
( ) [ ]325.26911930.5 25.88409 Volts
2O dcVπ
= − = −
• Average DC Load Current
( )( ) 25.884094
0.51768 Amps50
O dc
O dcL
VI
R
−= = = −
• Average & RMS Thyristor Currents
Im
iT1
π 2πα (2 + )π α
3π
α αωt
Fig.: Thyristor Current Waveform Referring to the thyristor current waveform of a single phase half-wave ac voltage
controller circuit, we can calculate the average thyristor current ( )T AvgI as
( ) ( )1sin .
2 mT AvgI I t d tπ
α
ω ωπ
= ∫
( ) ( )sin .2
mT Avg
II t d t
π
α
ω ωπ
= ∫
20
( ) ( )cos2
mT Avg
II t
π
α
ωπ
= −
( ) ( )cos cos2
mT Avg
II π α
π= − +
( ) [ ]1 cos2
mT Avg
II α
π= +
Where, mm
L
VI
R= = Peak thyristor current = Peak load current.
2 230
50mI×=
6.505382 AmpsmI =
( ) [ ]1 cos2
mT Avg
L
VI
Rα
π= +
( ) ( )02 2301 cos 60
2 50T AvgIπ
× = + ×
( ) [ ]2 2301 0.5
100T AvgIπ
×= +
( ) 1.5530 AmpsT AvgI =
• RMS thyristor current ( )T RMSI can be calculated by using the expression
( ) ( )2 21sin .
2 mT RMSI I t d tπ
α
ω ωπ
= ∫
( )( ) ( )
2 1 cos 2.
2 2m
T RMS
tII d t
π
α
ωω
π −
= ∫
( ) ( ) ( )2
cos 2 .4
mT RMS
II d t t d t
π π
α α
ω ω ωπ
= − ∫ ∫
( ) ( )1 sin 224mT RMS
tI I t
π π
α α
ωωπ = −
21
( ) ( )1 sin 2 sin 2
4 2mT RMSI Iπ απ α
π − = − −
( ) ( )1 sin 2
4 2mT RMSI Iαπ α
π = − +
( ) ( )1 sin 2
2 22m
T RMS
II
απ απ = − +
( )( )0sin 1206.50538 1
2 3 22T RMSI
πππ
= − +
( )1 2 0.8660254
4.62 3 2T RMSI
ππ = +
( ) 4.6 0.6342 2.91746T RMSI A= × =
( ) 2.91746 AmpsT RMSI =
SINGLE PHASE FULL WAVE AC VOLTAGE CONTROLLER (AC REGULATOR) OR RMS VOLTAGE CONTROLLER WITH RESISTIVE LOAD Single phase full wave ac voltage controller circuit using two SCRs or a single
triac is generally used in most of the ac control applications. The ac power flow to the
load can be controlled in both the half cycles by varying the trigger angle' 'α .
The RMS value of load voltage can be varied by varying the trigger angle ' 'α .
The input supply current is alternating in the case of a full wave ac voltage controller and
due to the symmetrical nature of the input supply current waveform there is no dc
component of input supply current i.e., the average value of the input supply current is
zero.
A single phase full wave ac voltage controller with a resistive load is shown in the
figure below. It is possible to control the ac power flow to the load in both the half cycles
by adjusting the trigger angle' 'α . Hence the full wave ac voltage controller is also
referred to as to a bi-directional controller.
22
Fig.: Single phase full wave ac voltage controller (Bi-directional Controller) using
SCRs
The thyristor 1T is forward biased during the positive half cycle of the input
supply voltage. The thyristor 1T is triggered at a delay angle of ' 'α ( )0 radiansα π≤ ≤ .
Considering the ON thyristor 1T as an ideal closed switch the input supply voltage
appears across the load resistor LR and the output voltage O Sv v= during tω α= to
π radians. The load current flows through the ON thyristor 1T and through the load
resistor LR in the downward direction during the conduction time of 1T from tω α= to
π radians. At tω π= , when the input voltage falls to zero the thyristor current (which is
flowing through the load resistor LR ) falls to zero and hence 1T naturally turns off . No
current flows in the circuit during tω π= to ( )π α+ .
The thyristor 2T is forward biased during the negative cycle of input supply and
when thyristor 2T is triggered at a delay angle ( )π α+ , the output voltage follows the
negative halfcycle of input from ( )tω π α= + to 2π . When 2T is ON, the load current
flows in the reverse direction (upward direction) through 2T during ( )tω π α= + to
2π radians. The time interval (spacing) between the gate trigger pulses of 1T and 2T is
kept at π radians or 1800. At 2tω π= the input supply voltage falls to zero and hence the load current also falls to zero and thyristor 2T turn off naturally.
Instead of using two SCR’s in parallel, a Triac can be used for full wave ac voltage
control.
Fig.: Single phase full wave ac voltage controller (Bi-directional Controller) using
TRIAC
23
Fig: Waveforms of single phase full wave ac voltage controller EQUATIONS
Input supply voltage sin 2 sinS m Sv V t V tω ω= = ;
Output voltage across the load resistor LR ;
sinO L mv v V tω= = ;
for to tω α π= and ( ) to 2tω π α π= +
Output load current
sinsinO m
O mL L
v V ti I t
R R
ω ω= = = ;
for to tω α π= and ( ) to 2tω π α π= +
TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF OUTPUT (LOAD) VOLTAGE
The RMS value of output voltage (load voltage) can be found using the expression
( ) ( ) ( )2
2 2 2
0
1
2 LO RMS L RMSV V v d tπ
ωπ
= = ∫ ;
24
For a full wave ac voltage controller, we can see that the two half cycles of output voltage waveforms are symmetrical and the output pulse time period (or output pulse repetition time) is π radians. Hence we can also calculate the RMS output voltage by using the expression given below.
( )2 2 2
0
1sin .mL RMSV V t d t
π
ω ωπ
= ∫
( ) ( )2
2 2
0
1.
2 LL RMSV v d tπ
ωπ
= ∫ ;
sinL O mv v V tω= = ; For to tω α π= and ( ) to 2tω π α π= +
Hence,
( ) ( ) ( ) ( ) ( )2
2 22 1sin sin
2 m mL RMSV V t d t V t d tπ π
α π α
ω ω ω ωπ +
= +
∫ ∫
( ) ( )2
2 2 2 21sin . sin .
2 m mV t d t V t d tπ π
α π α
ω ω ω ωπ +
= +
∫ ∫
( ) ( )22 1 cos 2 1 cos 2
2 2 2mV t t
d t d tπ π
α π α
ω ωω ωπ +
− −= + ∫ ∫
( ) ( ) ( ) ( )2 22
cos 2 . cos 2 .2 2
mVd t t d t d t t d t
π π π π
α α π α π α
ω ω ω ω ω ωπ + +
= − + − ×
∫ ∫ ∫ ∫
( ) ( )2 22 sin 2 sin 2
4 2 2mV t t
t tπ π π π
α π α α π α
ω ωω ωπ + +
= + − −
( ) ( ) ( ) ( )( )2 1 1
sin 2 sin 2 sin 4 sin 24 2 2
mV π α π α π α π π απ = − + − − − − − +
( ) ( ) ( )( )2 1 1
2 0 sin 2 0 sin 24 2 2
mV π α α π απ = − − − − − +
( ) ( )2 sin 2sin 22
4 2 2mV π ααπ απ
+ = − + +
( ) ( )2 sin 2 2sin 22
4 2 2mV π ααπ απ
+ = − + +
25
( ) ( )2 sin 2 1
2 sin 2 .cos 2 cos 2 .sin 24 2 2
mV απ α π α π απ = − + + +
sin 2 0 & cos 2 1π π= =
Therefore,
( ) ( )2
2 sin 2 sin 22
4 2 2m
L RMS
VV
α απ απ = − + +
( )2
2 sin 24
mV π α απ
= − +
( ) ( )2
2 2 2 sin 24
mL RMS
VV π α α
π= − +
Taking the square root, we get
( ) ( )2 2 sin 22
mL RMS
VV π α α
π= − +
( ) ( )2 2 sin 22 2
mL RMS
VV π α α
π= − +
( ) ( )12 2 sin 2
22m
L RMS
VV π α α
π= − +
( ) ( )1 sin 22
2 22m
L RMS
VV
απ απ = − +
( ) ( )1 sin 2
22m
L RMS
VV
απ απ = − +
( ) ( ) ( )1 sin 2
2L RMS i RMSV Vαπ α
π = − +
( ) ( )1 sin 2
2SL RMSV Vαπ α
π = − +
Maximum RMS voltage will be applied to the load when0α = , in that case the
full sine wave appears across the load. RMS load voltage will be the same as the RMS
supply voltage 2mV= . When α is increased the RMS load voltage decreases.
26
( ) ( )0
1 sin 2 00
22m
L RMS
VV
α
ππ=
× = − +
( ) ( )0
1 0
22m
L RMS
VV
α
ππ=
= +
( ) ( )0 2
mSL RMS i RMS
VV V V
α =
= = =
The output control characteristic for a single phase full wave ac voltage controller
with resistive load can be obtained by plotting the equation for ( )O RMSV
CONTROL CHARACTERISTIC OF SINGLE PHASE FULL-WAVE AC VOLTAGE CONTROLLER WITH RESISTIVE LOAD The control characteristic is the plot of RMS output voltage ( )O RMSV versus the
trigger angle α ; which can be obtained by using the expression for the RMS output voltage of a full-wave ac controller with resistive load.
( ) ( )1 sin 2
2SO RMSV Vαπ α
π = − +
;
Where 2m
S
VV = =RMS value of input supply voltage
Trigger angle α
in degrees Trigger angle α
in radians ( )O RMSV %
0 0 SV 100% SV
030 6π ( )1; 6
π 0.985477 SV 98.54% SV
060 3π ( )2; 6
π 0.896938 SV 89.69% SV
090 2π ( )3; 6
π 0.7071 SV 70.7% SV
0120 23
π ( )4; 6π 0.44215 SV 44.21% SV
0150 56
π ( )5; 6π 0.1698 SV 16.98% SV
0180 π ( )6; 6π 0 SV 0 SV
27
VO(RMS)
Trigger angle in degreesα
0 60 120 180
VS
0.2 VS
0.6VS
We can notice from the figure, that we obtain a much better output control characteristic by using a single phase full wave ac voltage controller. The RMS output voltage can be varied from a maximum of 100% SV at 0α = to a minimum of ‘0’ at
0180α = . Thus we get a full range output voltage control by using a single phase full wave ac voltage controller. Need For Isolation In the single phase full wave ac voltage controller circuit using two SCRs or Thyristors 1T and 2T in parallel, the gating circuits (gate trigger pulse generating circuits)
of Thyristors 1T and 2T must be isolated. Figure shows a pulse transformer with two
separate windings to provide isolation between the gating signals of 1T and 2T .
G1
K1
G2
K2
Gate
TriggerPulse
Generator
Fig.: Pulse Transformer SINGLE PHASE FULL-WAVE AC VOLTAGE CONTROLLER WITH COMMON CATHODE
It is possible to design a single phase full wave ac controller with a common cathode configuration by having a common cathode point for 1T and 2T & by adding two diodes in a full wave ac controller circuit as shown in the figure below
28
Fig.: Single phase full wave ac controller with common cathode
(Bidirectional controller in common cathode configuration)
Thyristor 1T and diode 1D are forward biased during the positive half cycle of
input supply. When thyristor 1T is triggered at a delay angle α , Thyristor 1T and diode
1D conduct together from tω α= to π during the positive half cycle.
The thyristor 2T and diode 2D are forward biased during the negative half cycle
of input supply, when trigged at a delay angle α , thyristor 2T and diode 2D conduct
together during the negative half cycle from ( )tω π α= + to 2π .
In this circuit as there is one single common cathode point, routing of the gate trigger pulses to the thyristor gates of 1T and 2T is simpler and only one isolation circuit is required. But due to the need of two power diodes the costs of the devices increase. As there are two power devices conducting at the same time the voltage drop across the ON devices increases and the ON state conducting losses of devices increase and hence the efficiency decreases. SINGLE PHASE FULL WAVE AC VOLTAGE CONTROLLER USING A SINGLE THYRISTOR
RL
T1
ACSupply
-
D1
D4
D3
D2
+
29
A single phase full wave ac controller can also be implemented with one thyristor and four diodes connected in a full wave bridge configuration as shown in the above figure. The four diodes act as a bridge full wave rectifier. The voltage across the thyristor
1T and current through thyristor 1T are always unidirectional. When 1T is triggered at
tω α= , during the positive half cycle ( )0 α π≤ ≤ , the load current flows through 1D , 1T ,
diode 2D and through the load. With a resistive load, the thyristor current (flowing
through the ON thyristor 1T ) , the load current falls to zero at tω π= , when the input
supply voltage decreases to zero at tω π= , the thyristor naturally turns OFF. In the negative half cycle, diodes 3 4&D D are forward biased during
to 2tω π π= radians. When 1T is triggered at ( )tω π α= + , the load current flows in the
opposite direction (upward direction) through the load, through 3D , 1T and 4D . Thus 3D ,
4D and 1T conduct together during the negative half cycle to supply the load power. When
the input supply voltage becomes zero at 2tω π= , the thyristor current (load current) falls to zero at 2tω π= and the thyristor 1T naturally turns OFF. The waveforms and the expression for the RMS output voltage are the same as discussed earlier for the single phase full wave ac controller. But however if there is a large inductance in the load circuit, thyristor 1T may not be turned OFF at the zero crossing points, in every half cycle of input voltage and this may result in a loss of output control. This would require detection of the zero crossing of the load current waveform in order to ensure guaranteed turn off of the conducting thyristor before triggering the thyristor in the next half cycle, so that we gain control on the output voltage. In this full wave ac controller circuit using a single thyristor, as there are three power devices conducting together at the same time there is more conduction voltage drop and an increase in the ON state conduction losses and hence efficiency is also reduced. The diode bridge rectifier and thyristor (or a power transistor) act together as a
bidirectional switch which is commercially available as a single device module and it has
relatively low ON state conduction loss. It can be used for bidirectional load current
control and for controlling the RMS output voltage.
SINGLE PHASE FULL WAVE AC VOLTAGE CONTROLLER (BIDIRECTIONAL CONTROLLER) WITH RL LOAD In this section we will discuss the operation and performance of a single phase full
wave ac voltage controller with RL load. In practice most of the loads are of RL type. For
example if we consider a single phase full wave ac voltage controller controlling the
speed of a single phase ac induction motor, the load which is the induction motor winding
is an RL type of load, where R represents the motor winding resistance and L represents
the motor winding inductance.
30
A single phase full wave ac voltage controller circuit (bidirectional controller)
with an RL load using two thyristors 1T and 2T ( 1T and 2T are two SCRs) connected in
parallel is shown in the figure below. In place of two thyristors a single Triac can be used
to implement a full wave ac controller, if a suitable Traic is available for the desired RMS
load current and the RMS output voltage ratings.
Fig: Single phase full wave ac voltage controller with RL load The thyristor 1T is forward biased during the positive half cycle of input supply.
Let us assume that 1T is triggered at tω α= , by applying a suitable gate trigger pulse to
1T during the positive half cycle of input supply. The output voltage across the load
follows the input supply voltage when 1T is ON. The load current Oi flows through the
thyristor 1T and through the load in the downward direction. This load current pulse
flowing through 1T can be considered as the positive current pulse. Due to the inductance
in the load, the load current Oi flowing through 1T would not fall to zero at tω π= , when
the input supply voltage starts to become negative.
The thyristor 1T will continue to conduct the load current until all the inductive
energy stored in the load inductor L is completely utilized and the load current through 1T
falls to zero at tω β= , where β is referred to as the Extinction angle, (the value of tω )
at which the load current falls to zero. The extinction angle β is measured from the point
of the beginning of the positive half cycle of input supply to the point where the load
current falls to zero.
31
The thyristor 1T conducts from tω α= to β . The conduction angle of 1T is
( )δ β α= − , which depends on the delay angle α and the load impedance angle φ . The
waveforms of the input supply voltage, the gate trigger pulses of 1T and 2T , the thyristor
current, the load current and the load voltage waveforms appear as shown in the figure
below.
Fig.: Input supply voltage & Thyristor current waveforms β is the extinction angle which depends upon the load inductance value.
Fig.: Gating Signals
32
Waveforms of single phase full wave ac voltage controller with RL load for α φ> .
Discontinuous load current operation occurs for α φ> and ( )β π α< + ;
i.e., ( )β α π− < , conduction angle π< .
Fig.: Waveforms of Input supply voltage, Load Current, Load Voltage and
Thyristor Voltage across 1T Note
• The RMS value of the output voltage and the load current may be varied by varying the trigger angle α .
• This circuit, AC RMS voltage controller can be used to regulate the RMS voltage across the terminals of an ac motor (induction motor). It can be used to control the temperature of a furnace by varying the RMS output voltage.
33
• For very large load inductance ‘L’ the SCR may fail to commutate, after it is triggered and the load voltage will be a full sine wave (similar to the applied input supply voltage and the output control will be lost) as long as the gating signals are applied to the thyristors 1T and 2T . The load current waveform will appear as a full continuous sine wave and the load current waveform lags behind the output sine wave by the load power factor angle φ.
TO DERIVE AN EXPRESSION FOR THE OUTPUT (INDUCTIVE L OAD) CURRENT, DURING to tω α β= WHEN THYRISTOR 1T CONDUCTS Considering sinusoidal input supply voltage we can write the expression for the supply voltage as sinS mv V tω= = instantaneous value of the input supply voltage. Let us assume that the thyristor 1T is triggered by applying the gating signal to 1T
at tω α= . The load current which flows through the thyristor 1T during tω α= to β can be found from the equation
sinOO m
diL Ri V t
dtω + =
;
The solution of the above differential equation gives the general expression for the output load current which is of the form
( ) 1sint
mO
Vi t A e
Zτω φ−
= − + ;
Where 2m SV V= = maximum or peak value of input supply voltage.
( )22Z R Lω= + = Load impedance.
1tanL
R
ωφ − =
= Load impedance angle (power factor angle of load).
L
Rτ = = Load circuit time constant.
Therefore the general expression for the output load current is given by the equation
( )
1sinR
tm L
O
Vi t A e
Zω φ
−
= − + ;
34
The value of the constant 1A can be determined from the initial condition. i.e.
initial value of load current 0Oi = , at tω α= . Hence from the equation for Oi equating
Oi to zero and substituting tω α= , we get
( )
10 sinR
tm L
O
Vi A e
Zα φ
−
= = − +
Therefore ( )1 sinR
tmL
VAe
Zα φ
− −= −
( )1
1sinm
Rt
L
VA
Ze
α φ−
− = −
( )1 sinR
tmL
VA e
Zα φ
+ − = −
( )
( )1 sinR t
mLV
A eZ
ωω α φ− = −
By substituting tω α= , we get the value of constant 1A as
( )
( )1 sinR
mLV
A eZ
αω α φ− = −
Substituting the value of constant 1A from the above equation into the expression for Oi , we obtain
( )( )
( )sin sinRR
tm mLL
O
V Vi t e e
Z Z
αωω φ α φ
− − = − + −
;
( )( ) ( )
( )sin sinR t R
m mL LO
V Vi t e e
Z Z
ω αω ωω φ α φ
− − = − + −
( ) ( ) ( )sin sinR
tm mL
O
V Vi t e
Z Z
ω αωω φ α φ− − − = − + −
Therefore we obtain the final expression for the inductive load current of a single
phase full wave ac voltage controller with RL load as
( ) ( ) ( )sin sin
Rt
m LO
Vi t e
Z
ω αωω φ α φ− −
= − − −
; Where tα ω β≤ ≤ .
35
The above expression also represents the thyristor current 1Ti , during the
conduction time interval of thyristor 1T from to tω α β= .
To Calculate Extinction Angle β The extinction angle β , which is the value of tω at which the load current
Oi falls to zero and 1T is turned off can be estimated by using the condition that
0Oi = , at tω β= By using the above expression for the output load current, we can write
( ) ( ) ( )0 sin sin
Rm L
O
Vi e
Z
β αωβ φ α φ− −
= = − − −
As 0mV
Z≠ we can write
( ) ( ) ( )sin sin 0
R
Leβ α
ωβ φ α φ− −
− − − =
Therefore we obtain the expression
( ) ( ) ( )sin sin
R
Leβ α
ωβ φ α φ− −
− = −
The extinction angle β can be determined from this transcendental equation by
using the iterative method of solution (trial and error method). After β is calculated, we
can determine the thyristor conduction angle ( )δ β α= − .
β is the extinction angle which depends upon the load inductance value. Conduction angle δ increases as α is decreased for a known value of β .
For δ π< radians, i.e., for ( )β α π− < radians, for ( )π α β+ > the load current
waveform appears as a discontinuous current waveform as shown in the figure. The output load current remains at zero during tω β= to ( )π α+ . This is referred to as
discontinuous load current operation which occurs for ( )β π α< + .
When the trigger angle α is decreased and made equal to the load impedance angle φ i.e., when α φ= we obtain from the expression for ( )sin β φ− ,
( )sin 0β φ− = ; Therefore ( )β φ π− = radians.
Extinction angle ( ) ( )β π φ π α= + = + ; for the case when α φ=
Conduction angle ( ) 0 radians 180δ β α π= − = = ; for the case when α φ=
Each thyristor conducts for 1800 ( radiansπ ) . 1T conducts from tω φ= to
( )π φ+ and provides a positive load current. 2T conducts from ( )π φ+ to ( )2π φ+ and
provides a negative load current. Hence we obtain a continuous load current and the
36
output voltage waveform appears as a continuous sine wave identical to the input supply voltage waveform for trigger angle α φ≤ and the control on the output is lost.
vO
π 2πφ
3π
φ
ωt
Vm
0
φ φ
Im
ωtφ
v =vO S
iO
Fig.: Output voltage and output current waveforms for a single phase full wave ac voltage controller with RL load for α φ≤
Thus we observe that for trigger angle α φ≤ , the load current tends to flow continuously and we have continuous load current operation, without any break in the load current waveform and we obtain output voltage waveform which is a continuous sinusoidal waveform identical to the input supply voltage waveform. We loose the control on the output voltage for α φ≤ as the output voltage becomes equal to the input supply voltage and thus we obtain
( ) 2m
SO RMS
VV V= = ; for α φ≤
Hence, RMS output voltage = RMS input supply voltage for α φ≤
TO DERIVE AN EXPRESSION FOR RMS OUTPUT VOLTAGE ( )O RMSV OF A
SINGLE PHASE FULL-WAVE AC VOLTAGE CONTROLLER WITH R L LOAD.
37
When Oα > , the load current and load voltage waveforms become discontinuous as shown in the figure above.
( ) ( )1
22 21sin .mO RMSV V t d t
β
α
ω ωπ
= ∫
Output sino mv V tω= , for to tω α β= , when 1T is ON.
( )( ) ( )
122 1 cos 2
2m
O RMS
tVV d t
β
α
ωω
π −
=
∫
( ) ( ) ( )122
cos 2 .2
mO RMS
VV d t t d t
β β
α α
ω ω ωπ
= − ∫ ∫
( ) ( )122 sin 2
22m
O RMS
V tV t
β β
α α
ωωπ
= −
( ) ( )1
2 2sin 2 sin 2
2 2 2m
O RMS
VV
β αβ απ
= − − +
( ) ( )121 sin 2 sin 2
2 2 2mO RMSV Vα ββ α
π = − + −
( ) ( )121 sin 2 sin 2
2 22m
O RMS
VV
α ββ απ = − + −
The RMS output voltage across the load can be varied by changing the trigger angle α . For a purely resistive load 0L = , therefore load power factor angle 0φ = .
1tan 0L
R
ωφ − = =
;
Extinction angle 0 radians 180β π= =
38
PERFORMANCE PARAMETERS OF A SINGLE PHASE FULL WAVE AC VOLTAGE CONTROLLER WITH RESISTIVE LOAD
• RMS Output Voltage ( ) ( )1 sin 2
22m
O RMS
VV
απ απ = − +
; 2m
S
VV= = RMS
input supply voltage.
• ( )( )O RMS
O RMSL
VI
R= = RMS value of load current.
• ( )S O RMSI I= = RMS value of input supply current.
• Output load power
( )2
O LO RMSP I R= ×
• Input Power Factor
( )
( )
( )2
L LO RMS O RMSO
S S S SO RMS
I R I RPPF
V I V I V
× ×= = =
× ×
( ) ( )1 sin 2
2O RMS
S
VPF
V
απ απ = = − +
• Average Thyristor Current,
Im
iT1
π 2πα (2 + )π α
3π
α αωt
Fig.: Thyristor Current Waveform
( ) ( ) ( )1 1sin .
2 2T mT AvgI i d t I t d tπ π
α α
ω ω ωπ π
= =∫ ∫
( ) ( )sin . cos2 2
m mT Avg
I II t d t t
π π
αα
ω ω ωπ π
= = −
∫
( ) [ ] [ ]cos cos 1 cos2 2
m mT Avg
I II π α α
π π= − + = +
39
• Maximum Average Thyristor Current, for 0α = ,
( )m
T Avg
II
π=
• RMS Thyristor Current
( ) ( )2 21sin .
2 mT RMSI I t d tπ
α
ω ωπ
= ∫
( ) ( )1 sin 2
2 22m
T RMS
II
απ απ = − +
• Maximum RMS Thyristor Current, for 0α = ,
( ) 2m
T RMS
II =
In the case of a single phase full wave ac voltage controller circuit using a Triac with resistive load, the average thyristor current ( ) 0T AvgI = . Because the Triac conducts in
both the half cycles and the thyristor current is alternating and we obtain a symmetrical thyristor current waveform which gives an average value of zero on integration. PERFORMANCE PARAMETERS OF A SINGLE PHASE FULL WAVE AC VOLTAGE CONTROLLER WITH R-L LOAD The Expression for the Output (Load) Current The expression for the output (load) current which flows through the thyristor, during to tω α β= is given by
( ) ( ) ( )1
sin sinR
tm L
O T
Vi i t e
Z
ω αωω φ α φ− −
= = − − −
; for tα ω β≤ ≤
Where,
2m SV V= = Maximum or peak value of input ac supply voltage.
( )22Z R Lω= + = Load impedance.
1tanL
R
ωφ − =
= Load impedance angle (load power factor angle).
α = Thyristor trigger angle = Delay angle.
β = Extinction angle of thyristor, (value of tω ) at which the thyristor (load) current falls to zero.
β is calculated by solving the equation
( ) ( ) ( )sin sin
R
Leβ α
ωβ φ α φ− −
− = −
40
Thyristor Conduction Angle ( )δ β α= −
Maximum thyristor conduction angle ( )δ β α π= − = radians = 1800 for α φ≤ .
RMS Output Voltage
( ) ( )1 sin 2 sin 2
2 22m
O RMS
VV
α ββ απ = − + −
The Average Thyristor Current
( ) ( )1
1
2 TT AvgI i d tβ
α
ωπ
= ∫
( ) ( ) ( ) ( ) ( )1sin sin
2
Rt
m LT Avg
VI t e d t
Z
βω α
ω
α
ω φ α φ ωπ
− − = − − −
∫
( ) ( ) ( ) ( ) ( ) ( )sin . sin2
Rt
m LT Avg
VI t d t e d t
Z
β βω α
ω
α α
ω φ ω α φ ωπ
− − = − − −
∫ ∫
Maximum value of ( )T AvgI occur at 0α = . The thyristors should be rated for
maximum ( )m
T Avg
II
π =
, where mm
VI
Z= .
RMS Thyristor Current ( )T RMSI
( ) ( )1
21
2 TT RMSI i d tβ
α
ωπ
=
∫
Maximum value of ( )T RMSI occurs at 0α = . Thyristors should be rated for
maximum ( ) 2m
T RMS
II
=
When a Triac is used in a single phase full wave ac voltage controller with RL
type of load, then ( ) 0T AvgI = and maximum ( ) 2m
T RMS
II =
41
PROBLEMS
1. A single phase full wave ac voltage controller supplies an RL load. The input supply voltage is 230V, RMS at 50Hz. The load has L = 10mH, R = 10Ω, the
delay angle of thyristors 1T and 2T are equal, where 1 2 3
πα α= = . Determine
a. Conduction angle of the thyristor 1T .
b. RMS output voltage. c. The input power factor.
Comment on the type of operation. Given 230sV V= , 50f Hz= , 10L mH= , 10R = Ω , 060α = ,
1 2 3
πα α α= = = radians, .
2 2 230 325.2691193 m SV V V= = × =
( ) ( ) ( )2 2 22 Load Impedance 10Z R L Lω ω= = + = +
( ) ( )32 2 50 10 10 3.14159L fLω π π π−= = × × × = = Ω
( ) ( )2 210 3.14159 109.8696 10.4818Z = + = = Ω
2 230
31.03179 10.4818
mm
VI A
Z
×= = =
Load Impedance Angle 1tanL
R
ωφ − =
( )1 1 0tan tan 0.314159 17.4405910
πφ − − = = =
Trigger Angle α φ> . Hence the type of operation will be discontinuous load current operation, we get
( )β π α< +
( )180 60β < + ; 0240β <
Therefore the range of β is from 180 degrees to 240 degrees.
( )0 0180 240β< <
42
Extinction Angle β is calculated by using the equation
( ) ( ) ( )sin sin
R
Leβ α
ωβ φ α φ− −
− = −
In the exponential term the value of α and β should be substituted in
radians. Hence
( ) ( ) ( )sin sin
Rad RadR
Leβ α
ωβ φ α φ− −
− = − ; 3Rad
πα =
( ) ( ) 060 17.44059 42.5594α φ− = − =
( ) ( ) ( )100 0sin 17.44 sin 42.5594e
β απβ
− −− =
( ) ( )0 3.183sin 17.44 0.676354e β αβ − −− =
0180 radians, π→0
0180Rad
β πβ ×=
Assuming 0190β = ;
0 0
0
1903.3161
180 180Rad
β π πβ × ×= = =
L.H.S: ( ) ( )0sin 190 17.44 sin 172.56 0.129487− = =
R.H.S: 3.183 3.3161
430.676354 4.94 10eπ − − − × = ×
Assuming 0183β = ;
0 0
0
1833.19395
180 180Rad
β π πβ × ×= = =
( ) 3.19395 2.146753
πβ α − = − =
L.H.S: ( ) ( ) 0sin sin 183 17.44 sin165.56 0.24936β φ− = − = =
R.H.S: ( )3.183 2.14675 40.676354 7.2876 10e− −= × Assuming 0180β ≈
0 0
0
180
180 180Rad
β π πβ π× ×= = =
( ) 2
3 3
π πβ α π − = − =
43
L.H.S: ( ) ( )sin sin 180 17.44 0.2997β φ− = − =
R.H.S: 3.183
430.676354 8.6092 10eππ − − − = ×
Assuming 0196β =
0 0
0
1963.420845
180 180Rad
β π πβ × ×= = =
L.H.S: ( ) ( )sin sin 196 17.44 0.02513β φ− = − =
R.H.S: 3.183 3.420845
430.676354 3.5394 10eπ − − − = ×
Assuming 0197β =
0 0
0
1973.43829
180 180Rad
β π πβ × ×= = =
L.H.S: ( ) ( ) 3sin sin 197 17.44 7.69 7.67937 10β φ −− = − = = ×
R.H.S: 3.183 3.43829
430.676354 4.950386476 10eπ − − − = ×
Assuming 0197.42β =
0
0
197.423.4456
180 180Rad
β π πβ × ×= = =
L.H.S: ( ) ( ) 4sin sin 197.42 17.44 3.4906 10β φ −− = − = ×
R.H.S: 3.183 3.4456
430.676354 3.2709 10eπ − − − = ×
Conduction Angle ( ) ( )0 0 0197.42 60 137.42δ β α= − = − =
RMS Output Voltage
( ) ( )1 sin 2 sin 2
2 2SO RMSV Vα ββ α
π = − + −
( )( ) ( )0 0sin 2 60 sin 2 197.421
230 3.44563 2 2O RMSVπ
π
= − + −
( ) ( )1230 2.39843 0.4330 0.285640O RMSV
π= + −
( ) 230 0.9 207.0445 VO RMSV = × =
44
Input Power Factor
O
S S
PPF
V I=
×
( )( ) 207.0445
19.7527 A10.4818
O RMS
O RMS
VI
Z= = =
( ) ( )22 19.7527 10 3901.716 WO LO RMSP I R= × = × =
( )230 , 19.7527S S O RMSV V I I= = =
3901.716
0.8588230 19.7527
O
S S
PPF
V I= = =
× ×
2. A single phase full wave controller has an input voltage of 120 V (RMS) and a
load resistance of 6 ohm. The firing angle of thyristor is2π . Find a. RMS output voltage b. Power output c. Input power factor d. Average and RMS thyristor current.
Solution
090 , 120 V, 62 SV Rπα = = = = Ω
RMS Value of Output Voltage
1
21 sin 2
2O SV Vαπ α
π = − +
1
21 sin180120
2 2OVππ
π = − +
84.85 VoltsOV =
RMS Output Current
84.85
14.14 A6
OO
VI
R= = =
Load Power 2
O OP I R= ×
( )214.14 6 1200 wattsOP = × =
45
Input Current is same as Load Current Therefore 14.14 AmpsS OI I= =
Input Supply Volt-Amp 120 14.14 1696.8 S SV I VA= = × = Therefore
Input Power Factor = ( )Load Power 12000.707
Input Volt-Amp 1696.8lag= =
Each Thyristor Conducts only for half a cycle Average thyristor current ( )T AvgI
( ) ( )1sin .
2 mT AvgI V t d tR
π
α
ω ωπ
= ∫
( ) m1 cos ; V 22
mS
VV
Rα
π= + =
[ ]2 1201 cos90 4.5 A
2 6π×= + =×
RMS thyristor current ( )T RMSI
( ) ( )2 2
2
sin1
2m
T RMS
V tI d t
R
π
α
ω ωπ
= ∫
( ) ( )
2
2
1 cos 2
2 2m
tVd t
R
π
α
ωω
π−
= ∫
1
21 sin 2
2 2mV
R
απ απ = − +
1
22 1 sin 2
2 2SV
R
απ απ = − +
1
22 120 1 sin18010 Amps
2 6 2 2
πππ
× = − + = ×
46
3. A single phase half wave ac regulator using one SCR in anti-parallel with a diode
feeds 1 kW, 230 V heater. Find load power for a firing angle of 450.
Solution 045 , 230 V
4 SVπα = = = ; 1 1000OP KW W= =
At standard rms supply voltage of 230V, the heater dissipates 1KW of output power
Therefore
2
O O OO O O
V V VP V I
R R
×= × = =
Resistance of heater
( )22 230
52.91000
O
O
VR
P= = = Ω
RMS value of output voltage
1
21 sin 22
2 2O SV Vαπ α
π = − +
; for firing angle 045α =
1
21 sin 90230 2 224.7157 Volts
2 4 2OVππ
π = − + =
RMS value of output current
224.9
4.2479 Amps52.9
OO
VI
R= = =
Load Power
( )22 4.25 52.9 954.56 WattsO OP I R= × = × =
4. Find the RMS and average current flowing through the heater shown in figure.
The delay angle of both the SCRs is 450.
SCR2
SCR1 io+
1 kW, 220Vheater
1-220V
ac
φ
47
Solution
045 , 220 V4 SVπα = = =
Resistance of heater
( )22 220
48.41000
VR
R= = = Ω
Resistance value of output voltage
1 sin 2
2O SV Vαπ α
π = − +
1 sin90
2204 2OVππ
π = − +
1 1
220 209.769 Volts4 2OVππ
π = − + =
RMS current flowing through heater 209.769
4.334 Amps48.4
OV
R= = =
Average current flowing through the heater 0AvgI =
5. A single phase voltage controller is employed for controlling the power flow from
220 V, 50 Hz source into a load circuit consisting of R = 4 Ω and ωL = 6 Ω. Calculate the following
a. Control range of firing angle b. Maximum value of RMS load current c. Maximum power and power factor d. Maximum value of average and RMS thyristor current.
Solution For control of output power, minimum angle of firing angle αααα is equal to the load impedance angle θθθθ
, load angleα θ=
1 1 06tan tan 56.3
4
L
R
ωθ − − = = =
Maximum possible value of α is 0180 Therefore control range of firing angle is 0 056.3 180α< <
48
Maximum value of RMS load current occurs when 056.3α θ= = . At this value of α the Maximum value of RMS load current
2 2
22030.5085 Amps
4 6S
O
VI
Z= = =
+
Maximum Power ( )22 30.5085 4 3723.077 WO OP I R= = × =
Input Volt-Amp 220 30.5085 6711.87 WS OV I= = × =
Power Factor 3723.077
0.5547 6711.87
OP
Input VA= = =
Average thyristor current will be maximum when α θ= and conduction angle 0180γ = . Therefore maximum value of average thyristor current
( ) ( ) ( )1sin
2m
T Avg
VI t d t
Z
π α
α
ω θ ωπ
+
= −∫
Note: ( ) ( ) ( )1
sin sinR
tm L
O T
Vi i t e
Z
ω αωω θ α θ− −
= = − − −
At 0α = ,
( )1
sinmT O
Vi i t
Zω θ= = −
( ) ( )cos2
mT Avg
VI t
Z
π α
αω θ
π+
= − −
( ) ( ) ( )cos cos2
mT Avg
VI
Zπ α θ α θ
π= − + − + −
But α θ= ,
( ) ( ) ( ) [ ]cos cos 0 22 2
m m mT Avg
V V VI
Z Z Zπ
π π π= − + = =
( ) 2 2
2 22013.7336 Amps
4 6m
T Avg
VI
Zπ π×∴ = = =
+
Similarly, maximum RMS value occurs when 0α = and γ π= . Therefore maximum value of RMS thyristor current
( ) ( )2
1sin
2m
TM
VI t d t
Z
π α
α
ω θ ωπ
+ = −
∫
49
( ) ( )
2
2
1 cos 2 2
2 2m
TM
tVI d t
Z
π α
α
ω θω
π
+ − − =
∫
( )2
2
sin 2 2
4 2m
TM
tVI t
Z
π α
α
ω θω
π
+−
= −
[ ]2
20
4m
TM
VI
Zπ α α
π= + − −
2 2
2 22021.57277 Amps
2 2 4 6m
TM
VI
Z
×= = =+
50
CONTROLLED RECTIFIERS (Line Commutated AC to DC converters)
INTRODUCTION TO CONTROLLED RECTIFIERS Controlled rectifiers are line commutated ac to dc power converters which are used to convert a fixed voltage, fixed frequency ac power supply into variable dc output voltage.
LineCommutated
Converter
+
-
DC OutputV0(dc)
ACInput
Voltage
Type of input: Fixed voltage, fixed frequency ac power supply. Type of output: Variable dc output voltage
The input supply fed to a controlled rectifier is ac supply at a fixed rms voltage and at a fixed frequency. We can obtain variable dc output voltage by using controlled rectifiers. By employing phase controlled thyristors in the controlled rectifier circuits we can obtain variable dc output voltage and variable dc (average) output current by varying the trigger angle (phase angle) at which the thyristors are triggered. We obtain a uni-directional and pulsating load current waveform, which has a specific average value. The thyristors are forward biased during the positive half cycle of input supply and can be turned ON by applying suitable gate trigger pulses at the thyristor gate leads. The thyristor current and the load current begin to flow once the thyristors are triggered (turned ON) say at tω α= . The load current flows when the thyristors conduct from
tω α= to β . The output voltage across the load follows the input supply voltage through the conducting thyristor. At tω β= , when the load current falls to zero, the thyristors turn off due to AC line (natural) commutation. In some bridge controlled rectifier circuits the conducting thyristor turns off, when the other thyristor is (other group of thyristors are) turned ON. The thyristor remains reverse biased during the negative half cycle of input supply. The type of commutation used in controlled rectifier circuits is referred to AC line commutation or Natural commutation or AC phase commutation.
When the input ac supply voltage reverses and becomes negative during the negative half cycle, the thyristor becomes reverse biased and hence turns off. There are several types of power converters which use ac line commutation. These are referred to as line commutated converters. Different types of line commutated converters are
• Phase controlled rectifiers which are AC to DC converters. • AC to AC converters
AC voltage controllers, which convert input ac voltage into variable ac output voltage at the same frequency.
Cyclo converters, which give low output frequencies.
51
All these power converters operate from ac power supply at a fixed rms input supply voltage and at a fixed input supply frequency. Hence they use ac line commutation for turning off the thyristors after they have been triggered ON by the gating signals. DIFFERENCES BETWEEN DIODE RECTIFIERS AND PHASE CONT ROLLED RECTIFIERS The diode rectifiers are referred to as uncontrolled rectifiers which make use of power semiconductor diodes to carry the load current. The diode rectifiers give a fixed dc output voltage (fixed average output voltage) and each diode rectifying element conducts for one half cycle duration (T/2 seconds), that is the diode conduction angle = 1800 or π radians.
A single phase half wave diode rectifier gives (under ideal conditions) an average
dc output voltage ( )m
O dc
VV
π= and single phase full wave diode rectifier gives (under ideal
conditions) an average dc output voltage ( )2 m
O dc
VV
π= , where mV is maximum value of
the available ac supply voltage. Thus we note that we can not control (we can not vary) the dc output voltage or the average dc load current in a diode rectifier circuit. In a phase controlled rectifier circuit we use a high current and a high power thyristor device (silicon controlled rectifier; SCR) for conversion of ac input power into dc output power. Phase controlled rectifier circuits are used to provide a variable voltage output dc and a variable dc (average) load current. We can control (we can vary) the average value (dc value) of the output load voltage (and hence the average dc load current) by varying the thyristor trigger angle.
We can control the thyristor conduction angle δ from 1800 to 00 by varying the trigger angle α from 00 to 1800, where thyristor conduction angle ( )δ π α= −
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. • Reactor controls. • Portable hand tool drives. • Variable speed industrial drives. • Battery charges. • High voltage DC transmission. • Uninterruptible power supply systems (UPS).
Some years back ac to dc power conversion was achieved using motor generator
sets, mercury arc rectifiers, and thyratorn tubes. The modern ac to dc power converters are designed using high power, high current thyristors and presently most of the ac-dc power converters are thyristorised power converters. The thyristor devices are phase controlled to obtain a variable dc output voltage across the output load terminals. The
52
phase controlled thyristor converter uses ac line commutation (natural commutation) for commutating (turning off) the thyristors that have been turned ON.
The phase controlled converters are simple and less expensive and are widely used in industrial applications for industrial dc drives. These converters are classified as two quadrant converters if the output voltage can be made either positive or negative for a given polarity of output load current. There are also single quadrant ac-dc converters where the output voltage is only positive and cannot be made negative for a given polarity of output current. Of course single quadrant converters can also be designed to provide only negative dc output voltage.
The two quadrant converter operation can be achieved by using fully controlled bridge converter circuit and for single quadrant operation we use a half controlled bridge converter. CLASSIFICATION OF PHASE CONTROLLED RECTIFIERS The phase controlled rectifiers can be classified based on the type of input power supply as
• Single Phase Controlled Rectifiers which operate from single phase ac input power supply.
• Three Phase Controlled Rectifiers which operate from three phase ac input power supply.
DIFFERENT TYPES OF SINGLE PHASE CONTROLLED RECTIFIE RS Single Phase Controlled Rectifiers are further subdivided into different types
• Half wave controlled rectifier which uses a single thyristor device (which provides output control only in one half cycle of input ac supply, and it provides low dc output).
• Full wave controlled rectifiers (which provide higher dc output)
o Full wave controlled rectifier using a center tapped transformer (which requires two thyristors).
o Full wave bridge controlled rectifiers (which do not require a center tapped transformer) Single phase semi-converter (half controlled bridge converter,
using two SCR’s and two diodes, to provide single quadrant operation).
Single phase full converter (fully controlled bridge converter which requires four SCR’s, to provide two quadrant operation).
Three Phase Controlled Rectifiers are of different types
• Three phase half wave controlled rectifiers. • Three phase full wave controlled rectiriers.
o Semi converter (half controlled bridge converter). o Full converter (fully controlled bridge converter).
PRINCIPLE OF PHASE CONTROLLED RECTIFIER OPERATION The basic principle of operation of a phase controlled rectifier circuit is explained with reference to a single phase half wave phase controlled rectifier circuit with a resistive load shown in the figure.
53
Load ResistanceLR R= =
Fig.: Single Phase Half-Wave Thyristor Converter with a Resistive Load
A single phase half wave thyristor converter which is used for ac-dc power
conversion is shown in the above figure. The input ac supply is obtained from a main supply transformer to provide the desired ac supply voltage to the thyristor converter depending on the output dc voltage required. Pv represents the primary input ac supply
voltage. Sv represents the secondary ac supply voltage which is the output of the transformer secondary.
During the positive half cycle of input supply when the upper end of the transformer secondary is at a positive potential with respect to the lower end, the thyristor anode is positive with respect to its cathode and the thyristor is in a forward biased state. The thyristor is triggered at a delay angle of tω α= , by applying a suitable gate trigger pulse to the gate lead of thyristor. When the thyristor is triggered at a delay angle of tω α= , the thyristor conducts and assuming an ideal thyristor, the thyristor behaves as a closed switch and the input supply voltage appears across the load when the thyristor conducts from tω α= to π radians. Output voltage O Sv v= , when the thyristor
conducts from to tω α π= . For a purely resistive load, the load current Oi (output current) that flows when
the thyristor 1T is on, is given by the expression
, for OO
L
vi t
Rα ω π= ≤ ≤
The output load current waveform is similar to the output load voltage waveform during the thyristor conduction time from to α π . The output current and the output voltage waveform are in phase for a resistive load. The load current increases as the input
supply voltage increases and the maximum load current flows at 2
tπω = , when the input
supply voltage is at its maximum value. The maximum value (peak value) of the load current is calculated as
( )maxm
mOL
Vi I
R= = .
54
Note that when the thyristor conducts (1T is on) during to tω α π= , the thyristor
current 1Ti , the load current Oi through LR and the source current Si flowing through the transformer secondary winding are all one and the same. Hence we can write
1
sin ; for O m
S T O
v V ti i i t
R R
ω α ω π= = = = ≤ ≤
mI is the maximum (peak) value of the load current that flows through the
transformer secondary winding, through 1T and through the load resistor LR at the instant
2t
πω = , when the input supply voltage reaches its maximum value.
When the input supply voltage decreases the load current decreases. When the supply voltage falls to zero at tω π= , the thyristor and the load current also falls to zero at tω π= . Thus the thyristor naturally turns off when the current flowing through it falls to zero at tω π= .
During the negative half cycle of input supply when the supply voltage reverses and becomes negative during to 2tω π π= radians, the anode of thyristor is at a negative potential with respect to its cathode and as a result the thyristor is reverse biased and hence it remains cut-off (in the reverse blocking mode). The thyristor cannot conduct during its reverse biased state between to 2tω π π= . An ideal thyristor under reverse biased condition behaves as an open switch and hence the load current and load voltage are zero during to 2tω π π= . The maximum or peak reverse voltage that appears across the thyristor anode and cathode terminals is mV .
The trigger angle α (delay angle or the phase angle α ) is measured from the beginning of each positive half cycle to the time instant when the gate trigger pulse is applied. The thyristor conduction angle is from to α π , hence the conduction angle
( )δ π α= − . The maximum conduction angle is π radians (1800) when the trigger angle
0α = .
Fig: Quadrant Diagram
The waveforms shows the input ac supply voltage across the secondary winding
of the transformer which is represented as Sv , the output voltage across the load, the output (load) current, and the thyristor voltage waveform that appears across the anode and cathode terminals.
55
Fig: Waveforms of single phase half-wave controlled rectifier with resistive load
EQUATIONS sins mv V tω= = the ac supply voltage across the transformer secondary. mV = max. (peak) value of input ac supply voltage across transformer secondary.
2m
S
VV = = RMS value of input ac supply voltage across transformer secondary.
O Lv v= = the output voltage across the load ; O Li i= = output (load) current.
56
When the thyristor is triggered at tω α= (an ideal thyristor behaves as a closed switch) and hence the output voltage follows the input supply voltage. sinO L mv v V tω= = ; for to tω α π= , when the thyristor is on.
OO L
vi i
R= = = Load current for to tω α π= , when the thyristor is on.
TO DERIVE AN EXPRESSION FOR THE AVERAGE (DC) OUTPUT VOLTAGE ACROSS THE LOAD
If mV is the peak input supply voltage, the average output voltage dcV can be found from
( ) ( )1.
2dc OO dcV V v d tπ
α
ωπ
= = ∫
( ) ( )1sin .
2dc mO dcV V V t d tπ
α
ω ωπ
= = ∫
( ) ( )1sin .
2 mO dcV V t d tπ
α
ω ωπ
= ∫
( ) ( )sin .2
mO dc
VV t d t
π
α
ω ωπ
= ∫
( ) cos2
mO dc
VV t
π
α
ωπ
= −
( ) [ ]cos cos2
mO dc
VV π α
π= − + ; cos 1π = −
( ) [ ]1 cos2
mO dc
VV α
π= + ; 2m SV V=
The maximum average (dc) output voltage is obtained when 0α = and the
maximum dc output voltage ( )maxm
dmdc
VV V
π= = .
The average dc output voltage can be varied by varying the trigger angle α from 0 to a maximum of ( )0180 radiansπ .
We can plot the control characteristic, which is a plot of dc output voltage versus the trigger angle α by using the equation for ( )O dcV .
57
CONTROL CHARACTERISTIC OF SINGLE PHASE HALF WAVE PH ASE CONTROLLED RECTIFIER WITH RESISTIVE LOAD The average dc output voltage is given by the expression
( ) [ ]1 cos2
mO dc
VV α
π= +
We can obtain the control characteristic by plotting the expression for the dc output voltage as a function of trigger angle α
Trigger angle α in degrees ( )O dcV %
0 mdm
VV
π= 100% dmV
030 0.933 dmV 93.3 % dmV 060 0.75 dmV 75 % dmV 090 0.5 dmV 50 % dmV 0120 0.25 dmV 25 % dmV 0150 0.06698 dmV 6.69 % dmV 0180 0 0
( )maxm
dm dc
VV V
π= =
VO(dc)
Trigger angle in degreesα
0 60 120 180
Vdm
0.2 Vdm
0.6Vdm
Fig.: Control characteristic Normalizing the dc output voltage with respect to dmV , the normalized output voltage
( )
( )
max
O dc dcdcn
dmdc
V VV
V V= =
58
( )1 cos
2m
dcdcn n
mdm
VV
V VVV
απ
π
+= = =
( )11 cos
2dc
n dcndm
VV V
Vα= = + =
TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF OUTPUT VOLTAGE OF A SINGLE PHASE HALF WAVE CONTROLLED RECT IFIER WITH RESISTIVE LOAD The rms output voltage is given by
( ) ( )2
2
0
1.
2 OO RMSV v d tπ
ωπ
=
∫
Output voltage sin ; for to O mv V t tω ω α π= =
( ) ( )1
22 21sin .
2 mO RMSV V t d tπ
α
ω ωπ
=
∫
By substituting 2 1 cos 2sin
2
tt
ωω −= , we get
( )( ) ( )
1
22 1 cos 21
.2 2mO RMS
tV V d t
π
α
ωω
π −
=
∫
( ) ( ) ( )1
2 2
1 cos 2 .4
mO RMS
VV t d t
π
α
ω ωπ
= −
∫
( ) ( ) ( )1
2 2
cos 2 .4
mO RMS
VV d t t d t
π π
α α
ω ω ωπ
= −
∫ ∫
( ) ( )1
21 sin 222
mO RMS
V tV t
π π
α α
ωωπ = −
( ) ( ) ( )1
2sin 2 sin 21
2 2m
O RMS
VV
π απ α
π −
= − −
; sin 2 0π =
59
Hence we get,
( ) ( )1
21 sin 2
2 2m
O RMS
VV
απ απ = − +
( ) ( )1
2sin 2
22m
O RMS
VV
απ απ = − +
PERFORMANCE PARAMETERS OF PHASE CONTROLLED RECTIFIE RS Output dc power (average or dc output power delivered to the load)
( ) ( ) ( )O dc O dc O dcP V I= × ; i.e., dc dc dcP V I= ×
Where
( ) dcO dcV V= = average or dc value of output (load) voltage.
( ) dcO dcI I= = average or dc value of output (load) current.
Output ac power ( ) ( ) ( )O ac O RMS O RMSP V I= ×
Efficiency of Rectification (Rectification Ratio)
( )
( )Efficiency
O dc
O ac
P
Pη = ; ( )
( )% Efficiency 100
O dc
O ac
P
Pη = ×
The output voltage can be considered as being composed of two components
• The dc component ( )O dcV = DC or average value of output voltage.
• The ac component or the ripple component ( )ac r rmsV V= =RMS value of all
the ac ripple components. The total RMS value of output voltage is given by
( ) ( ) ( )2 2
O RMS O dc r rmsV V V= +
Therefore
( ) ( ) ( )2 2
ac r rms O RMS O dcV V V V= = −
60
Form Factor (FF) which is a measure of the shape of the output voltage is given by
( )
( )
( )( )
RMS output load voltage
DC output load voltageO RMS
O dc
VFF
V= =
The Ripple Factor (RF) which is a measure of the ac ripple content in the output voltage waveform. The output voltage ripple factor defined for the output voltage waveform is given by
( )
( )
r rms acv
dcO dc
V Vr RF
V V= = =
( ) ( )
( )
( )
( )
22 2
1O RMS O dc O RMS
v
O dc O dc
V V Vr
V V
− = = −
Therefore
2 1vr FF= −
Current Ripple Factor defined for the output (load) current waveform is given by
( )
( )
r rms aci
dcO dc
I Ir
I I= =
Where ( ) ( ) ( )2 2
acr rms O RMS O dcI I I I= = −
Some times the peak to peak output ripple voltage is also considered to express
the peak to peak output ripple voltage as
( ) peak to peak ac ripple output voltager ppV =
The peak to peak ac ripple load current is the difference between the maximum and the minimum values of the output load current. ( ) ( ) ( )max minr pp O OI I I= −
Transformer Utilization Factor (TUF)
( )O dc
S S
PTUF
V I=
×
Where
SV = RMS value of transformer secondary output voltage (RMS supply voltage at the secondary)
61
SI = RMS value of transformer secondary current (RMS line or supply current).
Supply voltage at the transformer secondary sideSv = .
Input supply current (transformer secondary winding current)Si = .
1 Fundamental component of the input supply currentSi = .
Peak value of the input supply currentPI = .
φ = Phase angle difference between (sine wave components) the fundamental components of input supply current and the input supply voltage. φ = Displacement angle (phase angle) For an RL load Displacement angle = Load impedance angleφ =
1 tan for an RL loadL
R
ωφ − ∴ =
Displacement Factor (DF) or Fundamental Power Factor
DF Cosφ= Harmonic Factor (HF) or Total Harmonic Distortion Factor (THD) The harmonic factor is a measure of the distortion in the output waveform and is also referred to as the total harmonic distortion (THD)
112 22 2 2
121 1
1S S S
S S
I I IHF
I I
− = = −
Where SI = RMS value of input supply current.
1SI = RMS value of fundamental component of the input supply current.
φ
62
Input Power Factor (PF)
1 1cos cosS S S
S S S
V I IPF
V I Iφ φ= =
The Crest Factor (CF)
( ) Peak input supply current
RMS input supply currentS peak
S
ICF
I= =
For an Ideal Controlled Rectifier 1FF = ; which means that ( ) ( )O RMS O dcV V= .
Efficiency 100%η = ; which means that ( ) ( )O dc O acP P= .
( ) 0ac r rmsV V= = ; so that 0vRF r= = ; Ripple factor = 0 (ripple free converter).
1TUF = ; which means that ( ) S SO dcP V I= ×
0HF THD= = ; which means that 1S SI I=
1PF DPF= = ; which means that 0φ =
SINGLE PHASE HALF WAVE CONTROLLED RECTIFIER WITH AN RL LOAD In this section we will discuss the operation and performance of a single phase
half wave controlled rectifier with RL load. In practice most of the loads are of RL type.
For example if we consider a single phase controlled rectifier controlling the speed of a
dc motor, the load which is the dc motor winding is an RL type of load, where R
represents the motor winding resistance and L represents the motor winding inductance.
A single phase half wave controlled rectifier circuit with an RL load using a
thyristor 1T ( 1T is an SCR) is shown in the figure below.
63
The thyristor 1T is forward biased during the positive half cycle of input supply.
Let us assume that 1T is triggered at tω α= , by applying a suitable gate trigger pulse to
1T during the positive half cycle of input supply. The output voltage across the load
follows the input supply voltage when 1T is ON. The load current Oi flows through the
thyristor 1T and through the load in the downward direction. This load current pulse
flowing through 1T can be considered as the positive current pulse. Due to the inductance
in the load, the load current Oi flowing through 1T would not fall to zero at tω π= , when
the input supply voltage starts to become negative. A phase shift appears between the
load voltage and the load current waveforms, due to the load inductance.
The thyristor 1T will continue to conduct the load current until all the inductive
energy stored in the load inductor L is completely utilized and the load current through 1T
falls to zero at tω β= , where β is referred to as the Extinction angle, (the value of tω )
at which the load current falls to zero. The extinction angle β is measured from the point
of the beginning of the positive half cycle of input supply to the point where the load
current falls to zero.
The thyristor 1T conducts from tω α= to β . The conduction angle of 1T is
( )δ β α= − , which depends on the delay angle α and the load impedance angle φ . The
waveforms of the input supply voltage, the gate trigger pulse of 1T , the thyristor current,
the load current and the load voltage waveforms appear as shown in the figure below.
Fig.: Input supply voltage & Thyristor current waveforms
1 O Si i i= =
64
β is the extinction angle which depends upon the load inductance value.
Fig.: Output (load) voltage waveform of a single phase half wave controlled
rectifier with RL load
From β to 2π , the thyristor remains cut-off as it is reverse biased and behaves as an open switch. The thyristor current and the load current are zero and the output voltage also remains at zero during the non conduction time interval between β to 2π . In the
next cycle the thyristor is triggered again at a phase angle of ( )2π α+ , and the same
operation repeats. TO DERIVE AN EXPRESSION FOR THE OUTPUT (INDUCTIVE L OAD) CURRENT, DURING to tω α β= WHEN THYRISTOR 1T CONDUCTS Considering sinusoidal input supply voltage we can write the expression for the supply voltage as sinS mv V tω= = instantaneous value of the input supply voltage. Let us assume that the thyristor 1T is triggered by applying the gating signal to 1T
at tω α= . The load current which flows through the thyristor 1T during tω α= to β can be found from the equation
sinOO m
diL Ri V t
dtω + =
;
The solution of the above differential equation gives the general expression for the output load current which is of the form
( ) 1sint
mO
Vi t Ae
Zτω φ−
= − + ;
Where 2m SV V= = maximum or peak value of input supply voltage.
( )22Z R Lω= + = Load impedance.
65
1tanL
R
ωφ − =
= Load impedance angle (power factor angle of load).
L
Rτ = = Load circuit time constant.
Therefore the general expression for the output load current is given by the equation
( )
1sinR
tm L
O
Vi t Ae
Zω φ
−
= − + ;
The value of the constant 1A can be determined from the initial condition. i.e.
initial value of load current 0Oi = , at tω α= . Hence from the equation for Oi equating
Oi to zero and substituting tω α= , we get
( )
10 sinR
tm L
O
Vi A e
Zα φ
−
= = − +
Therefore ( )1 sinR
tmL
VAe
Zα φ
− −= −
( )1
1sinm
Rt
L
VA
Ze
α φ−
− = −
( )1 sinR
tmL
VA e
Zα φ
+ − = −
( )
( )1 sinR t
mLV
A eZ
ωω α φ− = −
By substituting tω α= , we get the value of constant 1A as
( )
( )1 sinR
mLV
A eZ
αω α φ− = −
Substituting the value of constant 1A from the above equation into the expression for Oi , we obtain
( )( )
( )sin sinRR
tm mLL
O
V Vi t e e
Z Z
αωω φ α φ
− − = − + −
;
( )( ) ( )
( )sin sinR t R
m mL LO
V Vi t e e
Z Z
ω αω ωω φ α φ
− − = − + −
66
( ) ( ) ( )sin sinR
tm mL
O
V Vi t e
Z Z
ω αωω φ α φ− − − = − + −
Therefore we obtain the final expression for the inductive load current of a single
phase half wave controlled rectifier with RL load as
( ) ( ) ( )sin sin
Rt
m LO
Vi t e
Z
ω αωω φ α φ− −
= − − −
; Where tα ω β≤ ≤ .
The above expression also represents the thyristor current 1Ti , during the
conduction time interval of thyristor 1T from to tω α β= .
TO CALCULATE EXTINCTION ANGLE β The extinction angle β , which is the value of tω at which the load current
Oi falls to zero and 1T is turned off can be estimated by using the condition that
0Oi = , at tω β= By using the above expression for the output load current, we can write
( ) ( ) ( )0 sin sin
Rm L
O
Vi e
Z
β αωβ φ α φ− −
= = − − −
As 0mV
Z≠ , we can write
( ) ( ) ( )sin sin 0
R
Leβ α
ωβ φ α φ− −
− − − =
Therefore we obtain the expression
( ) ( ) ( )sin sin
R
Leβ α
ωβ φ α φ− −
− = −
The extinction angle β can be determined from this transcendental equation by
using the iterative method of solution (trial and error method). After β is calculated, we
can determine the thyristor conduction angle ( )δ β α= − .
β is the extinction angle which depends upon the load inductance value.
Conduction angle δ increases as α is decreased for a specific value of β . Conduction angle ( )δ β α= − ; for a purely resistive load or for an RL load
when the load inductance L is negligible the extinction angle β π= and the conduction
angle ( )δ π α= −
67
Equations sin Input supply voltages mv V tω= = sin Output load voltage for to O L mv v V t tω ω α β= = = = ,
when the thyristor 1T conducts (1T is on).
Expression for the load current (thyristor current): for to tω α β=
( ) ( ) ( )sin sin
Rt
m LO
Vi t e
Z
ω αωω φ α φ− −
= − − −
; Where tα ω β≤ ≤ .
Extinction angle β can be calculated using the equation
( ) ( ) ( )sin sin
R
Leβ α
ωβ φ α φ− −
− = −
TO DERIVE AN EXPRESSION FOR AVERAGE (DC) LOAD VOLTA GE
( ) ( )2
0
1.
2L OO dcV V v d tπ
ωπ
= = ∫
( ) ( ) ( ) ( )2
0
1. . .
2L O O OO dcV V v d t v d t v d tβα π
α β
ω ω ωπ
= = + + ∫ ∫ ∫ ;
0 for 0 to & for to 2Ov t tω α ω β π= = = ;
( ) ( )1. ; sin for to
2L O O mO dcV V v d t v V t tβ
α
ω ω ω α βπ
∴ = = = = ∫
( ) ( )1sin .
2L mO dcV V V t d tβ
α
ω ωπ
= = ∫
( ) ( )cos cos cos2 2
m mLO dc
V VV V t
β
α
ω α βπ π
= = − = −
( ) ( )cos cos2
mLO dc
VV V α β
π∴ = = −
Note: During the period to tω π β= , we can see from the output load voltage waveform that the instantaneous output voltage is negative and this reduces the average or the dc output voltage when compared to a purely resistive load.
68
Average DC Load Current
( ) ( )( ) ( )cos cos
2O dc m
O dc L AvgL L
V VI I
R Rα β
π= = = −
SINGLE PHASE HALF WAVE CONTROLLED RECTIFIER WITH RL LOAD AND FREE WHEELING DIODE
V0
i0T
R
L
Vs ~+
+
−
−
FWD
Fig. : Single Phase Half Wave Controlled Rectifier with RL Load and Free
Wheeling Diode (FWD) With a RL load it was observed that the average output voltage reduces. This
disadvantage can be overcome by connecting a diode across the load as shown in figure. The diode is called as a Free Wheeling Diode (FWD). The waveforms are shown below.
2π
2π
3π
3π
π
π
α
α
α
α
0
0
0
0
Vs
iG
VO
ωt
ωt
ωt
ωt
Supply voltage
Load current
Load voltage
ωt=β
2π+α
Gate pulses
Vm
-Vm
α
π ββ 2π
iO
69
At tω π= , the source voltage Sv falls to zero and as Sv becomes negative, the
free wheeling diode is forward biased. The stored energy in the inductance maintains the load current flow through R, L, and the FWD. Also, as soon as the FWD is forward biased, at tω π= , the SCR becomes reverse biased, the current through it becomes zero and the SCR turns off. During the period to tω π β= , the load current flows through FWD (free wheeling load current) and decreases exponentially towards zero at tω β= .
Also during this free wheeling time period the load is shorted by the conducting FWD and the load voltage is almost zero, if the forward voltage drop across the conducting FWD is neglected. Thus there is no negative region in the load voltage wave form. This improves the average output voltage.
The average output voltage [ ]1 cos2
mdc
VV α
π= + , which is the same as that of a
purely resistive load. The output voltage across the load appears similar to the output voltage of a purely resistive load.
The following points are to be noted.
• If the inductance value is not very large, the energy stored in the inductance is able to maintain the load current only upto tω β= , where
2π β π< < , well before the next gate pulse and the load current tends to become discontinuous.
• During the conduction period to α π , the load current is carried by the SCR and during the free wheeling period to π β , the 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. Generally 2π β π< < .
If the value of the inductance is very large, the load current does not decrease to
zero during the free wheeling time interval and the load current waveform appears as shown in the figure.
2π 3ππα0 ωt
2π+α
t1
i0
SCR SCRFWD FWD
t3t2 t4
Fig. : Waveform of Load Current in Single Phase Half Wave Controlled Rectifier
with a Large Inductance and FWD
70
During the periods 1 3, ,.....t t the SCR carries the load current and during the periods
2 4, ,.....t t the FWD carries the load current.
It is to be noted that
• The load current becomes continuous and the load current does not fall to zero for large value of load inductance.
• The ripple in the load current waveform (the amount of variation in the output load current) decreases.
SINGLE PHASE HALF WAVE CONTROLLED RECTIFIER WITH A GENERAL LOAD A general load consists of R, L and a DC source ‘E’ in the load circuit
R
vS~+
− L
E+−
vO
iO
In the half wave controlled rectifier circuit shown in the figure, the load circuit consists of a dc source ‘E’ in addition to resistance and inductance. When the thyristor is in the cut-off state, the current in the circuit is zero and the cathode will be at a voltage equal to the dc voltage in the load circuit i.e. the cathode potential will be equal to ‘E’. The thyristor will be forward biased for anode supply voltage greater than the load dc voltage.
When the supply voltage is less than the dc voltage ‘E’ in the circuit the thyristor is reverse biased and hence the thyristor cannot conduct for supply voltage less than the load circuit dc voltage.
The value of tω at which the supply voltage increases and becomes equal to the load circuit dc voltage can be calculated by using the equation sinmV t Eω = . If we
assume the value of tω is equal to γ then we can write sinmV Eγ = . Therefore γ is
calculated as 1sinm
E
Vγ −
=
.
For trigger angle α γ< , the thyristor conducts only from to tω γ β= . For trigger angle α γ> , the thyristor conducts from to tω α β= . The waveforms appear as shown in the figure
71
0
0
iO
ωt
ωt
Load current
E
vO
Load voltage
γ απ 2π
2π+α
Vm
Im
α β 2π+α 2π+β
α
β
δ
Equations sin Input supply voltageS mv V tω= = . sin Output load voltage for to O mv V t tω ω α β= = = for 0 to & for to 2Ov E t tω α ω β π= = = Expression for the Load Current When the thyristor is triggered at a delay angle of α , the equation for the circuit can be written as
sin +E ; Om O
diV t i R L t
dtω α ω β = × + ≤ ≤
The general expression for the output load current can be written as
( )sint
mO
V Ei t Ae
Z Rτω φ−
= − − +
Where
( )22 = Load ImpedanceZ R Lω= +
1tan Load impedance angleL
R
ωφ − = =
Load circuit time constantL
Rτ = =
The general expression for the output load current can be written as
72
( )sinR
tm L
O
V Ei t Ae
Z Rω φ
−
= − − +
To find the value of the constant ‘A’ apply the initial condition at tω α= , load current 0Oi = . Equating the general expression for the load current to zero at tω α= , we get
( )0 sinR
m LO
V Ei Ae
Z R
αωα φ
− ×= = − − +
We obtain the value of constant ‘A’ as
( )sinR
m LVE
A eR Z
αωα φ = − −
Substituting the value of the constant ‘A’ in the expression for the load current, we get the complete expression for the output load current as
( ) ( ) ( )sin sin
Rt
m m LO
V VE Ei t e
Z R R Z
ω αωω φ α φ− − = − − + − −
The Extinction angle β can be calculated from the final condition that the output
current 0Oi = at tω β= . By using the above expression we get,
( ) ( ) ( )0 sin sin
Rm m L
O
V VE Ei e
Z R R Z
β αωβ φ α φ− − = = − − + − −
To derive an expression for the average or dc load voltage
( ) ( )2
0
1.
2 OO dcV v d tπ
ωπ
= ∫
( ) ( ) ( ) ( )2
0
1. . .
2 O O OO dcV v d t v d t v d tβα π
α β
ω ω ωπ
= + + ∫ ∫ ∫
sin Output load voltage for to O mv V t tω ω α β= = =
for 0 to & for to 2Ov E t tω α ω β π= = =
( ) ( ) ( )2
0
1. sin .
2 mO dcV E d t V t E d tβα π
α β
ω ω ωπ
= + + ∫ ∫ ∫
( ) ( ) ( ) ( )2
0
1cos
2 mO dcV E t V t E tπα β
βα
ω ω ωπ
= + − +
73
( ) ( ) ( ) ( )10 cos cos 2
2 mO dcV E V Eα β α π βπ
= − − − + −
( ) ( ) ( )cos cos 22 2
mO dc
V EV α β π β α
π π= − + − +
( ) ( ) ( )2cos cos
2 2m
O dc
VV E
π β αα β
π π− −
= − +
Conduction angle of thyristor ( )δ β α= −
RMS Output Voltage can be calculated by using the expression
( ) ( )2
2
0
1.
2 OO RMSV v d tπ
ωπ
= ∫
DISADVANTAGES OF SINGLE PHASE HALF WAVE CONTROLLED RECTIFIERS Single phase half wave controlled rectifier gives
• Low dc output voltage. • Low dc output power and lower efficiency. • Higher ripple voltage & ripple current. • Higher ripple factor. • Low transformer utilization factor. • The input supply current waveform has a dc component which can result in dc
saturation of the transformer core.
Single phase half wave controlled rectifiers are rarely used in practice as they give low dc output and low dc output power. They are only of theoretical interest.
The above disadvantages of a single phase half wave controlled rectifier can be over come by using a full wave controlled rectifier circuit. Most of the practical converter circuits use full wave controlled rectifiers.
SINGLE PHASE FULL WAVE CONTROLLED RECTIFIERS Single phase full wave controlled rectifier circuit combines two half wave controlled rectifiers in one single circuit so as to provide two pulse output across the load. Both the half cycles of the input supply are utilized and converted into a uni-directional output current through the load so as to produce a two pulse output waveform. Hence a full wave controlled rectifier circuit is also referred to as a two pulse converter. Single phase full wave controlled rectifiers are of various types
• Single phase full wave controlled rectifier using a center tapped transformer (two pulse converter with mid point configuration).
• Single phase full wave bridge controlled rectifier Half controlled bridge converter (semi converter). Fully controlled bridge converter (full converter).
74
SINGLE PHASE FULL WAVE CONTROLLED RECTIFIER USING A CENTER TAPPED TRANSFORMER
ACSupply
O
A
B
T1
T2
R L
vO
+
FWD
iO
iS
vS
Sv = Supply Voltage across the upper half of the transformer secondary winding
sinS AO mv v V tω= =
sinBO AO mv v V tω= − = − = supply voltage across the lower half of the transformer
secondary winding. This type of full wave controlled rectifier requires a center tapped transformer and two thyristors 1T and 2T . The input supply is fed through the mains supply transformer, the primary side of the transformer is connected to the ac line voltage which is available (normally the primary supply voltage is 230V RMS ac supply voltage at 50Hz supply frequency in India). The secondary side of the transformer has three lines and the center point of the transformer (center line) is used as the reference point to measure the input and output voltages. The upper half of the secondary winding and the thyristor 1T along with the load act as a half wave controlled rectifier, the lower half of the secondary winding and the thyristor 2T with the common load act as the second half wave controlled rectifier so as to produce a full wave load voltage waveform. There are two types of operations possible.
Discontinuous load current operation, which occurs for a purely resistive load or an RL load with low inductance value.
Continuous load current operation which occurs for an RL type of load with large load inductance.
Discontinuous Load Current Operation (for low value of load inductance)
Generally the load current is discontinuous when the load is purely resistive or when the RL load has a low value of inductance.
During the positive half cycle of input supply, when the upper line of the secondary winding is at a positive potential with respect to the center point ‘O’ the thyristor 1T is forward biased and it is triggered at a delay angle of α. The load current
75
flows through the thyristor 1T , through the load and through the upper part of the
secondary winding, during the period to α β , when the thyristor 1T conducts. The output voltage across the load follows the input supply voltage that appears
across the upper part of the secondary winding from to tω α β= . The load current
through the thyristor 1T decreases and drops to zero at tω β= , where β π> for RL type
of load and the thyristor 1T naturally turns off at tω β= .
vOVm
0 α
2π 3πα π β( )π+α ( )π+β
αβ
iO
ωt
ωt0
Fig.: Waveform for Discontinuous Load Current Operation without FWD During the negative half cycle of the input supply the voltage at the supply line
‘A’ becomes negative whereas the voltage at line ‘B’ (at the lower side of the secondary winding) becomes positive with respect to the center point ‘O’. The thyristor 2T is forward biased during the negative half cycle and it is triggered at a delay angle of
( )π α+ . The current flows through the thyristor 2T , through the load, and through the
lower part of the secondary winding when 2T conducts during the negative half cycle the
load is connected to the lower half of the secondary winding when 2T conducts.
For purely resistive loads when L = 0, the extinction angle β π= . The load current falls to zero at tω β π= = , when the input supply voltage falls to zero at tω π= . The load current and the load voltage waveforms are in phase and there is no phase shift between the load voltage and the load current waveform in the case of a purely resistive load. For low values of load inductance the load current would be discontinuous and the extinction angle β π> but ( )β π α< + .
For large values of load inductance the load current would be continuous and does not fall to zero. The thyristor 1T conducts from ( ) to α π α+ , until the next thyristor 2T
is triggered. When 2T is triggered at ( )tω π α= + , the thyristor 1T will be reverse biased
and hence 1T turns off.
76
TO DERIVE AN EXPRESSION FOR THE DC OUTPUT VOLTAGE O F A SINGLE PHASE FULL WAVE CONTROLLED RECTIFIER WITH RL LOAD (WITHOUT FREE WHEELING DIODE (FWD)) The average or dc output voltage of a full-wave controlled rectifier can be calculated by finding the average value of the output voltage waveform over one output
cycle (i.e., π radians) and note that the output pulse repetition time is 2
T seconds where T
represents the input supply time period and 1
Tf
= ; where f = input supply frequency.
Assuming the load inductance to be small so that β π> , ( )β π α< + we obtain
discontinuous load current operation. The load current flows through 1T form
to tω α β= , where α is the trigger angle of thyristor 1T and β is the extinction angle
where the load current through 1T falls to zero at tω β= . Therefore the average or dc output voltage can be obtained by using the expression
( ) ( )2.
2dc OO dct
V V v d tβ
ω α
ωπ =
= = ∫
( ) ( )1.dc OO dc
t
V V v d tβ
ω α
ωπ =
= = ∫
( ) ( )1sin .dc mO dcV V V t d t
β
α
ω ωπ
= = ∫
( ) cosmdcO dc
VV V t
β
α
ωπ
= = −
( ) ( )cos cosmdcO dc
VV V α β
π= = −
Therefore ( ) ( )cos cosmO dc
VV α β
π= − , for discontinuous load current operation,
( )π β π α< < + .
When the load inductance is small and negligible that is 0L ≈ , the extinction angle radiansβ π= . Hence the average or dc output voltage for resistive load is obtained as
( ) ( )cos cosmO dc
VV α π
π= − ; cos 1π = −
( ) ( )( )cos 1mO dc
VV α
π= − −
77
( ) ( )1 cosmO dc
VV α
π= + ; for resistive load, when 0L ≈
THE EFFECT OF LOAD INDUCTANCE Due to the presence of load inductance the output voltage reverses and becomes negative during the time period to tω π β= . This reduces the dc output voltage. To prevent this reduction of dc output voltage due to the negative region in the output load voltage waveform, we can connect a free wheeling diode across the load. The output voltage waveform and the dc output voltage obtained would be the same as that for a full wave controlled rectifier with resistive load. When the Free wheeling diode (FWD) is connected across the load When 1T is triggered at tω α= , during the positive half cycle of the input supply
the FWD is reverse biased during the time period to tω α π= . FWD remains reverse biased and cut-off from to tω α π= . The load current flows through the conducting thyristor 1T , through the RL load and through upper half of the transformer secondary
winding during the time period to α π . At tω π= , when the input supply voltage across the upper half of the secondary winding reverses and becomes negative the FWD turns-on. The load current continues to flow through the FWD from to tω π β= .
vOVm
0 α
2π 3πα π β( )π+α ( )π+β
αβ
iO
ωt
ωt0
Fig.: Waveform for Discontinuous Load Current Operation with FWD EXPRESSION FOR THE DC OUTPUT VOLTAGE OF A SINGLE PH ASE FULL WAVE CONTROLLED RECTIFIER WITH RL LOAD AND FWD
( ) ( )0
1.dc OO dc
t
V V v d tπ
ω
ωπ =
= = ∫
Thyristor 1T is triggered at tω α= . 1T conducts from to tω α π=
78
Output voltage sin ; for O mv V t t toω ω α π= = FWD conducts from to tω π β= and 0Ov ≈ during discontinuous load current
Therefore ( ) ( )1sin .dc mO dcV V V t d t
π
α
ω ωπ
= = ∫
( ) cosmdcO dc
VV V t
π
α
ωπ
= = −
( ) [ ]cos cos ; cos 1mdcO dc
VV V π α π
π= = − + = −
Therefore ( ) ( )1 cosmdcO dc
VV V α
π= = +
The DC output voltage dcV is same as the DC output voltage of a single phase full
wave controlled rectifier with resistive load. Note that the dc output voltage of a single phase full wave controlled rectifier is two times the dc output voltage of a half wave controlled rectifier. CONTROL CHARACTERISTICS OF A SINGLE PHASE FULL WAVE CONTROLLED RECTIFIER WITH R LOAD OR RL LOAD WITH FW D The control characteristic can be obtained by plotting the dc output voltage dcV
versus the trigger angle α . The average or dc output voltage of a single phase full wave controlled rectifier circuit with R load or RL load with FWD is calculated by using the equation
( ) ( )1 cosmdcO dc
VV V α
π= = +
dcV can be varied by varying the trigger angle α from 00 to 180 . (i.e., the range
of trigger angle α is from 0 to π radians). Maximum dc output voltage is obtained when 0α =
( ) ( )max
21 cos0m m
dcdc
V VV V
π π= = + =
Therefore ( )max
2 mdcdc
VV V
π= = for a single phase full wave controlled rectifier.
Normalizing the dc output voltage with respect to its maximum value, we can write the normalized dc output voltage as
79
( )max
dc dcdcn n
dmdc
V VV V
V V= = =
( )
( )1 cos 1
1 cos2 2
m
dcn nm
V
V VV
απ α
π
+= = = +
Therefore ( )11 cos
2dc
dcn ndm
VV V
Vα= = + =
( )11 cos
2dc dmV Vα= +
Trigger angle α
in degrees ( )O dcV Normalized dc output voltage Vn
0 2
0.636619mdm m
VV V
π= = 1
030 0.593974 mV 0.9330 060 0.47746 mV 0.75 090 0.3183098 mV 0.5 0120 0.191549 mV 0.25 0150 0.04264 mV 0.06698 0180 0 0
VO(dc)
Trigger angle in degreesα
0 60 120 180
Vdm
0.2 Vdm
0.6Vdm
Fig.: Control characteristic of a single phase full wave controlled rectifier with R load or RL load with FWD
80
CONTINUOUS LOAD CURRENT OPERATION (WITHOUT FWD) For large values of load inductance the load current flows continuously without decreasing and falling to zero and there is always a load current flowing at any point of time. This type of operation is referred to as continuous current operation. Generally the load current is continuous for large load inductance and for low trigger angles. The load current is discontinuous for low values of load inductance and for large values of trigger angles. The waveforms for continuous current operation are as shown.
vOVm
0
2π 3πα π( )π+α
α
iO
ωt
ωt0
α α α
( )2π+α
T ON1 T ON2 T ON1
Fig.: Load voltage and load current waveform of a single phase full wave controlled rectifier with RL load & without FWD for continuous load current operation
In the case of continuous current operation the thyristor 1T which is triggered at a
delay angle of α , conducts from ( ) to tω α π α= + . Output voltage follows the input
supply voltage across the upper half of the transformer secondary winding sinO AO mv v V tω= = .
The next thyristor 2T is triggered at ( )tω π α= + , during the negative half cycle
input supply. As soon as 2T is triggered at ( )tω π α= + , the thyristor 1T will be reverse
biased and 1T turns off due to natural commutation (ac line commutation). The load
current flows through the thyristor 2T from ( ) ( ) to 2tω π α π α= + + . Output voltage
across the load follows the input supply voltage across the lower half of the transformer secondary winding sinO BO mv v V tω= = − .
Each thyristor conducts for ( )0 radians 180π in the case of continuous current
operation.
81
TO DERIVE AN EXPRESSION FOR THE AVERAGE OR DC OUTPU T VOLTAGE OF SINGLE PHASE FULL WAVE CONTROLLED RECTIF IER WITH LARGE LOAD INDUCTANCE ASSUMING CONTINUOUS LOAD CURRENT OPERATION.
( ) ( )( )
1.dc OO dc
t
V V v d tπ α
ω α
ωπ
+
=
= = ∫
( ) ( )( )
1sin .dc mO dcV V V t d t
π α
α
ω ωπ
+ = =
∫
( )
( )cosm
dcO dc
VV V t
π α
α
ωπ
+ = = −
( ) ( )cos cosmdcO dc
VV V α π α
π= = − + ; ( )cos cosπ α α+ = −
( ) [ ]cos cosmdcO dc
VV V α α
π= = +
( )2
cosmdcO dc
VV V α
π∴ = =
The above equation can be plotted to obtain the control characteristic of a single phase full wave controlled rectifier with RL load assuming continuous load current operation.
Normalizing the dc output voltage with respect to its maximum value, the normalized dc output voltage is given by
( )
( )
max
2cos
cos2
m
dcdcn n
mdc
VV
V VVV
απ α
π
= = = =
Therefore cosdcn nV V α= =
82
Trigger angle α in degrees ( )O dcV Remarks
0 2 m
dm
VV
π =
Maximum dc output voltage
( )max
2 mdmdc
VV V
π = =
030 0.866 dmV 060 0.5 dmV 090 0 dmV 0120 -0.5 dmV 0150 -0.866 dmV
0180 2 m
dm
VV
π − = −
VO(dc)
Trigger angle in degreesα
030 60 90
Vdm
0.2 Vdm
0.6Vdm
-0.6 Vdm
-0.2Vdm
-Vdm
α
120 150 180
Fig.: Control Characteristic
We notice from the control characteristic that by varying the trigger angle α we can vary the output dc voltage across the load. Thus it is possible to control the dc output voltage by changing the trigger angle α . For trigger angle α in the range of 0 to 90 degrees ( )0. ., 0 90i e α≤ ≤ , dcV is positive and the circuit operates as a controlled
rectifier to convert ac supply voltage into dc output power which is fed to the load. For trigger angle 090 ,cosα α> becomes negative and as a result the average dc
output voltage dcV becomes negative, but the load current flows in the same positive direction. Hence the output power becomes negative. This means that the power flows from the load circuit to the input ac source. This is referred to as line commutated inverter operation. During the inverter mode operation for 090α > the load energy can be fed back from the load circuit to the input ac source.
83
TO DERIVE AN EXPRESSION FOR RMS OUTPUT VOLTAGE The rms value of the output voltage is calculated by using the equation
( ) ( )( )
1
222.
2 OO RMSV v d tπ α
α
ωπ
+ =
∫
( ) ( )( )
1
22 21sin .mO RMSV V t d t
π α
α
ω ωπ
+ =
∫
( ) ( )( )
1
222sin .m
O RMS
VV t d t
π α
α
ω ωπ
+ =
∫
( )( ) ( )
( )1
22 1 cos 2.
2m
O RMS
tVV d t
π α
α
ωω
π
+ −=
∫
( ) ( ) ( )( )( )
1
21
cos 2 .2mO RMSV V d t t d t
π α π α
α α
ω ω ωπ
+ + = −
∫ ∫
( ) ( )( ) ( )
1
21 sin 222mO RMS
tV V t
π α π α
α α
ωωπ
+ + = −
( ) ( ) ( )1
2sin 2 sin 21
2 2mO RMSV Vπ α α
π α απ
+ − = + − −
( ) ( )1
21 sin 2 cos 2 cos 2 sin 2 sin 2
2 2mO RMSV Vπ α π α απ
π × + × − = −
( ) ( )1
21 0 sin 2 sin 2
2 2mO RMSV Vα απ
π + − = −
( ) ( )1
21
2 2m
mO RMS
VV V π
π = =
Therefore
( ) 2m
O RMS
VV = ; The rms output voltage is same as the input rms supply voltage.
84
SINGLE PHASE SEMICONVERTERS
Errata: Consider diode 2D as 1D in the figure and diode 1D as 2D
Single phase semi-converter circuit is a full wave half controlled bridge converter which uses two thyristors and two diodes connected in the form of a full wave bridge configuration. The two thyristors are controlled power switches which are turned on one after the other by applying suitable gating signals (gate trigger pulses). The two diodes are uncontrolled power switches which turn-on and conduct one after the other as and when they are forward biased. The circuit diagram of a single phase semi-converter (half controlled bridge converter) is shown in the above figure with highly inductive load and a dc source in the load circuit. When the load inductance is large the load current flows continuously and we can consider the continuous load current operation assuming constant load current, with negligible current ripple (i.e., constant and ripple free load current operation). The ac supply to the semiconverter is normally fed through a mains supply transformer having suitable turns ratio. The transformer is suitably designed to supply the required ac supply voltage (secondary output voltage) to the converter. During the positive half cycle of input ac supply voltage, when the transformer secondary output line ‘A’ is positive with respect to the line ‘B’ the thyristor 1T and the
diode 1D are both forward biased. The thyristor 1T is triggered at tω α= ; ( )0 α π≤ ≤
by applying an appropriate gate trigger signal to the gate of 1T . The current in the circuit
flows through the secondary line ‘A’, through 1T , through the load in the downward
direction, through diode 1D back to the secondary line ‘B’.
1T and 1D conduct together from to tω α π= and the load is connected to the input ac supply. The output load voltage follows the input supply voltage (the secondary output voltage of the transformer) during the period to tω α π= . At tω π= , the input supply voltage decreases to zero and becomes negative during the period ( ) to tω π π α= + . The free wheeling diode mD across the load
becomes forward biased and conducts during the period ( ) to tω π π α= + .
85
Fig:. Waveforms of single phase semi-converter for RLE load and constant load
current for αααα > 900
86
The load current is transferred from 1T and 1D to the FWD mD . 1T and 1D are
turned off. The load current continues to flow through the FWD mD . The load current free wheels (flows continuously) through the FWD during the free wheeling time period
( ) to π π α+ .
During the negative half cycle of input supply voltage the secondary line ‘A’ becomes negative with respect to line ‘B’. The thyristor 2T and the diode 2D are both
forward biased. 2T is triggered at ( )tω π α= + , during the negative half cycle. The FWD
is reverse biased and turns-off as soon as 2T is triggered. The load current continues to
flow through 2T and 2D during the period ( ) to 2tω π α π= +
TO DERIVE AN EXPRESSION FOR THE AVERAGE OR DC OUTPU T VOLTAGE OF A SINGLE PHASE SEMI-CONVERTER
The average output voltage can be found from
( )2sin .
2dc mV V t d tπ
α
ω ωπ
= ∫
[ ]2cos
2m
dc
VV t
π
αωπ
= −
[ ]cos cos ; cos 1mdc
VV π α π
π= − + = −
Therefore [ ]1 cosmdc
VV α
π= +
dcV can be varied from 2 mV
π to 0 by varying α from 0 to π .
The maximum average output voltage is
( )max
2 mdmdc
VV V
π= =
Normalizing the average output voltage with respect to its maximum value
( )0.5 1 cosdcdcn n
dm
VV V
Vα= = = +
The output control characteristic can be plotted by using the expression for dcV
87
TO DERIVE AN EXPRESSION FOR THE RMS OUTPUT VOLTAGE OF A SINGLE PHASE SEMI-CONVERTER
The rms output voltage is found from
( ) ( )1
22 22sin .
2 mO RMSV V t d tπ
α
ω ωπ
=
∫
( ) ( ) ( )1
2 2
1 cos 2 .2
mO RMS
VV t d t
π
α
ω ωπ
= −
∫
( )
1
21 sin 2
22m
O RMS
VV
απ απ = − +
SINGLE PHASE FULL CONVERTER (FULLY CONTROLLED BRIDG E CONVERTER)
The circuit diagram of a single phase fully controlled bridge converter is shown in the figure with a highly inductive load and a dc source in the load circuit so that the load current is continuous and ripple free (constant load current operation). The fully controlled bridge converter consists of four thyristors 1T , 2T , 3T and 4T connected in the form of full wave bridge configuration as shown in the figure. Each thyristor is controlled and turned on by its gating signal and naturally turns off when a reverse voltage appears across it. During the positive half cycle when the upper line of the transformer secondary winding is at a positive potential with respect to the lower end the thyristors 1T and 2T are forward biased during the time interval 0 to tω π= . The
thyristors 1T and 2T are triggered simultaneously ( ) ; 0tω α α π= ≤ ≤ , the load is
connected to the input supply through the conducting thyristors 1T and 2T . The output voltage across the load follows the input supply voltage and hence output voltage
sinO mv V tω= . Due to the inductive load 1T and 2T will continue to conduct beyond
tω π= , even though the input voltage becomes negative. 1T and 2T conduct together
88
during the time period ( ) to α π α+ , for a time duration of π radians (conduction angle
of each thyristor = 0180 ) During the negative half cycle of input supply voltage for to 2tω π π= the thyristors 3T and 4T are forward biased. 3T and 4T are triggered at ( )tω π α= + . As
soon as the thyristors 3T and 4T are triggered a reverse voltage appears across the
thyristors 1T and 2T and they naturally turn-off and the load current is transferred from
1T and 2T to the thyristors 3T and 4T . The output voltage across the load follows the
supply voltage and sinO mv V tω= − during the time period ( ) ( ) to 2tω π α π α= + + . In
the next positive half cycle when 1T and 2T are triggered, 3T and 4T are reverse biased and they turn-off. The figure shows the waveforms of the input supply voltage, the output load voltage, the constant load current with negligible ripple and the input supply current.
89
During the time period to tω α π= , the input supply voltage Sv and the input
supply current Si are both positive and the power flows from the supply to the load. The
converter operates in the rectification mode during to tω α π= . During the time period ( ) to tω π π α= + , the input supply voltage Sv is negative
and the input supply current Si is positive and there will be reverse power flow from the load circuit to the input supply. The converter operates in the inversion mode during the time period ( ) to tω π π α= + and the load energy is fed back to the input source.
The single phase full converter is extensively used in industrial applications up to about 15kW of output power. Depending on the value of trigger angle α , the average output voltage may be either positive or negative and two quadrant operation is possible. TO DERIVE AN EXPRESSION FOR THE AVERAGE (DC) OUTPUT VOLTAGE The average (dc) output voltage can be determined by using the expression
( ) ( )2
0
1. ;
2dc OO dcV V v d tπ
ωπ
= = ∫
The output voltage waveform consists of two output pulses during the input
supply time period between 0 & 2 radiansπ . In the continuous load current operation of a single phase full converter (assuming constant load current) each thyristor conduct for π radians (1800) after it is triggered. When thyristors 1T and 2T are triggered at tω α=
1T and 2T conduct from ( ) to α π α+ and the output voltage follows the input supply
voltage. Therefore output voltage sinO mv V tω= ; for ( ) to tω α π α= +
Hence the average or dc output voltage can be calculated as
( ) ( )2sin .
2dc mO dcV V V t d tπ α
α
ω ωπ
+ = =
∫
( ) ( )1sin .dc mO dcV V V t d t
π α
α
ω ωπ
+ = =
∫
( ) ( )sin .mdcO dc
VV V t d t
π α
α
ω ωπ
+ = =
∫
( ) [ ]cosmdcO dc
VV V t
π α
αωπ
+= = −
( ) ( )cos cosmdcO dc
VV V π α α
π= = − + + ; ( )cos cosπ α α+ = −
Therefore ( )2
cosmdcO dc
VV V α
π= =
90
The dc output voltage dcV can be varied from a maximum value of 02 for 0 to mV α
π= a
minimum value of 02 for radians 180mV α π
π− = =
The maximum average dc output voltage is calculated for a trigger angle 00α = and is obtained as
( ) ( )max
2 2cos 0m m
dmdc
V VV V
π π= = × =
Therefore ( )max
2 mdmdc
VV V
π= =
The normalized average output voltage is given by
( )
( )max
O dc dcdcn n
dmdc
V VV V
V V= = =
2cos
cos2
m
dcn nm
V
V VV
απ α
π
= = =
Therefore cosdcn nV V α= = ; for a single phase full converter assuming continuous and constant load current operation. CONTROL CHARACTERISTIC OF SINGLE PHASE FULL CONVERT ER The dc output control characteristic can be obtained by plotting the average or dc output voltage dcV versus the trigger angle α
For a single phase full converter the average dc output voltage is given by the
equation ( )2
cosmdcO dc
VV V α
π= =
Trigger angle α
in degrees ( )O dcV Remarks
0 2 m
dm
VV
π =
Maximum dc output voltage
( )max
2 mdmdc
VV V
π = =
030 0.866 dmV 060 0.5 dmV 090 0 dmV 0120 -0.5 dmV 0150 -0.866 dmV
0180 2 m
dm
VV
π − = −
91
VO(dc)
Trigger angle in degreesα
030 60 90
Vdm
0.2 Vdm
0.6Vdm
-0.6 Vdm
-0.2Vdm
-Vdm
α
120 150 180
Fig.: Control Characteristic
We notice from the control characteristic that by varying the trigger angle α we can vary the output dc voltage across the load. Thus it is possible to control the dc output voltage by changing the trigger angle α . For trigger angle α in the range of 0 to 90 degrees ( )0. ., 0 90i e α≤ ≤ , dcV is positive and the average dc load current dcI is also
positive. The average or dc output power dcP is positive, hence the circuit operates as a controlled rectifier to convert ac supply voltage into dc output power which is fed to the load. For trigger angle 090 ,cosα α> becomes negative and as a result the average dc
output voltage dcV becomes negative, but the load current flows in the same positive
direction i.e., dcI is positive . Hence the output power becomes negative. This means that the power flows from the load circuit to the input ac source. This is referred to as line commutated inverter operation. During the inverter mode operation for 090α > the load energy can be fed back from the load circuit to the input ac source TWO QUADRANT OPERATION OF A SINGLE PHASE FULL CONVE RTER
92
The above figure shows the two regions of single phase full converter operation in the dcV versus dcI plane. In the first quadrant when the trigger angle α is less than 900,
and dc dcV I are both positive and the converter operates as a controlled rectifier and converts the ac input power into dc output power. The power flows from the input source to the load circuit. This is the normal controlled rectifier operation where dcP is positive.
When the trigger angle is increased above 900 , dcV becomes negative but dcI is
positive and the average output power (dc output power) dcP becomes negative and the power flows from the load circuit to the input source. The operation occurs in the fourth quadrant where dcV is negative and dcI is positive. The converter operates as a line commutated inverter.
TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF THE OU TPUT VOLTAGE The rms value of the output voltage is calculated as
( ) ( )2
2
0
1.
2 OO RMSV v d tπ
ωπ
= ∫
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
( ) ( )22.
2 OO RMSV v d tπ α
α
ωπ
+ =
∫
( ) ( )2 21sin .mO RMSV V t d t
π α
α
ω ωπ
+ =
∫
( ) ( )2
2sin .mO RMS
VV t d t
π α
α
ω ωπ
+ =
∫
( )( ) ( )
2 1 cos 2.
2m
O RMS
tVV d t
π α
α
ωω
π
+ −=
∫
( ) ( ) ( )2
cos 2 .2
mO RMS
VV d t t d t
π α π α
α α
ω ω ωπ
+ + = −
∫ ∫
( ) ( )2 sin 2
22m
O RMS
V tV t
π α π α
α α
ωωπ
+ + = −
93
( ) ( ) ( )2 sin 2 sin 2
2 2m
O RMS
VV
π α απ α α
π + −
= + − −
( ) ( ) ( ) ( )2 sin 2 2 sin 2
; sin 2 2 sin 22 2
mO RMS
VV
π α απ π α α
π + −
= − + =
( ) ( )2 sin 2 sin 2
2 2m
O RMS
VV
α αππ − = −
( ) ( )2 2
02 2 2
m m mO RMS
V V VV π
π= − = =
Therefore ( ) 2m
SO RMS
VV V= =
Hence the rms output voltage is same as the rms input supply voltage
The rms thyristor current can be calculated as Each thyristor conducts for π radians or 0180 in a single phase full converter operating at continuous and constant load current.
Therefore rms value of the thyristor current is calculated as
( ) ( ) ( )1
2 2T RMS O RMS O RMSI I Iππ
= =
( )( )
2
O RMS
T RMS
II =
The average thyristor current can be calculated as
( ) ( ) ( )1
2 2T Avg O dc O dcI I Iππ
= × = ×
( )( )
2O dc
T Avg
II =
94
SINGLE PHASE DUAL CONVERTER
95
We have seen in the case of a single phase full converter with inductive loads the converter can operate in two different quadrants in the versus dc dcV I operating diagram. If two single phase full converters are connected in parallel and in opposite direction (connected in back to back) across a common load four quadrant operation is possible. Such a converter is called as a dual converter which is shown in the figure. The dual converter system will provide four quadrant operation and is normally used in high power industrial variable speed drives. The converter number 1 provides a positive dc output voltage and a positive dc load current, when operated in the rectification mode.
The converter number 2 provides a negative dc output voltage and a negative dc load current when operated in the rectification mode. We can thus have bi-directional load current and bi-directional dc output voltage. The magnitude of output dc load voltage and the dc load current can be controlled by varying the trigger angles 1 2 & α α of the converters 1 and 2 respectively.
Fig.: Four quadrant operation of a dual converter
There are two modes of operations possible for a dual converter system.
• Non circulating current mode of operation (circulating current free mode of operation).
• Circulating current mode of operation. NON CIRCULATING CURRENT MODE OF OPERATION (CIRCULAT ING CURRENT FREE MODE OF OPERATION) In this mode of operation only one converter is switched on at a time while the second converter is switched off. When the converter 1 is switched on and the gate trigger signals are released to the gates of thyristors in converter 1, we get an average output voltage across the load, which can be varied by adjusting the trigger angle 1α of the
converter 1. If 1α is less than 900, the converter 1 operates as a controlled rectifier and
converts the input ac power into dc output power to feed the load. dcV and dcI are both positive and the operation occurs in the first quadrant. The average output power
dc dc dcP V I= × is positive. The power flows from the input ac supply to the load. When 1α
is increased above 900 converter 1 operates as a line commutated inverter and dcV
becomes negative while dcI is positive and the output power dcP becomes negative. The power is fed back from the load circuit to the input ac source through the converter 1. The load current falls to zero when the load energy is utilized completely. The second converter 2 is switched on after a small delay of about 10 to 20 mill seconds to allow all the thyristors of converter 1 to turn off completely. The gate signals
96
are released to the thyristor gates of converter 2 and the trigger angle 2α is adjusted such
that 020 90α≤ ≤ so that converter 2 operates as a controlled rectifier. The dc output
voltage dcV and dcI are both negative and the load current flows in the reverse direction.
The magnitude of dcV and dcI are controlled by the trigger angle 2α . The operation
occurs in the third quadrant where dcV and dcI are both negative and output power dcP is positive and the converter 2 operates as a controlled rectifier and converts the ac supply power into dc output power which is fed to the load.
When we want to reverse the load current flow so that dcI is positive we have to
operate converter 2 in the inverter mode by increasing the trigger angle 2α above 090 .
When 2α is made greater than 090 , the converter 2 operates as a line commutated inverter and the load power (load energy) is fed back to ac mains. The current falls to zero when all the load energy is utilized and the converter 1 can be switched on after a short delay of 10 to 20 milli seconds to ensure that the converter 2 thyristors are completely turned off.
The advantage of non circulating current mode of operation is that there is no circulating current flowing between the two converters as only one converter operates and conducts at a time while the other converter is switched off. Hence there is no need of the series current limiting inductors between the outputs of the two converters. The current rating of thyristors is low in this mode.
But the disadvantage is that the load current tends to become discontinuous and the transfer characteristic becomes non linear. The control circuit becomes complex and the output response is sluggish as the load current reversal takes some time due to the time delay between the switching off of one converter and the switching on of the other converter. Hence the output dynamic response is poor. Whenever a fast and frequent reversal of the load current is required, the dual converter is operated in the circulating current mode.
CIRCULATING CURRENT MODE OF OPERATION In this mode of operation both the converters 1 and 2 are switched on and operated simultaneously and both the converters are in a state of conduction. If converter 1 is operated as a controlled rectifier by adjusting the trigger angle 1α between 0 to 900 the second converter 2 is operated as a line commutated inverter by increasing its trigger angle 2α above 900. The trigger angles 1α and 2α are adjusted such that they produce the same average dc output voltage across the load terminals.
The average dc output voltage of converter 1 is
1 1
2cosm
dc
VV α
π=
The average dc output voltage of converter 2 is
2 2
2cosm
dc
VV α
π=
97
In the dual converter operation one converter is operated as a controlled rectifier with 0
1 90α < and the second converter is operated as a line commutated inverter in the
inversion mode with 02 90α > .
1 2dc dcV V= −
( )1 2 2
2 2 2cos cos cosm m mV V Vα α α
π π π−= = −
Therefore ( )1 2 2 1 1cos cos or cos cos cosα α α α π α= − = − = −
Therefore ( )2 1α π α= − or ( )1 2 radiansα α π+ =
Which gives ( )2 1α π α= −
When the trigger angle 1α of converter 1 is set to some value the trigger angle 2α
of the second converter is adjusted such that ( )02 1180α α= − . Hence for circulating
current mode of operation where both converters are conducting at the same time
( ) 01 2 180α α+ = so that they produce the same dc output voltage across the load.
When 01 90α < (say 0
1 30α = ) the converter 1 operates as a controlled rectifier
and converts the ac supply into dc output power and the average load current dcI is positive. At the same time the converter 2 is switched on and operated as a line
commutated inverter, by adjusting the trigger angle 2α such that ( )02 1180α α= − , which
is equal to 1500 , when 01 30α = . The converter 2 will operate in the inversion mode and
feeds the load energy back to the ac supply. When we want to reverse the load current flow we have to switch the roles of the two converters. When converter 2 is operated as a controlled rectifier by adjusting the trigger angle 2α such that 0
2 90α < , the first converter1 is operated as a line commutated
inverter, by adjusting the trigger angle 1α such that 01 90α > . The trigger angle 1α is
adjusted such that ( )01 2180α α= − for a set value of 2α .
In the circulating current mode a current builds up between the two converters even when the load current falls to zero. In order to limit the circulating current flowing between the two converters, we have to include current limiting reactors in series between the output terminals of the two converters. The advantage of the circulating current mode of operation is that we can have faster reversal of load current as the two converters are in a state of conduction simultaneously. This greatly improves the dynamic response of the output giving a faster dynamic response. The output voltage and the load current can be linearly varied by adjusting the trigger angles 1 2&α α to obtain a smooth and linear output control. The control circuit becomes relatively simple. The transfer characteristic between the output voltage and the trigger angle is linear and hence the output response is very fast. The load current is free to flow in either direction at any time. The reversal of the load current can be done in a faster and smoother way.
98
The disadvantage of the circulating current mode of operation is that a current flows continuously in the dual converter circuit even at times when the load current is zero. Hence we should connect current limiting inductors (reactors) in order to limit the peak circulating current within specified value. The circulating current flowing through the series inductors gives rise to increased power losses, due to dc voltage drop across the series inductors which decreases the efficiency. Also the power factor of operation is low. The current limiting series inductors are heavier and bulkier which increases the cost and weight of the dual converter system. The current flowing through the converter thyristors is much greater than the dc load current. Hence the thyristors should be rated for a peak thyristor current of
( ) ( ) ( )max max maxT dc rI I i= + , where ( )maxdcI is the maximum dc load current and ( )maxri is the
maximum value of the circulating current. TO CALCULATE THE CIRCULATING CURRENT
Fig.: Waveforms of dual converter
99
As the instantaneous output voltages of the two converters are out of phase, there will be an instantaneous voltage difference and this will result in circulating current between the two converters. In order to limit the circulating current, current limiting reactors are connected in series between the outputs of the two converters. This circulating current will not flow through the load and is normally limited by the current reactor Lr . If vO1 and vO2 are the instantaneous output voltages of the converters 1 and 2, respectively the circulating current 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.
( )( )
( )1
1 2
2
1. ;
t
r r r O Or
i v d t v v vL
ω
π α
ωω −
= = −
∫
As the output voltage 2Ov is negative ( )1 2r O Ov v v= +
Therefore ( ) ( )( )1
1 2
2
1. ;
t
r O Or
i v v d tL
ω
π α
ωω −
= +
∫
( )1 1sin for 2 to O mv V t tω π α ω= − −
( ) ( )( )( )1 12 2
sin . sin .t t
mr
r
Vi t d t t d t
L
ω ω
π α π α
ω ω ω ωω − −
= − −
∫ ∫
( )( )
( )( )1 12 2
cos cost t
mr
r
Vi t t
L
ω ω
π α π α
ω ωω − −
= +
( ) ( ) ( ) ( )1 1cos cos 2 cos cos 2mr
r
Vi t t
Lω π α ω π α
ω= − − + − −
( )12cos 2cos 2mr
r
Vi t
Lω π α
ω= − −
( )1
2cos cosm
rr
Vi t
Lω α
ω= −
The instantaneous value of the circulating current depends on the delay angle.
100
For trigger angle (delay angle) α1 = 0, its magnitude becomes minimum when , 0, 2,4,....t n nω π= = and magnitude becomes maximum when , 1,3,5,....t n nω π= =
If the peak load current is pI , one of the converters that controls the power flow
may carry a peak current of 4 m
pr
VI
Lω+ ,
Where ( ) ( )max max
4, & m m
p L rL r
V VI I i
R Lω= = =
Problems
1. What will be the average power in the load for the circuit shown, when 4
πα = .
Assume SCR to be ideal. Supply voltage is 330 sin314t. Also calculate the RMS power and the rectification efficiency.
T
100ΩR330Sin314t ~
+
−
The circuit is that of a single phase half wave controlled rectifier with a resistive load
( )1 cos ; 2 4
mdc
VV radians
πα απ
= + =
330
1 cos2 4dcV
ππ = +
89.66 VoltsdcV =
Average Power 2 289.66
80.38 Watts100
dcV
R= = =
89.66
0.8966 Amps100
dcdc
VI
R= = =
1
21 sin 2
2 2m
RMS
VV
απ απ = − +
101
1
2
sin 2330 1 42 4 2RMSV
πππ
π
× = − +
157.32 RMSV V=
RMS Power (AC power)
2 2157.32247.50 Watts
100RMSV
R= = =
Rectification Efficiency Average power
RMS power=
80.38
0.3248247.47
= =
2. In the circuit shown find out the average voltage across the load assuming that the
conduction drop across the SCR is 1 volt. Take α = 450.
100ΩR330Sin314t ~
+
−
VAK
The wave form of the load voltage is shown below (not to scale).
2π 3ππα
βγ0
Vm
VAK
ωt
Voltage acrossresistanceL
oad
volta
ge
It is observed that the SCR turns off when tω β= , where ( )β π γ= − because the
SCR turns-off for anode supply voltage below 1 Volt.
sin 1 volt (given)AK mV V γ= =
102
Therefore ( )1 1 01sin sin 0.17 0.003 radians
330AK
m
V
Vγ − − = = =
( )0180β γ= − ; By symmetry of the curve.
0179.83β = ; 3.138 radians.
( ) ( )1sin
2dc m AKV V t V d tβ
α
ω ωπ
= −∫
( ) ( )1sin .
2dc m AKV V t d t V d tβ β
α α
ω ω ωπ
= − ∫ ∫
( ) ( )1cos
2dc m AKV V t V tβ β
α α
ω ωπ
= − −
( ) ( )1cos cos
2dc m AKV V Vα β β απ
= − − −
( ) ( )0 01330 cos 45 cos179.83 1 3.138 0.003
2dcVπ = − − −
89.15 VoltsdcV =
Note: β and α values should be in radians
3. In the figure find out the battery charging current when 4
πα = . Assume ideal
SCR.
24V(V )B
200 V50 Hz ~
+
−
10Ω
R
Solution It is obvious that the SCR cannot conduct when the instantaneous value of the supply voltage is less than 24 V, the battery voltage. The load voltage waveform is as shown (voltage across ion).
103
2π 3ππ
αβγ
0
Vm
VB
ωt
Voltage acrossresistance
sinB mV V γ=
24 200 2 sinγ=
1 024sin 4.8675 0.085 radians
200 2γ − = = = ×
3.056 radiansβ π γ= − = Average value of voltage across 10Ω
( ) ( )1sin .
2 m BV t V d tβ
α
ω ωπ
= − ∫
(The integral gives the shaded area)
( ) ( )3.056
4
1200 2 sin 24 .
2t d t
π
ω ωπ
= × −
∫
1
200 2 cos cos3.056 24 3.0562 4 4
π ππ = − − −
68 Vots= Therefore charging current
Average voltage across R
R=
68
6.8 Amps10
= =
Note: If value of γ is more than α , then the SCR will trigger only at tω γ= , (assuming that the gate signal persists till then), when it becomes forward biased.
104
Therefore ( ) ( )1sin .
2dc m BV V t V d tβ
γ
ω ωπ
= − ∫
4. In a single phase full wave rectifier supply is 200 V AC. The load resistance is
10Ω , 060α = . Find the average voltage across the load and the power consumed in the load. Solution In a single phase full wave rectifier
( )1 cosmdc
VV α
π= +
( )0200 21 cos60dcV
π×= +
135 VoltsdcV = Average Power
2 2135
1.823 10
dcVkW
R= = =
5. In the circuit shown find the charging current if the trigger angle 090α = .
R = 10 Ω
+
−
200 V50 Hz ~
+
−10V(V )B
Solution With the usual notation sinB mV V γ=
10 200 2 sinγ=
Therefore 1 10sin 0.035 radians
200 2γ − = = ×
105
090 radians2
πα = = ; ( ) 3.10659β π γ= − =
Average voltage across ( ) ( )210 sin .
2 m BV t V d tβ
α
ω ωπ
Ω = − ∫
( )1cosm BV t V t
β
αω ω
π= − −
( ) ( )1cos cosm BV Vα β β α
π= − − −
1
200 2 cos cos3.106 10 3.1062 2
π ππ = × − − −
85 V =
Note that the values of & α β are in radians.
Charging current dc voltage across resistance
resistance=
85
8.5 Amps10
= =
6. A single phase full wave controlled rectifier is used to supply a resistive load of
10 Ω from a 230 V, 50 Hz, supply and firing angle of 900. What is its mean load voltage? If a large inductance is added in series with the load resistance, what will be the new output load voltage?
Solution For a single phase full wave controlled rectifier with resistive load,
( )1 cosmdc
VV α
π= +
230 2
1 cos2dcVπ
π× = +
103.5 VoltsdcV = When a large inductance is added in series with the load, the output voltage wave form will be as shown below, for trigger angle 090α = .
106
2π
3π
π0
V0
ωt
α
2
cosmdc
VV α
π=
Since 2
πα = ; cos 02
cosπα = =
Therefore 0dcV = and this is evident from the waveform also.
7. The figure shows a battery charging circuit using SCRs. The input voltage to the
circuit is 230 V RMS. Find the charging current for a firing angle of 450. If any one of the SCR is open circuited, what is the charging current? Solution
10Ω
+
−
Vs
VL
~+
−
100V
With the usual notations sinS mV V tω=
2 230sinSV tω= × sinm BV Vγ = , the battery voltage
2 230sin 100γ× =
107
Therefore 1 100sin
2 230γ − = ×
017.9 or 0.312 radiansγ = ( ) ( )0.312β π γ π= − = −
2.829 radiansβ = Average value of voltage across load resistance
( ) ( )2sin
2 m BV t V d tβ
α
ω ωπ
= − ∫
( )1cosm BV t V t
β
αω ω
π= − −
( ) ( )1cos cosm BV Vα β β α
π= − − −
1
230 2 cos cos 2.829 100 2.8294 4
π ππ = × − − −
( )1230 2 0.707 0.9517 204.36
π = × + −
106.68 Volts=
Charging current Voltage across resistance
R=
106.68
10.668 Amps10
= =
If one of the SCRs is open circuited, the circuit behaves like a half wave rectifier. The average voltage across the resistance and the charging current will be half of that of a full wave rectifier.
Therefore Charging Current 10.668
5.334 Amps2
= =
108
THREE PHASE CONTROLLED RECTIFIERS INTRODUCTION TO 3-PHASE CONTROLLED RECTIFIERS Single phase half controlled bridge converters & fully controlled bridge converters are used extensively in industrial applications up to about 15kW of output power. The
single phase controlled rectifiers provide a maximum dc output of ( )max
2 mdc
VV
π= .
The output ripple frequency is equal to the twice the ac supply frequency. The single phase full wave controlled rectifiers provide two output pulses during every input supply cycle and hence are referred to as two pulse converters.
Three phase converters are 3-phase controlled rectifiers which are used to convert ac input power supply into dc output power across the load. Features of 3-phase controlled rectifiers are
• Operate from 3 phase ac supply voltage. • They provide higher dc output voltage and higher dc output power. • Higher output voltage ripple frequency. • Filtering requirements are simplified for smoothing out load voltage and load
current
Three phase controlled rectifiers are extensively used in high power variable speed industrial dc drives. 3-PHASE HALF WAVE CONVERTER
Three single phase half-wave converters are connected together to form a three phase half-wave converter as shown in the figure.
109
THEE PHASE SUPPLY VOLTAGE EQUATIONS We define three line neutral voltages (3 phase voltages) as follows sin ;RN an mv v V tω= = Max. Phase VoltagemV =
2
sin3YN bn mv v V tπω = = −
( )0sin 120YN bn mv v V tω= = −
2
sin3BN cn mv v V tπω = = +
( )0sin 120BN cn mv v V tω= = +
( )0sin 240BN cn mv v V tω= = −
VAN
VCN
VBN
1200
1200
1200
Vector diagram of 3-phase supply voltages
110
The 3-phase half wave converter combines three single phase half wave controlled rectifiers in one single circuit feeding a common load. The thyristor 1T in series with one
of the supply phase windings ' 'a n− acts as one half wave controlled rectifier. The second thyristor 2T in series with the supply phase winding ' 'b n− acts as the second half
wave controlled rectifier. The third thyristor 3T in series with the supply phase winding
' 'c n− acts as the third half wave controlled rectifier. The 3-phase input supply is applied through the star connected supply transformer as shown in the figure. The common neutral point of the supply is connected to one end of the load while the other end of the load connected to the common cathode point.
When the thyristor 1T is triggered at ( )0306
tπω α α = + = +
, the phase voltage
anv appears across the load when 1T conducts. The load current flows through the supply
phase winding ' 'a n− and through thyristor 1T as long as 1T conducts.
When thyristor 2T is triggered at ( )05150
6t
πω α α = + = +
, 1T becomes reverse
biased and turns-off. The load current flows through the thyristor 2T and through the
supply phase winding ' 'b n− . When 2T conducts the phase voltage bnv appears across the
load until the thyristor 3T is triggered .
When the thyristor 3T is triggered at ( )03270
2t
πω α α = + = +
, 2T is reversed
biased and hence 2T turns-off. The phase voltage cnv appears across the load when 3T conducts.
When 1T is triggered again at the beginning of the next input cycle the thyristor 3T
turns off as it is reverse biased naturally as soon as 1T is triggered. The figure shows the 3-phase input supply voltages, the output voltage which appears across the load, and the load current assuming a constant and ripple free load current for a highly inductive load and the current through the thyristor 1T .
For a purely resistive load where the load inductance ‘L = 0’ and the trigger angle
6
πα >
, the load current appears as discontinuous load current and each thyristor is
naturally commutated when the polarity of the corresponding phase supply voltage reverses. The frequency of output ripple frequency for a 3-phase half wave converter is 3 Sf , where Sf is the input supply frequency.
The 3-phase half wave converter is not normally used in practical converter
systems because of the disadvantage that the supply current waveforms contain dc components (i.e., the supply current waveforms have an average or dc value).
111
TO DERIVE AN EXPRESSION FOR THE AVERAGE OUTPUT VOLT AGE OF A 3-PHASE HALF WAVE CONVERTER FOR CONTINUOUS LOAD C URRENT
The reference phase voltage is sinRN an mv v V tω= = . The trigger angle α is
measured from the cross over points of the 3-phase supply voltage waveforms. When the phase supply voltage anv begins its positive half cycle at 0tω = , the first cross over point
appears at 0 306
t radiansπω = =
.
The trigger angle α for the thyristor 1T is measured from the cross over point at 030tω = . The thyristor 1T is forward biased during the period 0 030 to 150tω = , when the
phase supply voltage anv has a higher amplitude than the other phase supply voltages.
Hence 1T can be triggered between 0 030 to 150. When the thyristor 1T is triggered at a
trigger angle α , the average or dc output voltage for continuous load current is calculated using the equation
( )5
6
6
3.
2dc OV v d t
π α
π α
ωπ
+
+
=
∫
Output voltage ( ) ( )0 0sin for 30 to 150O an mv v V t tω ω α α= = = + +
( )5
6
6
3sin .
2dc mV V t d t
π α
π α
ω ωπ
+
+
=
∫
As the output load voltage waveform has three output pulses during the input cycle of 2π radians
( )5
6
6
3sin .
2m
dc
VV t d t
π α
π α
ω ωπ
+
+
=
∫
( )5
6
6
3cos
2m
dc
VV t
π α
π α
ωπ
+
+
= −
112
3 5cos cos
2 6 6m
dc
VV
π πα απ = − + + +
Note from the trigonometric relationship
( ) ( )cos cos .cos sin .sinA B A B A B+ = −
( ) ( ) ( ) ( )3 5 5cos cos sin sin cos .cos sin sin
2 6 6 6 6m
dc
VV
π π π πα α α απ = − + + −
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0 0 0 03cos 150 cos sin 150 sin cos 30 .cos sin 30 sin
2m
dc
VV α α α α
π = − + + −
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0 0 0 0 0 03cos 180 30 cos sin 180 30 sin cos 30 .cos sin 30 sin
2m
dc
VV α α α α
π = − − + − + −
Note: ( ) ( )0 0 0cos 180 30 cos 30− = −
( ) ( )0 0 0sin 180 30 sin 30− =
Therefore
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0 0 0 03cos 30 cos sin 30 sin cos 30 .cos sin 30 sin
2m
dc
VV α α α α
π = + + + −
( ) ( )032cos 30 cos
2m
dc
VV α
π =
( )3 32 cos
2 2m
dc
VV α
π
= ×
( ) ( )3 3 33 cos cos
2 2m m
dc
V VV α α
π π = =
113
( )3cos
2Lm
dc
VV α
π=
Where
3 Max. line to line supply voltageLm mV V= = for a 3-phase star connected transformer.
The maximum average or dc output voltage is obtained at a delay angle α = 0 and is given by
( )max
3 3
2m
dmdc
VV V
π= =
Where mV is the peak phase voltage. And the normalized average output voltage is
cosdcdcn n
dm
VV V
Vα= = =
TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF THE OU TPUT VOLTAGE OF A 3-PHASE HALF WAVE CONVERTER FOR CONTIN UOUS LOAD CURRENT The rms value of output voltage is found by using the equation
( ) ( )
15 26
2 2
6
3sin .
2 mO RMSV V t d t
π α
π α
ω ωπ
+
+
=
∫
and we obtain
( )
1
21 33 cos 2
6 8mO RMSV V απ
= +
114
3 PHASE HALF WAVE CONTROLLED RECTIFIER OUTPUT VOLTA GE WAVEFORMS FOR DIFFERENT TRIGGER ANGLES WITH RL LOAD
0
0
0
300
300
300
600
600
600
900
900
900
1200
1200
1200
1500
1500
1500
1800
1800
1800
2100
2100
2100
2400
2400
2400
2700
2700
2700
3000
3000
3000
3300
3300
3300
3600
3600
3600
3900
3900
3900
4200
4200
4200
Van
↑V 0
↑V 0
↑V 0
Van
Van
α
α
α
α=300
α=600
α=900
Vbn
Vbn
Vbn
Vcn
Vcn
Vcn
ωt
ωt
ωt
115
3 PHASE HALF WAVE CONTROLLED RECTIFIER OUTPUT VOLTA GE WAVEFORMS FOR DIFFERENT TRIGGER ANGLES WITH R LOAD
0
0
0
0
300
300
300
300
600
600
600
600
900
900
900
900
1200
1200
1200
1200
1500
1500
1500
1500
1800
1800
1800
1800
2100
2100
2100
2100
2400
2400
2400
2400
2700
2700
2700
2700
3000
3000
3000
3000
3300
3300
3300
3300
3600
3600
3600
3600
3900
3900
3900
3900
4200
4200
4200
4200
Vs
V0
Van
α
α
α
α=0
α=150
Vbn Vcn
ωt
Van Vbn Vcn
ωt
V0
α=300
Van Vbn Vcn
ωt
V0
α=600
VanVbn Vcn
ωt
116
TO DERIVE AN EXPRESSION FOR THE AVERAGE OR DC OUTPU T VOLTAGE OF A 3 PHASE HALF WAVE CONVERTER WITH RESIS TIVE LOAD OR RL LOAD WITH FWD. In the case of a three-phase half wave controlled rectifier with resistive load, the
thyristor 1T is triggered at ( )030tω α= + and 1T conducts up to 0180tω π= = radians.
When the phase supply voltage anv decreases to zero at tω π= , the load current falls to
zero and the thyristor 1T turns off. Thus 1T conducts from ( ) ( )0 030 to 180tω α= + .
Hence the average dc output voltage for a 3-pulse converter (3-phase half wave controlled rectifier) is calculated by using the equation
( )0
0
180
30
3.
2dc OV v d tα
ωπ +
=
∫
( ) ( )0 0sin ; for 30 to 180O an mv v V t tω ω α= = = +
( )0
0
180
30
3sin .
2dc mV V t d tα
ω ωπ +
=
∫
( )0
0
180
30
3sin .
2m
dc
VV t d t
α
ω ωπ +
=
∫
0
0
180
30
3cos
2m
dc
VV t
α
ωπ
+
= −
( )0 03cos180 cos 30
2m
dc
VV α
π = − + +
Since 0cos180 1,= −
We get ( )031 cos 30
2m
dc
VV α
π = + +
117
THREE PHASE SEMICONVERTERS 3-phase semi-converters are three phase half controlled bridge controlled rectifiers which employ three thyristors and three diodes connected in the form of a bridge configuration. Three thyristors are controlled switches which are turned on at appropriate times by applying appropriate gating signals. The three diodes conduct when they are forward biased by the corresponding phase supply voltages. 3-phase semi-converters are used in industrial power applications up to about 120kW output power level, where single quadrant operation is required. The power factor of 3-phase semi-converter decreases as the trigger angle α increases. The power factor of a 3-phase semi-converter is better than three phase half wave converter. The figure shows a 3-phase semi-converter with a highly inductive load and the load current is assumed to be a constant and continuous load current with negligible ripple.
Thyristor 1T is forward biased when the phase supply voltage anv is positive and
greater than the other phase voltages bnv and cnv . The diode 1D is forward biased when
the phase supply voltage cnv is more negative than the other phase supply voltages.
Thyristor 2T is forward biased when the phase supply voltage bnv is positive and
greater than the other phase voltages. Diode 2D is forward biased when the phase supply
voltage anv is more negative than the other phase supply voltages.
Thyristor 3T is forward biased when the phase supply voltage cnv is positive and
greater than the other phase voltages. Diode 3D is forward biased when the phase supply
voltage bnv is more negative than the other phase supply voltages. The figure shows the waveforms for the three phase input supply voltages, the
output voltage, the thyristor and diode current waveforms, the current through the free wheeling diode mD and the supply current ai . The frequency of the output supply
waveform is 3 Sf , where Sf is the input ac supply frequency. The trigger angle α can be
varied from 0 00 to 180 .
During the time period 7
6 6t
π πω ≤ ≤
i.e., for 0 030 210tω≤ ≤ , thyristor 1T is
forward biased. If 1T is triggered at 6
tπω α = +
, 1T and 1D conduct together and the
118
line to line supply voltage acv appears across the load. At 7
6t
πω =
, acv starts to
become negative and the free wheeling diode mD turns on and conducts. The load current
continues to flow through the free wheeling diode mD and thyristor 1T and diode 1D are turned off.
If the free wheeling diode mD is not connected across the load, then 1T would
continue to conduct until the thyristor 2T is triggered at 5
6t
πω α = +
and the free
wheeling action is accomplished through 1T and 2D , when 2D turns on as soon as anv
becomes more negative at 7
6t
πω =
. If the trigger angle 3
πα ≤
each thyristor
conducts for 2
3
π radians ( )0120 and the free wheeling diode mD does not conduct. The
waveforms for a 3-phase semi-converter with 3
πα ≤
is shown in figure
119
120
We define three line neutral voltages (3 phase voltages) as follows sin ;RN an mv v V tω= = Max. Phase VoltagemV =
2
sin3YN bn mv v V tπω = = −
( )0sin 120YN bn mv v V tω= = −
2
sin3BN cn mv v V tπω = = +
( )0sin 120BN cn mv v V tω= = +
( )0sin 240BN cn mv v V tω= = −
The corresponding line-to-line voltages are
( ) 3 sin6RB ac an cn mv v v v V tπω = = − = −
( ) 53 sin
6YR ba bn an mv v v v V tπω = = − = −
( ) 3 sin2BY cb cn bn mv v v v V tπω = = − = +
( ) 3 sin6RY ab an bn mv v v v V tπω = = − = +
Where mV is the peak phase voltage of a star (Y) connected source. TO DERIVE AN EXPRESSION FOR THE AVERAGE OUTPUT VOLT AGE OF
THREE PHASE SEMICONVERTER FOR 3
πα >
AND DISCONTINUOUS
OUTPUT VOLTAGE
For 3
πα ≥ and discontinuous output voltage: the average output voltage is found
from
( )7
6
6
3.
2dc acV v d t
π
πα
ωπ
+
= ∫
121
( )7
6
6
33 sin
2 6dc mV V t d t
π
πα
πω ωπ
+
= −
∫
( )3 31 cos
2m
dc
VV α
π= +
( )31 cos
2mL
dc
VV α
π= +
The maximum average output voltage that occurs at a delay angle of 0α = is
3 3 mdm
VV
π=
The normalized average output voltage is
( )0.5 1 cosdcn
dm
VV
Vα= = +
The rms output voltage is found from
( ) ( )
17 26
2 2
6
33 sin
2 6mO RMSV V t d t
π
πα
πω ωπ
+
= −
∫
( )
1
23 13 sin 2
4 2mO RMSV V π α απ
= − +
For 3
πα ≤ , and continuous output voltage
Output voltage 3 sin6O ab mv v V tπω = = +
; for to
6 2t
π πω α = +
Output voltage 3 sin6O ac mv v V tπω = = −
; for
5 to
2 6t
π πω α = +
The average or dc output voltage is calculated by using the equation
( ) ( )5
62
6 2
3. .
2dc ab acV v d t v d t
ππα
π πα
ω ωπ
+
+
= + ∫ ∫
122
( )3 31 cos
2m
dc
VV α
π= +
( )0.5 1 cosdcn
dm
VV
Vα= = +
The RMS value of the output voltage is calculated by using the equation
( ) ( ) ( )
15 262
2 2
6 2
3. .
2 ab acO RMSV v d t v d t
ππα
π πα
ω ωπ
+
+
= +
∫ ∫
( )
1
223 2
3 3 cos4 3mO RMSV V
π απ
= +
THREE PHASE FULL CONVERTER Three phase full converter is a fully controlled bridge controlled rectifier using six thyristors connected in the form of a full wave bridge configuration. All the six thyristors are controlled switches which are turned on at a appropriate times by applying suitable gate trigger signals. The three phase full converter is extensively used in industrial power applications upto about 120kW output power level, where two quadrant operation is required. The figure shows a three phase full converter with highly inductive load. This circuit is also known as three phase full wave bridge or as a six pulse converter.
The thyristors are triggered at an interval of 3
π
radians (i.e. at an interval of
060 ). The frequency of output ripple voltage is 6 Sf and the filtering requirement is less
than that of three phase semi and half wave converters.
123
At 6
tπω α = +
, thyristor 6T is already conducting when the thyristor 1T is
turned on by applying the gating signal to the gate of 1T . During the time period
to 6 2
tπ πω α α = + +
, thyristors 1T and 6T conduct together and the line to line
supply voltage abv appears across the load.
At 2
tπω α = +
, the thyristor 2T is triggered and 6T is reverse biased
immediately and 6T turns off due to natural commutation. During the time period
5 to
2 6t
π πω α α = + +
, thyristor 1T and 2T conduct together and the line to line
supply voltage acv appears across the load. The thyristors are numbered in the circuit diagram corresponding to the order in
which they are triggered. The trigger sequence (firing sequence) of the thyristors is 12, 23, 34, 45, 56, 61, 12, 23, and so on. The figure shows the waveforms of three phase input supply voltages, output voltage, the thyristor current through 1T and 4T , the supply current through the line ‘a’. We define three line neutral voltages (3 phase voltages) as follows
sin ; Max. Phase VoltageRN an m mv v V t Vω= = =
( )02sin sin 120
3YN bn m mv v V t V tπω ω = = − = −
( ) ( )0 02sin sin 120 sin 240
3BN cn m m mv v V t V t V tπω ω ω = = + = + == −
Where mV is the peak phase voltage of a star (Y) connected source.
The corresponding line-to-line voltages are
( ) 3 sin6RY ab an bn mv v v v V tπω = = − = +
( ) 3 sin2YB bc bn cn mv v v v V tπω = = − = −
( ) 3 sin2BR ca cn an mv v v v V tπω = = − = +
124
iG1
iG2
iG3
iG4
iG5
iG6
(30 + )0 α
600
600
600
600
600
(360 +30 + )0 0 α
T1 T2 T3 T4 T5 T6 T1 T2T6
t
t
t
t
t
t
Gating (Control) Signals of 3-phase full converter
125
TO DERIVE AN EXPRESSION FOR THE AVERAGE OUTPUT VOLT AGE OF THREE PHASE FULL CONVERTER WITH HIGHLY INDUCTIVE LO AD ASSUMING CONTINUOUS AND CONSTANT LOAD CURRENT
The output load voltage consists of 6 voltage pulses over a period of 2π radians, hence the average output voltage is calculated as
( )
2
6
6. ;
2dc OO dcV V v d t
π α
π α
ωπ
+
+
= = ∫
3 sin6O ab mv v V tπω = = +
2
6
33 sin .
6dc mV V t d t
π α
π α
πω ωπ
+
+
= +
∫
3 3 3cos cosm mL
dc
V VV α α
π π= =
Where mLV 3 Max. line-to-line supply voltagemV= = The maximum average dc output voltage is obtained for a delay angle α = 0,
( )max
3 3 3m mLdmdc
V VV V
π π= = =
The normalized average dc output voltage is
cosdcdcn n
dm
VV V
Vα= = =
The rms value of the output voltage is found from
( ) ( )
1
22
2
6
6.
2 OO rmsV v d t
π α
π α
ωπ
+
+
=
∫
126
( ) ( )
1
22
2
6
6.
2 abO rmsV v d t
π α
π α
ωπ
+
+
=
∫
( ) ( )
1
22
2 2
6
33 sin .
2 6mO rmsV V t d t
π α
π α
πω ωπ
+
+
= +
∫
( )
1
21 3 33 cos 2
2 4mO rmsV V απ
= +
127
THREE PHASE DUAL CONVERTERS In many variable speed drives, the four quadrant operation is generally required and three phase dual converters are extensively used in applications up to the 2000 kW level. Figure shows three phase dual converters where two three phase full converters are connected back to back across a common load. We have seen that due to the instantaneous voltage differences between the output voltages of converters, a circulating current flows through the converters. The circulating current is normally limited by circulating reactor, rL . The two converters are controlled in such a way that if 1α is the
delay angle of converter 1, the delay angle of converter 2 is ( )2 1α π α= − .
The operation of a three phase dual converter is similar that of a single phase dual converter system. The main difference being that a three phase dual converter gives much higher dc output voltage and higher dc output power than a single phase dual converter system. But the drawback is that the three phase dual converter is more expensive and the design of control circuit is more complex.
The figure below shows the waveforms for the input supply voltages, output voltages of converter1 and conveter2 , and the voltage across current limiting reactor (inductor) rL . The operation of each converter is identical to that of a three phase full converter.
During the interval 16
π α +
to 12
π α +
, the line to line voltage abv appears
across the output of converter 1 and bcv appears across the output of converter 2 We define three line neutral voltages (3 phase voltages) as follows
sin ;RN an mv v V tω= = Max. Phase VoltagemV =
( )02sin sin 120
3YN bn m mv v V t V tπω ω = = − = −
( )02sin sin 120
3BN cn m mv v V t V tπω ω = = + = +
( )0sin 240mV tω= −
128
The corresponding line-to-line supply voltages are
( ) 3 sin6RY ab an bn mv v v v V tπω = = − = +
( ) 3 sin2YB bc bn cn mv v v v V tπω = = − = −
( ) 3 sin2BR ca cn an mv v v v V tπω = = − = +
129
TO OBTAIN AN EXPRESSION FOR THE CIRCULATING CURRENT If 1Ov and 2Ov are the output voltages of converters 1 and 2 respectively, the
instantaneous voltage across the current limiting inductor during the interval
1 16 2t
π πα ω α + ≤ ≤ +
is
( ) ( )1 2r O O ab bcv v v v v= + = −
3 sin sin6 2r mv V t tπ πω ω = + − −
3 cos6r mv V tπω = −
The circulating current can be calculated by using the equation
( ) ( )16
1.
t
r rr
i t v d tL
ω
π α
ωω
+
= ∫
( ) ( )16
13 cos .
6
t
r mr
i t V t d tL
ω
π α
πω ωω
+
= −
∫
( ) 1
3sin sin
6m
rr
Vi t t
L
πω αω
= − −
( )max
3 mr
r
Vi
Lω= = maximum value of the circulating current.
There are two different modes of operation of a three phase dual converter system.
• Circulating current free (non circulating) mode of operation • Circulating current mode of operation
CIRCULATING CURRENT FREE (NON-CIRCULATING) MODE OF OPERATION In this mode of operation only one converter is switched on at a time when the converter number 1 is switched on and the gate signals are applied to the thyristors the average output voltage and the average load current are controlled by adjusting the trigger angle α1 and the gating signals of converter 1 thyristors. The load current flows in the downward direction giving a positive average load current when the converter 1 is switched on. For 0
1 90α < the converter 1 operates in the
rectification mode dcV is positive, dcI is positive and hence the average load power dcP is positive.
130
The converter 1 converts the input ac supply and feeds a dc power to the load. Power flows from the ac supply to the load during the rectification mode. When the trigger angle 1α is increased above 090 , dcV becomes negative where as dcI is positive because the thyristors of converter 1 conduct in only one direction and reversal of load current through thyristors of converter 1 is not possible.
For 01 90α > converter 1 operates in the inversion mode & the load energy is
supplied back to the ac supply. The thyristors are switched-off when the load current decreases to zero & after a short delay time of about 10 to 20 milliseconds, the converter 2 can be switched on by releasing the gate control signals to the thyristors of converter 2. We obtain a reverse or negative load current when the converter 2 is switched ON. The average or dc output voltage and the average load current are controlled by adjusting the trigger angle 2α of the gate trigger pulses supplied to the thyristors of converter 2.
When 2α is less than 090 , converter 2 operates in the rectification mode and converts the input ac supply in to dc output power which is fed to the load. When 2α is less than 090 for converter 2, dcV is negative & dcI is negative, converter 2 operates as a controlled rectifier & power flows from the ac source to the load circuit. When 2α is increased above 900, the converter 2 operates in the inversion mode
with dcV positive and dcI negative and hence dcP is negative, which means that power flows from the load circuit to the input ac supply. The power flow from the load circuit to the input ac source is possible if the load circuit has a dc source of appropriate polarity. When the load current falls to zero the thyristors of converter 2 turn-off and the converter 2 can be turned off. CIRCULATING CURRENT MODE OF OPERATION Both the converters are switched on at the same time in the mode of operation. One converter operates in the rectification mode while the other operates in the inversion mode. Trigger angles 1α & 2α are adjusted such that ( ) 0
1 2 180α α+ =
When 01 90α < , converter 1 operates as a controlled rectifier. When 2α is made
greater than 090 , converter 2 operates in the inversion mode. dcV , dcI , dcP are positive.
When 02 90α < , converter 2 operates as a controlled rectifier. When 1α is made
greater than 090 , converter 1 operates as an Inverter. dcV and dcI are negative while dcP is positive.
131
Problems
1. A 3 phase fully controlled bridge rectifier is operating from a 400 V, 50 Hz
supply. The thyristors are fired at 4
πα = . There is a FWD across the load. Find
the average output voltage for 045α = and 075α = . Solution
For 045α = , 3
cosmdc
VV α
π=
03 2 400cos 45 382 VoltsdcV
π× ×= =
For 075α = , ( )061 cos 60
2m
dc
VV α
π = + +
( )0 06 2 4001 cos 60 75
2dcVπ
× × = + +
158.4 VoltsdcV =
2. A 6 pulse converter connected to 415 V ac supply is controlling a 440 V dc motor. Find the angle at which the converter must be triggered so that the voltage drop in the circuit is 10% of the motor rated voltage. Solution
+
+
−−
440 V
44V
RaLa
3 phaseFull WaveRectifier
A
B
C
484V=V0
aR - Armature resistance of motor.
aL - Armature Inductance. If the voltage across the armature has to be the rated voltage i.e., 440 V, then the output voltage of the rectifier should be 440 + drop in the motor That is 440 01 440 484 Volts+ × = .
132
Therefore 3 cos
484mO
VV
απ
= =
That is 3 2 415 cos
484α
π× × × =
Therefore 030.27α =
3. A 3 phase half controlled bridge rectifier is feeding a RL load. If input voltage is
400 sin314t and SCR is fired at 4
πα = . Find average load voltage. If any one
supply line is disconnected what is the average load voltage.
Solution
4
πα = radians which is less than 3
π
Therefore [ ]31 cos
2m
dc
VV α
π= +
03 4001 cos 45
2dcVπ
× = +
326.18 VoltsdcV = If any one supply line is disconnected, the circuit behaves like a single phase half controlled rectifies with RL load.
[ ]1 cosmdc
VV α
π= +
04001 cos 45dcV
π = +
217.45 VoltsdcV =
133
EDUSAT PROGRAMME
LECTURE NOTES
ON
POWER ELECTRONICS
BY
PROF. T.K. ANANTHA KUMAR
DEPARTMENT OF
ELECTRICAL & ELECTRONICS ENGG.
M.S. RAMAIAH INSTITUTE OF TECHNOLOGY
BANGALORE – 560 054
134
THYRISTOR COMMUTATION TECHNIQUES INTRODUCTION
In practice it becomes necessary to turn off a conducting thyristor. (Often thyristors are used as switches to turn on and off power to the load). The process of turning off a conducting thyristor is called commutation. The principle involved is that either the anode should be made negative with respect to cathode (voltage commutation) or the anode current should be reduced below the holding current value (current commutation).
The reverse voltage must be maintained for a time at least equal to the turn-off time of SCR otherwise a reapplication of a positive voltage will cause the thyristor to conduct even without a gate signal. On similar lines the anode current should be held at a value less than the holding current at least for a time equal to turn-off time otherwise the SCR will start conducting if the current in the circuit increases beyond the holding current level even without a gate signal. Commutation circuits have been developed to hasten the turn-off process of Thyristors. The study of commutation techniques helps in understanding the transient phenomena under switching conditions.
The reverse voltage or the small anode current condition must be maintained for a time at least equal to the TURN OFF time of SCR; Otherwise the SCR may again start conducting. The techniques to turn off a SCR can be broadly classified as
• Natural Commutation • Forced Commutation.
NATURAL COMMUTATION (CLASS F)
This type of commutation takes place when supply voltage is AC, because a negative voltage will appear across the SCR in the negative half cycle of the supply voltage and the SCR turns off by itself. Hence no special circuits are required to turn off the SCR. That is the reason that this type of commutation is called Natural or Line Commutation. Figure 1.1 shows the circuit where natural commutation takes place and figure 1.2 shows the related waveforms. ct is the time offered by the circuit within which
the SCR should turn off completely. Thus ct should be greater than qt , the turn off time
of the SCR. Otherwise, the SCR will become forward biased before it has turned off completely and will start conducting even without a gate signal.
~
T
+
−
vovsR↑ ↑
Fig. 1.1: Circuit for Natural Commutation
135
Fig. 1.2: Natural Commutation – Waveforms of Supply and Load Voltages
(Resistive Load)
This type of commutation is applied in ac voltage controllers, phase controlled rectifiers and cyclo converters. FORCED COMMUTATION
When supply is DC, natural commutation is not possible because the polarity of the supply remains unchanged. Hence special methods must be used to reduce the SCR current below the holding value or to apply a negative voltage across the SCR for a time interval greater than the turn off time of the SCR. This technique is called FORCED COMMUTATION and is applied in all circuits where the supply voltage is DC - namely, Choppers (fixed DC to variable DC), inverters (DC to AC). Forced commutation techniques are as follows:
• Self Commutation • Resonant Pulse Commutation • Complementary Commutation • Impulse Commutation • External Pulse Commutation. • Load Side Commutation.
ωt
ωt
ωt
ωt
Supply voltage vs Sinusoidal
Voltage across SCR
Load voltage vo
Turn offoccurs here
0
0
π
π
2π
2π
3π
3π
α
tc
136
• Line Side Commutation. SELF COMMUTATION OR LOAD COMMUTATION OR CLASS A COMMUTATION: (COMMUTATION BY RESONATING THE LOAD)
In this type of commutation the current through the SCR is reduced below the holding current value by resonating the load. i.e., the load circuit is so designed that even though the supply voltage is positive, an oscillating current tends to flow and when the current through the SCR reaches zero, the device turns off. This is done by including an inductance and a capacitor in series with the load and keeping the circuit under-damped. Figure 1.3 shows the circuit.
This type of commutation is used in Series Inverter Circuit.
V
R L V (0)c
C
T i
Load
+ -
Fig. 1.3: Circuit for Self Commutation EXPRESSION FOR CURRENT
At 0t = , when the SCR turns ON on the application of gate pulse assume the current in the circuit is zero and the capacitor voltage is ( )0CV .
Writing the Laplace Transformation circuit of figure 1.3 the following circuit is
obtained when the SCR is conducting.
VS
R sL1
CS
V (0)SC
C
T I(S) + +- -
Fig.: 1.4.
137
( )( )0
1
C
S
V V
SI SR sL
C
−
=+ +
( )
2
0
1
S CC V V
SRCs s LC
−
=+ +
( )2
0
1CC V V
RLC s s
L LC
− = + +
( )
2
0
1
CV V
LR
s sL LC
−
=+ +
( )( )2 2
2
0
12 2
CV V
LR R R
s sL LC L L
−
= + + + −
( )( )2
2 2
0
12 2
CV V
L
R Rs
L LC L
−
= + + −
( )2 2
A
s δ ω=
+ +,
Where
( )( ) 20 1
, ,2 2
CV V R RA
L L LC Lδ ω
− = = = −
ω is called the natural frequency
( )( )2 2
AI S
s
ωω δ ω
=+ +
138
Taking inverse Laplace transforms
( ) sintAi t e tδ ω
ω−=
Therefore expression for current
( ) ( )2
0sin
RtC L
V Vi t e t
Lω
ω
−−=
Peak value of current ( )( )0CV V
Lω−
=
Expression for voltage across capacitor at the time of turn off Applying KVL to figure 1.3
c R Lv V v V= − −
c
div V iR L
dt= − −
Substituting for i,
sin sint tc
A d Av V R e t L e t
dtδ δω ω
ω ω− − = − −
( )sin cos sint t tc
A Av V R e t L e t e tδ δ δω ω ω δ ω
ω ω− − −= − − −
[ ]sin cos sintc
Av V e R t L t L tδ ω ω ω δ ω
ω−= − + −
sin cos sin2
tc
A Rv V e R t L t L t
Lδ ω ω ω ω
ω− = − + −
sin cos2
tc
A Rv V e t L tδ ω ω ω
ω− = − +
Substituting for A,
( ) ( )( )0sin cos
2C t
c
V V Rv t V e t L t
Lδ ω ω ω
ω−− = − +
( ) ( )( )0sin cos
2C t
c
V V Rv t V e t t
Lδ ω ω ω
ω−− = − +
139
SCR turns off when current goes to zero. i.e., at tω π= . Therefore at turn off
( )( ) ( )0
0 cosC
c
V Vv V e
δπω ω π
ω
−−= − +
( )0c Cv V V V eδπω
−
= + −
Therefore ( ) 20R
Lc Cv V V V e
πω
−
= + −
Note: For effective commutation the circuit should be under damped.
That is 2
1
2
R
L LC <
• With R = 0, and the capacitor initially uncharged that is ( )0 0CV =
sinV t
iL LCω
=
But 1
LCω =
Therefore sin sinV t C t
i LC VL LLC LC
= =
and capacitor voltage at turn off is equal to 2V.
• Figure 1.5 shows the waveforms for the above conditions. Once the SCR turns off voltage across it is negative voltage.
• Conduction time of SCR πω
= .
140
Current i
Capacitor voltage
Gate pulse
Voltage across SCR
0 ππ/2ωt
ωt
ωt
ωt
V
−V
2V
CV
L
Fig. 1.5: Self Commutation – Wave forms of Current and Capacitors Voltage Problem 1.1 : Calculate the conduction time of SCR and the peak SCR current that flows in the circuit employing series resonant commutation (self commutation or class A commutation), if the supply voltage is 300 V, C = 1µF, L = 5 mH and RL = 100 Ω.
Assume that the circuit is initially relaxed.
V=300V
RLL
1 Fµ100 Ω 5 mH
CT+ −
Fig. 1.6.
141
Solution:
2
1
2LR
LC Lω = −
2
3 6 3
1 100
5 10 1 10 2 5 10ω − − −
= − × × × × ×
10,000 rad/secω =
Since the circuit is initially relaxed, initial voltage across capacitor is zero as also
the initial current through L and the expression for current i is
sintVi e t
Lδ ω
ω−= , where
2
R
Lδ = ,
Therefore peak value of V
iLω
=
3
3006
10000 5 10i A−= =
× ×
Conducting time of SCR 0.314msec10000
π πω
= = =
Problem 1.2 : Figure 1.7 shows a self commutating circuit. The inductance carries an initial current of 200 A and the initial voltage across the capacitor is V, the supply voltage. Determine the conduction time of the SCR and the capacitor voltage at turn off.
V=100V
L
50 Fµ
10 HµC
T
+−
i(t)
→ IO
VC(0)=V
Fig. 1.7.
142
Solution : The transformed circuit of figure 1.7 is shown in figure 1.8.
sL
VS
V (0)SC
I LO
+
+
+
−
−
−
1CS
I(S)=V
Fig.1.8: Transformed Circuit of Fig. 1.7
The governing equation is
( ) ( ) ( )0 1CO
VVI S sL I L I S
s s Cs= − + +
Therefore ( )( )0
1
CO
VVI L
s sI SsL
Cs
− +=
+
( )
( )
2 2
0
1 1
C
O
VVCs
s s I LCsI S
s LC s LC
−
= ++ +
( ) ( )2 2
0
1 1C O
V V C I LCsI S
LC s LC sLC LC
− = + + +
( ) ( )2 22 2
0C OV V sI
I SsL s ωω
−= +
+ +
( ) ( )2 22 2
0C OV V sI
I SsL s
ωωω ω
− = ++ +
where 1
LCω =
Taking inverse LT
( ) ( )0 sin cosC O
Ci t V V t I t
Lω ω= − +
143
The capacitor voltage is given by
( ) ( ) ( )0
10
t
c Cv t i t dt VC
= +∫
( ) ( ) ( )0
10 sin cos 0
t
c C O C
Cv t V V t I t dt V
C Lω ω
= − + + ∫
( ) ( )( ) ( ) ( ) ( )01
cos sin 0C Oc C
V V t tICv t t t V
o oC Lω ω
ω ω −
= − + +
( ) ( )( ) ( ) ( ) ( )01
1 cos sin 0C Oc C
V V ICv t t t V
C Lω ω
ω ω −
= − + +
( ) ( )( ) ( ) ( )1sin 0 1 cos 0O
c C C
I Cv t LC t V V LC t V
C C Lω ω= × + − − +
( ) ( ) ( ) ( )sin cos 0 0 cos 0c O C C C
Lv t I t V V t V V t V
Cω ω ω= + − − + +
( ) ( )( )sin 0 cosc O C
Lv t I t V V t V
Cω ω= − − +
In this problem ( )0CV V=
Therefore we get, ( ) cosOi t I tω= and
( ) sinc O
Lv t I t V
Cω= +
144
The waveforms are as shown in figure 1.9
I0i(t)
π/2
π/2
ωt
ωt
vc(t)
V
Fig.: 1.9
Turn off occurs at a time to so that 2Otπω =
Therefore 0.5
0.5Ot LCπ π
ω= =
6 60.5 10 10 50 10Ot π − −= × × × ×
60.5 10 500 35.1 secondsOt π µ−= × × =
and the capacitor voltage at turn off is given by
( ) sinc O O O
Lv t I t V
Cω= +
( )6
06
10 10200 sin 90 100
50 10c Ov t−
−
×= +×
( ) 35.12200 0.447 sin 100
22.36c Ov t = × × +
( ) 89.4 100 189.4 c Ov t V= + =
145
Problem 1.3: In the circuit shown in figure 1.10. V = 600 volts, initial capacitor voltage is zero, L = 20 µH, C = 50µF and the current through the inductance at the time of SCR triggering is Io = 350 A. Determine (a) the peak values of capacitor voltage and current
(b) the conduction time of T1.
V
L
i(t)→ I0
C
T1
Fig. 1.10
Solution: (Refer to problem 1.2).
The expression for ( )i t is given by
( ) ( )0 sin cosC O
Ci t V V t I t
Lω ω= − +
It is given that the initial voltage across the capacitor, ( )CV O is zero.
Therefore ( ) sin cosO
Ci t V t I t
Lω ω= +
( )i t can be written as
( ) ( )2 2 sinO
Ci t I V t
Lω α= + +
where 1tanO
LI
CV
α −=
and 1
LCω =
The peak capacitor current is
2 2O
CI V
L+
Substituting the values, the peak capacitor current
6
2 26
50 10350 600 1011.19
20 10A
−
−
×= + × =×
146
The expression for capacitor voltage is
( ) ( )( )sin 0 cosc O C
Lv t I t V V t V
Cω ω= − − +
with ( ) ( )0 0, sin cosC c O
LV v t I t V t V
Cω ω= = − +
This can be rewritten as
( ) ( )2 2 sinc O
Lv t V I t V
Cω β= + − +
Where 1tanO
CV
LI
β −=
The peak value of capacitor voltage is
2 2O
LV I V
C= + +
Substituting the values, the peak value of capacitor voltage
6
2 26
20 10600 350 600
50 10
−
−
×= + × +×
639.5 600 1239.5V= + = To calculate conduction time of 1T
The waveform of capacitor current is shown in figure 1.11. When the capacitor current becomes zero the SCR turns off.
ωt
Capacitorcurrent
π − α
0α
Fig. 1.11.
147
Therefore conduction time of SCR π α
ω−=
1tan
1
O
LI
CV
LC
π −
− =
Substituting the values
1tanO
LI
CV
α −
=
6
16
350 20 10tan
600 50 10α
−−
−
×=×
020.25 i.e., 0.3534 radiansα =
6 6
1 131622.8 rad/sec
20 10 50 10LCω
− −= = =
× × ×
Therefore conduction time of SCR
0.3534
88.17 sec31622.8
π µ−= =
RESONANT PULSE COMMUTATION (CLASS B COMMUTATION)
The circuit for resonant pulse commutation is shown in figure 1.12.
L
C
VLoad
FWD
ab
iT
IL
Fig. 1.12: Circuit for Resonant Pulse Commutation
148
This is a type of commutation in which a LC series circuit is connected across the SCR. Since the commutation circuit has negligible resistance it is always under-damped i.e., the current in LC circuit tends to oscillate whenever the SCR is on.
Initially the SCR is off and the capacitor is charged to V volts with plate ‘a’ being positive. Referring to figure 1.13 at 1t t= the SCR is turned ON by giving a gate pulse. A
current LI flows through the load and this is assumed to be constant. At the same time SCR short circuits the LC combination which starts oscillating. A current ‘i’ starts flowing in the direction shown in figure. As ‘i’ reaches its maximum value, the capacitor voltage reduces to zero and then the polarity of the capacitor voltage reverses ‘b’ becomes positive). When ‘i’ falls to zero this reverse voltage becomes maximum, and then direction of ‘i’ reverses i.e., through SCR the load current LI and ‘i’ flow in opposite
direction. When the instantaneous value of ‘i’ becomes equal to LI , the SCR current becomes zero and the SCR turns off. Now the capacitor starts charging and its voltage reaches the supply voltage with plate a being positive. The related waveforms are shown in figure 1.13.
Gate pulseof SCR
Capacitor voltagevab
πt1
V
t
t
t
t
t
Ip i
Voltage acrossSCR
IL
tC
∆t
πω
ISCR
Fig. 1.13: Resonant Pulse Commutation – Various Waveforms
149
EXPRESSION FOR ct , THE CIRCUIT TURN OFF TIME
Assume that at the time of turn off of the SCR the capacitor voltage abv V≈ − and
load current LI is constant. ct is the time taken for the capacitor voltage to reach 0 volts from – V volts and is derived as follows.
0
1 ct
LV I dtC
= ∫
L cI t
VC
=
secondscL
VCt
I=
For proper commutation ct should be greater than qt , the turn off time of T. Also,
the magnitude of pI , the peak value of i should be greater than the load current LI and
the expression for i is derived as follows The LC circuit during the commutation period is shown in figure 1.14.
i
L
C
T
+−
VC(0)=V
Fig. 1.14.
The transformed circuit is shown in figure 1.15.
I(S)
sL
T 1Cs
Vs
+
−
Fig. 1.15.
150
( )1
V
sI SsL
Cs
=+
( ) 2 1
VCs
sI S
s LC
=
+
( )2 1
VCI S
LC sLC
= +
( )2
11
VI S
L sLC
= ×+
( )2
11
1 1
V LCI SL s
LC LC
= × ×
+
( )2
1
1C LCI S VL s
LC
= ×
+
Taking inverse LT
( ) sinC
i t V tL
ω=
Where 1
LCω =
Or ( ) sin sinp
Vi t t I t
Lω ω
ω= =
Therefore ampsp
CI V
L= .
151
EXPRESSION FOR CONDUCTION TIME OF SCR For figure 1.13 (waveform of i), the conduction time of SCR
tπω
= + ∆
1sin L
p
I
Iπω ω
− = +
ALTERNATE CIRCUIT FOR RESONANT PULSE COMMUTATION
The working of the circuit can be explained as follows. The capacitor C is assumed to be charged to ( )0CV with polarity as shown, 1T is conducting and the load
current LI is a constant. To turn off1T , 2T is triggered. L, C, 1T and 2T forms a resonant
circuit. A resonant current ( )ci t flows in the direction shown, i.e., in a direction opposite
to that of load current LI .
( )ci t = sinpI tω (refer to the previous circuit description). Where ( )0p C
CI V
L=
& and the capacitor voltage is given by
( ) ( )
( ) ( )
( ) ( )
1.
10 sin .
0 cos
c C
c C
c C
v t i t dtC
Cv t V t dt
C L
v t V t
ω
ω
=
=
= −
∫
∫ .
VLOADFWD
L
T1 IL
T3
T2
iC(t)
iC(t)
VC(0)
a b
+−
C
Fig. 1.16: Resonant Pulse Commutation – An Alternate Circuit
152
When ( )ci t becomes equal to LI (the load current), the current through 1T
becomes zero and 1T turns off. This happens at time 1t such that
1sinL p
tI I
LC=
( )0p C
CI V
L=
( )
11 sin
0L
C
I Lt LC
V C−
=
and the corresponding capacitor voltage is ( ) ( )1 1 10 cosc Cv t V V tω= − = −
Once the thyristor 1T turns off, the capacitor starts charging towards the supply
voltage through 2T and load. As the capacitor charges through the load capacitor current
is same as load currentLI , which is constant. When the capacitor voltage reaches V, the supply voltage, the FWD starts conducting and the energy stored in L charges C to a still higher voltage. The triggering of 3T reverses the polarity of the capacitor voltage and the
circuit is ready for another triggering of 1T . The waveforms are shown in figure 1.17.
EXPRESSION FOR ct
Assuming a constant load current LI which charges the capacitor
1 secondscL
CVt
I=
Normally ( )1 0CV V≈
For reliable commutation ct should be greater thanqt , the turn off time of SCR1T .
It is to be noted that ct depends upon LI and becomes smaller for higher values of load current.
153
t
t
tC
t1
V1
V
VC(0)
Capacitorvoltage vab
Current iC(t)
Fig. 1.17: Resonant Pulse Commutation – Alternate Circuit – Various Waveforms
RESONANT PULSE COMMUTATION WITH ACCELERATING DIODE
VLOADFWD
LCT1
IL
T3
T2iC(t)
VC(0)+-
D2 iC(t)
Fig. 1.17(a)
154
Fig. 1.17(b)
A diode 2D is connected as shown in the figure 1.17(a) to accelerate the
discharging of the capacitor ‘C’. When thyristor 2T is fired a resonant current ( )Ci t
flows through the capacitor and thyristor1T . At time 1t t= , the capacitor current ( )Ci t
equals the load current LI and hence current through 1T is reduced to zero resulting in
turning off of 1T . Now the capacitor current ( )Ci t continues to flow through the diode 2D
until it reduces to load current level LI at time 2t . Thus the presence of 2D has accelerated the discharge of capacitor ‘C’. Now the capacitor gets charged through the load and the charging current is constant. Once capacitor is fully charged 2T turns off by
itself. But once current of thyristor 1T reduces to zero the reverse voltage appearing across
1T is the forward voltage drop of 2D which is very small. This makes the thyristor recovery process very slow and it becomes necessary to provide longer reverse bias time. From figure 1.17(b)
2 1t LC tπ= − ( ) ( )2 2cosC CV t V O tω= −
Circuit turn-off time 2 1Ct t t= − Problem 1.4 : The circuit in figure 1.18 shows a resonant pulse commutation circuit. The initial capacitor voltage ( ) 200C OV V= , C = 30µF and L = 3µH. Determine the circuit
turn off time ct , if the load currentLI is (a) 200 A and (b) 50 A.
IL
0
VC
0
V1V (O)C
iC
t
tt1 t2
tC
155
VLOADFWD
LCT1
IL
T3
T2iC(t)
VC(0)+−
Fig. 1.18. Solution
(a) When 200LI A=
Let 2T be triggered at 0t = .
The capacitor current ( )ci t reaches a value LI at 1t t= , when 1T turns off
( )1
1 sin0
L
C
I Lt LC
V C−
=
6
6 6 11 6
200 3 103 10 30 10 sin
200 30 10t
−− − −
−
×= × × × ×
1 3.05 sect µ= .
6 6
1 1
3 10 30 10LCω
− −= =
× × ×
60.105 10 / secradω = × . At 1t t= , the magnitude of capacitor voltage is ( )1 10 cosCV V tω=
That is 6 6
1 200cos0.105 10 3.05 10V −= × × × 1 200 0.9487V = × 1 189.75 VoltsV =
and 1c
L
CVt
I=
156
630 10 189.75
28.46 sec200ct µ−× ×= = .
(b) When 50LI A=
66 6 1
1 6
50 3 103 10 30 10 sin
200 30 10t
−− − −
−
×= × × × ×
1 0.749 sect µ= .
6 61 200cos0.105 10 0.749 10V −= × × ×
1 200 1 200 VoltsV = × = .
1c
L
CVt
I=
630 10 200
120 sec50ct µ
−× ×= = .
It is observed that as load current increases the value of ct reduces.
Problem 1.4a : Repeat the above problem for 200LI A= , if an antiparallel diode 2D is
connected across thyristor 1T as shown in figure 1.18a.
VLOADFWD
LCT1
IL
T3
T2iC(t)
VC(0)+-
D2 iC(t)
Fig. 1.18(a)
157
Solution
200LI A=
Let 2T be triggered at 0t = .
Capacitor current ( )Ci t reaches the value 1 1 at , when turns offLI t t T=
Therefore ( )
11 sin L
C
I Lt LC
V O C−
=
6
6 6 11 6
200 3 103 10 30 10 sin
200 30 10t
−− − −
−
×= × × × ×
` 1 3.05 sect µ= .
6 6
1 1
3 10 30 10LCω
− −= =
× × ×
6 0.105 10 radians/secω = × 1 At t t=
( ) ( )1 1 1cosC CV t V V O tω= = −
( ) ( )6 61 200cos 0.105 10 3.05 10CV t −= − × × ×
( )1 189.75CV t V= −
( )Ci t flows through diode 2D after 1T turns off.
( )Ci t current falls back to 2 at LI t
2 1t LC tπ= −
6 6 62 3 10 30 10 3.05 10t π − − −= × × × − ×
2 26.75 sect µ= .
6 6
1 1
3 10 30 10LCω
− −= =
× × ×
60.105 10 rad/sec.ω = ×
158
2 At t t=
( ) 6 62 2 200cos0.105 10 26.75 10CV t V + −= = − × × ×
( )2 2 189.02 CV t V V= =
Therefore 6 6
2 1 26.75 10 3.05 10Ct t t − −= − = × − ×
23.7 secsCt µ= Problem 1.5: For the circuit shown in figure 1.19 calculate the value of L for proper commutation of SCR. Also find the conduction time of SCR.
V=30V L
i
4 Fµ
RL
IL30 Ω
Fig. 1.19. Solution:
The load current 30
1 Amp30L
L
VI
R= = =
For proper SCR commutationpI , the peak value of resonant current i, should be
greater thanLI ,
Let 2p LI I= , Therefore 2 AmpspI = .
Also 1p
V V CI V
L LLLC
ω= = =
×
Therefore 64 10
2 30L
−×= ×
Therefore 0.9L mH= .
3 6
1 116666 rad/sec
0.9 10 4 10LCω
− −= = =
× × ×
159
Conduction time of SCR =
1sin L
p
I
Iπω ω
− +
1 1sin
216666 16666
π− = +
0.523
radians16666
π +=
0.00022 seconds= 0.22 msec=
Problem 1.6: For the circuit shown in figure 1.20 given that the load current to be commutated is 10 A, turn off time required is 40µsec and the supply voltage is 100 V. Obtain the proper values of commutating elements.
V=100V L i IL
IL
C
Fig. 1.20.
Solution
pI peak value of C
i VL
= and this should be greater than LI . Let 1.5p LI I= .
Therefore ( )1.5 10 100 ...C
aL
× =
Also, assuming that at the time of turn off the capacitor voltage is approximately
equal to V, (and referring to waveform of capacitor voltage in figure 1.13) and the load current linearly charges the capacitor
secondscL
CVt
I=
and this ct is given to be 40 µsec.
Therefore 6 10040 10
10C−× = ×
160
Therefore 4C Fµ=
Substituting this in equation (a)
64 10
1.5 10 100L
−×× =
4 6
2 2 10 4 101.5 10
L
−× ×× =
Therefore 41.777 10L H−= × 0.177L mH= . Problem 1.7 : In a resonant commutation circuit supply voltage is 200 V. Load current is 10 A and the device turn off time is 20µs. The ratio of peak resonant current to load current is 1.5. Determine the value of L and C of the commutation circuit. Solution
Given 1.5p
L
I
I=
Therefore 1.5 1.5 10 15p LI I A= = × = .
That is ( )15 ...p
CI V A a
L= =
It is given that the device turn off time is 20 µsec. Thereforect , the circuit turn off
time should be greater than this,
Let 30 secct µ= .
And cL
CVt
I=
Therefore 6 20030 10
10
C− ×× =
Therefore 1.5C Fµ= . Substituting in (a)
61.5 10
15 200L
−×=
161
6
2 2 1.5 1015 200
L
−×= ×
Therefore 0.2666 mHL = COMPLEMENTARY COMMUTATION (CLASS C COMMUTATION, PARALLEL CAPACITOR COMMUTATION)
In complementary commutation the current can be transferred between two loads.
Two SCRs are used and firing of one SCR turns off the other. The circuit is shown in figure 1.21.
V
R1 R2
T1 T2
IL
iC
C
a b
Fig. 1.21: Complementary Commutation
The working of the circuit can be explained as follows. Initially both 1T and 2T are off; Now, 1T is fired. Load current LI flows
through 1R . At the same time, the capacitor C gets charged to V volts through 2R and 1T (‘b’ becomes positive with respect to ‘a’). When the capacitor gets fully charged, the capacitor current ci becomes zero.
To turn off 1T , 2T is fired; the voltage across C comes across 1T and reverse biases
it, hence 1T turns off. At the same time, the load current flows through 2R and 2T . The
capacitor ‘C’ charges towards V through 1R and 2T and is finally charged to V volts with ‘a’ plate positive. When the capacitor is fully charged, the capacitor current becomes zero. To turn off 2T , 1T is triggered, the capacitor voltage (with ‘a’ positive) comes across
2T and 2T turns off. The related waveforms are shown in figure 1.22. EXPRESSION FOR CIRCUIT TURN OFF TIME ct
From the waveforms of the voltages across 1T and capacitor, it is obvious that ct is the time taken by the capacitor voltage to reach 0 volts from – V volts, the time constant being RC and the final voltage reached by the capacitor being V volts. The equation for capacitor voltage ( )cv t can be written as
162
( ) ( ) tc f i fv t V V V e τ−= + −
Where fV is the final voltage, iV is the initial voltage and τ is the time constant.
At ct t= , ( ) 0cv t = ,
1R Cτ = , fV V= , iV V= − ,
Therefore ( ) 10ct
R CV V V e−
= + − −
10 2ct
R CV Ve−
= −
Therefore 12ct
R CV Ve−
=
10.5ct
R Ce−
=
Taking natural logarithms on both sides
1
ln 0.5 ct
R C
−=
10.693ct R C=
This time should be greater than the turn off time qt of 1T .
Similarly when 2T is commutated
20.693ct R C=
And this time should be greater than qt of 2T .
Usually 1 2R R R= =
163
Gate pulseof T1
Gate pulseof T2
Current through R1
p
ILV
t
t
t
t
t
t
Current through T1
Voltage acrosscapacitor vab
Voltage across T1
Current through T2
tC tC
tC
V
- V
2
2
VR
2
1
V
R
V
R1
V
R2
2
1
V
RV
R1
Fig. 1.22
164
Problem 1.8 : In the circuit shown in figure 1.23 the load resistances 1 2 5R R R= = = Ω
and the capacitance C = 7.5 µF, V = 100 volts. Determine the circuit turn off time ct .
V
R1 R2
T1 T2
C
Fig. 1.23. Solution
The circuit turn-off time 0.693 RC secondsct = 60.693 5 7.5 10ct
−= × × ×
26 secct µ= .
Problem 1.9: Calculate the values of LR and C to be used for commutating the main SCR
in the circuit shown in figure 1.24. When it is conducting a full load current of 25 A flows. The minimum time for which the SCR has to be reverse biased for proper commutation is 40µsec. Also find 1R , given that the auxiliary SCR will undergo natural commutation
when its forward current falls below the holding current value of 2 mA.
V=100V
R1 RL
MainSCR
AuxiliarySCR
iC
C
ILi1
Fig. 1.24. Solution
In this circuit only the main SCR carries the load and the auxiliary SCR is used to turn off the main SCR. Once the main SCR turns off the current through the auxiliary SCR is the sum of the capacitor charging current ci and the current 1i through 1R , ci
reduces to zero after a time ct and hence the auxiliary SCR turns off automatically after a
time ct , 1i should be less than the holding current.
165
Given 25LI A=
That is 100
25L L
VA
R R= =
Therefore 4LR = Ω
40 sec 0.693c Lt R Cµ= =
That is 640 10 0.693 4 C−× = × ×
Therefore 640 10
4 0.693C
−×=×
14.43C Fµ=
11
Vi
R= should be less than the holding current of auxiliary SCR.
Therefore 1
100
R should be < 2mA.
Therefore 1 3
100
2 10R −>
×
That is 1 50R K> Ω
IMPULSE COMMUTATION (CLASS D COMMUTATION) The circuit for impulse commutation is as shown in figure 1.25.
VLOADFWD
C
T1
T3
IL
T2
VC(O)+
−
L
Fig. 1.25: Circuit for Impulse Commutation
166
The working of the circuit can be explained as follows. It is assumed that initially the capacitor C is charged to a voltage ( )CV O with polarity as shown. Let the thyristor 1T
be conducting and carry a load current LI . If the thyristor 1T is to be turned off, 2T is
fired. The capacitor voltage comes across 1T , 1T is reverse biased and it turns off. Now
the capacitor starts charging through 2T and the load. The capacitor voltage reaches V with top plate being positive. By this time the capacitor charging current (current through
2T ) would have reduced to zero and 2T automatically turns off. Now 1T and 2T are both
off. Before firing 1T again, the capacitor voltage should be reversed. This is done by
turning on 3T , C discharges through 3T and L and the capacitor voltage reverses. The waveforms are shown in figure 1.26.
Gate pulseof T2
Gate pulseof T3
Voltage across T1
Capacitorvoltage
Gate pulseof T1
VS
t
t
t
tC
VC
VC
Fig. 1.26: Impulse Commutation – Waveforms of Capacitor Voltage, Voltage across 1T .
167
EXPRESSION FOR CIRCUIT TURN OFF TIME (AVAILABLE TUR N OFF TIME) ct
ct depends on the load current LI and is given by the expression
0
1 ct
C LV I dtC
= ∫
(assuming the load current to be constant)
L cC
I tV
C=
secondsCc
L
V Ct
I=
For proper commutation ct should be > qt , turn off time of 1T .
Note:
• 1T is turned off by applying a negative voltage across its terminals. Hence this is voltage commutation.
• ct depends on load current. For higher load currents ct is small. This is a disadvantage of this circuit.
• When 2T is fired, voltage across the load is CV V+ ; hence the current through load shoots up and then decays as the capacitor starts charging.
AN ALTERNATIVE CIRCUIT FOR IMPULSE COMMUTATION
Is shown in figure 1.27.
V
C
D
T1
IT1
IL
i
T2
L
RL
VC(O)+
_
Fig. 1.27: Impulse Commutation – An Alternate Circuit
168
The working of the circuit can be explained as follows: Initially let the voltage across the capacitor be ( )CV O with the top plate positive.
Now 1T is triggered. Load current flows through 1T and load. At the same time, C
discharges through 1T , L and D (the current is ‘i’) and the voltage across C reverses i.e., the bottom plate becomes positive. The diode D ensures that the bottom plate of the capacitor remains positive.
To turn off 1T , 2T is triggered; the voltage across the capacitor comes across 1T .
1T is reverse biased and it turns off (voltage commutation). The capacitor now starts
charging through 2T and load. When it charges to V volts (with the top plate positive), the
current through 2T becomes zero and 2T automatically turns off. The related waveforms are shown in figure 1.28.
Gate pulseof T1
Gate pulseof T2
Current through SCR
Load current
This is due to i
Voltage across T1
Capacitorvoltage
t
t
t
t
t
tC
tC
VC
IL
IL
V
IT1
−V
VRL
2VRL
Fig. 1.28: Impulse Commutation – (Alternate Circuit) – Various Waveforms
169
Problem 1.10: An impulse commutated thyristor circuit is shown in figure 1.29. Determine the available turn off time of the circuit if V = 100 V, R = 10 Ω and C = 10 µF. Voltage across capacitor before 2T is fired is V volts with polarity as shown.
C
T1
T2
V (0)C
V +
+
-
-
R
Fig. 1.29.
Solution
When 2T is triggered the circuit is as shown in figure 1.30.
C
i(t)
T2V
++ -
-
R
V (O)C
Fig. 1.30. Writing the transform circuit, we obtain
Vs
V (0)s
C
+
+
−
−
I(s)
1Cs
R
Fig. 1.31.
170
We have to obtain an expression for capacitor voltage. It is done as follows:
( )( )( )10
1
CV VsI S
RCs
+=
+
( ) ( )( )0
1CC V V
I SRCs
+=
+
( ) ( )( )0
1CV V
I SR s
RC
+=
+
Voltage across capacitor ( ) ( ) ( )01 CC
VV s I s
Cs s= −
( ) ( ) ( )0 011
C CC
V V VV s
RCs ss
RC
+= −
+
( ) ( ) ( ) ( )0 0 0
1C C C
C
V V V V VV s
s ss
RC
+ += − −
+
( ) ( )01 1
CC
VV VV s
s s sRC RC
= − −+ +
( ) ( ) ( )1 0t tRC RC
c Cv t V e V e− −
= − −
In the given problem ( )0CV V=
Therefore ( ) ( )1 2tRC
cv t V e−
= −
The waveform of ( )cv t is shown in figure 1.32.
171
t
tC
V
V (0)C
v (t)C
Fig. 1.32.
At ct t= , ( ) 0cv t =
Therefore 0 1 2ct
RCV e− = −
1 2ct
RCe−
=
1
2
ctRCe
−=
Taking natural logarithms
1
log2
ce
t
RC
− =
( )ln 2ct RC=
( )610 10 10 ln 2ct
−= × ×
69.3 secct µ= . Problem 1.11 : In the commutation circuit shown in figure 1.33. C = 20 µF, the input voltage V varies between 180 and 220 V and the load current varies between 50 and 200 A. Determine the minimum and maximum values of available turn off time ct .
C
T1
T2 I0
I0
V (0)=C V+
−
V
Fig. 1.33.
172
Solution It is given that V varies between 180 and 220 V and OI varies between 50 and
200 A. The expression for available turn off time ct is given by
cO
CVt
I=
ct is maximum when V is maximum and OI is minimum.
Therefore maxmax
minc
O
CVt
I=
6max
22020 10 88 sec
50ct µ−= × × =
and minmin
maxc
O
CVt
I=
6min
18020 10 18 sec
200ct µ−= × × =
EXTERNAL PULSE COMMUTATION (CLASS E COMMUTATION)
VS VAUX
L
C
T1 T3T2
RL 2VAUX
+
−
Fig. 1.34: External Pulse Commutation
In this type of commutation an additional source is required to turn-off the conducting thyristor. Figure 1.34 shows a circuit for external pulse commutation. SV is
the main voltage source and AUXV is the auxiliary supply. Assume thyristor 1T is
conducting and load LR is connected across supplySV . When thyristor 3T is turned ON at
0t = , AUXV , 3T , L and C from an oscillatory circuit. Assuming capacitor is initially
uncharged, capacitor C is now charged to a voltage 2 AUXV with upper plate positive at
t LCπ= . When current through 3T falls to zero, 3T gets commutated. To turn-off the
173
main thyristor 1T , thyristor 2T is turned ON. Then 1T is subjected to a reverse voltage
equal to 2S AUXV V− . This results in thyristor 1T being turned-off. Once 1T is off capacitor
‘C’ discharges through the load LR
LOAD SIDE COMMUTATION In load side commutation the discharging and recharging of capacitor takes place through the load. Hence to test the commutation circuit the load has to be connected. Examples of load side commutation are Resonant Pulse Commutation and Impulse Commutation. LINE SIDE COMMUTATION In this type of commutation the discharging and recharging of capacitor takes place through the supply.
LOAD
FWD
Lr
CT3
IL
L
T2
VS
T1
+
_
+
_
Fig.: 1.35 Line Side Commutation Circuit
Figure 1.35 shows line side commutation circuit. Thyristor 2T is fired to charge
the capacitor ‘C’. When ‘C’ charges to a voltage of 2V, 2T is self commutated. To
reverse the voltage of capacitor to -2V, thyristor 3T is fired and 3T commutates by itself.
Assuming that 1T is conducting and carries a load current LI thyristor 2T is fired to turn
off 1T . The turning ON of 2T will result in forward biasing the diode (FWD) and applying
a reverse voltage of 2V across 1T . This turns off 1T , thus the discharging and recharging of capacitor is done through the supply and the commutation circuit can be tested without load.
174
DC CHOPPERS
INTRODUCTION A chopper is a static device which is used to obtain a variable dc voltage from a
constant dc voltage source. A chopper is also known as dc-to-dc converter. The thyristor converter offers greater efficiency, faster response, lower maintenance, smaller size and smooth control. Choppers are widely used in trolley cars, battery operated vehicles, traction motor control, control of large number of dc motors, etc….. They are also used in regenerative braking of dc motors to return energy back to supply and also as dc voltage regulators.
Choppers are of two types
• Step-down choppers • Step-up choppers.
In step-down choppers, the output voltage will be less than the input voltage whereas in step-up choppers output voltage will be more than the input voltage. PRINCIPLE OF STEP-DOWN CHOPPER
V
i0
V0
Chopper
R
+
−
Fig. 2.1: Step-down Chopper with Resistive Load
Figure 2.1 shows a step-down chopper with resistive load. The thyristor in the circuit acts as a switch. When thyristor is ON, supply voltage appears across the load and when thyristor is OFF, the voltage across the load will be zero. The output voltage and current waveforms are as shown in figure 2.2.
175
Vdc
v0
V
V/R
i0
Idc
t
t
tON
T
tOFF
Fig. 2.2: Step-down choppers — output voltage and current waveforms
dcV = average value of output or load voltage
dcI = average value of output or load current
ONt = time interval for which SCR conducts
OFFt = time interval for which SCR is OFF.
ON OFFT t t= + = period of switching or chopping period
1
fT
= = frequency of chopper switching or chopping frequency.
Average output voltage
( )... 2.1ONdc
ON OFF
tV V
t t
= +
( ). ... 2.2ONdc
tV V V d
T = =
but ( )duty cycle ... 2.3ONtd
t = =
Average output current,
( )... 2.4dcdc
VI
R=
( )... 2.5ONdc
tV VI d
R T R = =
176
RMS value of output voltage
2
0
1 ONt
O oV v dtT
= ∫
But during ,ON ot v V= Therefore RMS output voltage
2
0
1 ONt
OV V dtT
= ∫
( )2
. ... 2.6ONO ON
tVV t V
T T= =
( ). ... 2.7OV d V=
Output power O O OP V I=
But OO
VI
R=
Therefore output power 2
OO
VP
R=
( )2
... 2.8O
dVP
R=
Effective input resistance of chopper
( )... 2.9idc
VR
I=
( )... 2.10i
RR
d=
The output voltage can be varied by varying the duty cycle. METHODS OF CONTROL
The output dc voltage can be varied by the following methods. • Pulse width modulation control or constant frequency operation. • Variable frequency control.
PULSE WIDTH MODULATION
In pulse width modulation the pulse width ( )ONt of the output waveform is varied
keeping chopping frequency ‘f’ and hence chopping period ‘T’ constant. Therefore output voltage is varied by varying the ON time, ONt . Figure 2.3 shows the output voltage waveforms for different ON times.
177
V0
V
V
V0
t
ttON
tON tOFF
tOFF
T
Fig. 2.3: Pulse Width Modulation Control
VARIABLE FREQUENCY CONTROL
In this method of control, chopping frequency f is varied keeping either ONt or
OFFt constant. This method is also known as frequency modulation.
Figure 2.4 shows the output voltage waveforms for a constant ONt and variable chopping period T.
In frequency modulation to obtain full output voltage, range frequency has to be varied over a wide range. This method produces harmonics in the output and for large
OFFt load current may become discontinuous.
v0
V
V
v0
t
t
tON
tON
T
T
tOFF
tOFF
Fig. 2.4: Output Voltage Waveforms for Time Ratio Control
178
STEP-DOWN CHOPPER WITH R-L LOAD Figure 2.5 shows a step-down chopper with R-L load and free wheeling diode.
When chopper is ON, the supply is connected across the load. Current flows from the supply to the load. When chopper is OFF, the load current Oi continues to flow in the same direction through the free-wheeling diode due to the energy stored in the inductor L. The load current can be continuous or discontinuous depending on the values of L and duty cycle, d. For a continuous current operation the load current is assumed to vary between two limits minI and maxI .
Figure 2.6 shows the output current and output voltage waveforms for a continuous current and discontinuous current operation.
V
i0
V0
Chopper
R
LFWD
E
+
−
Fig. 2.5: Step Down Chopper with R-L Load
Outputvoltage
Outputcurrent
v0
V
i0
Imax
Imin
t
t
tON
T
tOFF
Continuouscurrent
Outputcurrent
t
Discontinuouscurrent
i0
Fig. 2.6: Output Voltage and Load Current Waveforms (Continuous Current)
179
When the current exceeds maxI the chopper is turned-off and it is turned-on when current reduces tominI . EXPRESSIONS FOR LOAD CURRENT Oi FOR CONTINUOUS CURRENT
OPERATION WHEN CHOPPER IS ON ( )0 ONt t≤ ≤
V
i0
V0
R
L
E
+
-
Fig. 2.5 (a) Voltage equation for the circuit shown in figure 2.5(a) is
( )... 2.11OO
diV i R L E
dt= + +
Taking Laplace Transform
( ) ( ) ( ) ( ). 0 ... 2.12O O O
V ERI S L S I S i
S S− = + − +
At 0t = , initial current ( ) min0Oi I− =
( ) ( )min ... 2.13O
IV EI S
RR SLS SLL
−= + ++
Taking Inverse Laplace Transform
( ) ( )min1 ... 2.14R R
t tL L
O
V Ei t e I e
R
− −
−= − +
This expression is valid for 0 ONt t≤ ≤ . i.e., during the period chopper is ON. At the instant the chopper is turned off, load current is ( ) maxO ONi t I=
180
When Chopper is OFF ( )0 OFFt t≤ ≤
i0
R
L
E
Fig. 2.5 (b)
Voltage equation for the circuit shown in figure 2.5(b) is
( )0 ... 2.15OO
diRi L E
dt= + +
Taking Laplace transform
( ) ( ) ( )0 0O O O
ERI S L SI S i
S− = + − +
Redefining time origin we have at 0t = , initial current ( ) max0Oi I− =
Therefore ( ) maxO
I EI S
R RS LS SL L
= − + +
Taking Inverse Laplace Transform
( ) ( )max 1 ... 2.16R R
t tL L
O
Ei t I e e
R
− − = − −
The expression is valid for 0 OFFt t≤ ≤ , i.e., during the period chopper is OFF. At
the instant the chopper is turned ON or at the end of the off period, the load current is ( ) minO OFFi t I=
181
TO FIND maxI AND minI
From equation (2.14), At ( ) max, ON Ot t dT i t I= = =
Therefore ( )max min1 ... 2.17dRT dRT
L LV EI e I e
R
− − −= − +
From equation (2.16), At ( ) min, OFF ON Ot t T t i t I= = − =
( )1OFFt t d T= = −
Therefore ( ) ( )
( )1 1
min max 1 ... 2.18d RT d RT
L LEI I e e
R
− −− −
= − −
Substituting for minI in equation (2.17) we get,
( )max
1... 2.19
1
dRT
L
RT
L
V e EI
R Re
−
−
− = −
−
Substituting for maxI in equation (2.18) we get,
( )min
1... 2.20
1
dRT
L
RT
L
V e EI
R Re
− = −
−
( )max minI I− is known as the steady state ripple.
Therefore peak-to-peak ripple current max minI I I∆ = − Average output voltage ( ). ... 2.21dcV d V=
Average output current
( ) ( )max min ... 2.222dc approx
I II
+=
182
Assuming load current varies linearly from minI to maxI instantaneous load current is given by
( ) ( )min
. 0O ON
I ti I for t t dT
dT
∆= + ≤ ≤
( )max minmin ... 2.23O
I Ii I t
dT
− = +
RMS value of load current
( )20
0
1 dT
O RMSI i dtdT
= ∫
( )( ) 2
max minmin
0
1 dT
O RMS
I I tI I dt
dT dT
− = +
∫
( )( )2
min max min2 2max minmin
0
21 dT
O RMS
I I I tI II I t dt
dT dT dT
−− = + +
∫
RMS value of output current
( )( ) ( ) ( )
12 2
max min2min min max min ... 2.24
3O RMS
I II I I I I
−= + + −
RMS chopper current
20
0
1 dT
CHI i dtT
= ∫
2
max minmin
0
1 dT
CH
I II I t dt
T dT
− = +
∫
( ) ( )
12 2
max min2min min max min3CH
I II d I I I I
−= + + −
( ) ( )... 2.25CH O RMSI d I=
Effective input resistance is
iS
VR
I=
183
Where SI = Average source current S dcI dI=
Therefore ( )... 2.26idc
VR
dI=
PRINCIPLE OF STEP-UP CHOPPER
+
−
VOV
Chopper
CLOAD
DLI
+ −
Fig. 2.13: Step-up Chopper
Figure 2.13 shows a step-up chopper to obtain a load voltage OV higher than the input voltage V. The values of L and C are chosen depending upon the requirement of output voltage and current. When the chopper is ON, the inductor L is connected across the supply. The inductor current ‘I’ rises and the inductor stores energy during the ON time of the chopper, ONt . When the chopper is off, the inductor current I is forced to flow
through the diode D and load for a period, OFFt . The current tends to decrease resulting in reversing the polarity of induced EMF in L. Therefore voltage across load is given by
( ) . ., ... 2.27O O
dIV V L i e V V
dt= + >
If a large capacitor ‘C’ is connected across the load then the capacitor will provide a continuous output voltage OV . Diode D prevents any current flow from capacitor to the source. Step up choppers are used for regenerative braking of dc motors. EXPRESSION FOR OUTPUT VOLTAGE
Assume the average inductor current to be I during ON and OFF time of Chopper.
When Chopper is ON Voltage across inductor L V=
184
Therefore energy stored in inductor = ( ). . ... 2.28ONV I t ,
where ONt ON= period of chopper.
When Chopper is OFF (energy is supplied by inductor to load)
Voltage across OL V V= − Energy supplied by inductor ( )O OFFL V V It= − , where OFFt OFF= period of
Chopper. Neglecting losses, energy stored in inductor L = energy supplied by inductor L Therefore ( )ON O OFFVIt V V It= −
[ ]ON OFF
OOFF
V t tV
t
+=
OON
TV V
T t
= −
Where T = Chopping period or period of switching. ON OFFT t t= +
1
1O
ON
V Vt
T
= −
Therefore ( )1... 2.29
1OV Vd
= −
Where duty cyleONtd
T= =
For variation of duty cycle ‘d’ in the range of 0 1d< < the output voltage OV will vary
in the range OV V< < ∞ .
PERFORMANCE PARAMETERS
The thyristor requires a certain minimum time to turn ON and turn OFF. Hence duty cycle d can be varied only between a minimum and a maximum value, limiting the minimum and maximum value of the output voltage. Ripple in the load current depends inversely on the chopping frequency, f. Therefore to reduce the load ripple current, frequency should be as high as possible.
185
CLASSIFICATION OF CHOPPERS Choppers are classified as follows
• Class A Chopper • Class B Chopper • Class C Chopper • Class D Chopper • Class E Chopper
CLASS A CHOPPER
V
Chopper
FWD
+
−
v0
v0
i0
i0
LOAD
V
Fig. 2.14: Class A Chopper and O Ov i− Characteristic
Figure 2.14 shows a Class A Chopper circuit with inductive load and free-wheeling diode. When chopper is ON, supply voltage V is connected across the load i.e.,
Ov V= and current i0 flows as shown in figure. When chopper is OFF, v0 = 0 and the
load current Oi continues to flow in the same direction through the free wheeling diode.
Therefore the average values of output voltage and current i.e., Ov and Oi are always positive. Hence, Class A Chopper is a first quadrant chopper (or single quadrant chopper). Figure 2.15 shows output voltage and current waveforms for a continuous load current.
186
Output current
Thyristorgate pulse
Output voltage
ig
i0
v0
t
t
ttON
T
CH ON
FWD Conducts
Fig. 2.15: First quadrant Chopper - Output Voltage and Current Waveforms Class A Chopper is a step-down chopper in which power always flows from
source to load. It is used to control the speed of dc motor. The output current equations obtained in step down chopper with R-L load can be used to study the performance of Class A Chopper.
CLASS B CHOPPER
V
Chopper
+
−
v0
v0
−i0
i0
L
E
R
D
Fig. 2.16: Class B Chopper Fig. 2.16 shows a Class B Chopper circuit. When chopper is ON, 0Ov = and E
drives a current Oi through L and R in a direction opposite to that shown in figure 2.16. During the ON period of the chopper, the inductance L stores energy. When Chopper is OFF, diode D conducts, Ov V= and part of the energy stored in inductor L is returned to
the supply. Also the current Oi continues to flow from the load to source. Hence the average output voltage is positive and average output current is negative. Therefore Class
187
B Chopper operates in second quadrant. In this chopper, power flows from load to source. Class B Chopper is used for regenerative braking of dc motor. Figure 2.17 shows the output voltage and current waveforms of a Class B Chopper.
The output current equations can be obtained as follows. During the interval diode ‘D’ conducts (chopper is off) voltage equation is given by
V
i0
V0
R
L
E
+
-
D Conducting
OO
LdiV Ri E
dt= + +
For the initial condition i.e., ( ) minOi t I= at 0t = .
The solution of the above equation is obtained along similar lines as in step-down chopper with R-L load
Therefore ( ) min1 0R R
t tL L
O OFF
V Ei t e I e t t
R
− − −= − + < <
At OFFt t= ( ) ( ) maxOi t I=
max min1OFF OFF
R Rt t
L LV EI e I e
R
− − −= − +
During the interval chopper is ON voltage equation is given by
i0
V0
R
L
E
+
-
ChopperON
0 OO
LdiRi E
dt= + +
188
Redefining the time origin, at 0t = ( ) maxOi t I= .
The solution for the stated initial condition is
( ) max 1 0R R
t tL L
O ON
Ei t I e e t t
R
− − = − − < <
At ( ) minON Ot t i t I= =
Therefore min max 1ON ON
R Rt t
L LE
I I e eR
− − = − −
Output current
D conductsChopper
conducts
Thyristorgate pulse
Output voltage
ig
i0
v0
t
t
t
Imin
Imax
T
tONtOFF
Fig. 2.17: Class B Chopper - Output Voltage and Current Waveforms CLASS C CHOPPER
Class C Chopper is a combination of Class A and Class B Choppers. Figure 2.18 shows a Class C two quadrant Chopper circuit. For first quadrant operation, 1CH is ON
or 2D conducts and for second quadrant operation, 2CH is ON or 1D conducts. When
1CH is ON, the load current Oi is positive. i.e., Oi flows in the direction as shown in figure 2.18.
The output voltage is equal to ( )OV v V= and the load receives power from the
source.
189
V
Chopper
+
−
v0
D1
D2CH2
CH1
v0i0
i0
L
E
R
Fig. 2.18: Class C Chopper When 1CH is turned OFF, energy stored in inductance L forces current to flow
through the diode 2D and the output voltage 0Ov = , but Oi continues to flow in positive
direction. When 2CH is triggered, the voltage E forces Oi to flow in opposite direction
through L and 2CH . The output voltage 0Ov = . On turning OFF 2CH , the energy stored
in the inductance drives current through diode1D and the supply; output voltage Ov V= the input current becomes negative and power flows from load to source.
Thus the average output voltage Ov is positive but the average output current
Oi can take both positive and negative values. Choppers 1CH and 2CH should not be turned ON simultaneously as it would result in short circuiting the supply. Class C Chopper can be used both for dc motor control and regenerative braking of dc motor. Figure 2.19 shows the output voltage and current waveforms.
Gate pulseof CH2
Gate pulseof CH1
Output current
Output voltage
ig1
ig2
i0
V0
t
t
t
t
D1 D1D2 D2CH1 CH2 CH1 CH2
ON ON ON ON
Fig. 2.19: Class C Chopper - Output Voltage and Current Waveforms
190
CLASS D CHOPPER
V+ −v0
D2
D1 CH2
CH1
v0
i0
L ER i0
Fig. 2.20: Class D Chopper
Figure 2.20 shows a class D two quadrant chopper circuit. When both 1CH and
2CH are triggered simultaneously, the output voltage Ov V= and output current Oi flows
through the load in the direction shown in figure 2.20. When 1CH and 2CH are turned
OFF, the load current Oi continues to flow in the same direction through load, 1D and 2D ,
due to the energy stored in the inductor L, but output voltage Ov V= − . The average load
voltage Ov is positive if chopper ON-time ( )ONt is more than their OFF-time ( )OFFt and
average output voltage becomes negative if ON OFFt t< . Hence the direction of load current is always positive but load voltage can be positive or negative. Waveforms are shown in figures 2.21 and 2.22.
Gate pulseof CH2
Gate pulseof CH1
Output current
Output voltage
Average v0
ig1
ig2
i0
v0
V
t
t
t
t
CH ,CHON1 2 D1,D2 Conducting
Fig. 2.21: Output Voltage and Current Waveforms for ON OFFt t>
191
Gate pulseof CH2
Gate pulseof CH1
Output current
Output voltage
Average v0
ig1
ig2
i0
v0
V
t
t
t
t
CHCH
1
2
D , D1 2
Fig. 2.22: Output Voltage and Current Waveforms for ON OFFt t< CLASS E CHOPPER
V
v0
i0L ER
CH2 CH4D2 D4
D1 D3CH1 CH3
+ −
Fig. 2.23: Class E Chopper
192
v0
i0
CH - CH ONCH - D Conducts
1 4
4 2
D D2 3 - ConductsCH - D Conducts4 2
CH - CH ONCH - D Conducts
3 2
2 4
CH - D ConductsD - D Conducts
2 4
1 4
Fig. 2.23(a): Four Quadrant Operation Figure 2.23 shows a class E 4 quadrant chopper circuit. When 1CH and 4CH are
triggered, output current Oi flows in positive direction as shown in figure 2.23 through
1CH and 4CH , with output voltage Ov V= . This gives the first quadrant operation. When
both 1CH and 4CH are OFF, the energy stored in the inductor L drives Oi through 3D
and 2D in the same direction, but output voltage Ov V= − . Therefore the chopper operates in the fourth quadrant. For fourth quadrant operation the direction of battery must be reversed. When 2CH and 3CH are triggered, the load current Oi flows in
opposite direction and output voltageOv V= − .
Since both Oi and Ov are negative, the chopper operates in third quadrant. When
both 2CH and 3CH are OFF, the load current Oi continues to flow in the same direction
through 1D and 4D and the output voltageOv V= . Therefore the chopper operates in
second quadrant as Ov is positive but Oi is negative. Figure 2.23(a) shows the devices which are operative in different quadrants.
EFFECT OF SOURCE AND LOAD INDUCTANCE
In choppers, the source inductance should be as small as possible to limit the transient voltage. Usually an input filter is used to overcome the problem of source inductance. Also source inductance may cause commutation problem for the chopper. The load ripple current is inversely proportional to load inductance and chopping frequency. Therefore the peak load current depends on load inductance. To limit the load ripple current, a smoothing inductor is connected in series with the load. Problem 2.1 : For the first quadrant chopper shown in figure 2.24, express the following variables as functions of V, R and duty cycle ‘d’ in case load is resistive.
• Average output voltage and current • Output current at the instant of commutation • Average and rms free wheeling diode current. • RMS value of output voltage • RMS and average thyristor currents.
193
V
i0
v0
Chopper
FWD
+
−
LOAD
Fig. 6.24.
Solution
• Average output voltage, ONdc
tV V dV
T = =
Average output current, dcdc
V dVI
R R= =
• The thyristor is commutated at the instant ONt t= .
Therefore output current at the instant of commutation is V
R, since V is the output
voltage at that instant.
• Free wheeling diode (FWD) will never conduct in a resistive load. Therefore average and RMS free wheeling diode currents are zero.
• RMS value of output voltage
( )20
0
1 ONt
O RMSV v dtT
= ∫
But Ov V= during ONt
( )2
0
1 ONt
O RMSV V dtT
= ∫
( )2 ON
O RMS
tV V
T =
( )O RMSV dV=
Where duty cycle, ONtd
T=
194
• RMS value of thyristor current = RMS value of load current
( )O RMSV
R=
dV
R=
Average value of thyristor current = Average value of load current
dV
R=
Problem 2.2 : A Chopper circuit is operating on TRC at a frequency of 2 kHz on a 460 V supply. If the load voltage is 350 volts, calculate the conduction period of the thyristor in each cycle. Solution
V = 460 V, dcV = 350 V, f = 2 kHz
Chopping period 1
Tf
=
3
10.5 sec
2 10T m−= =
×
Output voltage ONdc
tV V
T =
Conduction period of thyristor
dcON
T Vt
V
×=
30.5 10 350
460ONt−× ×=
0.38 msecONt =
Problem 2.3 : Input to the step up chopper is 200 V. The output required is 600 V. If the conducting time of thyristor is 200 µssec. Compute
• Chopping frequency, • If the pulse width is halved for constant frequency of operation, find the new
output voltage.
195
Solution V = 200 V, 200ONt sµ= , 600dcV V=
dcON
TV V
T t
= −
6
600 200200 10
T
T − = − ×
Solving for T 300T sµ=
• Chopping frequency
1
fT
=
6
13.33
300 10f KHz−= =
×
• Pulse width is halved
Therefore 6200 10
1002ONt sµ
−×= =
Frequency is constant
Therefore 3.33f KHz=
1
300T sf
µ= =
Therefore output voltage = ON
TV
T t
−
( )
6
6
300 10200 300 Volts
300 100 10
−
−
×= = −
Problem 2.4: A dc chopper has a resistive load of 20Ω and input voltage 220SV V= .
When chopper is ON, its voltage drop is 1.5 volts and chopping frequency is 10 kHz. If the duty cycle is 80%, determine the average output voltage and the chopper on time.
196
Solution 220SV V= , 20R = Ω , f = 10 kHz
0.80ONtd
T= =
chV = Voltage drop across chopper = 1.5 volts Average output voltage
( )ONdc S ch
tV V V
T = −
( )0.80 220 1.5 174.8 VoltsdcV = − =
Chopper ON time, ONt dT=
Chopping period, 1
Tf
=
33
10.1 10 secs 100 µsecs
10 10T −= = × =
×
Chopper ON time, ONt dT=
30.80 0.1 10ONt −= × ×
30.08 10 80 µsecsONt −= × =
Problem 2.5: In a dc chopper, the average load current is 30 Amps, chopping frequency is 250 Hz. Supply voltage is 110 volts. Calculate the ON and OFF periods of the chopper if the load resistance is 2 ohms. Solution
30 AmpsdcI = , f = 250 Hz, V = 110 V, 2R = Ω
Chopping period, 31 14 10 4 msecs
250T
f−= = = × =
dcdc
VI
R= and dcV dV=
Therefore dc
dVI
R=
197
30 2
0.545110
dcI Rd
V
×= = =
Chopper ON period, 30.545 4 10 2.18 msecsONt dT −= = × × =
Chopper OFF period, OFF ONt T t= − 3 34 10 2.18 10OFFt − −= × − ×
31.82 10 1.82 msecOFFt −= × =
Problem 2.6: A dc chopper in figure 2.25 has a resistive load of 10R = Ω and input voltage of V = 200 V. When chopper is ON, its voltage drop is 2 V and the chopping frequency is 1 kHz. If the duty cycle is 60%, determine
• Average output voltage • RMS value of output voltage • Effective input resistance of chopper • Chopper efficiency.
V
i0
Chopper
+
−
R v0
Fig. 2.25
Solution
V = 200 V, 10R = Ω , Chopper voltage drop, 2chV V= , d = 0.60, f = 1 kHz.
• Average output voltage ( )dc chV d V V= −
[ ]0.60 200 2 118.8 VoltsdcV = − =
• RMS value of output voltage
( )O chV d V V= −
( )0.6 200 2 153.37 VoltsOV = − =
198
• Effective input resistance of chopper is
iS dc
V VR
I I= =
118.8
11.88 Amps10
dcdc
VI
R= = =
200
16.8311.88i
S dc
V VR
I I= = = = Ω
• Output power is
20
0
1 dT
O
vP dt
T R= ∫
( )2
0
1 dTch
O
V VP dt
T R
−= ∫
( )2
chO
d V VP
R
−=
[ ]2
0.6 200 22352.24 watts
10OP−
= =
• Input power, 0
1 dT
i OP Vi dtT
= ∫
( )
0
1 dTch
O
V V VP dt
T R
−= ∫
( ) [ ]0.6 200 200 2
2376 watts10
chO
dV V VP
R
× −−= = =
• Chopper efficiency,
100O
i
P
Pη = ×
2352.24
100 99%2376
η = × =
Problem 2.7: A chopper is supplying an inductive load with a free-wheeling diode. The load inductance is 5 H and resistance is 10Ω. The input voltage to the chopper is 200
199
volts and the chopper is operating at a frequency of 1000 Hz. If the ON/OFF time ratio is 2:3. Calculate
• Maximum and minimum values of load current in one cycle of chopper operation. • Average load current
Solution:
L = 5 H, R = 10 Ω, f = 1000 Hz, V = 200 V, : 2 :3ON OFFt t =
Chopping period, 1 1
1 msecs1000
Tf
= = =
2
3ON
OFF
t
t=
2
3ON OFFt t=
ON OFFT t t= +
2
3 OFF OFFT t t= +
5
3 OFFT t=
3
5OFFt T=
331 10 0.6 msec
5T −= × × =
ON OFFt T t= −
( ) 31 0.6 10 0.4 msecONt −= − × =
Duty cycle, 3
3
0.4 100.4
1 10ONt
dT
−
−
×= = =×
• Refer equations (2.19) and (2.20) for expressions of maxI and minI .
Maximum value of load current [equation (2.19)] is
max
1
1
dRT
L
RT
L
V e EI
R Re
−
−
− = −
−
200
Since there is no voltage source in the load circuit, E = 0
Therefore max
1
1
dRT
L
RT
L
V eI
Re
−
−
− =
−
3
3
0.4 10 1 10
5
max 10 1 10
5
200 1
101
eI
e
−
−
× × ×−
× ×−
− =
−
3
3
0.8 10
max 2 10
120
1
eI
e
−
−
− ×
− ×
−= −
max 8.0047AI =
Minimum value of load current from equation (2.20) with E = 0 is
min
1
1
dRT
L
RT
L
V eI
Re
− =
−
3
3
0.4 10 1 10
5
min 10 1 10
5
200 17.995 A
101
eI
e
−
−
× × ×
× ×
− = =
−
• Average load current
max min
2dc
I II
+=
8.0047 7.995
8 A2dcI+= ≈
Problem 2.8 : A chopper feeding on RL load is shown in figure 2.26. With V = 200 V, R = 5Ω, L = 5 mH, f = 1 kHz, d = 0.5 and E = 0 V. Calculate
• Maximum and minimum values of load current • Average value of load current • RMS load current • Effective input resistance as seen by source • RMS chopper current.
Solution
V = 200 V, R = 5 Ω, L = 5 mH, f = 1kHz, d = 0.5, E = 0
201
Chopping period is 33
1 11 10 secs
1 10T
f−= = = ×
×
i0
v0
Chopper
R
LFWD
E
+
− Fig.: 2.26
Refer equations (2.19) and (2.20) for expressions of maxI and minI . Maximum value of load current
max
1
1
dRT
L
RT
L
V e EI
R Re
−
−
− = −
−
3
3
3
3
0.5 5 1 10
5 10
max 5 1 10
5 10
200 10
51
eI
e
−
−
−
× × ×−×
× ×−×
− = −
−
0.5
max 1
140 24.9 A
1
eI
e
−
−
−= = −
Minimum value of load current is
min
1
1
dRT
L
RT
L
V e EI
R Re
− = −
−
3
3
3
3
0.5 5 1 10
5 10
min 5 1 10
5 10
200 10
51
eI
e
−
−
−
−
× × ××
× ××
− = −
−
0.5
min 1
140 15.1 A
1
eI
e
−= = −
Average value of load current is
1 2
2dc
I II
+= for linear variation of currents
202
Therefore 24.9 15.1
20 A2dcI+= =
Refer equations (2.24) and (2.25) for RMS load current and RMS chopper current. RMS load current from equation (2.24) is
( )( ) ( )
12 2
max min2min min max min3O RMS
I II I I I I
−= + + −
( )( ) ( )
12 2
2 24.9 15.115.1 15.1 24.9 15.1
3O RMSI −
= + + −
( )
1
296.04228.01 147.98 20.2 A
3O RMSI = + + =
RMS chopper current from equation is (2.25) is
( ) 0.5 20.2 14.28 Ach O RMSI d I= = × =
Effective input resistance is
iS
VR
I=
SI = Average source current S dcI dI= 0.5 20 10 ASI = × = Therefore effective input resistance is
200
2010i
S
VR
I= = = Ω
Problem 2.9: A 200 V dc motor fed by a chopper, runs at 1000 rpm with a duty ratio of 0.8. What must be the ON time of the chopper if the motor has to run at 800 rpm. The chopper operates at 100 Hz. Solution
Speed of motor 1N = 1000 rpm
Duty ratio 1 0.8d = , f = 100 Hz
203
We know that back EMF of motor bE is given by
60b
ZNPE
A
φ=
Where N = speed in rpm
φ = flux/pole in wbs Z = Number of Armature conductors P = Number of poles A = Number of parallel paths Therefore bE Nα φ
if flux is constantbE Nα φ
V
Ia
Chopper
Ra
Eb
+
+
−−
Vdc
M
Fig. 2.27 b dc a aE V I R= −
where aI = Armature current
aR = Armature Resistance Since aR is not given, a aI R drop is neglected.
Therefore 1 1
200 voltsb dcE V= =
1 1dcV d V=
Supply, 1
1
dcVV
d=
200
0.8V =
250 VoltsV =
204
1 1 bE Nα
( )200 1000 ... 2.30α
Now speed changes hence ‘d’ also changes. Given 2 800N = rpm
2?bE =
2 2 bE Nα ( )
2 800 ... 2.31bE α
Dividing equation (2.30) by equation (2.31) we get
2
200 1000
800bE=
2
800 200160 V
1000bE =× =
But
2 2 2b dcE V d V= =
2
2
1600.64
250dcV
dV
= = =
Chopping frequency f = 100 Hz
1 1
0.01 sec100
Tf
= = =
10 msecsT =
2ONt
dT
=
ON time of chopper 2ONt d T= 30.64 10 10ONt −= × × 6.4 msecsONt =
205
IMPULSE COMMUTATED CHOPPER Impulse commutated choppers are widely used in high power circuits where load fluctuation is not large. This chopper is also known as parallel capacitor turn-off chopper or voltage commutated chopper or classical chopper. Fig. 2.28 shows an impulse commutated chopper with two thyristors T1 and T2. We shall assume that the load current remains constant at a value IL during the commutation process.
LOAD
L
C
IL
LS
VS
+
_
+
_
T2
T1
D1
a
biC
iT1
vO
+
_
FWD
Fig. 2.28
To start the circuit, capacitor ‘C’ is initially charged with polarity (with plate ‘a’
positive) as shown in the fig. 2.28 by triggering the thyristor T2. Capacitor ‘C’ gets charged through ‘VS’, ‘C’, T2 and load. As the charging current decays to zero thyristor T2 will be turned-off. With capacitor charged with plate ‘a’ positive the circuit is ready for operation. For convenience the chopper operation is divided into five modes. MODE – 1 Thyristor T1 is fired at t = 0. The supply voltage comes across the load. Load current IL flows through T1 and load. At the same time capacitor discharges through T1, D1, L1, and ‘C’ and the capacitor reverses its voltage. This reverse voltage on capacitor is held constant by diode D1. Fig. 2.29 shows the equivalent circuit of Mode 1.
LOADL
C
IL
LS
VS
+
_
+
_
T1
D1
VC iC
Fig. 2.29
206
Capacitor Discharge Current
( ) sinC
Ci t V t
Lω=
( ) sinC Pi t I tω= ; where P
CI V
L=
Where 1
LCω =
& Capacitor Voltage ( ) cosCV t V tω=
MODE – 2 Thyristor T2 is now fired to commutate thyristor T1. When T2 is ON capacitor voltage reverse biases T1 and turns it off. Now the capacitor discharges through the load from –VS to 0 and the discharge time is known as circuit turn-off time.
Circuit turn-off time is given by
CC
L
V Ct
I
×=
Where IL is load current. Since tC depends on load current, it must be designed for the worst case condition which occur at the maximum value of load current and minimum value of capacitor voltage. Then the capacitor recharges back to the supply voltage (with plate ‘a’ positive). This time is called the recharging time and is given by
Sd
L
V Ct
I
×=
The total time required for the capacitor to discharge and recharge is called the commutation time and it is given by r C dt t t= + At the end of Mode-2 capacitor has recharged to ‘VS’ and the free wheeling diode starts conducting. The equivalent circuit for Mode-2 is shown in fig. 2.30.
207
LOAD
C
LS
VS+
_+
_
T2
VC
IL
IL
Fig. 2.30. MODE – 3 Free wheeling diode FWD starts conducting and the load current decays. The energy stored in source inductance LS is transferred to capacitor. Instantaneous current is
( ) cosLi t I tω= Hence capacitor charges to a voltage higher than supply voltage. 2T
naturally turns-off. The instantaneous capacitor voltage is
( ) sinSC S L S
LV t V I t
Cω= +
Where 1
S
SL Cω =
Fig. 2.31 shows the equivalent circuit of Mode – 3.
LOAD
C
LS
VS
+_
+
_
T2VS
FWD
IL
IL
Fig. 2.31 MODE – 4 Since the capacitor has been overcharged i.e. its voltage is above supply voltage it starts discharging in reverse direction. Hence capacitor current becomes negative. The capacitor discharges through LS, VS, FWD, D1 and L. When this current reduces to zero D1 will stop conducting and the capacitor voltage will be same as the supply voltage fig. 2.32 shows in equivalent circuit of Mode – 4.
208
LOAD
C
LS
VS
+_
+
_
D1
LFWD
IL
VC
Fig. 2.32 MODE – 5 In mode 5 both thyristors are off and the load current flows through the free wheeling diode (FWD). This mode will end once thyristor T1 is fired. The equivalent circuit for mode 5 is shown in fig. 2.33
LOAD
IL
FWD
Fig. 2.33
Fig. 2.34 shows the current and voltage waveforms for a voltage commutated chopper.
209
Capacitor CurrentIL
t
t
t
t
t
Ip Current through T1
Voltage across T1
Output Voltage
Capacitor Voltage
tctd
ic
0Ip
iT1
0vT1
Vc
0vo
Vs c+V
Vs
vc
Vc
-Vc
IL
Fig. 2.34
Though voltage commutated chopper is a simple circuit it has the following disadvantages.
• A starting circuit is required and the starting circuit should be such that it triggers thyristor T2 first.
• Load voltage jumps to twice the supply voltage when the commutation is initiated. • The discharging and charging time of commutation capacitor are dependent on the
load current and this limits high frequency operation, especially at low load current.
• Chopper cannot be tested without connecting load. • Thyristor T1 has to carry load current as well as resonant current resulting in
increasing its peak current rating.
210
Jone’s Chopper
C
D
+
−
V
+
−
LFWD
R
T1
T2
L2
L1
v0
Fig. 2.35: Jone’s Chopper
Figure 2.35 shows a Jone’s Chopper circuit for an inductive load with free
wheeling diode. Jone’s Chopper is an example of class D commutation. Two thyristors are used, T1 is the main thyristor and T2 is the auxiliary thyristor. Commutating circuit
for T1 consists of thyristor T2, capacitor C, diode D and autotransformer (L1 and L2).
Initially thyristor T2 is turned ON and capacitor C is charged to a voltage V with a
polarity as shown in figure 2.35. As C charges, the charging current through thyristor T2
decays exponentially and when current falls below holding current level, thyristor T2
turns OFF by itself. When thyristor T1 is triggered, load current flows through thyristor
T1, L2 and load. The capacitor discharges through thyristor T1, L1 and diode D. Due to
resonant action of the auto transformer inductance L2 and capacitance C, the voltage
across the capacitor reverses after some time. It is to be noted that the load current in L1 induces a voltage in L2 due to
autotransformer action. Due to this voltage in L2 in the reverse direction, the capacitor
charges to a voltage greater than the supply voltage. (The capacitor now tries to discharge in opposite direction but it is blocked by diode D and hence capacitor maintains the reverse voltage across it). When thyristor T1 is to be commutated, thyristor T2 is turned
ON resulting in connecting capacitor C directly across thyristor T1. Capacitor voltage
reverse biases thyristor T1 and turns it off. The capacitor again begins to charge through
thyristor T2 and the load for the next cycle of operation.
The various waveforms are shown in figure 2.36
211
Gate pulse of T2 Gate pulse of T1 Gate pulse of T2
Capacitor Voltage
Capacitordischarge current
Current of T1
Voltage across T1
Auto transformer action
Resonant action
Ig
IL
IL
VC
+V
−V
t
t
t
t
tC
tC
