02 Power Electronics
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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
1
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 ‘ ’ V0(RMS) AC Input Voltage fs
Vs fs
AC Voltage Controller
Variable AC RMSO/P Voltage fS
There are two different types of thyristor control used in practice to control the ac power flow OnOff control Phase control These are the two ac output voltage control techniques. In OnOff 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 ‘ ’ (delay angle), the output RMS voltage across the load can be controlled. The trigger delay angle ‘ ’ is defined as the phase angle (the value of t) 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
2
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 Unidirectional or half wave ac controller. Bidirectional or full wave ac controller. In brief different types of ac voltage controllers are Single phase half wave ac voltage controller (unidirectional controller). Single phase full wave ac voltage controller (bidirectional controller). Three phase half wave ac voltage controller (unidirectional controller). Three phase full wave ac voltage controller (bidirectional 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 ONOFF CONTROL TECHNIQUE (INTEGRAL CYCLE CONTROL) The basic principle of onoff control technique is explained with reference to a single phase full wave ac voltage controller circuit shown below. The thyristor switches T1 and T2 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 tON . The 3
thyristor switches T1 and T2 are turned off by blocking the gate trigger pulses for ‘m’ number of input cycles during the time interval tOFF . The ac controller ON time tON usually consists of an integral number of input cycles.
R RL = Load Resistance Fig.: Single phase full wave AC voltage controller circuit
Vs
n
m
wt
Vo io wt
ig1
Gate pulse of T1 wt
ig2
Gate pulse of T2 wt
Fig.: Waveforms Example Referring to the waveforms of ONOFF control technique in the above diagram, n Two input cycles. Thyristors are turned ON during tON for two input cycles.
4
m
One input cycle. Thyristors are turned OFF during tOFF for one input cycle
Fig.: Power Factor Thyristors are turned ON precisely at the zero voltage crossings of the input supply. The thyristor T1 is turned on at the beginning of each positive half cycle by applying the gate trigger pulses to T1 as shown, during the ON time tON . The load current flows in the positive direction, which is the downward direction as shown in the circuit diagram when T1 conducts. The thyristor T2 is turned on at the beginning of each negative half cycle, by applying gating signal to the gate of T2 , during tON . The load current flows in the reverse direction, which is the upward direction when T2 conducts. Thus we obtain a bidirectional 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, vs Vm sin t 2VS sin t V VS RMS value of input ac supply = m = RMS phase supply voltage. 2 If the input ac supply is connected to load for ‘n’ number of input cycles and disconnected for ‘m’ number of input cycles, then tON
n T,
tOFF
m T
1 = input cycle time (time period) and f f = input supply frequency. tON = controller on time = n T . tOFF = controller off time = m T .
Where T
TO = Output time period = tON
tOFF
nT mT .
5
We can show that, Output RMS voltage VO RMS
tON TO
Vi RMS
VS
tON TO
Where Vi RMS is the RMS input supply voltage = VS . TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF OUTPUT VOLTAGE, FOR ONOFF CONTROL METHOD.
Output RMS voltage VO RMS
VO RMS
VO RMS
VO RMS
VO RMS
Now
Sin2
Vm 2 Sin 2 t.d
t
t 0
tON
Vm 2 TO
VO RMS
Substituting for
tON
1 TO
Sin 2 t.d
t
0
1 Cos 2 2
Vm 2 TO
Vm 2 2 TO
tON
1 Cos 2 t d 2
0
tON
t
tON
d
t
0
Cos 2 t.d
t
0
tON
Vm 2 2 TO
t
Vm 2 2 TO
tON
0
0
Sin 2 t 2
sin 2 tON 2
tON
0
sin 0
tON = An integral number of input cycles; Hence tON
T , 2T ,3T , 4T ,5T ,..... &
tON
2 , 4 , 6 ,8 ,10 ,......
Where T is the input supply time period (T = input cycle time period). Thus we note that sin 2 tON 0 VO RMS
Vm 2 tON 2 TO
Vm
tON 2 TO
6
VO RMS
Where Vi RMS tON TO
VO RMS
Vm 2
tON TO
Vi RMS
tON TO
VS
VS = RMS value of input supply voltage;
tON tON tOFF
nT nT mT
n
VS
n
k = duty cycle (d).
n m
VS k
m n
PERFORMANCE PARAMETERS OF AC VOLTAGE CONTROLLERS RMS Output (Load) Voltage
VO RMS
2
VO RMS
Vm 2
VO RMS
Vi RMS
Where VS
2
Vm sin
t.d
2
t
0
n m n
k
2
Vi RMS
k
VS k
VS k
Vi RMS = RMS value of input supply voltage.
Duty Cycle t k ON TO Where, k
1
2
n n m
tON tON tOFF
n m n
nT m n T
= duty cycle (d).
RMS Load Current
IO RMS
VO RMS
VO RMS
Z
RL
;
for a resistive load Z
RL .
Output AC (Load) Power
PO
IO2 RMS
RL
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Input Power Factor PO VA
PF
output load power input supply volt amperes
I O2 RMS
PF
Vi RMS
RL
;
I in RMS
IS
PO VS I S
RMS input supply current.
Iin RMS
The input supply current is same as the load current I in Hence, RMS supply current = RMS load current; Iin RMS
IO2 RMS
PF
Vi RMS
PF
k
RL
VO RMS
Iin RMS
Vi RMS
Vi RMS Vi RMS
k
IO
IL
IO RMS .
k
n m n
The Average Current of Thyristor IT
Avg
Waveform of Thyristor Current iT
m
n
Im
0
2 IT
IT
IT
IT
Avg
Avg
Avg
Avg
3
n m n
0
nI m 2 m n
0
2
t
I m sin t.d
sin t.d
nI m 2 m n
cos t
nI m 2 m n
cos
t
t
0
cos 0
8
IT
Avg
IT
Avg
IT
Avg
k
IT
Where I m
nI m 2 m n
1
n 2Im m n
2
Imn m n
k .I m
tON tON tOFF
duty cycle
Avg
1
Imn m n
k .I m
n n m
,
Vm = maximum or peak thyristor current. RL
RMS Current of Thyristor IT
RMS 1
IT
IT
IT
RMS
RMS
RMS
2
n n m
nI m2 2 n m nI m2 2 n m
I m2 sin 2 t.d
2
t
0
1
sin
2
t.d
2
t
0
1
1 cos 2 t d 2
0
2
t
1
IT
RMS
nI m2 4 n m
d
t
cos 2 t.d
0
t
0
1
IT
IT
RMS
RMS
nI m2 4 n m nI m2 4 n m
2
t 0
sin 2 t 2
2
0
1
0
sin 2
sin 0
2
2
9
IT
RMS
IT
RMS
IT
RMS
IT
RMS
1
nI m2 4 n m 1
nI m2 4 n m Im 2
2
0 0
2
nI m2 4 n m
1
2
Im k 2
n m n
Im k 2
PROBLEM 1. A single phase full wave ac voltage controller working on ONOFF control technique has supply voltage of 230V, RMS 50Hz, load = 50 . 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.
230V ,
Vin RMS T
1 f
1 50 Hz
Vm 0.02sec ,
2 230V T
325.269 V, Vm
325.269V ,
20ms .
n = number of input cycles during which controller is ON; n 30 . m
number of input cycles during which controller is OFF; m 40 . tON
n T
30 20ms
600ms
tON
n T
0.6sec = controller ON time.
tOFF
m T
40 20ms
tOFF
m T
0.8sec = controller OFF time.
Duty cycle k
n m n
30 40 30
800ms
0.6sec
0.8sec
0.4285
RMS output voltage VO RMS
Vi RMS
n m n
10
230V
VO RMS
230V 0.42857
VO RMS
150.570V
I O RMS
VO RMS
VO RMS
Z
RL
IO2 RMS
PO Input Power Factor P.F
230
3 7
230 0.65465
150.570V 50
3.0114 A
3.01142 50 453.426498W
RL
k
n
PF
30 70
m n
PF
0.4285
0.654653
Average Thyristor Current Rating Im IT Avg
where
30 30 40
VO RMS
k Im
n m n
Im
Vm RL
Im
6.505382 A = Peak (maximum) thyristor current.
2 230 50
6.505382
IT
Avg
IT
Avg
RMS
3 7
0.88745 A
RMS Current Rating of Thyristor Im IT RMS 2
IT
325.269 50
n m n
Im k 2
6.505382 2
3 7
2.129386 A
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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 T1 is connected to the cathode of diode D1 and the cathode of T1 is connected to the anode of D1 . The output voltage across the load resistor ‘R’ and hence the ac power flow to the load is controlled by varying the trigger angle ‘ ’. The trigger angle or the delay angle ‘ ’ refers to the value of t or the instant at which the thyristor T1 is triggered to turn it ON, by applying a suitable gate trigger pulse between the gate and cathode lead. The thyristor T1 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 T1 is triggered it conducts and the load current flows through the thyristor T1 , the load and through the transformer secondary winding. By assuming T1 as an ideal thyristor switch it can be considered as a closed switch when it is ON during the period t to radians. The output voltage across the load follows the input supply voltage when the thyristor T1 is turnedon and when it conducts from t to radians. When the input supply voltage decreases to zero at t , for a resistive load the load current also falls to zero at t and hence the thyristor T1 turns off at t . Between the time period t to 2 , when the supply voltage reverses and becomes negative the diode D1 becomes forward biased and hence turns ON and conducts. The load current flows in the opposite direction during t to 2 radians when D1 is ON and the output voltage follows the negative half cycle of input supply.
Fig.: Halfwave AC phase controller (Unidirectional Controller)
12
Equations Input AC Supply Voltage across the Transformer Secondary Winding. vs
Vm sin t
VS
Vin RMS
Vm = RMS value of secondary supply voltage. 2
Output Load Voltage vo
vL
0 ; for
vo
vL
Vm sin t ; for
t
0 to
t
to 2 .
Output Load Current io
iL
vo RL
Vm sin t ; for RL
io
iL
0 ; for
t
0 to
to 2 .
t
.
TO DERIVE AN EXPRESSION FOR RMS OUTPUT VOLTAGE VO RMS
VO RMS
1 2
VO RMS
Vm 2 2
2
Vm 2 sin 2 t.d
2
t
1 cos 2 t .d 2
t
13
VO RMS
VO RMS
VO RMS
VO RMS
VO RMS
VO RMS
VO RMS
2
Vm 2 4
1 cos 2 t .d
2
Vm
2
d
2
t
2
Vm
t
2
Vm 2
Vm 2
Vm
2
sin 4 2
VO RMS
Vi RMS
VO RMS
VS
1 2
1 2
2
sin 2 2
;sin 4
0
sin 2 2
2 2
Vm 2
2
sin 2 t 2
sin 2 t 2
Vm
VO RMS
cos 2 t.d t
2
2
2
t
1 2
2
2
sin 2 2
2
sin 2 2
2
sin 2 2 sin 2 2
Vm = RMS value of input supply voltage (across the 2 transformer secondary winding).
Where, Vi RMS
VS
Note: Output RMS voltage across the load is controlled by changing ' ' as indicated by the expression for VO RMS
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PLOT OF VO RMS VERSUS TRIGGER ANGLE
FOR A SINGLE PHASE HALF
WAVE AC VOLTAGE CONTROLLER (UNIDIRECTIONAL CONTROLLER)
VO RMS
Vm 2
VO RMS
VS
1 2 1 2
sin 2 2
2
sin 2 2
2
By using the expression for VO RMS we can obtain the control characteristics, which is the plot of RMS output voltage VO RMS versus the trigger angle
. A typical
control characteristic of single phase halfwave phase controlled ac voltage controller is as shown below Trigger angle in degrees
Trigger angle in radians
0
VS
0
300
6
600
3
900
2
1200
2
1500
5
3 6
;1
; 2 ; 3 ; 4 ; 5 ; 6
1800
VO RMS Vm 2
6
0.992765 VS
6
0.949868 VS
6
0.866025 VS
6
0.77314 VS
6
0.717228 VS
6
0.707106 VS
VO(RMS) 70.7% VS
100% VS
60% VS
20% VS 0
60
120
Trigger angle
180 in degrees
Fig.: Control characteristics of single phase halfwave phase controlled ac voltage controller 15
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 VS to 70.7% of VS when we vary the trigger angle 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 2 1 VO dc Vm sin t.d t 2 2
VO dc
Vm 2
VO dc
Vm 2
cos t
VO dc
Vm 2
cos 2
cos
1
; Vm
Vdc
Hence Vdc
sin t.d
t
2
Vm cos 2 2VS cos 2
; cos 2
1
2VS
1
When ' ' is varied from 0 to
. Vdc varies from 0 to
Vm
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 VS at a trigger angle 0 to a low of 70.7% of VS at Radians . These drawbacks of single phase half wave ac voltage controller can be over come by using a single phase full wave ac voltage controller.
16
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 halfwave ac voltage controller has a load resistance R 50 , input ac supply voltage is 230V RMS at 50Hz. The input supply transformer has a turns ratio of 1:1. If the thyristor T1 is triggered at 600 . Calculate RMS output voltage. Output power. RMS load current and average load current. Input power factor. Average and RMS thyristor current.
Given,
Vp
230V , RMS primary supply voltage.
f RL
Input supply frequency = 50Hz. 50 600
Therefore
VS
radians. 3 RMS secondary voltage.
Vp
Np
VS
NS
Vp
VS
1 1 1 230V
Where, N p = Number of turns in the primary winding. N S = Number of turns in the secondary winding.
17
RMS Value of Output (Load) Voltage VO RMS 2
1 2
VO RMS
Vm2 sin 2 t.d
t
We have obtained the expression for VO RMS as 1 2
sin 2 2
VO RMS
VS
2
VO RMS
230
1 2
VO RMS
230
1 5.669 2
VO RMS
218.4696 V
2
sin1200 2
3
230 0.94986
218.47 V
RMS Load Current IO RMS
IO RMS
VO RMS
218.46966 50
RL
4.36939 Amps
Output Load Power PO
PO
IO2 RMS
PO
0.9545799 KW
RL
4.36939
2
50 954.5799 Watts
Input Power Factor PF
PO VS I S
VS = RMS secondary supply voltage = 230V. I S = RMS secondary supply current = RMS load current.
IS PF
IO RMS
4.36939 Amps
954.5799 W 230 4.36939 W
0.9498
18
Average Output (Load) Voltage
VO dc
1 2
2
Vm sin t.d
t
We have obtained the expression for the average / DC output voltage as, VO dc
VO dc
VO dc
Vm cos 2
1
2 230 cos 600 2
325.2691193 2
1
0.5
325.2691193 0.5 1 2
25.88409 Volts
Average DC Load Current
IO dc
VO dc RL
25.884094 50
0.51768 Amps
Average & RMS Thyristor Currents
iT1 Im 3
2 (2 + )
t
Fig.: Thyristor Current Waveform Referring to the thyristor current waveform of a single phase halfwave ac voltage controller circuit, we can calculate the average thyristor current IT Avg as
IT
Avg
IT
Avg
1 2
I m sin t.d
Im 2
sin t.d
t
t
19
IT
Avg
IT
Avg
IT
Avg
Where, I m
cos t
Im 2
cos
Im 1 cos 2
2 230 50 6.505382 Amps
IT
Avg
IT
Avg
IT
Avg
IT
Avg
Vm 1 cos 2 RL 2 230 1 cos 600 2 50 2 230 1 0.5 100
1.5530 Amps
RMS thyristor current IT
IT
IT
cos
Vm = Peak thyristor current = Peak load current. RL
Im Im
Im 2
RMS
RMS
IT
RMS
IT
RMS
1 2
I m2 sin 2 t.d
I m2 2
t
1 cos 2 t .d 2
I m2 4
Im
can be calculated by using the expression
RMS
d
1 4
t
t
t
cos 2 t.d
t
sin 2 t 2
20
IT
RMS
Im
1 4
IT
RMS
Im
1 4
IT
RMS
Im 2
sin 2 2
sin 2 2
1 2
sin 2 2 sin 1200
6.50538 1 2 2
IT
RMS
IT
RMS
4.6
IT
RMS
4.6 0.6342 2.91746 A
IT
RMS
2.91746 Amps
1 2
sin 2
2 3
3
2
0.8660254 2
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 ' ' . The RMS value of load voltage can be varied by varying the trigger angle ' ' . 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 ' ' . Hence the full wave ac voltage controller is also referred to as to a bidirectional controller.
21
Fig.: Single phase full wave ac voltage controller (Bidirectional Controller) using SCRs The thyristor T1 is forward biased during the positive half cycle of the input supply voltage. The thyristor T1 is triggered at a delay angle of ' ' 0
radians .
Considering the ON thyristor T1 as an ideal closed switch the input supply voltage appears across the load resistor RL and the output voltage vO vS during t to radians. The load current flows through the ON thyristor T1 and through the load resistor RL in the downward direction during the conduction time of T1 from t to radians. At t , when the input voltage falls to zero the thyristor current (which is flowing through the load resistor RL ) falls to zero and hence T1 naturally turns off . No current flows in the circuit during
t
to
.
The thyristor T2 is forward biased during the negative cycle of input supply and when thyristor T2 is triggered at a delay angle negative halfcycle of input from
t
, the output voltage follows the
to 2 . When T2 is ON, the load current
flows in the reverse direction (upward direction) through T2 during
t
to
2 radians. The time interval (spacing) between the gate trigger pulses of T1 and T2 is kept at radians or 1800. At t 2 the input supply voltage falls to zero and hence the load current also falls to zero and thyristor T2 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 (Bidirectional Controller) using TRIAC
22
Fig: Waveforms of single phase full wave ac voltage controller EQUATIONS Input supply voltage vS Vm sin t
2VS sin t ;
Output voltage across the load resistor RL ; vO vL Vm sin t ; for
t
to
Output load current vO iO RL for
and
Vm sin t RL
t
to
and
t
to 2
I m sin t ;
t
to 2
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
VO2 RMS
V 2 L RMS
1 2
2
vL 2 d
t ;
0
23
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 radians. Hence we can also calculate the RMS output voltage by using the expression given below.
V 2 L RMS
1
Vm 2 sin 2 t.d t 0
V
2
vL
L RMS
vO
1 2
2
vL 2 .d
t ;
0
Vm sin t ; For
t
and
to
t
to 2
Hence,
VL2 RMS
1 2
2 2
Vm sin t d
Vm 2
t
2
1 cos 2 t d 2
t
sin 2 t.d
t
1 cos 2 t d 2
t
t
2
Vm 2 2 2 Vm 2 4
Vm sin t d 2
1 Vm 2 sin 2 t.d 2 Vm 2 2
2
t
d
t
cos 2 t.d
2
t
2
d
sin 2 t 2
t
Vm 2 4
t
1 sin 2 2
Vm 2 2 4
1 0 sin 2 2
Vm 2 2 4
sin 2 2
sin 2
Vm 2 2 4
sin 2 2
sin 2
t
cos 2 t.d
sin 2 t 2
sin 2
t
2
1 sin 4 2
sin 2
1 0 sin 2 2
2 2 2
24
Vm 2 2 4
sin 2
sin 2 2
0 & cos 2
1 sin 2 .cos 2 2
cos 2 .sin 2
1
Therefore, VL2 RMS
V
2 L RMS
Vm 2 2 4
sin 2 2
Vm 2 2 4
sin 2
Vm 2 4
2
2
sin 2 2
sin 2
Taking the square root, we get VL RMS
VL RMS
VL RMS
VL RMS
Vm 2
2
2
Vm 2 2
Vm
2
1 2
2 Vm
2
sin 2
2
sin 2
2
sin 2
1 2 2
2
VL RMS
Vm 2
VL RMS
Vi RMS
VL RMS
VS
1
1
sin 2 2
sin 2 2 1
sin 2 2
sin 2 2
Maximum RMS voltage will be applied to the load when 0 , in that case the full sine wave appears across the load. RMS load voltage will be the same as the RMS Vm supply voltage . When is increased the RMS load voltage decreases. 2
25
1
0
Vm 2
1
0
Vm 2
0 2
Vi RMS
VS
VL RMS
VL RMS
Vm
VL RMS 0
2
sin 2 0 2
0
The output control characteristic for a single phase full wave ac voltage controller with resistive load can be obtained by plotting the equation for VO RMS CONTROL CHARACTERISTIC OF SINGLE PHASE FULLWAVE AC VOLTAGE CONTROLLER WITH RESISTIVE LOAD The control characteristic is the plot of RMS output voltage VO RMS versus the trigger angle ; which can be obtained by using the expression for the RMS output voltage of a fullwave ac controller with resistive load.
VO RMS
VS
Where VS Trigger angle in degrees 0
1
sin 2 2
Vm 2
RMS value of input supply voltage
Trigger angle in radians 0
300
6
600
3
900
2
1200
2
1500
5
1800
;
3 6
;1 ; 2 ; 3
; 4 ; 5 ; 6
6 6 6
6 6 6
VO RMS VS
% 100% VS
0.985477 VS
98.54% VS
0.896938 VS
89.69% VS
0.7071 VS
70.7% VS
0.44215 VS
44.21% VS
0.1698 VS
16.98% VS
0 VS
0 VS
26
VO(RMS) VS
0.6VS
0.2 VS 0
60
120
Trigger angle
180 in degrees
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% VS at 0 to a minimum of ‘0’ at 1800 . 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 T1 and T2 in parallel, the gating circuits (gate trigger pulse generating circuits) of Thyristors T1 and T2 must be isolated. Figure shows a pulse transformer with two separate windings to provide isolation between the gating signals of T1 and T2 .
G1
Gate Trigger Pulse Generator
K1 G2 K2
Fig.: Pulse Transformer
SINGLE PHASE FULLWAVE COMMON CATHODE
AC
VOLTAGE
CONTROLLER
WITH
It is possible to design a single phase full wave ac controller with a common cathode configuration by having a common cathode point for T1 and T2 & by adding two diodes in a full wave ac controller circuit as shown in the figure below
27
Fig.: Single phase full wave ac controller with common cathode (Bidirectional controller in common cathode configuration) Thyristor T1 and diode D1 are forward biased during the positive half cycle of input supply. When thyristor T1 is triggered at a delay angle , Thyristor T1 and diode D1 conduct together from t to during the positive half cycle. The thyristor T2 and diode D2 are forward biased during the negative half cycle of input supply, when trigged at a delay angle , thyristor T2 and diode D2 conduct together during the negative half cycle from t to 2 . In this circuit as there is one single common cathode point, routing of the gate trigger pulses to the thyristor gates of T1 and T2 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
D1
D3
+ T1 D4
AC Supply
D2 RL

28
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 T1 and current through thyristor T1 are always unidirectional. When T1 is triggered at , during the positive half cycle 0 , the load current flows through D1 , T1 , t diode D2 and through the load. With a resistive load, the thyristor current (flowing through the ON thyristor T1 ) , the load current falls to zero at t , when the input supply voltage decreases to zero at t , the thyristor naturally turns OFF. In the negative half cycle, diodes D3 & D4 are forward biased during
t
to 2 radians. When T1 is triggered at
t
, the load current flows in the
opposite direction (upward direction) through the load, through D3 , T1 and D4 . Thus D3 , D4 and T1 conduct together during the negative half cycle to supply the load power. When the input supply voltage becomes zero at t 2 , the thyristor current (load current) falls to zero at t 2 and the thyristor T1 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 T1 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 (BIDIRECTIONAL CONTROLLER) WITH RL LOAD
CONTROLLER
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.
29
A single phase full wave ac voltage controller circuit (bidirectional controller) with an RL load using two thyristors T1 and T2 ( T1 and T2 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 T1 is forward biased during the positive half cycle of input supply. Let us assume that T1 is triggered at
t
, by applying a suitable gate trigger pulse to
T1 during the positive half cycle of input supply. The output voltage across the load
follows the input supply voltage when T1 is ON. The load current iO flows through the thyristor T1 and through the load in the downward direction. This load current pulse flowing through T1 can be considered as the positive current pulse. Due to the inductance in the load, the load current iO flowing through T1 would not fall to zero at
t
, when
the input supply voltage starts to become negative. The thyristor T1 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 T1 falls to zero at
t
, where
is referred to as the Extinction angle, (the value of
at which the load current falls to zero. The extinction angle
t)
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.
30
The thyristor T1 conducts from
t
, which depends on the delay angle
to
. The conduction angle of T1 is and the load impedance angle
. The
waveforms of the input supply voltage, the gate trigger pulses of T1 and T2 , 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 is the extinction angle which depends upon the load inductance value.
Fig.: Gating Signals
31
Waveforms of single phase full wave ac voltage controller with RL load for Discontinuous load current operation occurs for i.e.,
, conduction angle
and
.
;
.
Fig.: Waveforms of Input supply voltage, Load Current, Load Voltage and Thyristor Voltage across T1 Note The RMS value of the output voltage and the load current may be varied by varying the trigger angle . 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. 32
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 T1 and T2 . 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 . TO DERIVE AN EXPRESSION FOR THE OUTPUT (INDUCTIVE LOAD) CURRENT, DURING t to WHEN THYRISTOR T1 CONDUCTS Considering sinusoidal input supply voltage we can write the expression for the supply voltage as vS
Vm sin t = instantaneous value of the input supply voltage.
Let us assume that the thyristor T1 is triggered by applying the gating signal to T1 at t . The load current which flows through the thyristor T1 during t to can be found from the equation L
diO dt
RiO
Vm sin t ;
The solution of the above differential equation gives the general expression for the output load current which is of the form iO
Where Vm Z
Vm sin Z
t
t
A1e
;
2VS = maximum or peak value of input supply voltage. R2
tan
1
L
L R
2
= Load impedance.
= Load impedance angle (power factor angle of load).
L = Load circuit time constant. R
Therefore the general expression for the output load current is given by the equation R t Vm iO sin t A1e L ; Z
33
The value of the constant A1 can be determined from the initial condition. i.e. initial value of load current iO 0 , at t . Hence from the equation for iO equating iO to zero and substituting t , we get Vm 0 sin Z
iO
R
Therefore
A1e L
Vm sin Z
t
e A1
e
R t L
R t L
R
A1 By substituting
Vm sin Z
1
A1
e
Vm sin Z
t L
Vm sin Z
, we get the value of constant A1 as
t R
A1
A1e
R t L
e
L
Vm sin Z
Substituting the value of constant A1 from the above equation into the expression for iO , we obtain R R t Vm Vm L ; iO sin t e e L sin Z Z R
iO
Vm sin Z
t
e
iO
Vm sin Z
t
e
t L
R L
R
e
t
Vm sin Z
L
Vm sin Z
Therefore we obtain the final expression for the inductive load current of a single phase full wave ac voltage controller with RL load as
iO
Vm sin Z
t
sin
e
R L
t
;
Where
t
.
34
The above expression also represents the thyristor current iT 1 , during the conduction time interval of thyristor T1 from t to . To Calculate Extinction Angle The extinction angle , which is the value of t at which the load current iO falls to zero and T1 is turned off can be estimated by using the condition that iO 0 , at t By using the above expression for the output load current, we can write
iO As
Vm Z
0
Vm sin Z
sin
e
R L
0 we can write
sin
sin
e
sin
e
R L
0
Therefore we obtain the expression
sin
R L
The extinction angle can be determined from this transcendental equation by using the iterative method of solution (trial and error method). After is calculated, we can determine the thyristor conduction angle
.
is the extinction angle which depends upon the load inductance value. Conduction angle increases as is decreased for a known value of . For radians, i.e., for radians, for the load current waveform appears as a discontinuous current waveform as shown in the figure. The output load current remains at zero during t to . This is referred to as discontinuous load current operation which occurs for
.
When the trigger angle is decreased and made equal to the load impedance angle i.e., when we obtain from the expression for sin ,
sin
0 ; Therefore
Extinction angle Conduction angle
radians.
; for the case when
radians 1800 ; for the case when
Each thyristor conducts for 1800 (
radians ) . T1 conducts from
t
to
and provides a positive load current. T2 conducts from to 2 and provides a negative load current. Hence we obtain a continuous load current and the
35
output voltage waveform appears as a continuous sine wave identical to the input supply voltage waveform for trigger angle and the control on the output is lost. vO
vO=vS
Vm
0
2
3 t
iO Im t
Fig.: Output voltage and output current waveforms for a single phase full wave ac voltage controller with RL load for Thus we observe that for trigger angle , 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 as the output voltage becomes equal to the input supply voltage and thus we obtain Vm VO RMS VS ; for 2 Hence, RMS output voltage = RMS input supply voltage for TO DERIVE AN EXPRESSION FOR RMS OUTPUT VOLTAGE VO RMS OF A SINGLE PHASE FULLWAVE AC VOLTAGE CONTROLLER WITH RL LOAD.
36
O , the load current and load voltage waveforms become discontinuous When as shown in the figure above.
VO RMS Output vo
1 2
1
Vm 2 sin 2 t.d
Vm sin t , for
t
t
to
, when T1 is ON. 1
VO RMS
Vm 2
1 cos 2 t d 2
2
t
1
VO RMS
Vm 2 2
VO RMS
Vm 2 2
VO RMS
Vm 2 2
d
t
cos 2 t.d
t
1
t
sin 2 t 2
2
1
sin 2 2
2
sin 2 2
2
1
VO RMS
Vm
1 2
sin 2 2
sin 2 2
Vm
1
sin 2 2
sin 2 2
1
VO RMS
2
2
2
The RMS output voltage across the load can be varied by changing the trigger angle . For a purely resistive load L 0 , therefore load power factor angle 0. L tan 1 0 ; R radians 1800 Extinction angle
37
PERFORMANCE PARAMETERS OF A SINGLE PHASE FULL WAVE AC VOLTAGE CONTROLLER WITH RESISTIVE LOAD
Vm 2
RMS Output Voltage VO RMS
1
sin 2 2
Vm 2
;
VS = RMS
input supply voltage.
IO RMS
IS
VO RMS RL
= RMS value of load current.
IO RMS = RMS value of input supply current.
Output load power
IO2 RMS
PO
RL
Input Power Factor PF
PF
PO VS I S
I O2 RMS
RL
I O RMS
VS I O RMS
VO RMS
RL
VS
1
sin 2 2
VS
Average Thyristor Current, iT1 Im 3
2
t
(2 + )
Fig.: Thyristor Current Waveform
IT
Avg
IT
Avg
IT
Avg
1 2
iT d
t
Im sin t.d 2 Im 2
cos
1 2
I m sin t.d
t
Im 2
cos
t
cos t
Im 1 cos 2
38
Maximum Average Thyristor Current, for Im IT Avg
0,
RMS Thyristor Current IT
RMS
IT
RMS
1 2
Im 2
I m2 sin 2 t.d
1 2
t
sin 2 2
Maximum RMS Thyristor Current, for
0,
Im 2 In the case of a single phase full wave ac voltage controller circuit using a Triac with resistive load, the average thyristor current IT Avg 0 . Because the Triac conducts in IT
RMS
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 RL LOAD The Expression for the Output (Load) Current The expression for the output (load) current which flows through the thyristor, during t to is given by
iO
Vm sin Z
iT1
t
sin
e
R L
t
;
for
t
Where,
Vm Z
2VS = Maximum or peak value of input ac supply voltage. R2
tan
1
L L R
2
= Load impedance.
= Load impedance angle (load power factor angle).
= Thyristor trigger angle = Delay angle. = Extinction angle of thyristor, (value of current falls to zero. is calculated by solving the equation
sin
sin
e
t ) at which the thyristor (load)
R L
39
Thyristor Conduction Angle radians = 1800 for
Maximum thyristor conduction angle
.
RMS Output Voltage
Vm 2
VO RMS
1
sin 2 2
sin 2 2
The Average Thyristor Current IT
IT
IT
Avg
Avg
1 2
iT1 d
1 2
Vm sin Z
Vm 2 Z
Avg
Maximum value of IT maximum IT
Im Avg
RMS
t
sin
.d
t
e
sin
R L
t
d
e
R L
t
t
d
t
0 . The thyristors should be rated for
Vm . Z
RMS
RMS
Maximum value of IT maximum IT
t
occur at
Avg
, where I m
RMS Thyristor Current IT
IT
sin
t
1 2
RMS
iT21 d
t
occurs at
0 . Thyristors should be rated for
Im 2
When a Triac is used in a single phase full wave ac voltage controller with RL Im type of load, then IT Avg 0 and maximum IT RMS 2
40
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 = 10 , the delay angle of thyristors T1 and T2 are equal, where
1
a. Conduction angle of the thyristor T1 . b. RMS output voltage. c. The input power factor. Comment on the type of operation. Given Vs 230V , f 50Hz , L 10mH , 1
2
3
2
3
. Determine
600 ,
R 10 ,
radians, .
Vm
2VS
Z
2 230 325.2691193 V R2
Load Impedance
L Z
Im
2 fL 10
Vm Z
2
50 10 10
3.14159
2 230 10.4818 tan
Load Impedance Angle
tan
2
1
10
L
10
2
L
2
3.14159
109.8696 10.4818
31.03179 A L R
1
tan
2
3
2
1
0.314159
17.440590
Trigger Angle . Hence the type of operation will be discontinuous load current operation, we get
180 60 ; Therefore
180
0
240
the
range
of
2400
is
from
180 degrees
to
240
degrees.
0
41
Extinction Angle
is calculated by using the equation
sin
sin
e
R L
In the exponential term the value of radians. Hence
sin
sin
e
60 17.44059
R L
and
Rad
Rad
;
should be substituted in
Rad
3
42.55940 10
sin
17.44
sin
17.44
0
sin 42.55940 e
0
0.676354e
3.183
0
1800
Assuming
radians,
Rad
1900 ; 0 Rad
1900 180
1800
L.H.S: sin 190 17.44
0
3.183 3.3161
3
0.129487
4.94 10
4
1830 ; 0 Rad
1800
1830 180
3.19395
L.H.S: sin
3.19395
3
2.14675
sin 183 17.44
R.H.S: 0.676354e Assuming
3.3161
sin 172.56
R.H.S: 0.676354 e Assuming
1800
3.183 2.14675
sin165.560
7.2876 10
0.24936
4
1800 0 Rad
1800
1800 180
3
2 3
42
L.H.S: sin
sin 180 17.44 3.183
3
R.H.S: 0.676354e Assuming
8.6092 10
Rad
1960 180
1800
L.H.S: sin
3.420845
sin 196 17.44 3.183 3.420845
3
R.H.S: 0.676354e
0.02513
3.5394 10
4
197 0 0 Rad
1970 180
1800
L.H.S: sin
3.43829
7.69 7.67937 10 3
sin 197 17.44 3.183 3.43829
R.H.S: 0.676354e
Assuming
4
1960 0
Assuming
0.2997
3
4.950386476 10
4
197.420 0 Rad
180
197.42 180
0
L.H.S: sin
3.4456
sin 197.42 17.44 3.183 3.4456
R.H.S: 0.676354e
3
3.2709 10
197.420 600
Conduction Angle
3.4906 10
4
4
137.420
RMS Output Voltage
VO RMS
VS
1
sin 2 2 1
VO RMS
230
VO RMS
230
VO RMS
230 0.9 207.0445 V
1
3.4456
2.39843
3
sin 2 2 sin 2 600
sin 2 197.420
2
2
0.4330 0.285640
43
Input Power Factor PO
PF
VS I S VO RMS
I O RMS
Z
PO
IO2 RMS
VS
230V ,
PF
207.0445 19.7527 A 10.4818
RL
19.7527
IS
PO
IO RMS
2
10 3901.716 W
19.7527
3901.716 230 19.7527
VS I S
0.8588
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 is 2 . Find a. RMS output voltage b. Power output c. Input power factor d. Average and RMS thyristor current. Solution 2
900 , VS
120 V,
R 6
RMS Value of Output Voltage
VO
VS
VO 120 VO
1
sin 2 2
1 2
1 2
sin180 2
1 2
84.85 Volts
RMS Output Current IO
VO R
PO
I O2 R
PO
14.14
84.85 14.14 A 6
Load Power
2
6 1200 watts
44
Input Current is same as Load Current Therefore I S I O 14.14 Amps Input Supply VoltAmp VS I S 120 14.14 1696.8 VA Therefore Input Power Factor =
Load Power Input VoltAmp
1200 1696.8
0.707 lag
Each Thyristor Conducts only for half a cycle Average thyristor current IT
IT
1 Avg
2 R
Avg
Vm sin t.d
Vm 1 cos 2 R
t
;
Vm
2 120 1 cos 90 2 6
RMS thyristor current IT
IT
RMS
2VS
4.5 A
RMS
1 2
Vm2 sin 2 t d R2
Vm2 2 R2
t
1 cos 2 t d 2
Vm 1 2R 2VS 1 2R
2 120 1 2 6
1 2
sin 2 2 sin 2 2
2
t
1 2
sin180 2
1 2
10 Amps
45
3. A single phase half wave ac regulator using one SCR in antiparallel with a diode feeds 1 kW, 230 V heater. Find load power for a firing angle of 450. Solution 450
, VS 230 V ; PO 1KW 1000W 4 At standard rms supply voltage of 230V, the heater dissipates 1KW of output
power Therefore PO
VO I O
VO2 R
VO VO R
Resistance of heater R
VO2 PO
2
230 1000
52.9
RMS value of output voltage
1 2
VO
VS
VO
1 230 2
sin 2 2
2
2
RMS value of output current VO 224.9 IO R 52.9 Load Power
PO
IO2 R
4
1 2
; for firing angle
450
1 2
sin 90 2
224.7157 Volts
4.2479 Amps
4.25
2
52.9 954.56 Watts
4. Find the RMS and average current flowing through the heater shown in figure. The delay angle of both the SCRs is 450.
io
SCR1 +
1220V ac
SCR2
1 kW, 220V heater
46
Solution 450
4
, VS
220 V
Resistance of heater 2
V2 R R
220 1000
48.4
Resistance value of output voltage 1
VO
VS
VO
220
VO
220
sin 2 2 1 4
sin 90 2
4
1 2
1
209.769 Volts
VO R
209.769 48.4
Average current flowing through the heater
I Avg
RMS current flowing through heater
4.334 Amps 0
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 and L = 6 . 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 load impedance angle
is equal to the
, load angle tan
1
L R
tan
1
6 4
56.30
Maximum possible value of is 1800 Therefore control range of firing angle is 56.30
1800
47
56.30 . At this value
Maximum value of RMS load current occurs when of the Maximum value of RMS load current VS Z
IO
220
30.5085 Amps
42 62
2
Maximum Power PO
IO2 R
30.5085
Input VoltAmp
VS I O
220 30.5085 6711.87 W
PO Input VA
Power Factor
3723.077 6711.87
4 3723.077 W
0.5547
Average thyristor current will be maximum when 1800 . angle Therefore maximum value of average thyristor current Vm 1 IT Avg sin t d t 2 Z
Note: At
But
iO
iT1
Vm sin Z
iT1
iO
Vm sin Z
t
sin
e
R L
and conduction
t
0,
IT
Avg
IT
Avg
IT
Avg
t
Vm 2 Z
cos
Vm 2 Z
cos
Vm 2 Z
cos
t
cos
,
IT
Vm Z
Avg
Vm 2 2 Z
cos 0
2 220 42 62
Vm Z
13.7336 Amps
Similarly, maximum RMS value occurs when
0 and
.
Therefore maximum value of RMS thyristor current
ITM
1 2
Vm sin Z
2
t
d
t
48
ITM
Vm2 2 Z2
ITM
Vm2 4 Z2
ITM
Vm2 4 Z2
ITM
Vm 2Z
1 cos 2 t 2 2
t
d
t
sin 2 t 2 2
0
2 220 2 42 62
21.57277 Amps
49
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.
+ AC Input Voltage
Line Commutated Converter
DC Output V0(dc ) 
Type of input: Fixed voltage, fixed frequency ac power supply. Type of output: Variable dc output voltage 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 unidirectional 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 t . The load current flows when the thyristors conduct from to . The output voltage across the load follows the input supply voltage through t the conducting thyristor. At t , 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. 50
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 CONTROLLED 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 radians. A single phase half wave diode rectifier gives (under ideal conditions) an average Vm dc output voltage VO dc and single phase full wave diode rectifier gives (under ideal conditions) an average dc output voltage VO dc
2Vm
, where Vm 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 from 1800 to 00 by varying the trigger angle from 00 to 1800, where thyristor conduction angle 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. Electrochemical and electrometallurgical 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 acdc 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
51
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 acdc 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 RECTIFIERS 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 semiconverter (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.
52
R
RL
Load Resistance
Fig.: Single Phase HalfWave Thyristor Converter with a Resistive Load A single phase half wave thyristor converter which is used for acdc 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. v P represents the primary input ac supply voltage. v S 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 t , by applying a suitable gate trigger pulse to the gate lead of thyristor. When the thyristor is triggered at a delay angle of t , 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 t to radians. Output voltage vO vS , when the thyristor conducts from t to . For a purely resistive load, the load current iO (output current) that flows when the thyristor T1 is on, is given by the expression vO iO , for t RL The output load current waveform is similar to the output load voltage waveform during the thyristor conduction time from to . 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
t
2
, when the input
supply voltage is at its maximum value. The maximum value (peak value) of the load current is calculated as Vm iO max I m . RL
53
Note that when the thyristor conducts ( T1 is on) during t to , the thyristor current iT 1 , the load current iO through RL and the source current iS flowing through the transformer secondary winding are all one and the same. Hence we can write vO Vm sin t iS iT 1 iO ; for t R R I m is the maximum (peak) value of the load current that flows through the transformer secondary winding, through T1 and through the load resistor RL at the instant t
2
, 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 t , the thyristor and the load current also falls to zero at t . Thus the thyristor naturally turns off when the current flowing through it falls to zero at t . During the negative half cycle of input supply when the supply voltage reverses and becomes negative during t to 2 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 cutoff (in the reverse blocking mode). The thyristor cannot conduct during its reverse biased state between t to 2 . An ideal thyristor under reverse biased condition behaves as an open switch and hence the load current and load voltage are zero during t to 2 . The maximum or peak reverse voltage that appears across the thyristor anode and cathode terminals is Vm . The trigger angle (delay angle or the phase angle ) 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 , hence the conduction angle . The maximum conduction angle is radians (1800) when the trigger angle
0.
Fig: Quadrant Diagram The waveforms shows the input ac supply voltage across the secondary winding of the transformer which is represented as v S , the output voltage across the load, the output (load) current, and the thyristor voltage waveform that appears across the anode and cathode terminals.
54
Fig: Waveforms of single phase halfwave controlled rectifier with resistive load EQUATIONS vs Vm sin t the ac supply voltage across the transformer secondary. Vm
max. (peak) value of input ac supply voltage across transformer secondary.
VS
Vm 2
vO
vL
RMS value of input ac supply voltage across transformer secondary. the output voltage across the load ; iO
iL
output (load) current.
55
When the thyristor is triggered at t (an ideal thyristor behaves as a closed switch) and hence the output voltage follows the input supply voltage. vO
vL
Vm sin t ; for
iO
iL
vO = Load current for R
t
to
, when the thyristor is on.
t
to
, when the thyristor is on.
TO DERIVE AN EXPRESSION FOR THE AVERAGE (DC) OUTPUT VOLTAGE ACROSS THE LOAD If Vm is the peak input supply voltage, the average output voltage Vdc can be found from 1 VO dc Vdc vO .d t 2
VO dc
Vdc
1 2
VO dc
1 2
Vm sin t.d
VO dc
Vm sin t.d 2
VO dc
Vm 2
cos t
VO dc
Vm 2
cos
VO dc
Vm 1 cos 2
Vm sin t.d
t
t
t
; cos
cos
; Vm
1
2VS
The maximum average (dc) output voltage is obtained when Vm maximum dc output voltage Vdc max Vdm .
0 and the
The average dc output voltage can be varied by varying the trigger angle from 0 0 to a maximum of 180 radians . We can plot the control characteristic, which is a plot of dc output voltage versus the trigger angle by using the equation for VO dc .
56
CONTROL CHARACTERISTIC OF SINGLE PHASE HALF WAVE PHASE CONTROLLED RECTIFIER WITH RESISTIVE LOAD The average dc output voltage is given by the expression Vm 1 cos 2
VO dc
We can obtain the control characteristic by plotting the expression for the dc output voltage as a function of trigger angle Trigger angle in degrees
VO dc Vdm
0
%
Vm
100% Vdm
300
0.933 Vdm
93.3 % Vdm
600
0.75 Vdm
75 % Vdm
900
0.5 Vdm
50 % Vdm
1200
0.25 Vdm
25 % Vdm
1500 1800
0.06698 Vdm 0
Vdm
Vm
Vdc max
6.69 % Vdm 0
VO(dc) Vdm
0.6Vdm
0.2 Vdm 0
60
120
Trigger angle
180 in degrees
Fig.: Control characteristic Normalizing the dc output voltage with respect to Vdm , the normalized output voltage Vdcn
VO ( dc ) Vdc max
Vdc Vdm
57
Vdcn
Vn
Vn
Vdc Vdm
Vm 1 cos 2 Vm
Vdc Vdm
1 1 cos 2
Vdcn
TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF OUTPUT VOLTAGE OF A SINGLE PHASE HALF WAVE CONTROLLED RECTIFIER WITH RESISTIVE LOAD The rms output voltage is given by
VO RMS Output voltage vO
1 2
2
vO2 .d
Vm sin t ; for
VO RMS
By substituting sin 2 t
t
0
t
to 1 2
1 2
Vm2 sin 2 t.d
t
1 cos 2 t , we get 2
VO RMS
VO RMS
VO RMS
1 2
Vm2
2 m
V 4
2 m
V 4
VO RMS
Vm 1 2
VO RMS
Vm 1 2
1 2
1 cos 2 t .d 2
t 1 2
1 cos 2 t .d
t 1 2
d
t
t
cos 2 t.d
t
1 2
sin 2 t 2
sin 2
sin 2 2
1 2
; sin 2
0
58
Hence we get,
VO RMS
VO RMS
Vm 1 2
1 2
sin 2 2
Vm
sin 2 2
2
1 2
PERFORMANCE PARAMETERS OF PHASE CONTROLLED RECTIFIERS Output dc power (average or dc output power delivered to the load)
PO dc
VO dc
IO dc ; i.e., Pdc
Vdc I dc
VO dc
Vdc
average or dc value of output (load) voltage.
IO dc
I dc
Where
average or dc value of output (load) current.
Output ac power
PO ac
VO RMS
IO RMS
Efficiency of Rectification (Rectification Ratio) PO dc
Efficiency
PO ac
% Efficiency
;
PO dc PO ac
100
The output voltage can be considered as being composed of two components The dc component VO dc = DC or average value of output voltage. The ac component or the ripple component Vac
Vr rms
RMS value of all
the ac ripple components. The total RMS value of output voltage is given by
VO RMS
VO2 dc
Vr2rms
Therefore Vac
Vr rms
VO2 RMS
VO2 dc
59
Form Factor (FF)
which is a measure of the shape of the output voltage is given by VO RMS
FF
RMS output load voltage DC output load voltage
VO dc
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 rv
Vr rms
RF
VO dc
VO2 RMS
rv
Vac Vdc
VO2 dc
VO dc
VO RMS VO dc
2
1
Therefore
FF 2 1
rv
Current Ripple Factor defined for the output (load) current waveform is given by I r rms
ri
I O dc
I r rms
Where
I ac I dc
I ac
I O2 RMS
I O2 dc
Some times the peak to peak output ripple voltage is also considered to express the peak to peak output ripple voltage as
Vr
pp
peak to peak ac ripple output voltage
The peak to peak ac ripple load current is the difference between the maximum and the minimum values of the output load current.
Ir
pp
IO max
IO min
Transformer Utilization Factor (TUF)
TUF
PO dc VS I S
Where VS
RMS value of transformer secondary output voltage (RMS supply voltage at the secondary)
60
IS
RMS value of transformer secondary current (RMS line or supply current).
vS
Supply voltage at the transformer secondary side .
iS
Input supply current (transformer secondary winding current) .
iS 1
Fundamental component of the input supply current .
IP
Peak value of the input supply current . Phase angle difference between (sine wave components) the fundamental components of input supply current and the input supply voltage. Displacement angle (phase angle)
For an RL load
Displacement angle = Load impedance angle tan
1
L for an RL load R
Displacement Factor (DF) or Fundamental Power Factor DF
Cos
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)
HF
I
2 S
I I
2 S1
2 S1
1 2
IS I S1
1 2
2
1
Where IS I S1
RMS value of input supply current. RMS value of fundamental component of the input supply current.
61
Input Power Factor (PF) PF
VS I S 1 cos VS I S
I S1 cos IS
The Crest Factor (CF)
CF
IS
peak
IS
Peak input supply current RMS input supply current
For an Ideal Controlled Rectifier
FF 1 ; which means that VO RMS Efficiency
Vac Vr rms
VO dc .
100% ; which means that PO dc
0 ; so that RF
rv
TUF 1 ; which means that PO dc
0 ; Ripple factor = 0 (ripple free converter).
VS I S
HF THD 0 ; which means that I S
PF
DPF 1 ; which means that
PO ac .
I S1
0
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 T1 ( T1 is an SCR) is shown in the figure below.
62
The thyristor T1 is forward biased during the positive half cycle of input supply. Let us assume that T1 is triggered at
, by applying a suitable gate trigger pulse to
t
T1 during the positive half cycle of input supply. The output voltage across the load
follows the input supply voltage when T1 is ON. The load current iO flows through the thyristor T1 and through the load in the downward direction. This load current pulse flowing through T1 can be considered as the positive current pulse. Due to the inductance in the load, the load current iO flowing through T1 would not fall to zero at
, when
t
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 T1 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 T1 falls to zero at
t
, where
is referred to as the Extinction angle, (the value of
at which the load current falls to zero. The extinction angle
t)
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 T1 conducts from
t
, which depends on the delay angle
to
. The conduction angle of T1 is and the load impedance angle
. The
waveforms of the input supply voltage, the gate trigger pulse of T1 , the thyristor current, the load current and the load voltage waveforms appear as shown in the figure below.
i1
iO
iS
Fig.: Input supply voltage & Thyristor current waveforms 63
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 to 2 , the thyristor remains cutoff 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 to 2 . In the next cycle the thyristor is triggered again at a phase angle of 2 operation repeats.
, and the same
TO DERIVE AN EXPRESSION FOR THE OUTPUT (INDUCTIVE LOAD) CURRENT, DURING t to WHEN THYRISTOR T1 CONDUCTS Considering sinusoidal input supply voltage we can write the expression for the supply voltage as vS
Vm sin t = instantaneous value of the input supply voltage.
Let us assume that the thyristor T1 is triggered by applying the gating signal to T1 at t . The load current which flows through the thyristor T1 during t to can be found from the equation L
diO dt
RiO
Vm sin t ;
The solution of the above differential equation gives the general expression for the output load current which is of the form iO
Where Vm Z
Vm sin Z
t
t
A1e
;
2VS = maximum or peak value of input supply voltage. R2
L
2
= Load impedance.
64
tan
L R
1
= Load impedance angle (power factor angle of load).
L = Load circuit time constant. R
Therefore the general expression for the output load current is given by the equation R t Vm L iO sin t A1e ; Z The value of the constant A1 can be determined from the initial condition. i.e. initial value of load current iO 0 , at t . Hence from the equation for iO equating iO to zero and substituting t , we get iO
R
Therefore
A1e L
1 e
A1
e
R t L
R t L
R
A1 By substituting
e
t
t L
Vm sin Z Vm sin Z
Vm sin Z
, we get the value of constant A1 as
t R
A1
A1e L
Vm sin Z
t
A1
R
Vm sin Z
0
e
L
Vm sin Z
Substituting the value of constant A1 from the above equation into the expression for iO , we obtain R R t Vm Vm L ; iO sin t e e L sin Z Z
iO
Vm sin Z
R
t
e
t L
R
e
L
Vm sin Z
65
iO
Vm sin Z
t
R L
e
t
Vm sin Z
Therefore we obtain the final expression for the inductive load current of a single phase half wave controlled rectifier with RL load as
iO
Vm sin Z
t
sin
R L
e
t
;
Where
t
.
The above expression also represents the thyristor current iT 1 , during the conduction time interval of thyristor T1 from t to .
TO CALCULATE EXTINCTION ANGLE The extinction angle , which is the value of t at which the load current iO falls to zero and T1 is turned off can be estimated by using the condition that iO 0 , at t By using the above expression for the output load current, we can write
iO As
Vm Z
0
Vm sin Z
sin
e
R L
0 , we can write
sin
sin
e
R L
0
Therefore we obtain the expression
sin
sin
e
R L
The extinction angle can be determined from this transcendental equation by using the iterative method of solution (trial and error method). After is calculated, we can determine the thyristor conduction angle
.
is the extinction angle which depends upon the load inductance value. Conduction angle increases as is decreased for a specific value of . Conduction angle
; for a purely resistive load or for an RL load
when the load inductance L is negligible the extinction angle angle
and the conduction
66
Equations vs
Vm sin t
vO
vL
Input supply voltage
Vm sin t
Output load voltage for t
to
,
when the thyristor T1 conducts ( T1 is on). Expression for the load current (thyristor current): for
iO Extinction angle
Vm sin Z
t
sin
e
R L
t
t
;
to
Where
t
.
can be calculated using the equation
sin
sin
e
R L
TO DERIVE AN EXPRESSION FOR AVERAGE (DC) LOAD VOLTAGE
VO dc
VO dc
vO
VL
VL
1 2 1 2
0 for t
2
vO .d
t
0
2
vO .d
t
vO .d
t
vO .d
t
;
0
0 to
& for t
to 2
VO dc
VL
1 2
vO .d
VO dc
VL
1 2
Vm sin t.d
t
VO dc
VL
Vm 2
cos t
Vm cos 2
VO dc
VL
Vm cos 2
t ; vO
;
Vm sin t for t
to
cos
cos
to , we can see from the output load voltage waveform Note: During the period t that the instantaneous output voltage is negative and this reduces the average or the dc output voltage when compared to a purely resistive load.
67
Average DC Load Current
IO dc
I L Avg
VO dc
Vm cos 2 RL
RL
cos
SINGLE PHASE HALF WAVE CONTROLLED RECTIFIER WITH RL LOAD AND FREE WHEELING DIODE T i0 + V0 R + Vs FWD
~
L 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. Vs
Vm Supply voltage
0
iG
t
Gate pulses
V m
0
iO
t
Load current t=
0
t 2
VO
0
Load voltage
t
68
At t , the source voltage v S falls to zero and as v S 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 t , the SCR becomes reverse biased, the current through it becomes zero and the SCR turns off. During the period t to , the load current flows through FWD (free wheeling load current) and decreases exponentially towards zero at t . 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. Vm The average output voltage Vdc 1 cos , which is the same as that of a 2 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 t , where 2 , well before the next gate pulse and the load current tends to become discontinuous. During the conduction period to , the load current is carried by the SCR and during the free wheeling period to , the load current is carried by the free wheeling diode. The value of depends on the value of R and L and the forward resistance of the FWD. Generally 2 . 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. i0
0
t1
t2
t3
t4
SCR
FWD
SCR
FWD
t
2
Fig. : Waveform of Load Current in Single Phase Half Wave Controlled Rectifier with a Large Inductance and FWD
69
During the periods t1 , t3 ,..... the SCR carries the load current and during the periods t2 , t4 ,..... 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
iO
R
+
~
vS
vO
L + E
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 cutoff 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 t at which the supply voltage increases and becomes equal to the load circuit dc voltage can be calculated by using the equation Vm sin t E . If we E . Therefore assume the value of t is equal to then we can write Vm sin is calculated as
sin
1
E . Vm
For trigger angle
, the thyristor conducts only from
For trigger angle
, the thyristor conducts from
t
t
to to
.
.
The waveforms appear as shown in the figure
70
Vm
vO
Load voltage E t
0
iO Im Load current 0
t
Equations vS Vm sin t
Input supply voltage .
vO
Vm sin t
Output load voltage for t
vO
E for t
0 to
& for t
to
to 2
Expression for the Load Current When the thyristor is triggered at a delay angle of can be written as di Vm sin t iO R L O +E ; t dt
, the equation for the circuit
The general expression for the output load current can be written as iO
Vm sin Z
t
Z
R2
L
E R
t
Ae
Where
tan
L R
1
L R
2
= Load Impedance
Load impedance angle
Load circuit time constant
The general expression for the output load current can be written as
71
iO
Vm sin Z
E R
t
Ae
R t L
To find the value of the constant ‘A’ apply the initial condition at t current iO 0 . Equating the general expression for the load current to zero at t get R Vm E iO 0 sin Ae L Z R
, load , we
We obtain the value of constant ‘A’ as A
E Vm sin R Z
e
R L
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 iO
Vm sin Z
E Vm sin R Z
E R
t
R L
e
t
The Extinction angle can be calculated from the final condition that the output current iO 0 at t . By using the above expression we get, iO
0
Vm sin Z
E Vm sin R Z
E R
e
R L
To derive an expression for the average or dc load voltage
VO dc
VO dc
1 2 1 2
2
vO .d
t
0
2
vO .d
t
vO .d
t
vO .d
vO
Vm sin t
Output load voltage for t
vO
E for t
0 to
VO dc
VO dc
1 2
1 2
t
0
& for t
to
to 2 2
E.d
t
Vm sin t
E.d
t
0
2
E
t
Vm
cos t
E
t
0
72
VO dc
1 2
VO dc
Vm 2
VO dc
Vm cos 2
E
0
cos
Vm cos
cos
E 2
E 2 2
cos
2
cos
2
E
Conduction angle of thyristor RMS Output Voltage can be calculated by using the expression
VO RMS
1 2
2
vO2 .d
t
0
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 unidirectional 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).
73
SINGLE PHASE FULL WAVE CONTROLLED RECTIFIER USING A CENTER TAPPED TRANSFORMER
iS T1
A
+ vO
vS AC Supply
O
L
R
iO
FWD
B
T2
v S = Supply Voltage across the upper half of the transformer secondary winding vS
vAO
Vm sin t
vBO vAO Vm sin t secondary winding.
supply voltage across the lower half of the transformer
This type of full wave controlled rectifier requires a center tapped transformer and two thyristors T1 and T2 . 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 T1 along with the load act as a half wave controlled rectifier, the lower half of the secondary winding and the thyristor T2 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 T1 is forward biased and it is triggered at a delay angle of . The load current 74
flows through the thyristor T1 , through the load and through the upper part of the secondary winding, during the period to , when the thyristor T1 conducts. The output voltage across the load follows the input supply voltage that appears across the upper part of the secondary winding from t to . The load current through the thyristor T1 decreases and drops to zero at t , where for RL type of load and the thyristor T1 naturally turns off at t .
vO
Vm
t 0 iO t 0 (
)
(
)
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 T2 is forward biased during the negative half cycle and it is triggered at a delay angle of . The current flows through the thyristor T2 , through the load, and through the lower part of the secondary winding when T2 conducts during the negative half cycle the load is connected to the lower half of the secondary winding when T2 conducts. For purely resistive loads when L = 0, the extinction angle . The load current falls to zero at t , when the input supply voltage falls to zero at t . 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 but . For large values of load inductance the load current would be continuous and does not fall to zero. The thyristor T1 conducts from to , until the next thyristor T2 is triggered. When T2 is triggered at
t
, the thyristor T1 will be reverse biased
and hence T1 turns off.
75
TO DERIVE AN EXPRESSION FOR THE DC OUTPUT VOLTAGE OF A SINGLE PHASE FULL WAVE CONTROLLED RECTIFIER WITH RL LOAD (WITHOUT FREE WHEELING DIODE (FWD)) The average or dc output voltage of a fullwave controlled rectifier can be calculated by finding the average value of the output voltage waveform over one output T cycle (i.e., radians) and note that the output pulse repetition time is seconds where T 2 1 represents the input supply time period and T ; where f = input supply frequency. f Assuming the load inductance to be small so that , we obtain discontinuous load current operation. The load current flows through T1 form is the trigger angle of thyristor T1 and is the extinction angle t to , where where the load current through T1 falls to zero at t . Therefore the average or dc output voltage can be obtained by using the expression
VO dc
VO dc
Vdc
Vdc
2 2
vO .d
t
t
1
vO .d
t
t
Therefore
VO dc
Vdc
VO dc
Vdc
VO dc
Vdc
VO dc
Vm
1
Vm
Vm
cos
Vm sin t.d
t
cos t
cos
cos
cos
, for discontinuous load current operation,
. When the load inductance is small and negligible that is L 0 , the extinction angle radians . Hence the average or dc output voltage for resistive load is obtained as VO dc
VO dc
Vm
cos
cos
Vm
cos
1
; cos
1
76
VO dc
Vm
1 cos
; for resistive load, when L 0
THE EFFECT OF LOAD INDUCTANCE Due to the presence of load inductance the output voltage reverses and becomes negative during the time period t to . 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 T1 is triggered at t , during the positive half cycle of the input supply the FWD is reverse biased during the time period t to . FWD remains reverse biased and cutoff from t to . The load current flows through the conducting thyristor T1 , through the RL load and through upper half of the transformer secondary winding during the time period to . At t , when the input supply voltage across the upper half of the secondary winding reverses and becomes negative the FWD turnson. The load current continues to flow through the FWD from t to .
vO
Vm
t 0 iO t 0 (
)
(
)
Fig.: Waveform for Discontinuous Load Current Operation with FWD EXPRESSION FOR THE DC OUTPUT VOLTAGE OF A SINGLE PHASE FULL WAVE CONTROLLED RECTIFIER WITH RL LOAD AND FWD
VO dc
Vdc
1
vO .d
t
t 0
Thyristor T1 is triggered at
t
. T1 conducts from
t
to
77
Output voltage vO
Vm sin t ; for t
FWD conducts from
t
Therefore
VO dc
Vdc
VO dc
Vdc
VO dc
Vdc
VO dc
Vdc
Therefore
to
1
to
and vO
0 during discontinuous load current
Vm sin t.d
Vm
cos t
Vm
cos
Vm
1 cos
t
cos
; cos
1
The DC output voltage Vdc 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 FWD The control characteristic can be obtained by plotting the dc output voltage Vdc versus the trigger angle . 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 VO dc
Vdc
Vm
1 cos
Vdc can be varied by varying the trigger angle of trigger angle is from 0 to radians). Maximum dc output voltage is obtained when
Therefore
Vdc max
Vdc
Vdc max
Vdc
Vm
2Vm
1 cos 0
from 0 to 1800 . (i.e., the range
0
2Vm
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
78
Vdcn
Vdc
Vn
Vdc max Vm
Therefore
Vdc Vdm
Vdcn
Vn
Vdcn
Vn
Vdc
1 1 cos 2
1 cos
1 1 cos 2
2Vm
Vdc Vdm
1 1 cos 2
Vdm
Trigger angle in degrees
Normalized dc output voltage Vn
VO dc Vdm
2Vm
0.636619Vm
1
300
0.593974 Vm
0.9330
0
0.47746 Vm
0.75
0
0.3183098 Vm
0.5
0.191549 Vm
0.25
0.04264 Vm 0
0.06698 0
0
60
90
1200 0
150 1800
VO(dc) Vdm
0.6Vdm
0.2 Vdm 0
60
120
Trigger angle
180 in degrees
Fig.: Control characteristic of a single phase full wave controlled rectifier with R load or RL load with FWD
79
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.
vO
Vm
t 0 iO
T2 ON
T1 ON
T1 ON
t
0 (
)
(
)
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 T1 which is triggered at a delay angle of , conducts from t . Output voltage follows the input to supply voltage across the upper half of the transformer secondary winding vO vAO Vm sin t . The next thyristor T2 is triggered at input supply. As soon as T2 is triggered at
t t
, during the negative half cycle , the thyristor T1 will be reverse
biased and T1 turns off due to natural commutation (ac line commutation). The load current flows through the thyristor T2 from t . Output voltage to 2 across the load follows the input supply voltage across the lower half of the transformer Vm sin t . secondary winding vO vBO Each thyristor conducts for
radians 1800
in the case of continuous current
operation.
80
TO DERIVE AN EXPRESSION FOR THE AVERAGE OR DC OUTPUT VOLTAGE OF SINGLE PHASE FULL WAVE CONTROLLED RECTIFIER WITH LARGE LOAD INDUCTANCE ASSUMING CONTINUOUS LOAD CURRENT OPERATION. VO dc
Vdc
1
vO .d
t
t
VO dc
Vdc
VO dc
Vdc
VO dc
Vdc
VO dc
Vdc
VO dc
Vdc
1
Vm
Vm sin t.d
t
cos t
Vm
cos
cos
Vm
cos
cos
2Vm
cos
;
cos
cos
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
2Vm
Therefore
Vdcn
Vn
Vdcn
Vn
Vdc Vdc max
cos 2Vm
cos
cos
81
Trigger angle in degrees
VO dc Vdm
0
Remarks
300
0.866 Vdm
600
0.5 Vdm
900
0 Vdm
1200
0.5 Vdm
150
Maximum dc output voltage 2Vm Vdc max Vdm
2Vm
0.866 Vdm
0
2Vm
Vdm
1800 V O(dc) Vdm 0.6Vdm 0.2 Vdm 0 0.2Vdm
30
60
90
120
150
180
0.6 V dm Vdm Trigger angle
in degrees
Fig.: Control Characteristic We notice from the control characteristic that by varying the trigger angle 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 . For trigger angle in the range of 0 to 90 0 degrees i.e., 0 90 , Vdc 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. 900 , cos becomes negative and as a result the average dc For trigger angle output voltage Vdc 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 900 the load energy can be fed operation. During the inverter mode operation for back from the load circuit to the input ac source.
82
TO DERIVE AN EXPRESSION FOR RMS OUTPUT VOLTAGE The rms value of the output voltage is calculated by using the equation VO RMS
VO RMS
VO RMS
VO RMS
VO RMS
VO RMS
VO RMS
VO RMS
VO RMS
VO RMS
1 2
2 2
vO2 .d
1 2
1
Vm2 sin 2 t.d
V
sin 2 t.d
2 m
t
1 2
1 cos 2 t .d 2
V
Vm
t
1 2
2 m
Vm
t
1 2
1 2
1 2
t
d
t
cos 2 t.d
t
1 2
sin 2 t 2
t
sin 2
Vm
1 2
Vm
1 2
sin 2
Vm
1 2
0 sin 2
Vm
1 2
sin 2
1 2
2
cos 2
sin 2
cos 2 2
sin 2
sin 2
1 2
1 2
2 1 2
Vm 2
Therefore VO RMS
Vm ; The rms output voltage is same as the input rms supply voltage. 2
83
SINGLE PHASE SEMICONVERTERS
Errata: Consider diode D2 as D1 in the figure and diode D1 as D2 Single phase semiconverter 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 turnon and conduct one after the other as and when they are forward biased. The circuit diagram of a single phase semiconverter (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 T1 and the diode D1 are both forward biased. The thyristor T1 is triggered at
; 0
t
by applying an appropriate gate trigger signal to the gate of T1 . The current in the circuit flows through the secondary line ‘A’, through T1 , through the load in the downward direction, through diode D1 back to the secondary line ‘B’. T1 and D1 conduct together from t to 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 t to . At t , the input supply voltage decreases to zero and becomes negative during the period t . The free wheeling diode Dm across the load to becomes forward biased and conducts during the period
t
to
.
84
Fig:. Waveforms of single phase semiconverter for RLE load and constant load current for > 900
85
The load current is transferred from T1 and D1 to the FWD Dm . T1 and D1 are turned off. The load current continues to flow through the FWD Dm . The load current free wheels (flows continuously) through the FWD during the free wheeling time period . to During the negative half cycle of input supply voltage the secondary line ‘A’ becomes negative with respect to line ‘B’. The thyristor T2 and the diode D2 are both forward biased. T2 is triggered at
, during the negative half cycle. The FWD
t
is reverse biased and turnsoff as soon as T2 is triggered. The load current continues to flow through T2 and D2 during the period
t
to 2
TO DERIVE AN EXPRESSION FOR THE AVERAGE OR DC OUTPUT VOLTAGE OF A SINGLE PHASE SEMICONVERTER The average output voltage can be found from
Vdc
2 Vm sin t.d 2
Vdc
2Vm 2
Vdc
Therefore
Vdc
cos t
Vm
cos
Vm
1 cos
Vdc can be varied from
t
2Vm
cos
; cos
to 0 by varying
1
from 0 to
.
The maximum average output voltage is Vdc max
Vdm
2Vm
Normalizing the average output voltage with respect to its maximum value Vdcn
Vn
Vdc Vdm
0.5 1 cos
The output control characteristic can be plotted by using the expression for Vdc
86
TO DERIVE AN EXPRESSION FOR THE RMS OUTPUT VOLTAGE OF A SINGLE PHASE SEMICONVERTER The rms output voltage is found from
VO RMS
VO RMS
VO RMS
2 2
1 2
Vm2 sin 2 t.d
2 m
V 2
Vm 1 2
t 1 2
1 cos 2 t .d
sin 2 2
t
1 2
SINGLE PHASE FULL CONVERTER (FULLY CONTROLLED BRIDGE 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 T1 , T2 , T3 and T4 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 T1 and T2 are forward biased during the time interval t 0 to . The thyristors T1 and T2 are triggered simultaneously
t
; 0
, the load is
connected to the input supply through the conducting thyristors T1 and T2 . The output voltage across the load follows the input supply voltage and hence output voltage vO Vm sin t . Due to the inductive load T1 and T2 will continue to conduct beyond , even though the input voltage becomes negative. T1 and T2 conduct together t
87
during the time period
to
, for a time duration of
radians (conduction angle
0
of each thyristor = 180 ) During the negative half cycle of input supply voltage for t thyristors T3 and T4 are forward biased. T3 and T4 are triggered at t
to 2
the . As
soon as the thyristors T3 and T4 are triggered a reverse voltage appears across the thyristors T1 and T2 and they naturally turnoff and the load current is transferred from T1 and T2 to the thyristors T3 and T4 . The output voltage across the load follows the supply voltage and vO
Vm sin t during the time period
t
to 2
. In
the next positive half cycle when T1 and T2 are triggered, T3 and T4 are reverse biased and they turnoff. 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.
88
During the time period t to , the input supply voltage v S and the input supply current iS are both positive and the power flows from the supply to the load. The converter operates in the rectification mode during t to . During the time period t , the input supply voltage v S is negative to and the input supply current iS 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 t and the load energy is fed back to the input source. to 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 , 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
VO dc
Vdc
1 2
2
vO .d
t
;
0
The output voltage waveform consists of two output pulses during the input supply time period between 0 & 2 radians . In the continuous load current operation of a single phase full converter (assuming constant load current) each thyristor conduct for radians (1800) after it is triggered. When thyristors T1 and T2 are triggered at t T1 and T2 conduct from
and the output voltage follows the input supply
to
voltage. Therefore output voltage vO Vm sin t ; for t to Hence the average or dc output voltage can be calculated as
Therefore
VO dc
Vdc
VO dc
Vdc
VO dc
Vdc
VO dc
Vdc
VO dc
Vdc
VO dc
Vdc
2 2 1
Vm
Vm sin t.d
Vm sin t.d
sin t.d
Vm
cos t
Vm
cos
2Vm
t
t
t
cos
; cos
cos
cos
89
The dc output voltage Vdc can be varied from a maximum value of 2Vm
minimum value of
for
radians
2Vm
for
00 to a
1800
The maximum average dc output voltage is calculated for a trigger angle and is obtained as 2Vm 2Vm Vdc max Vdm cos 0
Therefore
Vdc max
00
2Vm
Vdm
The normalized average output voltage is given by VO dc Vdc Vdcn Vn Vdc max Vdm
2Vm Vdcn
Vn
cos cos
2Vm
Vdcn Vn cos Therefore ; for a single phase full converter assuming continuous and constant load current operation.
CONTROL CHARACTERISTIC OF SINGLE PHASE FULL CONVERTER The dc output control characteristic can be obtained by plotting the average or dc output voltage Vdc versus the trigger angle For a single phase full converter the average dc output voltage is given by the 2Vm equation VO dc Vdc cos Trigger angle in degrees
VO dc Vdm
0
Remarks
2Vm
300
0.866 Vdm
600
0.5 Vdm
900
0 Vdm
120
0
1500 1800
Maximum dc output voltage 2Vm Vdc max Vdm
0.5 Vdm 0.866 Vdm Vdm
2Vm
90
V O(dc) Vdm 0.6Vdm 0.2 Vdm 0 0.2Vdm
30
60
90
120
150
180
0.6 V dm Vdm Trigger angle
in degrees
Fig.: Control Characteristic We notice from the control characteristic that by varying the trigger angle 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 . For trigger angle in the range of 0 to 90 0 degrees i.e., 0 90 , Vdc is positive and the average dc load current I dc is also positive. The average or dc output power Pdc 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. 900 , cos becomes negative and as a result the average dc For trigger angle output voltage Vdc becomes negative, but the load current flows in the same positive direction i.e., I dc 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 900 the load commutated inverter operation. During the inverter mode operation for energy can be fed back from the load circuit to the input ac source TWO QUADRANT OPERATION OF A SINGLE PHASE FULL CONVERTER
91
The above figure shows the two regions of single phase full converter operation in the Vdc versus I dc plane. In the first quadrant when the trigger angle is less than 900, Vdc and I dc 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 Pdc is positive. When the trigger angle is increased above 900 , Vdc becomes negative but I dc is positive and the average output power (dc output power) Pdc becomes negative and the power flows from the load circuit to the input source. The operation occurs in the fourth quadrant where Vdc is negative and I dc is positive. The converter operates as a line commutated inverter. TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF THE OUTPUT VOLTAGE The rms value of the output voltage is calculated as
VO RMS
1 2
2
vO2 .d
t
0
The single phase full converter gives two output voltage pulses during the input supply time period and hence the single phase full converter is referred to as a two pulse converter. The rms output voltage can be calculated as
VO RMS
VO RMS
VO RMS
VO RMS
2 2
1
Vm2
vO2 .d
t
Vm2 sin 2 t.d
sin 2 t.d
Vm2
t
t
1 cos 2 t .d 2
VO RMS
Vm2 2
d
VO RMS
Vm2 2
t
t
t
cos 2 t.d
t
sin 2 t 2
92
Therefore
VO RMS
Vm2 2
VO RMS
Vm2 2
sin 2
2 2
VO RMS
Vm2 2
sin 2
sin 2
VO RMS
Vm2 2 Vm 2
VO RMS
sin 2
sin 2 2 sin 2
; sin 2
2
sin 2
2
Vm2 2
0
Vm 2
VS
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 radians or 1800 in a single phase full converter operating at continuous and constant load current. Therefore rms value of the thyristor current is calculated as IT
IT
RMS
I O RMS
2
I O RMS
1 2
IO RMS RMS
2
The average thyristor current can be calculated as IT
Avg
IT
Avg
I O dc
2
I O dc
1 2
I O dc 2
93
SINGLE PHASE DUAL CONVERTER
94
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 Vdc versus I dc 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 bidirectional load current and bidirectional 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 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 (CIRCULATING 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 1 of the converter 1. If 1 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. Vdc and I dc are both positive and the operation occurs in the first quadrant. The average output power Pdc Vdc I dc is positive. The power flows from the input ac supply to the load. When 1 is increased above 900 converter 1 operates as a line commutated inverter and Vdc becomes negative while I dc is positive and the output power Pdc 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
95
are released to the thyristor gates of converter 2 and the trigger angle
2
is adjusted such
0
90 so that converter 2 operates as a controlled rectifier. The dc output that 0 2 voltage Vdc and I dc are both negative and the load current flows in the reverse direction. The magnitude of Vdc and I dc are controlled by the trigger angle 2 . The operation occurs in the third quadrant where Vdc and I dc are both negative and output power Pdc 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 I dc is positive we have to
operate converter 2 in the inverter mode by increasing the trigger angle
2
above 900 .
When 2 is made greater than 900 , 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 1 between 0 to 900 the second converter 2 is operated as a line commutated inverter by increasing its trigger angle 2 above 900. The trigger angles 1 and 2 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 Vdc1
2Vm
cos
1
The average dc output voltage of converter 2 is Vdc 2
2Vm
cos
2
96
with
In the dual converter operation one converter is operated as a controlled rectifier 900 and the second converter is operated as a line commutated inverter in the 1
inversion mode with Vdc1
2Vm
90 0 .
2
Vdc 2
cos
2Vm
Therefore
cos
Therefore
2
1
Which gives
2
1
1
cos
When the trigger angle
2
2Vm
cos
1
2
or cos
or
1
1
cos
2
cos
2
1
cos
1
radians
2
of converter 1 is set to some value the trigger angle
of the second converter is adjusted such that
0
2
180
1
2
. Hence for circulating
current mode of operation where both converters are conducting at the same time 1800 so that they produce the same dc output voltage across the load. 1 2 When 1 900 (say 1 300 ) the converter 1 operates as a controlled rectifier and converts the ac supply into dc output power and the average load current I dc is positive. At the same time the converter 2 is switched on and operated as a line commutated inverter, by adjusting the trigger angle 2 such that 2 1800 1 , which is equal to 1500 , when 1 300 . 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 2 such that 2 90 0 , the first converter1 is operated as a line commutated inverter, by adjusting the trigger angle adjusted such that
1
1800
2
1
such that
for a set value of
1
2
900 . The trigger angle
1
is
.
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 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. 97
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 IT max I dc max ir max , where I dc max is the maximum dc load current and ir max is the maximum value of the circulating current. TO CALCULATE THE CIRCULATING CURRENT
Fig.: Waveforms of dual converter 98
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 t = (2  1). As the two average output voltages during the interval t = ( + 1) to (2  1) are equal and opposite their contribution to the instantaneous circulating current ir is zero.
ir
1 Lr
t
vr .d 2
t
;
vr
vO1 vO 2
1
As the output voltage vO 2 is negative
Therefore
vr
vO1 vO 2
ir
1 Lr
vO1
ir
t
vO1 vO 2 .d 2
1
Vm sin t for 2 Vm Lr
1
to t
t
t
sin t.d 2
t
sin t.d 2
1
t
1
t
t
Vm Lr
cos t
ir
Vm Lr
cos t
ir
Vm 2 cos t 2 cos 2 Lr
ir
2Vm cos t cos Lr
ir
t ;
cos t 2
2
1
cos 2
1
cos t
1
cos 2
1
1
1
The instantaneous value of the circulating current depends on the delay angle.
99
t
For trigger angle (delay angle) 1 = 0, its magnitude becomes minimum when n , n 0, 2, 4,.... and magnitude becomes maximum when t n , n 1,3,5,.... If the peak load current is I p , one of the converters that controls the power flow
may carry a peak current of
Where
Ip
4Vm , Lr
Ip
I L max
Vm , & ir max RL
4Vm Lr
Problems 1. What will be the average power in the load for the circuit shown, when
. 4 Assume SCR to be ideal. Supply voltage is 330 sin314t. Also calculate the RMS power and the rectification efficiency. T
+ 330 Sin314t
~
R
100
The circuit is that of a single phase half wave controlled rectifier with a resistive load
Average Power
Vdc
Vm 1 cos 2
Vdc
330 1 cos 2 4
Vdc
89.66 Volts
Vdc2 R
I dc
VRMS
89.662 100
Vdc R
89.66 100
Vm 1 2
;
4
radians
80.38 Watts
0.8966 Amps
sin 2 2
1 2
100
1 2
VRMS
330 1 2
VRMS
157.32 V
sin 2 4
4
2
RMS Power (AC power) 2 VRMS R
Rectification Efficiency
157.322 100
247.50 Watts
Average power RMS power
80.38 247.47
0.3248
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 = 450. VAK
+ 330 Sin314t
~
R
100
Load voltage
The wave form of the load voltage is shown below (not to scale). Vm Voltage across resistance
VAK 0
t
It is observed that the SCR turns off when t , where SCR turnsoff for anode supply voltage below 1 Volt. VAK
Vm sin
because the
1 volt (given)
101
Therefore
sin
VAK Vm
1
sin
1800
1 330
0.170 0.003 radians
; By symmetry of the curve.
179.830
Note:
1
; 3.138 radians.
Vdc
1 2
Vm sin t VAK d
Vdc
1 2
Vm sin t.d
Vdc
1 Vm 2
Vdc
1 Vm cos 2
Vdc
1 330 cos 450 cos179.830 2
Vdc
89.15 Volts
and
t
cos t
t
VAK d
VAK
cos
t
t
VAK
1 3.138 0.003
values should be in radians
3. In the figure find out the battery charging current when
4
. Assume ideal
SCR. R 10 + 200 V 50 Hz
~
24V (VB)
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).
102
Vm Voltage across resistance VB 0
t
VB
Vm sin
24
200 2 sin
sin
24 200 2
1
4.86750
0.085 radians
3.056 radians
Average value of voltage across 10 1 2
Vm sin t VB .d
t
(The integral gives the shaded area) 1 2
3.056
200
2 sin t 24 .d
t
4
1 200 2 cos cos3.056 2 4
24 3.056
4
68 Vots Therefore charging current Average voltage across R R 68 10
6.8 Amps
Note: If value of is more than , then the SCR will trigger only at t (assuming that the gate signal persists till then), when it becomes forward biased.
,
103
1 2
Vdc
Therefore
Vm sin t VB .d
t
4. In a single phase full wave rectifier supply is 200 V AC. The load resistance is 600 . Find the average voltage across the load and the power consumed 10 , in the load. Solution In a single phase full wave rectifier Vm
Vdc
1 cos
200
Vdc Vdc
2
1 cos 600
135 Volts
Average Power Vdc2 R
1352 10
1.823 kW
5. In the circuit shown find the charging current if the trigger angle
900 .
R = 10 + 200 V 50 Hz
~ +
10V (VB)
Solution With the usual notation
Therefore
VB
Vm sin
10
200 2 sin
sin
1
10 200 2
0.035 radians
104
900
radians ;
2
2 2
Average voltage across 10
1
1
1
3.10659
Vm sin t VB .d
Vm cos t VB
Vm cos
200
t
cos
2 cos
t
VB
cos3.106
2
10 3.106
2
85 V Note that the values of Charging current
are in radians.
&
dc voltage across resistance resistance 85 10
8.5 Amps
6. A single phase full wave controlled rectifier is used to supply a resistive load of 10 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,
Vdc
Vdc Vdc
Vm
1 cos
230
2
1 cos
2
103.5 Volts
When a large inductance is added in series with the load, the output voltage wave 900 . form will be as shown below, for trigger angle
105
V0
0
t
Vdc
Since
Therefore Vdc
cos
cos
;
2
2Vm
cos
2
0
0 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
10
VL
+ Vs
~ + 100V
With the usual notations VS
Vm sin t
VS Vm sin
2 230sin t VB , the battery voltage
2 230sin
100
106
Therefore
sin
100 2 230
1
17.90 or 0.312 radians
0.312 2.829 radians
Average value of voltage across load resistance 2 2
1
1
1
1
Vm sin t VB d
Vm cos t VB
Vm cos
cos
t
t
VB
230
2 cos
230
2 0.707 0.9517
4
cos 2.829
100 2.829
4
204.36
106.68 Volts Charging current
Voltage across resistance R 106.68 10.668 Amps 10
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 2
5.334 Amps
107
THREE PHASE CONTROLLED RECTIFIERS INTRODUCTION TO 3PHASE 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 2Vm single phase controlled rectifiers provide a maximum dc output of Vdc max . 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 3phase controlled rectifiers which are used to convert ac input power supply into dc output power across the load. Features of 3phase 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.
3PHASE HALF WAVE CONVERTER Three single phase halfwave converters are connected together to form a three phase halfwave converter as shown in the figure.
108
THEE PHASE SUPPLY VOLTAGE EQUATIONS We define three line neutral voltages (3 phase voltages) as follows vRN
van
Vm sin t ; Vm
vYN
vbn
Vm sin
t
vYN
vbn
Vm sin
t 1200
Max. Phase Voltage
VCN
2 3
0
vBN
vBN
vcn
vcn
Vm sin
Vm sin
t
2 3
120 120
VAN
0 0
120
t 1200
VBN vBN
vcn
Vm sin
t 2400
Vector diagram of 3phase supply voltages
109
The 3phase half wave converter combines three single phase half wave controlled rectifiers in one single circuit feeding a common load. The thyristor T1 in series with one of the supply phase windings ' a n ' acts as one half wave controlled rectifier. The second thyristor T2 in series with the supply phase winding ' b n ' acts as the second half wave controlled rectifier. The third thyristor T3 in series with the supply phase winding ' c n ' acts as the third half wave controlled rectifier. The 3phase 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 T1 is triggered at
van
t
300
, the phase voltage 6 appears across the load when T1 conducts. The load current flows through the supply
phase winding ' a n ' and through thyristor T1 as long as T1 conducts. 5 1500 , T1 becomes reverse 6 biased and turnsoff. The load current flows through the thyristor T2 and through the
When thyristor T2 is triggered at
t
supply phase winding ' b n ' . When T2 conducts the phase voltage vbn appears across the load until the thyristor T3 is triggered . 3 2700 , T2 is reversed 2 biased and hence T2 turnsoff. The phase voltage vcn appears across the load when T3 conducts.
When the thyristor T3 is triggered at
t
When T1 is triggered again at the beginning of the next input cycle the thyristor T3 turns off as it is reverse biased naturally as soon as T1 is triggered. The figure shows the 3phase 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 T1 . For a purely resistive load where the load inductance ‘L = 0’ and the trigger angle , the load current appears as discontinuous load current and each thyristor is 6 naturally commutated when the polarity of the corresponding phase supply voltage reverses. The frequency of output ripple frequency for a 3phase half wave converter is 3 f S , where f S is the input supply frequency. The 3phase 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).
110
TO DERIVE AN EXPRESSION FOR THE AVERAGE OUTPUT VOLTAGE OF A 3PHASE HALF WAVE CONVERTER FOR CONTINUOUS LOAD CURRENT The reference phase voltage is vRN van Vm sin t . The trigger angle is measured from the cross over points of the 3phase supply voltage waveforms. When the phase supply voltage van begins its positive half cycle at t 0 , the first cross over point appears at
t
6
The trigger angle t
300 .
radians
for the thyristor T1 is measured from the cross over point at
300 . The thyristor T1 is forward biased during the period
t
300 to 1500 , when the
phase supply voltage van has a higher amplitude than the other phase supply voltages. Hence T1 can be triggered between 300 to 1500 . When the thyristor T1 is triggered at a trigger angle , the average or dc output voltage for continuous load current is calculated using the equation
Vdc
3 2
5 6
vO .d
t
6
Output voltage
vO
van
Vdc
3 2
300
Vm sin t for t
to 1500
5 6
Vm sin t.d
t
6
As the output load voltage waveform has three output pulses during the input cycle of 2 radians
Vdc
3Vm 2
5 6
sin t.d
t
6
Vdc
3Vm 2
5 6
cos t 6
111
3Vm 2
Vdc
cos
5 6
cos
6
Note from the trigonometric relationship
cos A B
cos A.cos B sin A.sin B
Vdc
3Vm 2
cos
Vdc
3Vm 2
cos 1500 cos
Vdc
3Vm 2
cos 1800 300 cos
5 cos 6
Note: cos 1800 300
sin
5 sin 6
sin 1500 sin
cos
6
.cos
sin
cos 300 .cos
sin 1800 300 sin
6
sin
sin 300 sin
cos 300 .cos
sin 300 sin
cos 300
sin 1800 300
sin 300
Therefore Vdc
3Vm 2
Vdc
3Vm 2cos 300 cos 2
Vdc
3Vm 2 2
Vdc
3Vm 2
cos 300 cos
sin 300 sin
cos 300 .cos
sin 300 sin
3 cos 2
3 cos
3 3Vm cos 2
112
Vdc
3VLm cos 2
Where
VLm
3Vm
Max. line to line supply voltage for a 3phase star connected transformer.
The maximum average or dc output voltage is obtained at a delay angle is given by 3 3 Vm Vdc max Vdm 2
= 0 and
Where Vm is the peak phase voltage.
And the normalized average output voltage is
Vdcn
Vn
Vdc Vdm
cos
TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF THE OUTPUT VOLTAGE OF A 3PHASE HALF WAVE CONVERTER FOR CONTINUOUS LOAD CURRENT The rms value of output voltage is found by using the equation
VO RMS
3 2
1 2
5 6
Vm2 sin 2 t.d
t
6
and we obtain
VO RMS
3Vm
1 6
3 8
1 2
cos 2
113
3 PHASE HALF WAVE CONTROLLED RECTIFIER OUTPUT VOLTAGE WAVEFORMS FOR DIFFERENT TRIGGER ANGLES WITH RL LOAD Van
Vbn
Vcn
V0 =30 0 30
0
60
0
90
0
120
0
0
150
180
0
210
0
240
Van
0
270
0
300
0
330
0
360
Vbn
0
0
390
420
t
0
Vcn
V0
=60
0 30
0
60
0
90
0
120
0
0
150
180
0
210
0
240
0
270
0
300
0
330
0
360
Vbn
Van
0
0
0
390
420
0
t
0
Vcn
V0 =90 0 30
0
60
0
90
0
120
0
0
150
180
0
210
0
240
0
270
0
300
0
330
0
360
0
0
390
420
0
0
t
114
3 PHASE HALF WAVE CONTROLLED RECTIFIER OUTPUT VOLTAGE WAVEFORMS FOR DIFFERENT TRIGGER ANGLES WITH R LOAD Vbn
Van
Vcn
=0 Vs 0 30
0
60
0
90
0
0
120 150
0
0
0
0
180 210 240 270
0
0
0
300 330 360
0
Vbn
Van
0
390
420
t
0
Vcn
=150 V0
0 30
0
60
0
90
0
0
120 150
0
0
0
0
180 210 240 270
0
0
0
300 330 360
0
Vbn
Van
0
390
420
t
0
Vcn
=300 V0
0 30
0
60
0
90
0
0
120 150
0
0
0
0
180 210 240 270
0
0
0
300 330 360
0
Vbn
Van
0
390
420
t
0
Vcn
=60 V0
0 30
0
60
0
90
0
0
120 150
0
0
0
0
180 210 240 270
0
0
0
300 330 360
0
0
390
420
0
0
t
115
TO DERIVE AN EXPRESSION FOR THE AVERAGE OR DC OUTPUT VOLTAGE OF A 3 PHASE HALF WAVE CONVERTER WITH RESISTIVE LOAD OR RL LOAD WITH FWD. In the case of a threephase half wave controlled rectifier with resistive load, the thyristor T1 is triggered at t 300 and T1 conducts up to t 1800 radians. When the phase supply voltage van decreases to zero at
t
, the load current falls to
zero and the thyristor T1 turns off. Thus T1 conducts from
t
300
to 1800 .
Hence the average dc output voltage for a 3pulse converter (3phase half wave controlled rectifier) is calculated by using the equation Vdc
3 2
vO
van
Vdc
3 2
Vdc
Vdc
Vdc
3Vm 2
1800
vO .d
t
300
1800
Vm sin t.d 30
t
0
1800
sin t.d 30
t
0
1800
3Vm 2
cos t
3Vm 2
cos1800 cos
Since
cos1800
We get
Vdc
300 to 1800
Vm sin t; for t
300
300
1,
3Vm 1 cos 2
300
116
THREE PHASE SEMICONVERTERS 3phase semiconverters 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. 3phase semiconverters are used in industrial power applications up to about 120kW output power level, where single quadrant operation is required. The power factor of 3phase semiconverter decreases as the trigger angle increases. The power factor of a 3phase semiconverter is better than three phase half wave converter. The figure shows a 3phase semiconverter with a highly inductive load and the load current is assumed to be a constant and continuous load current with negligible ripple.
Thyristor T1 is forward biased when the phase supply voltage van is positive and greater than the other phase voltages vbn and vcn . The diode D1 is forward biased when the phase supply voltage vcn is more negative than the other phase supply voltages. Thyristor T2 is forward biased when the phase supply voltage vbn is positive and greater than the other phase voltages. Diode D2 is forward biased when the phase supply voltage van is more negative than the other phase supply voltages. Thyristor T3 is forward biased when the phase supply voltage vcn is positive and greater than the other phase voltages. Diode D3 is forward biased when the phase supply voltage vbn 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 Dm and the supply current ia . The frequency of the output supply waveform is 3 f S , where f S is the input ac supply frequency. The trigger angle can be varied from 00 to 1800 . During the time period
6
forward biased. If T1 is triggered at
t t
7 6 6
i.e., for 300
t
2100 , thyristor T1 is
, T1 and D1 conduct together and the
117
7 , vac starts to 6 become negative and the free wheeling diode Dm turns on and conducts. The load current
line to line supply voltage vac appears across the load. At
t
continues to flow through the free wheeling diode Dm and thyristor T1 and diode D1 are turned off.
If the free wheeling diode Dm is not connected across the load, then T1 would 5 and the free 6 wheeling action is accomplished through T1 and D2 , when D2 turns on as soon as van
continue to conduct until the thyristor T2 is triggered at
becomes more negative at conducts for
t
2 radians 1200 3
7 6
t
. If the trigger angle
3
each thyristor
and the free wheeling diode Dm does not conduct. The
waveforms for a 3phase semiconverter with
3
is shown in figure
118
119
We define three line neutral voltages (3 phase voltages) as follows vRN
van
Vm sin t ; Vm
Max. Phase Voltage
vYN
vbn
Vm sin
t
vYN
vbn
Vm sin
t 1200
vBN
vcn
Vm sin
t
vBN
vcn
Vm sin
t 1200
vBN
vcn
Vm sin
t 2400
2 3
2 3
The corresponding linetoline voltages are vRB
vac
van vcn
3Vm sin
t
vYR
vba
vbn van
3Vm sin
t
vBY
vcb
vcn vbn
3Vm sin
t
vRY
vab
van vbn
3Vm sin
t
6 5 6
2
6
Where Vm is the peak phase voltage of a star (Y) connected source. TO DERIVE AN EXPRESSION FOR THE AVERAGE OUTPUT VOLTAGE OF THREE PHASE SEMICONVERTER FOR
3
AND DISCONTINUOUS
OUTPUT VOLTAGE For
3
and discontinuous output voltage: the average output voltage is found
from 7
Vdc
3 2
6
vac .d
t
6
120
7
Vdc
6
3 2
3 Vm sin
t
6
d
t
6
Vdc
3 3Vm 1 cos 2
Vdc
3VmL 1 cos 2
The maximum average output voltage that occurs at a delay angle of
0 is
3 3Vm
Vdm
The normalized average output voltage is Vn
Vdc Vdm
0.5 1 cos
The rms output voltage is found from 7
VO RMS
3 2
1 2
6
3Vm2 sin 2
t
6
d
t
6
VO RMS
For
3
3Vm
3 4
1 sin 2 2
1 2
, and continuous output voltage
Output voltage vO
vab
3Vm sin
t
Output voltage vO
vac
3Vm sin
t
6
6
; for
t
; for
t
to
6
2
to
2
5 6
The average or dc output voltage is calculated by using the equation
Vdc
5
2
3 2
vab .d 6
6
t
vac .d
t
2
121
Vdc
Vn
3 3Vm 1 cos 2 Vdc Vdm
0.5 1 cos
The RMS value of the output voltage is calculated by using the equation
VO RMS
5
2
3 2
2 vab .d 6
VO RMS
3Vm
1 2
6
vac2 .d
t
t
2
3 4
2 3
3 cos
2
1 2
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
radians (i.e. at an interval of
600 ). The frequency of output ripple voltage is 6 f S and the filtering requirement is less than that of three phase semi and half wave converters.
122
, thyristor T6 is already conducting when the thyristor T1 is 6 turned on by applying the gating signal to the gate of T1 . During the time period At
t
t
to , thyristors T1 and T6 conduct together and the line to line 6 2 supply voltage vab appears across the load.
, the thyristor T2 is triggered and T6 is reverse biased 2 immediately and T6 turns off due to natural commutation. During the time period At
t
5 , thyristor T1 and T2 conduct together and the line to line 2 6 supply voltage vac 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 T1 and T4 , the supply current through the line ‘a’. t
to
We define three line neutral voltages (3 phase voltages) as follows vRN
van
Vm sin t
;
Vm
Max. Phase Voltage
vYN
vbn
Vm sin
t
2 3
Vm sin
t 1200
vBN
vcn
Vm sin
t
2 3
Vm sin
t 1200
Vm sin
t 2400
Where Vm is the peak phase voltage of a star (Y) connected source. The corresponding linetoline voltages are vRY
vab
van vbn
3Vm sin
t
vYB
vbc
vbn vcn
3Vm sin
t
vBR
vca
vcn van
3Vm sin
t
6
2
2
123
T6
T1
T2
T3
T4
T5
T6
T1
T2
iG1
t 0
iG2
0
(30 + ) 0 60
0
(360 +30 + ) t
iG3 iG4 iG5 iG6
0
60
t 0
60
t 0
60
t 0
60
t
Gating (Control) Signals of 3phase full converter
124
TO DERIVE AN EXPRESSION FOR THE AVERAGE OUTPUT VOLTAGE OF THREE PHASE FULL CONVERTER WITH HIGHLY INDUCTIVE LOAD ASSUMING CONTINUOUS AND CONSTANT LOAD CURRENT The output load voltage consists of 6 voltage pulses over a period of 2 radians, hence the average output voltage is calculated as
VO dc
6 2
Vdc
2
vO .d t
;
6
vO
vab
3
Vdc
3Vm sin
t
6
2
3Vm sin
t
6
.d t
6
Vdc
3 3Vm
Where VmL
3VmL
cos
3Vm
cos
Max. linetoline supply voltage
The maximum average dc output voltage is obtained for a delay angle Vdc max
3 3Vm
Vdm
= 0,
3VmL
The normalized average dc output voltage is Vdcn
Vn
Vdc Vdm
cos
The rms value of the output voltage is found from 1 2
VO rms
6 2
2
vO2 .d
t
6
125
1 2
VO rms
6 2
2 2 vab .d
t
6
1 2
VO rms
3 2
2
3Vm2 sin 2
t
6
.d
t
6
VO rms
3Vm
1 2
3 3 cos 2 4
1 2
126
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, Lr . The two converters are controlled in such a way that if 1 is the delay angle of converter 1, the delay angle of converter 2 is 2 1 . 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) Lr . The operation of each converter is identical to that of a three phase full converter. to 1 1 , the line to line voltage vab appears 6 2 across the output of converter 1 and vbc appears across the output of converter 2 During the interval
We define three line neutral voltages (3 phase voltages) as follows vRN
van
Vm sin t
; Vm
Max. Phase Voltage
vYN
vbn
Vm sin
t
2 3
Vm sin
t 1200
vBN
vcn
Vm sin
t
2 3
Vm sin
t 1200
Vm sin
t 2400
127
The corresponding linetoline supply voltages are vRY
vab
van vbn
3Vm sin
t
vYB
vbc
vbn vcn
3Vm sin
t
vBR
vca
vcn van
3Vm sin
t
6
2
2
128
TO OBTAIN AN EXPRESSION FOR THE CIRCULATING CURRENT If vO1 and vO 2 are the output voltages of converters 1 and 2 respectively, the instantaneous voltage across the current limiting inductor during the interval 6
1
t
is
1
2
vr
vO1 vO2
vab vbc
vr
3Vm sin
t
vr
3Vm cos
t
sin
6
t
2
6
The circulating current can be calculated by using the equation
ir t
t
1 Lr
vr .d 1
6
ir t
t
t
1 Lr
3Vm cos 6
t
6
.d
t
1
ir t
3Vm sin Lr
ir max
3Vm Lr
t
6
sin
1
= 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 (NONCIRCULATING) 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 1 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 1 900 the converter 1 operates in the rectification mode Vdc is positive, I dc is positive and hence the average load power Pdc is positive.
129
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 1 is increased above 900 , Vdc becomes negative where as I dc 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 1 900 converter 1 operates in the inversion mode & the load energy is supplied back to the ac supply. The thyristors are switchedoff 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 2 of the gate trigger pulses supplied to the thyristors of converter 2. When 2 is less than 900 , 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 2 is less than 900 for converter 2, Vdc is negative & I dc is negative, converter 2 operates as a controlled rectifier & power flows from the ac source to the load circuit. When 2 is increased above 900, the converter 2 operates in the inversion mode with Vdc positive and I dc negative and hence Pdc 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 turnoff 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 1 & 2 are adjusted such that 1 1800 2 When
1
900 , converter 1 operates as a controlled rectifier. When
2
is made
0
greater than 90 , converter 2 operates in the inversion mode. Vdc , I dc , Pdc are positive. When
2
90 0 , converter 2 operates as a controlled rectifier. When
1
is made
greater than 900 , converter 1 operates as an Inverter. Vdc and I dc are negative while Pdc is positive.
130
Problems 1. A 3 phase fully controlled bridge rectifier is operating from a 400 V, 50 Hz supply. The thyristors are fired at the average output voltage for
. There is a FWD across the load. Find 4 450 and 750 .
Solution For
450 , Vdc
Vdc
For
750 , Vdc
Vdc Vdc
3Vm
cos
3
2 400
cos 450
382 Volts
6Vm 1 cos 600 2 6
2 400 1 cos 600 750 2
158.4 Volts
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 44V A B
+ 3 phase Full Wave Rectifier
484 V=V0
Ra
La + 440 V
C
Ra  Armature resistance of motor.
La  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 .
131
Therefore
VO 3
That is
3Vm cos
484
2 415 cos
484
30.270
Therefore
3. A 3 phase half controlled bridge rectifier is feeding a RL load. If input voltage is
. Find average load voltage. If any one 4 supply line is disconnected what is the average load voltage. 400 sin314t and SCR is fired at
Solution
4
radians which is less than
Therefore
3
Vdc
3Vm 1 cos 2
Vdc
3 400 1 cos 450 2
Vdc
326.18 Volts
If any one supply line is disconnected, the circuit behaves like a single phase half controlled rectifies with RL load. Vdc
Vdc Vdc
Vm
1 cos
400
1 cos 450
217.45 Volts
132
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
133
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 turnoff 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 turnoff 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 turnoff 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. tc is the time offered by the circuit within which the SCR should turn off completely. Thus tc should be greater than tq , 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 +
vs
~
R
vo
Fig. 1.1: Circuit for Natural Commutation
134
Supply voltage vs
Sinusoidal
3 0
2
t
t Load voltage vo Turn off occurs here t
3 0
2
t
Voltage across SCR tc
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.
135
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 underdamped. Figure 1.3 shows the circuit. This type of commutation is used in Series Inverter Circuit.
T
i
L
R
Vc(0) + 
Load
C
V
Fig. 1.3: Circuit for Self Commutation
EXPRESSION FOR CURRENT At t 0 , when the SCR turns ON on the application of gate pulse assume the current in the circuit is zero and the capacitor voltage is VC 0 . Writing the Laplace Transformation circuit of figure 1.3 the following circuit is obtained when the SCR is conducting.
T
R I(S)
sL
1 VC(0) S CS +  + C
V S
Fig.: 1.4.
136
V VC 0 S
I S
1 CS
R sL
CS V VC 0 S RCs s 2 LC 1
C V VC 0 LC s 2
s
R L
1 LC
V VC 0 L R 1 2 s s L LC
V VC 0 L s
2
R s L
1 LC
R 2L
2
R 2L
2
V VC 0 L s
R 2L
2
1 LC
A s
2
2
R 2L
2
2
,
Where A
V VC 0 L
R , 2L
,
1 LC
R 2L
2
is called the natural frequency
I S
A s
2
2
137
Taking inverse Laplace transforms A
i t
t
e
sin t
Therefore expression for current V VC 0 2 RL t e sin t L
i t
V VC 0
Peak value of current
L
Expression for voltage across capacitor at the time of turn off Applying KVL to figure 1.3 vc
V
vR VL
vc
V iR L
di dt
Substituting for i, vc
V
R
vc
V
R
vc
V
vc
V
vc
V
A
A
A
A
A
e
t
sin t L
e
t
sin t L
d A e dt
A
e
t
t
sin t
cos t
e
e
t
R sin t
L cos t L sin t
e
t
R sin t
L cos t L
e
t
R sin t 2
L cos t
t
sin t
R sin t 2L
Substituting for A,
vc t
V
vc t
V
V VC 0 L V VC 0
e
t
e
t
R sin t 2
L cos t
R sin t 2L
cos t
138
SCR turns off when current goes to zero. i.e., at
.
t
Therefore at turn off
Therefore
vc
V
vc
V
vc
V
V VC 0
e
0
cos
V VC 0 e V VC 0 e
R 2L
Note: For effective commutation the circuit should be under damped.
R 2L
That is
2
1 LC
With R = 0, and the capacitor initially uncharged that is VC 0 i
But
Therefore
0
V t sin L LC
1 LC
i
V t LC sin L LC
V
C t sin L 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
.
139
V
C L
Current i
0
t
/2
2V Capacitor voltage
V
t
Gate pulse t
t V Voltage across SCR 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 = 1 F, L = 5 mH and RL = 100 . Assume that the circuit is initially relaxed.
T
RL 100
L 5 mH
+
C
1 F
V =300V
Fig. 1.6.
140
Solution: RL 2L
1 LC
2
1 5 10
3
1 10
100 2 5 10
6
2 3
10,000 rad/sec
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
Therefore peak value of
i
V e L
i
V L
i
Conducting time of SCR
t
R , 2L
sin t , where
300 10000 5 10
10000
3
6A
0.314msec
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.
L
T
i(t)
IO 10 H C 50 F
V =100V
+
VC(0)=V
Fig. 1.7.
141
Solution : The transformed circuit of figure 1.7 is shown in figure 1.8.
IOL
sL
+ +
I(S) + V S
VC(0) =V S 1 CS
Fig.1.8: Transformed Circuit of Fig. 1.7 The governing equation is V s
Therefore
I S
VC 0 s
V s
sL
I S
I S
I S
I S
1 Cs
I O LCs s 2 LC 1
2
s LC 1 V VC 0 C
I O LCs
1 LC
LC s 2
V VC 0 L s2
s
2
2
V VC 0 L s
1 LC
sI O
2
2
1 Cs
IO L
VC 0 Cs s
V s
LC s 2
I S
VC 0 s
I S sL I O L
sI O 2
s
2
2
where
1 LC
Taking inverse LT i t
V VC 0
C sin t I O cos t L
142
The capacitor voltage is given by t
vc t
1 i t dt VC 0 C0
vc t
1 C
Therefore we get,
C sin t I O cos t dt VC 0 L
V VC 0 0
vc t
1 C
V VC 0
C L
vc t
1 C
V VC 0
C 1 cos t L
vc t
In this problem VC 0
t
IO C
cos t
1 V VC 0 C
LC sin t
t o
IO
sin t
t VC 0 o
IO
sin t
VC 0
LC
vc t
IO
L sin t V V cos t VC 0 C
vc t
IO
L sin t C
C 1 cos t L
VC 0
VC 0 cos t VC 0
V VC 0 cos t V
V i t
vc t
IO cos t and IO
L sin t V C
143
The waveforms are as shown in figure 1.9
I0
i(t)
t
/2 vc(t)
V t
/2 Fig.: 1.9 Turn off occurs at a time to so that
Therefore
tO
0.5
0.5
tO
2
LC 6
50 10
6
tO
0.5
10 10
tO
0.5
10 6 500 35.1 seconds
and the capacitor voltage at turn off is given by L sin tO V C
vc tO
IO
vc tO
200
vc tO
200 0.447 sin
vc tO
89.4 100 189.4 V
10 10 6 sin 900 100 6 50 10 35.12 22.36
100
144
Problem 1.3: In the circuit shown in figure 1.10. V = 600 volts, initial capacitor voltage is zero, L = 20 H, C = 50 F 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. T1 L
I0
i(t)
V
C
Fig. 1.10 Solution: (Refer to problem 1.2). The expression for i t is given by i t
V VC 0
C sin t I O cos t L
It is given that the initial voltage across the capacitor, VC O is zero. Therefore
i t
C sin t I O cos t L
V
i t can be written as I O2 V 2
i t
and
1
tan
t
L C
IO
where
C sin L
V
1 LC
The peak capacitor current is I O2 V 2
C L
Substituting the values, the peak capacitor current
350
2
600
2
50 10 20 10
6 6
1011.19 A
145
The expression for capacitor voltage is vc t
with
L sin t C
IO
VC 0
0, vc t
IO
V VC 0 cos t V L sin t V cos t V C
This can be rewritten as V2
vc t
tan
L sin C
t
V
C L
V Where
I O2
1
IO
The peak value of capacitor voltage is V2
I O2
L C
V
Substituting the values, the peak value of capacitor voltage
6002 3502
20 10 6 50 10 6
600
639.5 600 1239.5V
To calculate conduction time of T1 The waveform of capacitor current is shown in figure 1.11. When the capacitor current becomes zero the SCR turns off. Capacitor current
t
0
Fig. 1.11.
146
Therefore conduction time of SCR
tan
L C
IO
1
V 1 LC
Substituting the values
L C
IO tan
1
V
350 20 10 tan 600 50 10
6
1
6
20.250 i.e., 0.3534 radians
1 LC
1 20 10
6
50 10
31622.8 rad/sec
6
Therefore conduction time of SCR
0.3534 31622.8
88.17 sec
RESONANT PULSE COMMUTATION (CLASS B COMMUTATION) The circuit for resonant pulse commutation is shown in figure 1.12.
L T
i a b
C
IL
V
Load FWD Fig. 1.12: Circuit for Resonant Pulse Commutation
147
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 underdamped 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 t t1 the SCR is turned ON by giving a gate pulse. A current I L 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 I L and ‘i’ flow in opposite direction. When the instantaneous value of ‘i’ becomes equal to I L , 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 pulse of SCR t t1
V
Capacitor voltage vab t tC
Ip
i t
IL
t
ISCR
t Voltage across SCR t
Fig. 1.13: Resonant Pulse Commutation – Various Waveforms
148
EXPRESSION FOR tc , THE CIRCUIT TURN OFF TIME V and Assume that at the time of turn off of the SCR the capacitor voltage vab load current I L is constant. tc is the time taken for the capacitor voltage to reach 0 volts from – V volts and is derived as follows. t
V
1 c I L dt C0
V
I L tc C
tc
VC seconds IL
For proper commutation tc should be greater than tq , the turn off time of T. Also, the magnitude of I p , the peak value of i should be greater than the load current I L and the expression for i is derived as follows The LC circuit during the commutation period is shown in figure 1.14.
L T
i C
+
VC(0) =V
Fig. 1.14. The transformed circuit is shown in figure 1.15.
I(S) sL T
1 Cs + V s Fig. 1.15.
149
V s
I S sL
1 Cs
V Cs s s 2 LC 1
I S
VC
I S
1 LC
LC s 2
V 1 L s2 1 LC
I S
1 V LC L s2 1 LC
I S
I S
V
C L
1 1 LC
1 LC 1 s2 LC
Taking inverse LT i t
C sin t L
V
1 LC
Where
Or
i t
Therefore
Ip
V sin t L V
I p sin t
C amps . L
150
EXPRESSION FOR CONDUCTION TIME OF SCR For figure 1.13 (waveform of i), the conduction time of SCR t
sin
1
IL Ip
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 VC 0 with polarity as shown, T1 is conducting and the load current I L is a constant. To turn off T1 , T2 is triggered. L, C, T1 and T2 forms a resonant circuit. A resonant current ic t flows in the direction shown, i.e., in a direction opposite to that of load current I L .
ic t = I p sin t (refer to the previous circuit description). Where I p
VC 0
C L
& and the capacitor voltage is given by
vc t
1 iC t .dt C
vc t
1 VC 0 C
vc t
C sin t.dt . L
VC 0 cos t
C
ab
T1
iC(t)
L
iC(t)
IL T2
+ VC(0) V
T3 FWD
L O A D
Fig. 1.16: Resonant Pulse Commutation – An Alternate Circuit
151
When ic t
becomes equal to I L (the load current), the current through T1
becomes zero and T1 turns off. This happens at time t1 such that IL
I p sin
t1 LC
Ip
VC 0
C L
t1
LC sin
1
IL VC 0
L C
and the corresponding capacitor voltage is
vc t1
V1
VC 0 cos t1
Once the thyristor T1 turns off, the capacitor starts charging towards the supply voltage through T2 and load. As the capacitor charges through the load capacitor current is same as load current I L , 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 T3 reverses the polarity of the capacitor voltage and the circuit is ready for another triggering of T1 . The waveforms are shown in figure 1.17. EXPRESSION FOR tc Assuming a constant load current I L which charges the capacitor tc
CV1 seconds IL
Normally V1 VC 0 For reliable commutation tc should be greater than tq , the turn off time of SCR T1 . It is to be noted that tc depends upon I L and becomes smaller for higher values of load current.
152
Current iC(t)
t
V
Capacitor voltage vab t
t1
V1 tC VC(0) Fig. 1.17: Resonant Pulse Commutation – Alternate Circuit – Various Waveforms RESONANT PULSE COMMUTATION WITH ACCELERATING DIODE
D2
C
T1 L
iC(t)
IL iC(t)
T2
+ VC(0) V
T3 FWD
L O A D
Fig. 1.17(a)
153
iC IL 0
t
VC 0
t1
t
t2
V1 VC(O)
tC Fig. 1.17(b)
A diode D2 is connected as shown in the figure 1.17(a) to accelerate the discharging of the capacitor ‘C’. When thyristor T2 is fired a resonant current iC t flows through the capacitor and thyristor T1 . At time t
t1 , the capacitor current iC t
equals the load current I L and hence current through T1 is reduced to zero resulting in turning off of T1 . Now the capacitor current iC t continues to flow through the diode D2 until it reduces to load current level I L at time t2 . Thus the presence of D2 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 T2 turns off by itself. But once current of thyristor T1 reduces to zero the reverse voltage appearing across T1 is the forward voltage drop of D2 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)
t2
LC t1
VC t2 Circuit turnoff time tC
VC O cos t2 t2 t1
Problem 1.4 : The circuit in figure 1.18 shows a resonant pulse commutation circuit. The initial capacitor voltage VC O 200V , C = 30 F and L = 3 H. Determine the circuit turn off time tc , if the load current I L is (a) 200 A and (b) 50 A.
154
IL
T1 L
C
T2
iC(t)
+ VC(0) T3
V
FWD
L O A D
Fig. 1.18. Solution (a) When I L 200 A Let T2 be triggered at t
0.
The capacitor current ic t reaches a value I L at t
t1
LC sin
t1
3 10
t1
IL VC 0
1
6
t1 , when T1 turns off
L C
30 10 6 sin
1
200 3 10 6 200 30 10 6
3.05 sec .
1 LC
1 3 10
6
30 10
6
0.105 106 rad / sec .
At t
t1 , the magnitude of capacitor voltage is V1 VC 0 cos t1
That is
V1
200 cos 0.105 106 3.05 10
V1
200 0.9487
6
V1 189.75 Volts
and
tc
CV1 IL
155
tc
(b)
When I L
30 10
6
189.75 200
28.46 sec .
50 A
t1
3 10
6
6
30 10 sin
1
50 3 10 6 200 30 10 6
t1
0.749 sec .
V1
200 cos 0.105 106 0.749 10
V1
200 1 200 Volts .
tc
CV1 IL
tc
30 10 6 200 120 sec . 50
6
It is observed that as load current increases the value of tc reduces. Problem 1.4a : Repeat the above problem for I L 200 A , if an antiparallel diode D2 is connected across thyristor T1 as shown in figure 1.18a.
D2
C
T1 L
iC(t)
IL iC(t)
T2
+ VC(0) V
T3 FWD
L O A D
Fig. 1.18(a)
156
Solution IL
Let
200 A T2 be triggered at t
0.
Capacitor current iC t reaches the value I L at t
Therefore
t1
LC sin
t1
3 10
t1
`
IL VC O
1
6
t1 , when T1 turns off
L C
30 10 6 sin
1
200 3 10 6 200 30 10 6
3.05 sec .
1 LC
1 3 10
6
30 10
6
0.105 106 radians/sec At t
t1
VC t1
V1
VC O cos t1
VC t1
200cos 0.105 106 3.05 10
VC t1
189.75V
6
iC t flows through diode D2 after T1 turns off. iC t current falls back to I L at t2 t2
LC t1
t2
3 10
t2
26.75 sec .
1 LC
6
30 10
6
3.05 10
6
1 3 10
6
30 10
6
0.105 106 rad/sec.
157
At t
t2
Therefore
200cos0.105 10 6 26.75 10
VC t2
V2
VC t2
V2 189.02 V
tC
t2 t1
26.75 10
tC
23.7 secs
6
3.05 10
6
6
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.
4 F V =30V
L RL
i IL
30
Fig. 1.19. Solution: 30 1 Amp 30 For proper SCR commutation I p , the peak value of resonant current i, should be
The load current I L
greater than I L , Let Also
V RL
Ip
2I L ,
Ip
V L
V
V
1 LC 4 10 L
Therefore
2 30
Therefore
L 0.9mH . 1 LC
Ip
Therefore
L
2 Amps .
C L
6
1 0.9 10
3
4 10
6
16666 rad/sec
158
sin
IL Ip
1
Conduction time of SCR =
1 2 16666
sin 16666
1
0.523 radians 16666
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 40 sec and the supply voltage is 100 V. Obtain the proper values of commutating elements.
C V =100V
L
i
IL IL
Fig. 1.20. Solution I p peak value of i V
C and this should be greater than I L . Let I p L
Therefore 1.5 10 100
C L
1.5I L .
... a
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 tc
CV seconds IL
and this tc is given to be 40 sec. Therefore
40 10
6
C
100 10
159
Therefore
C
4 F
Substituting this in equation (a)
4 10 1.5 10 100 L 1.5
Therefore
2
10
2
6
104 4 10 L
6
L 1.777 10 4 H
L 0.177mH . 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 20 s. 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
Ip IL
1.5
Therefore
Ip
1.5 I L
That is
Ip
V
1.5 10 15 A .
C L
15 A
... a
It is given that the device turn off time is 20 sec. Therefore tc , the circuit turn off time should be greater than this, Let
tc
30 sec .
And
tc
CV IL
Therefore
30 10
Therefore
C 1.5 F .
6
200 C 10
Substituting in (a)
15 200
1.5 10 L
6
160
152
Therefore
2002
1.5 10 L
6
L 0.2666 mH
COMPLEMENTARY COMMUTATION (CLASS PARALLEL CAPACITOR COMMUTATION)
C
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.
IL R1
R2 ab
iC
V C T1
T2
Fig. 1.21: Complementary Commutation The working of the circuit can be explained as follows. Initially both T1 and T2 are off; Now, T1 is fired. Load current I L flows through R1 . At the same time, the capacitor C gets charged to V volts through R2 and T1 (‘b’ becomes positive with respect to ‘a’). When the capacitor gets fully charged, the capacitor current ic becomes zero. To turn off T1 , T2 is fired; the voltage across C comes across T1 and reverse biases it, hence T1 turns off. At the same time, the load current flows through R2 and T2 . The capacitor ‘C’ charges towards V through R1 and T2 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 T2 , T1 is triggered, the capacitor voltage (with ‘a’ positive) comes across T2 and T2 turns off. The related waveforms are shown in figure 1.22. EXPRESSION FOR CIRCUIT TURN OFF TIME tc From the waveforms of the voltages across T1 and capacitor, it is obvious that tc 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 vc t can be written as
161
vc t
Vf
Vi V f e
t
Where V f is the final voltage, Vi is the initial voltage and At
t
tc , vc t
0,
R1C , V f
V , Vi
Therefore 0 V
V V e
0 V
Therefore
V
2Ve
2Ve
0.5 e
is the time constant.
V, tc R1C
tc R1C
tc R1C
tc R1C
Taking natural logarithms on both sides tc ln 0.5 R1C tc
0.693R1C
This time should be greater than the turn off time tq of T1 . Similarly when T2 is commutated tc
0.693R2C
And this time should be greater than tq of T2 . Usually
R1
R2
R
162
Gate pulse of T2
Gate pulse of T1 p
t
V
IL
Current through R1
V R1
2V R1
t 2V R2
Current through T 1 V R1
t Current through T2
2V R1
V R2
t
V Voltage across capacitor v ab t
V
tC
tC Voltage across T1
t
tC
Fig. 1.22
163
Problem 1.8 : In the circuit shown in figure 1.23 the load resistances R1
R2
R
5
and the capacitance C = 7.5 F, V = 100 volts. Determine the circuit turn off time tc .
R1
R2
V C T1
T2
Fig. 1.23. Solution The circuit turnoff time
tc
0.693 RC seconds
tc
0.693 5 7.5 10
tc
26 sec .
6
Problem 1.9: Calculate the values of RL 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 40 sec. Also find R1 , given that the auxiliary SCR will undergo natural commutation when its forward current falls below the holding current value of 2 mA.
i1
IL
R1
RL iC
V =100V
C Auxiliary SCR
Main SCR
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 ic and the current i1 through R1 , ic reduces to zero after a time tc and hence the auxiliary SCR turns off automatically after a time tc , i1 should be less than the holding current.
164
Given
IL
That is
25 A
Therefore
RL tc
25 A
V RL
100 RL
4 40 sec 0.693RLC
That is
40 10
Therefore
C
6
0.693 4 C
40 10 6 4 0.693
C 14.43 F V should be less than the holding current of auxiliary SCR. R1 100 Therefore should be < 2mA. R1 i1
Therefore
R1
100 2 10
That is
R1
50 K
3
IMPULSE COMMUTATION (CLASS D COMMUTATION) The circuit for impulse commutation is as shown in figure 1.25.
IL
T1 T3 V
VC(O) L
+
C T2 FWD
L O A D
Fig. 1.25: Circuit for Impulse Commutation
165
The working of the circuit can be explained as follows. It is assumed that initially the capacitor C is charged to a voltage VC O with polarity as shown. Let the thyristor T1 be conducting and carry a load current I L . If the thyristor T1 is to be turned off, T2 is fired. The capacitor voltage comes across T1 , T1 is reverse biased and it turns off. Now the capacitor starts charging through T2 and the load. The capacitor voltage reaches V with top plate being positive. By this time the capacitor charging current (current through T2 ) would have reduced to zero and T2 automatically turns off. Now T1 and T2 are both off. Before firing T1 again, the capacitor voltage should be reversed. This is done by turning on T3 , C discharges through T3 and L and the capacitor voltage reverses. The waveforms are shown in figure 1.26.
Gate pulse of T3
Gate pulse of T2
Gate pulse of T1 t
VS Capacitor voltage t
VC tC
Voltage across T1 t
VC
Fig. 1.26: Impulse Commutation – Waveforms of Capacitor Voltage, Voltage across T1 .
166
EXPRESSION FOR CIRCUIT TURN OFF TIME (AVAILABLE TURN OFF TIME) tc tc depends on the load current I L and is given by the expression t
VC
1 c I L dt C0
(assuming the load current to be constant) VC
tc
I L tc C VC C seconds IL
For proper commutation tc should be > tq , turn off time of T1 . Note: T1 is turned off by applying a negative voltage across its terminals. Hence this is voltage commutation. tc depends on load current. For higher load currents tc is small. This is a disadvantage of this circuit. When T2 is fired, voltage across the load is V VC ; 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.
i +
T1 VC(O)
IT 1 T2 V
_
C
D L
IL RL
Fig. 1.27: Impulse Commutation – An Alternate Circuit
167
The working of the circuit can be explained as follows: Initially let the voltage across the capacitor be VC O with the top plate positive. Now T1 is triggered. Load current flows through T1 and load. At the same time, C discharges through T1 , 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 T1 , T2 is triggered; the voltage across the capacitor comes across T1 . T1 is reverse biased and it turns off (voltage commutation). The capacitor now starts charging through T2 and load. When it charges to V volts (with the top plate positive), the current through T2 becomes zero and T2 automatically turns off. The related waveforms are shown in figure 1.28. Gate pulse of T2
Gate pulse of T1
t
VC Capacitor voltage t
V tC This is due to i IT 1
IL Current through SCR
V RL t
2V RL IL Load current t
V
Voltage across T1 t
tC
Fig. 1.28: Impulse Commutation – (Alternate Circuit) – Various Waveforms
168
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 and C = 10 F. Voltage across capacitor before T2 is fired is V volts with polarity as shown.
+ C V
T1

VC(0) T2
+
R
Fig. 1.29. Solution When T2 is triggered the circuit is as shown in figure 1.30.
VC(O)  +
+
i(t)
C T2
V
R
Fig. 1.30. Writing the transform circuit, we obtain
1 Cs
VC(0) s +
I(s)
+ R
V s
Fig. 1.31.
169
We have to obtain an expression for capacitor voltage. It is done as follows:
I S
I S
1 V VC 0 s 1 R Cs C V VC 0 1 RCs
V VC 0
I S
1 RC
R s
Voltage across capacitor
VC s
I s
1 V VC 0 1 RCs s RC
VC s
VC s
V VC 0 s
VC s
V s
vc t In the given problem VC 0 Therefore
vc t
1 VC 0 Cs s
V 1 RC
s
V 1 e
t
RC
VC 0 s
V VC 0 1 s RC
VC 0 s
VC 0 1 s RC
VC 0 e
t
RC
V V 1 2e
t
RC
The waveform of vc t is shown in figure 1.32.
170
V vC(t) t
VC(0) tC
Fig. 1.32. At t
tc , vc t
Therefore
0 0 V 1 2e
1 2e 1 2
e
tc
tc
tc
RC
RC
RC
Taking natural logarithms tc 1 log e 2 RC
tc
RC ln 2
tc 10 10 10 6 ln 2 tc
69.3 sec .
Problem 1.11 : In the commutation circuit shown in figure 1.33. C = 20 F, 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 tc . T1 I0
C V
+
VC(0)=V
T2
I0
Fig. 1.33.
171
Solution It is given that V varies between 180 and 220 V and I O varies between 50 and 200 A. The expression for available turn off time tc is given by CV IO
tc
tc is maximum when V is maximum and I O is minimum.
Therefore
and
tc max
CVmax I O min
tc max
20 10
tc min
CVmin I O max
tc min
20 10
6
6
220 50
88 sec
180 18 sec 200
EXTERNAL PULSE COMMUTATION (CLASS E COMMUTATION)
T1
VS
T2
RL
L
2VAUX
+ C
T3
VAUX
Fig. 1.34: External Pulse Commutation In this type of commutation an additional source is required to turnoff the conducting thyristor. Figure 1.34 shows a circuit for external pulse commutation. VS is the main voltage source and VAUX is the auxiliary supply. Assume thyristor T1 is conducting and load RL is connected across supply VS . When thyristor T3 is turned ON at t 0 , VAUX , T3 , L and C from an oscillatory circuit. Assuming capacitor is initially uncharged, capacitor C is now charged to a voltage 2VAUX with upper plate positive at t
LC . When current through T3 falls to zero, T3 gets commutated. To turnoff the
172
main thyristor T1 , thyristor T2 is turned ON. Then T1 is subjected to a reverse voltage equal to VS 2VAUX . This results in thyristor T1 being turnedoff. Once T1 is off capacitor ‘C’ discharges through the load RL 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.
L
T1
+
IL +
T3 VS
_C FWD Lr
L O A D
T2
_ Fig.: 1.35 Line Side Commutation Circuit Figure 1.35 shows line side commutation circuit. Thyristor T2 is fired to charge the capacitor ‘C’. When ‘C’ charges to a voltage of 2V, T2 is self commutated. To reverse the voltage of capacitor to 2V, thyristor T3 is fired and T3 commutates by itself. Assuming that T1 is conducting and carries a load current I L thyristor T2 is fired to turn off T1 . The turning ON of T2 will result in forward biasing the diode (FWD) and applying a reverse voltage of 2V across T1 . This turns off T1 , thus the discharging and recharging of capacitor is done through the supply and the commutation circuit can be tested without load.
173
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 dctodc 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 Stepdown choppers Stepup choppers. In stepdown choppers, the output voltage will be less than the input voltage whereas in stepup choppers output voltage will be more than the input voltage.
PRINCIPLE OF STEPDOWN CHOPPER
Chopper i0
V
+
R
V0
Fig. 2.1: Stepdown Chopper with Resistive Load
Figure 2.1 shows a stepdown 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.
174
v0 V Vdc t tON
tOFF
i0 V/R Idc t T Fig. 2.2: Stepdown choppers — output voltage and current waveforms Vdc
= average value of output or load voltage I dc = average value of output or load current tON = time interval for which SCR conducts tOFF = time interval for which SCR is OFF. T tON tOFF = period of switching or chopping period 1 frequency of chopper switching or chopping frequency. f T
Average output voltage
but
Vdc
V
tON tON tOFF
... 2.1
Vdc
V
tON T
V .d
... 2.2
duty cycle
... 2.3
tON t
d
Average output current, Vdc I dc R I dc
V tON R T
... 2.4 V d R
... 2.5
175
RMS value of output voltage VO
But during tON , vo
1 T
tON
vo2 dt 0
V
Therefore RMS output voltage VO
1 T
tON
V 2 dt 0
VO
V2 tON T
VO
d .V
Output power
PO
VO I O
But
IO
VO R
Therefore output power
PO
VO2 R
PO
dV 2 R
Effective input resistance of chopper V Ri I dc
tON .V T
... 2.6
... 2.7
... 2.8
... 2.9
R ... 2.10 d The output voltage can be varied by varying the duty cycle. Ri
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 tON 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, tON . Figure 2.3 shows the output voltage waveforms for different ON times. 176
V0 V tON
tOFF t T
V0 V
t tON
tOFF
Fig. 2.3: Pulse Width Modulation Control VARIABLE FREQUENCY CONTROL In this method of control, chopping frequency f is varied keeping either tON or tOFF constant. This method is also known as frequency modulation. Figure 2.4 shows the output voltage waveforms for a constant tON 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 tOFF load current may become discontinuous.
v0 V
tON
tOFF t T
v0 V tON
tOFF t T
Fig. 2.4: Output Voltage Waveforms for Time Ratio Control
177
STEPDOWN CHOPPER WITH RL LOAD Figure 2.5 shows a stepdown chopper with RL 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 iO continues to flow in the same direction through the freewheeling 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 I min and I max . Figure 2.6 shows the output current and output voltage waveforms for a continuous current and discontinuous current operation.
Chopper
i0
+ R
V
FWD
V0
L E
Fig. 2.5: Step Down Chopper with RL Load
v0
Output voltage
V tON i0
tOFF T
t
Imax
Output current
Imin
Continuous current
i0
t Output current Discontinuous current t
Fig. 2.6: Output Voltage and Load Current Waveforms (Continuous Current)
178
When the current exceeds current reduces to I min .
I max
the chopper is turnedoff and it is turnedon when
EXPRESSIONS FOR LOAD CURRENT iO FOR CONTINUOUS CURRENT OPERATION WHEN CHOPPER IS ON 0 t
tON
i0
+ R
V
V0
L E Fig. 2.5 (a)
Voltage equation for the circuit shown in figure 2.5(a) is
V
iO R L
diO dt
Taking Laplace Transform V RI O S S At t
0 , initial current iO 0
E
... 2.11
L S .I O S
E S
iO 0
... 2.12
I min V
IO S
I min R S L
E
LS S
R L
... 2.13
Taking Inverse Laplace Transform iO t
V
This expression is valid for 0 t
E R
1 e
R t L
I min e
R t L
... 2.14
tON . i.e., during the period chopper is ON.
At the instant the chopper is turned off, load current is
iO tON
I max
179
When Chopper is OFF
0 t tOFF
i0 R L E Fig. 2.5 (b) Voltage equation for the circuit shown in figure 2.5(b) is 0
RiO
L
diO dt
E
... 2.15
Taking Laplace transform 0
RIO S
L SI O S
Redefining time origin we have at t Therefore
IO S
iO 0
E S
0 , initial current iO 0
I max R S L
I max
E LS S
R L
Taking Inverse Laplace Transform
iO t
I max e
R t L
E 1 e R
R t L
... 2.16
The expression is valid for 0 t tOFF , 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
iO tOFF
I min
180
TO FIND I max AND I min From equation (2.14), At
t tON
Therefore
I max
dT , iO t
V
E
1 e
R
I max dRT L
I min e
dRT L
... 2.17
From equation (2.16), At
t tOFF t tOFF
Therefore
I min
T tON , iO t
I min
1 d T
I max e
1 d RT L
E 1 e R
1 d RT L
... 2.18
Substituting for I min in equation (2.17) we get,
I max
V 1 e R 1 e
dRT L RT L
E R
... 2.19
Substituting for I max in equation (2.18) we get, dRT
I min
I max
V e L 1 R RTL e 1
E R
... 2.20
I min is known as the steady state ripple.
Therefore peaktopeak ripple current I I max I min Average output voltage Vdc
d.V
... 2.21
Average output current I dc approx
I max
I min 2
... 2.22
181
Assuming load current varies linearly from I min to I max instantaneous load current is given by iO
I min
I .t for 0 t dT
iO
I min
I max I min t dT
tON dT
... 2.23
RMS value of load current I O RMS
I O RMS
I O RMS
1 dT
dT
1 dT
dT
1 dT
dT
i02 dt 0
I max
I min 0
I
I min t dT
I max I min dT
2 min
0
2
dt
2
t2
2 I min I max I min t dt dT
RMS value of output current 2 I min
I O RMS
I max
I min
1 2
2
I min I max
3
I min
... 2.24
RMS chopper current I CH
I CH
1 T
dT
1 T
dT
i02 dt 0
I min 0
I CH
2 d I min
ICH
d IO RMS
I max I min t dT
I max
I min 3
2
dt
1 2
2
I min I max
I min
... 2.25
Effective input resistance is Ri
V IS
182
Where I S = Average source current
Therefore
IS
dI dc
Ri
V dI dc
... 2.26
PRINCIPLE OF STEPUP CHOPPER
I
L
D +
+
C
V Chopper
L O A D
VO
Fig. 2.13: Stepup Chopper Figure 2.13 shows a stepup chopper to obtain a load voltage VO 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, tON . When the chopper is off, the inductor current I is forced to flow through the diode D and load for a period, tOFF . The current tends to decrease resulting in reversing the polarity of induced EMF in L. Therefore voltage across load is given by
VO
V
L
dI i.e., VO dt
V
... 2.27
If a large capacitor ‘C’ is connected across the load then the capacitor will provide a continuous output voltage VO . 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
183
Therefore energy stored in inductor = V .I .tON where tON
... 2.28 ,
ON period of chopper.
When Chopper is OFF (energy is supplied by inductor to load) Voltage across L VO V Energy supplied by inductor L Chopper.
VO V ItOFF , where tOFF
OFF period of
Neglecting losses, energy stored in inductor L = energy supplied by inductor L Therefore
VItON VO
VO Where
VO V ItOFF V tON
tOFF
tOFF V
T T tON
T = Chopping period or period of switching. T
tON
VO
V
Therefore
VO
V
Where
d
tON T
tOFF
1 t 1 ON T 1 1 d
... 2.29
duty cyle
For variation of duty cycle ‘d’ in the range of 0 d 1 the output voltage VO will vary in the range V VO . 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. 184
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 i0
+
v0
Chopper V FWD
L O A D
v0 V
i0
Fig. 2.14: Class A Chopper and vO iO Characteristic Figure 2.14 shows a Class A Chopper circuit with inductive load and freewheeling diode. When chopper is ON, supply voltage V is connected across the load i.e., vO V and current i0 flows as shown in figure. When chopper is OFF, v0 = 0 and the load current iO continues to flow in the same direction through the free wheeling diode. Therefore the average values of output voltage and current i.e., vO and iO 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.
185
ig
Thyristor gate pulse t
i0
Output current CH ON t FWD Conducts
v0
Output voltage
tON
t T
Fig. 2.15: First quadrant Chopper  Output Voltage and Current Waveforms
Class A Chopper is a stepdown 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 RL load can be used to study the performance of Class A Chopper. CLASS B CHOPPER D
i0
v0
+ R L v0
V Chopper
E
i0
Fig. 2.16: Class B Chopper Fig. 2.16 shows a Class B Chopper circuit. When chopper is ON, vO 0 and E drives a current iO 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, vO V and part of the energy stored in inductor L is returned to the supply. Also the current iO continues to flow from the load to source. Hence the average output voltage is positive and average output current is negative. Therefore Class 186
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
i0
+
D Conducting
R
V
V0
L E V
LdiO dt
For the initial condition i.e., iO t
RiO
E
I min at t 0 .
The solution of the above equation is obtained along similar lines as in stepdown chopper with RL load
V
Therefore
iO t
At t
iO t
tOFF
I max
E
1 e
R
R t L
I min e
R t L
0 t
tOFF
I max V
E R
1 e
R tOFF L
I min e
R tOFF L
During the interval chopper is ON voltage equation is given by
i0
+ R
Chopper ON
V0
L E 
0
LdiO dt
RiO
E
187
Redefining the time origin, at t
0
I max .
iO t
The solution for the stated initial condition is
iO t At t
tON
Therefore
I max e
iO t I min
R t L
E 1 e R
R t L
E 1 e R
R tON L
0 t
tON
I min
I max e
R tON L
ig
Thyristor gate pulse t
i0
tOFF
tON T Output current
Imax Imin v0
t
D conducts Chopper conducts
Output voltage
t
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, CH1 is ON or D2 conducts and for second quadrant operation, CH 2 is ON or D1 conducts. When CH1 is ON, the load current iO is positive. i.e., iO flows in the direction as shown in figure 2.18. The output voltage is equal to V vO V and the load receives power from the source. 188
CH1
D1 i0
v0
+ R
V CH2
L v0
D2 Chopper
i0
E
Fig. 2.18: Class C Chopper When CH1 is turned OFF, energy stored in inductance L forces current to flow through the diode D2 and the output voltage vO 0 , but iO continues to flow in positive direction. When CH 2 is triggered, the voltage E forces iO to flow in opposite direction through L and CH 2 . The output voltage vO 0 . On turning OFF CH 2 , the energy stored in the inductance drives current through diode D1 and the supply; output voltage vO V the input current becomes negative and power flows from load to source. Thus the average output voltage vO is positive but the average output current iO can take both positive and negative values. Choppers CH1 and CH 2 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.
ig1
Gate pulse of CH1 t
ig2
Gate pulse of CH2 t
i0 Output current t D1
V0
CH1 ON
D2
CH2 ON
D1
CH1 ON
D2
CH2 ON
Output voltage
t Fig. 2.19: Class C Chopper  Output Voltage and Current Waveforms
189
CLASS D CHOPPER v0 CH1
D2 R i0
L
E
V + D1
i0
v0 CH2
Fig. 2.20: Class D Chopper Figure 2.20 shows a class D two quadrant chopper circuit. When both CH1 and CH 2 are triggered simultaneously, the output voltage vO V and output current iO flows through the load in the direction shown in figure 2.20. When CH1 and CH 2 are turned OFF, the load current iO continues to flow in the same direction through load, D1 and D2 , V . The average load due to the energy stored in the inductor L, but output voltage vO voltage vO is positive if chopper ONtime tON is more than their OFFtime tOFF and average output voltage becomes negative if tON tOFF . 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.
ig1
Gate pulse of CH1 t
ig2
Gate pulse of CH2 t
i0 Output current
v0
CH1,CH2 ON
t D1,D2 Conducting Output voltage
V Average v0
Fig. 2.21: Output Voltage and Current Waveforms for tON
t tOFF
190
ig1
Gate pulse of CH1 t
ig2
Gate pulse of CH2 t
i0 Output current CH1 CH2
t D1, D2
v0
Output voltage
V Average v0 Fig. 2.22: Output Voltage and Current Waveforms for tON
t
tOFF
CLASS E CHOPPER
CH1
i0
V
+ CH2
CH3
D1 R
L
D3
E
v0 D2
CH4
D4
Fig. 2.23: Class E Chopper
191
v0 CH2  D4 Conducts D1  D4 Conducts
CH1  CH4 ON CH4  D2 Conducts i0
CH3  CH2 ON CH2  D4 Conducts
D2  D3 Conducts CH4  D2 Conducts
Fig. 2.23(a): Four Quadrant Operation Figure 2.23 shows a class E 4 quadrant chopper circuit. When CH1 and CH 4 are triggered, output current iO flows in positive direction as shown in figure 2.23 through CH1 and CH 4 , with output voltage vO V . This gives the first quadrant operation. When both CH1 and CH 4 are OFF, the energy stored in the inductor L drives iO through D3 V . Therefore the chopper and D2 in the same direction, but output voltage vO operates in the fourth quadrant. For fourth quadrant operation the direction of battery must be reversed. When CH 2 and CH 3 are triggered, the load current iO flows in V. opposite direction and output voltage vO Since both iO and vO are negative, the chopper operates in third quadrant. When both CH 2 and CH 3 are OFF, the load current iO continues to flow in the same direction through D1 and D4 and the output voltage vO V . Therefore the chopper operates in second quadrant as vO is positive but iO 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.
192
i0 Chopper V
+ L O A D
FWD
v0
Fig. 6.24. Solution Average output voltage, Vdc
tON V T
Vdc R
Average output current, I dc
dV
dV R
The thyristor is commutated at the instant t
tON .
Therefore output current at the instant of commutation is
V , since V is the output R
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 VO RMS
But
vO
tON
v02 dt 0
V during tON
VO RMS
Where duty cycle,
1 T
1 T
tON
VO RMS
V2
VO RMS
dV
d
V 2 dt 0
tON T
tON T
193
RMS value of thyristor current = RMS value of load current VO RMS 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, Vdc = 350 V, f = 2 kHz Chopping period
Output voltage
T
1 f
T
1 2 10
Vdc
3
0.5 m sec
tON V T
Conduction period of thyristor T Vdc tON V tON
0.5 10 3 350 460
tON
0.38 msec
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 ssec. Compute Chopping frequency, If the pulse width is halved for constant frequency of operation, find the new output voltage.
194
Solution V = 200 V,
tON
200 s , Vdc
Vdc
V
600
600V
T T tON
200
T 200 10
T
6
Solving for T T
300 s
Chopping frequency 1 f T f
1 300 10
6
3.33KHz
Pulse width is halved Therefore
200 10 2
tON
6
100 s
Frequency is constant Therefore
f
3.33KHz
T
1 f
Therefore output voltage
300 s
=V
T T tON
200
300 10 6 300 100 10
6
300 Volts
Problem 2.4: A dc chopper has a resistive load of 20 and input voltage VS 220V . 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.
195
Solution VS d
220V , R
tON T
20 , f = 10 kHz
0.80
Vch = Voltage drop across chopper = 1.5 volts
Average output voltage Vdc
Vdc
tON T
VS Vch
0.80 220 1.5
Chopper ON time,
tON
Chopping period,
T
1 f
T
1 10 103
174.8 Volts
dT
0.1 10 3 secs 100 μsecs
Chopper ON time, tON
dT
tON
0.80 0.1 10
tON
0.08 10
3
3
80 μsecs
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 I dc
30 Amps , f = 250 Hz, V = 110 V, R 2
Chopping period,
Therefore
T
1 f
1 250
4 10
I dc
Vdc and Vdc R
I dc
dV R
3
4 msecs
dV
196
I dc R V
d
30 2 110
Chopper ON period, tON
dT
Chopper OFF period, tOFF
T tON
0.545
0.545 4 10
3
tOFF
4 10
2.18 10
tOFF
1.82 10
3
3
2.18 msecs
3
1.82 msec
Problem 2.6: A dc chopper in figure 2.25 has a resistive load of R 10 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.
Chopper
i0
+
R v0
V
Fig. 2.25 Solution V = 200 V, R 10 , Chopper voltage drop, Vch
2V , d = 0.60, f = 1 kHz.
Average output voltage Vdc d V Vch
Vdc
0.60 200 2
118.8 Volts
RMS value of output voltage VO d V Vch
VO
0.6 200 2
153.37 Volts
197
Effective input resistance of chopper is V V Ri I S I dc I dc
Vdc R
Ri
V IS
118.8 11.88 Amps 10 V I dc
200 16.83 11.88
Output power is
PO
PO
PO
PO
Input power,
Pi
PO
PO
1 T
dT
1 T
dT
0
0
v02 dt R
d V Vch R
10
1 T
dt
2
0.6 200 2
1 T
2
V Vch R
2
2352.24 watts
dT
ViO dt 0
dT
0
V V Vch dt R
dV V Vch R
0.6 200 200 2 10
2376 watts
Chopper efficiency, PO 100 Pi 2352.24 100 99% 2376
Problem 2.7: A chopper is supplying an inductive load with a freewheeling diode. The load inductance is 5 H and resistance is 10 . The input voltage to the chopper is 200
198
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 Chopping period,
T
1 f 2 3
tON
2 tOFF 3
T
tON
T
2 tOFF 3
T
5 tOFF 3
tOFF
tOFF
3 T 5
T
3 1 10 5
tON
T tOFF
tON d
2:3
1 1 msecs 1000
tON tOFF
tOFF
Duty cycle,
, f = 1000 Hz, V = 200 V, tON : tOFF
3
0.6 msec
1 0.6 10 tON T
3
0.4 10 3 1 10 3
0.4 msec 0.4
Refer equations (2.19) and (2.20) for expressions of I max and I min . Maximum value of load current [equation (2.19)] is
I max
V 1 e R 1 e
dRT L RT L
E R
199
Since there is no voltage source in the load circuit, E = 0
Therefore
I max
V 1 e R 1 e
dRT L RT L
0.4 10 1 10 5
I max
200 1 e 10 1 e 1 e
I max
20
I max
8.0047A
1 e
10 1 10 5
0.8 10 2 10
3
3
3
3
Minimum value of load current from equation (2.20) with E = 0 is dRT L
I min
I min
V e 1 RT R eL 1
200 e 10
Average load current I max I dc
I dc
0.4 10 1 10 5
e
10 1 10 5
3
1
3
7.995 A
1
I min 2
8.0047 7.995 2
8A
Problem 2.8 : A chopper feeding on RL load is shown in figure 2.26. With V = 200 V, R = 5 , 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
, L = 5 mH, f = 1kHz, d = 0.5, E = 0
200
1 f
Chopping period is T
1 1 103
1 10
3
Chopper
secs
i0
+ R
FWD
L
v0
E Fig.: 2.26 Refer equations (2.19) and (2.20) for expressions of I max and I min . Maximum value of load current
I max
V 1 e R 1 e
dRT L
E R
RT L
0.5 5 1 103
I max
I max
200 1 e 5 1 e
5 10
3
5 1 10 5 10
1 e 0.5 40 1 e1
0
3
3
24.9 A
Minimum value of load current is dRT
I min
V e L R RTL e
1
E R
1
0.5 5 1 10
I min
I min
200 e 5
40
5 10
5 1 10
e
5 10
e0.5 1 e1 1
3
3
3
1 3
0
1
15.1 A
Average value of load current is I1 I 2 I dc for linear variation of currents 2
201
Therefore
24.9 15.1 20 A 2
I dc
Refer equations (2.24) and (2.25) for RMS load current and RMS chopper current. RMS load current from equation (2.24) is I max
2 I min
I O RMS
I min
1 2
2
I min I max
3
24.9 15.1 3
I min
1 2
2
I O RMS
15.12
I O RMS
96.04 228.01 147.98 3
15.1 24.9 15.1
1 2
20.2 A
RMS chopper current from equation is (2.25) is
I ch
d IO RMS
0.5 20.2 14.28 A
Effective input resistance is Ri
V IS
I S = Average source current IS
dI dc
IS
0.5 20 10 A
Therefore effective input resistance is Ri
V IS
200 10
20
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 N1 = 1000 rpm Duty ratio d1 0.8 , f = 100 Hz
202
We know that back EMF of motor Eb is given by ZNP Eb 60 A Where N = speed in rpm = flux/pole in wbs Z = Number of Armature conductors P = Number of poles A = Number of parallel paths Therefore
Eb
N
Eb
N if flux
is constant
Chopper
Ia
+
Ra Vdc
V
+ M Eb
Fig. 2.27 Eb
where I a = Ra =
Vdc
I a Ra
Armature current Armature Resistance
Since Ra is not given, I a Ra drop is neglected. Eb1 Vdc1 200 volts Therefore Vdc1
Supply,
V
d1V
Vdc1 d1
V
200 0.8
V
250 Volts
203
Eb1
N1
200
1000
... 2.30
Now speed changes hence ‘d’ also changes. Given N 2
800 rpm Eb2
?
Eb 2
N2
Eb2
800
... 2.31
Dividing equation (2.30) by equation (2.31) we get
200 Eb2 Eb2
But
Eb2
d2
1000 800 800 200 160 V 1000 Vdc2
d 2V
Vdc2
160 250
V
0.64
Chopping frequency f = 100 Hz 1 1 T 0.01 sec f 100
T 10 msecs tON T
d2
ON time of chopper tON d 2T tON
0.64 10 10
tON
6.4 msecs
3
204
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 turnoff 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.
T1
LS +
a
iT1 IL
+
C b _ iC
T2
FWD
VS
L
_
+ L O A D
D1
vO
_
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 turnedoff. 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.
T1
LS +
IL
+ VC
_C
iC
VS L
D1
L O A D
_ Fig. 2.29
205
Capacitor Discharge Current C sin t L
iC t
V
iC t
I P sin t ; where I P
V
C L
1 LC
Where & Capacitor Voltage
VC t
V cos t
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 turnoff time. Circuit turnoff time is given by VC C IL Where IL is load current. tC
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 td
VS C IL
The total time required for the capacitor to discharge and recharge is called the commutation time and it is given by tr
tC
td
At the end of Mode2 capacitor has recharged to ‘VS’ and the free wheeling diode starts conducting. The equivalent circuit for Mode2 is shown in fig. 2.30.
206
IL
_
LS
+
VC VS
IL C
+
L O A D
T2
_ 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 i t I L cos t Hence capacitor charges to a voltage higher than supply voltage. T2 naturally turnsoff. The instantaneous capacitor voltage is VC t
VS
S
1 LS C
Where
IL
LS sin C
S
t
Fig. 2.31 shows the equivalent circuit of Mode – 3.
IL +
LS VS
IL
+
_C
T2
VS FWD
L O A D
_ 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. 207
LS +
IL
+ VC
_C
L O A D
D1
VS L
FWD
_ 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
IL FWD
L O A D
Fig. 2.33 Fig. 2.34 shows the current and voltage waveforms for a voltage commutated chopper.
208
ic
Capacitor Current IL
0 Ip iT1 IL
t Ip
Current through T1 t
0 v T1
Voltage across T1
Vc
t 0 vo Vs+Vc
Output Voltage
Vs
t
vc Vc
t Capacitor Voltage
Vc
tc
td
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.
209
Jone’s Chopper +
T1
C
T2 D V
L2 L1 + R FWD
v0 L
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
210
Ig
Gate pulse of T2
Gate pulse of T1
Gate pulse of T2 t
VC +V
Capacitor Voltage t Resonant action
V Auto transformer action tC
Capacitor discharge current Current of T1 IL
t
IL
Voltage across T1
t
tC
211
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