l2. single phase ac voltage controllers
TRANSCRIPT
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EEL 744AC CONTROLLERS
Professor Bhim SinghDepartment of Electrical Engineering Indian Institute of Technology Delhi Hauz Khas, New Delhi-10016, India
email:[email protected],[email protected]
Ph.:011-2659-1045
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Lecture - 2
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1 – phase AC Voltage controller
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1 – phase AC Voltage controller• The power flow can be controlled by varying the rms value
of AC voltage applied to the load, a thyristor switch isconnected between the AC source and load.
• This type of power circuit is known as ac voltage controller
AC Voltage Controller
It converts fixed AC voltage directly to variable AC voltage without change in frequency.
The Power flow can be controlled by varying the rms voltage applied to the load
It has high efficiency, flexibility in Control, less maintenance, and compact size.
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AC voltage controllers can be divided in twobroad categories
Single phase;
Three phase.
These controllers can be used for an induction motoras:
Soft starters;
Energy saving controllers;
Solid state speed controllers.
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Salient features as soft starter
Step less control of motor voltage.Control flexibility due to low power controlcircuitry.Smooth acceleration and deceleration of the motor.Easy implementation of current control.Simple protection against single phasing orunbalanced operation in case of three phase motors.Absence of current inrush.Low maintenance for applications requiringfrequent starting and stopping.
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Soft start of single phase induction motors
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AC Voltage ControllersSalient features as energy saver:
In such applications, voltage control is used for reduction oflosses not for speed control;The motor losses primarily depend on three factors:o Loading on the motor;o Magnitude of applied voltage; ando Quality of motor construction.The most significant factor is motor loading;The motor running at light load has most savings;The applications with low duty cycles will allow moreenergy savings;
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AC Voltage Controllers
Following applications have significant no loadoperation and so voltage control can serve asenergy saver:
Gang Ripsaw (1.63 kW saving for 50 hp motor);Woodhog (1.2 kW saving on 16 kW motor);Air compressors : reciprocating type (12% saving of 200 hpmotor);Drill presses;Cutoff saws;Machine tools;Industrial sewing machines;
Gang ripsaws (1.63 kW saving for 50 hp motor)
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Applications 1- phase ac controller
Woodhog (1.2 kW saving on 16 kW motor)
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American Pulverizer wood hogs and shredders are designed toprovide a one-step operation reducing by impact rather than bycutting. No knives are used. The hammers are designed to producemaximum reduction of fibrous materials and are built to withstandthe impact of such foreign materials as nails, cleats etc.
Air compressors : reciprocating type (12% saving of 200 hp motor)
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Applications single phase ac controller
Drill presses
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Applications single phase ac controller
Cuttoff saw
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Applications single phase ac controller
CNC flat bed lathes
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Radial drilling machine
Machine tools
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Industrial sewing machine
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AC Voltage Controllers
The applied voltage is directly related with core losses;therefore, optimum voltage shall reduce the losses:
the motor operating near the distribution substation willhave higher voltage than the one at the end of thedistribution line;Therefore, the voltage reduction will allow energy savings;
A badly designed motor or a rewound motor with uneven airgap will draw more magnetizing current and will have highercore losses;
Reduced voltage operation of such motor will certainlyimprove energy utilization at all loads;
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Following applications use AC voltage controlleras speed controller:
Speed control of motors
Fans and blowers;
High pressure material handling fans;
Twist frames;
Light vacuum system;
Pumps (Single quadrant operation);
Crane drives (Four quadrant operation);
Cutting and forming machines;
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Following applications use AC voltage controlleras speed controller:
Grinders;
Plastic extruders;
Abrasive planer.
Lamp dimmer
Thyristor controlled reactor
Thyristor switched capacitor
Heating control
Heating chamber
Electric Boiler
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Speed control of 1- phase ac motors usingtriac based controller
Applications single phase ac controller
Fan regulator
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Applications single phase ac controller
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Applications single phase ac controller
Blowers
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Applications single phase ac controller
Twist frames
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Applications single phase ac controller
Light vacuum system
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Applications single phase ac controller
Pumps
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Applications single phase ac controller
Crane drives
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Applications single phase ac controller
Cutting machines
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Applications single phase ac controller
Forming machines
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Applications single phase ac controller
Grinders
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Applications single phase ac controller
Abrasive planer
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Applications single phase ac controller
Plastic extruder
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Lamp dimmer
Applications single phase ac controller
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Applications single phase ac controller
Thyristor controlled reactor
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Applications single phase ac controller
Thyristor switched capacitors
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Applications single phase ac controller
Heating control
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Applications single phase ac controller
Heating chamber for drying
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Electric boiler
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For power transferTwo types of control
1. On – off control2. Phase angle control
On – off control:- Thyristor switches are connect load to acsource for few cycles of input voltage and then disconnect forfew cycles . This type of control is also called as burst firing,zero voltage switching, cycle selection or integral cycleswitching
The Thyristors thus act as a high speed contactor (or highspeed ac switch).
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Phase angle control:- Thyristor switches connect the loadto the ac source for a portion of each cycle of input voltageor if each thyristor is triggered at some non-zero point onits respective anode voltage cycle, the load voltagewaveform is described as ‘phase angle controlled’
By controlling the phase angle or the trigger angle ‘α’(delay angle), the output RMS voltage across the load canbe controlled.The trigger delay angle ‘α’ is defined as the phase angle(the value of ωt) at which the thyristor turns on and theload current begins to flow.
The load voltage and current have identical positive andnegative alterations with frequency spectra containing onlyodd harmonics
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Phase control Thyristors which are relatively inexpensive,converter grade Thyristors which are slower than fastswitching inverter grade Thyristors are normally used.
For applications upto 400Hz, if Triacs are available tomeet the voltage and current ratings of a particularapplication, Triacs are more commonly used.
Due to ac line commutation or natural commutation, thereis no need of extra commutation circuitry or componentsand the circuits for ac voltage controllers are very simple.
Phase angle control
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Classification of ac voltage controllers
Single phase ac voltage controllers– Unidirectional or half-wave control– Bidirectional or full-wave control
Three phase ac voltage controllers– Unidirectional or half-wave control– Bidirectional or full-wave control
Phase control thyristors are relatively inexpensive and slower than fast switching thyristors are normally used
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Classification of ac voltage controllers
Single phase ac controllers operate with single phase acsupply voltage of 230V RMS at 50Hz in our country.
Three phase ac controllers operate with 3 phase ac supplyof 415V RMS at 50Hz supply frequency.
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If TRIACs are available for the given voltage and current rating then they are commonly used
The circuits of ac voltage controllers are very simple due to line or natural commutation
Due to the nature of output waveforms , the analysis for the derivations of explicit expressions for the performance parameters of circuits is not simple
For the sake of simplicity resistive loads are taken in many of the derivations.
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1
1
o
om
On – off control
Circuit waveforms
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n = Two input cycles. Thyristors are turned ON during for two input cycles.m = One input cycle. Thyristors are turned OFF during for one input cycle
For a sine wave input supply voltage,
sin 2 sin
RMS value of input ac supply = = RMS phase supply voltage.2
If the input ac supply is connected to load for 'n' number of input cycles and
s m S
mS
v V t V tVV
= ω = ω
=
disconnected for 'm' number of input cycles, then,
1Where = input cycle time (time period) and
ON OFFt n T t m T
Tf
= × = ×
=
On – off control
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( ) ( )
( ) ( )
= input supply frequency. = controller on time = .
= controller off time = . = Output time period = .We can show that,
Output RMS voltage
W
ON
OFF
O ON OFF
ON ONSO RMS i RMS
O O
ft n T
t m TT t t nT mT
t tV V VT T
×
×
+ = +
= =
( )here is the RMS input supply voltage = Si RMSV V
On – off control
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On – off control
Power factor
TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF OUTPUT VOLTAGE, FOR ON-OFF CONTROL METHOD.
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( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )
( )
2 2
0
22
0
2
2
0
2
0 0
1Output RMS voltage . .
.
1 2Substituting for 2
1 22
2 .2
ON
ON
ON
ON ON
t
mO RMSO t
tm
O RMSO
tm
O RMSO
t tm
O RMSO
O RMS
V V Sin t d tT
VV Sin t d tT
CosSin
V Cos tV d tT
VV d t Cos t d tT
V
ω
ω =
ω
ω
ω ω
= ω ωω
= ω ωω
− θθ =
− ω⎡ ⎤= ω⎢ ⎥ω ⎣ ⎦
⎡ ⎤= ω − ω ω⎢ ⎥
ω ⎢ ⎥⎣ ⎦
=
∫
∫
∫
∫ ∫
( )2
0 0
22 2
ON ONt tm
O
V Sin ttT
ω ω⎡ ⎤ωω −⎢ ⎥ω ⎣ ⎦
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( ) ( )2 sin 2 sin 0
02 2
Now = An integral number of input cycles; Hence , 2 ,3 , 4 ,5 ,.....& 2 , 4 ,6 ,8 ,10 ,......
Where T is the input supply time period (T = input cy
m ONONO RMS
O
ON
ON ON
V tV t
Tt
t T T T T T t
−⎡ ⎤= − −⎢ ⎥⎣ ⎦
= =
ωω
ω
ω π π π π π
( )
( ) ( )
( )
( )
2
cle time period). Thus we note that sin 2 0
2 2
Where = RMS value of input supply voltage;2
duty cycle (d
ON
m ON m ONO RMS
O O
ON ONSO RMS i RMS
O O
mSi RMS
ON ON
O ON OFF
t
V t V tV
T T
t tV V V
T TV
V V
t t nT n kT t t nT mT n m
=
= =
= =
= =
= = = = =+ + +
ω
ωω
( ) ( )
).
S SO RMSnV V V k
m n= =
+
Applications of on – off control or integral cycle control
Incandescent lighting control: An irritating flicker isnoticed even when only supply cycle is omitted from eachcontrol period of hundred cycles.
This form of control is unsuitable for normal lighting.
This type of control can be used for photographic andphotochemical applications where an exposure – timeprecision of not less than one supply period is needed
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Applications of on – off control or integral cycle control
Heating control: The temperature of a 75W element wasmonitored using a recording , thermocouple thermometer.
Control period of 10 cycles, a direct relationship wasfound between the heat energy developed and the powertransmitted, measured by the N/T ratio.
– Where N is the number of conducting cycles.– And T is the number of supply cycles.
Integral cycle control appears to be well suited to thisform of application.
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Applications of on – off control or integral cycle control
Control of ¼ HP universal motor : Control may beeffected at no load with fixed control period of 10 supplycycles and variable ON/OFF, N/T – N.
With N/T – N ≥ 1, the test motor ran smoothly and asmall degree of speed control was achieved.
With N/T – N < 1, when extinction interval exceeded theconduction interval, torque pulsations became visible andaudible and inching occurred during the conductionperiods.
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Applications of on – off control or integral cycle control
Speed control of a FHP, dc series motor: The speed of adc series motor can be controlled by use of field currentdiversion or by variation of applied voltage.
The scheme in the next slide was used to provide appliedvoltage variation using rectified integral cycle pulseswith constant control period to a 1/8 HP motor
Any integral number of consecutive conduction cycles upto full conduction could be applied so that this constitutesa form of pulse width modulation.
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55
Schematic drive for rectified, integral-cycle control of fractional horse power series dc motor
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The diode bridge rectifier provides a relaxation path for the motor current during excitation for the thyristors.
To provide smooth speed control, motor current variationsmust be kept to a minimum. For this reason the ON/OFFratio has a minimum value in which the maximumpermissible OFF time is determined by the electricalenergy storage capability of the motor.
When the energy recovered during OFF period isinsufficient, motor current decays to zero.
By the nature of the controller , the voltage and inputpower applied to the motor are pulsating.
Speed pulsating were obtained by means of a tachometer T
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PERFORMANCE PARAMETERS OF AC VOLTAGE CONTROLLERS
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( )
122
2 2
0
sin .2
2
Where = RM
RMS Output (Lo
S value of input sup
ad)
ply voltage
Voltage
.
mO RMS
mSO RMS i RMS
SO RMS i RMS
S i RMS
nV V t d tn m
V nV V k V km n
V V k V k
V V
π⎡ ⎤= ω ω⎢ ⎥π +⎣ ⎦
= = =+
= =
=
∫
58
PERFORMANCE PARAMETERS OF AC VOLTAGE CONTROLLERS
( ) ( )
( )
( )( ) ( )
( )2
Duty Cycle
RMS Load Current
Where, = duty cycle (d).
; for a resis
Output AC (Load) Power
tive load .
ON ON
O ON OFF
O RMS O RMSLO RMS
L
O LO RMS
t t nTkT t t m n T
nkm n
V VI Z R
Z R
P I R
= = =+ +
=+
= = =
= ×
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PERFORMANCE PARAMETERS OF AC VOLTAGE CONTROLLERS
( )
( ) ( )( )
2
output load powerinput supply volt amperes
; RMS input supply current.
The input supply current is same as the load cur
Input Power Fact
rent e
or
H
O O
S S
LO RMSS in RMS
i RMS in RMS
in O L
P PPFVA V I
I RPF I I
V I
I I I
= = =
×= = =
×
= =
( ) ( )
( )
( ) ( )
( )
( )
( )
( )
2
nce, RMS supply current = RMS load current; .in RMS O RMS
LO RMS O RMS i RMS
i RMS in RMS i RMS i RMS
I I
I R V V kPF k
V I V V
nPF km n
=
×= = = =
×
= =+
60
0 p 2 p 3p wt
Im
n miT
Waveform of Thyristor Current
( )The Average Current of Thyristor T AvgI
( ) ( ) ( )
( ) ( ) ( )
0
0
sin .2
sin .2
mT Avg
mT Avg
nI I t d tm n
nI I t d tm n
=+
=+
∫
∫
π
π
ω ωπ
ω ωπ
61
,
( ) ( )
( ) ( ) [ ]
( ) ( ) ( )
( ) ( ) [ ]
( ) ( )
( ) ( )
( ) ( )
0
cos2
cos cos 02
1 12
22
.
duty cycle
.
Where = maximum or peak thyristor cu
mT Avg
mT Avg
mT Avg
mT Avg
m mT Avg
ON
ON OFF
m mT Avg
mm
L
nII t
m n
nII
m nnI
Im nnI Im n
I n k II
m nt nk
t t n mI n k I
Im nV
IR
⎡ ⎤= −⎢ ⎥+ ⎣ ⎦
= − ++
= − − +⎡ ⎤⎣ ⎦+
=+
= =+
= = =+ +
= =+
=
π
ωπ
ππ
π
π
π π
π π
rrent.
62
,
( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )( ) ( )
( ) ( ) ( ) ( )
( )
12
2 2
0
122
2
0
122
0
122
0 0
2
RMS Cur
sin .2
sin .2
1 cos
rent of Thyr
22 2
cos 2 .
istor
4
mT RMS
mT RMS
mT RMS
mT RMS
T RMS
mT RMS
nI I t d tn m
nII t d t
n m
tnII d t
n m
nII d t t d t
n m
I
nII
⎡ ⎤= ⎢ ⎥+⎣ ⎦
⎡ ⎤= ⎢ ⎥+⎣ ⎦
⎡ ⎤−= ⎢ ⎥+⎣ ⎦
⎡ ⎤⎧ ⎫= −⎢ ⎥⎨ ⎬+⎢ ⎥⎩ ⎭⎣ ⎦
=
∫
∫
∫
∫ ∫
π
π
π
π π
ω ωπ
ω ωπ
ωω
π
ω ω ωπ
( ) ( )
( ) ( ) ( )
12
0 0
122
sin 24 2
sin 2 sin 004 2
mT RMS
ttn m
nII
n m
⎡ ⎤⎧ ⎫⎛ ⎞−⎢ ⎥⎨ ⎬⎜ ⎟+ ⎝ ⎠⎢ ⎥⎩ ⎭⎣ ⎦
⎡ ⎤⎧ − ⎫⎛ ⎞= − −⎨ ⎬⎢ ⎥⎜ ⎟+ ⎝ ⎠⎩ ⎭⎣ ⎦
π πωωπ
πππ
63
The RMS Current of Thyristor
,
( ) ( )
( ) ( ) ( )
( ) ( )
( )
122
1 12 22 2
0 04
4 4
2 2
2
mT RMS
m mT RMS
m mT RMS
mT RMS
nII
n m
nI nII
n m n m
I InI km n
II k
⎡ ⎤= − −⎢ ⎥+⎣ ⎦
⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦
= =+
=
ππ
ππ
64
Problem: A single phase full wave ac voltage controller working on ON-OFF control technique has supply voltage of 230V, RMS 50Hz, load =50Ω. The controller is ON for 30 cycles and off for 40 cycles. CalculateON & OFF time intervals.RMS output voltage.Input P.F.Average and RMS thyristor currents.
( ) 230 ; 2 230 325.269 ; 325.269
1 1 0.02 sec : 2050
n = num ber of input cycles during w hich controller is ON ; n = 30. m = num ber of inp
ut cycles during w hich cont
Solu
roller is
tion :
O FF
m min RM SV V V V V V V
T T msf Hz
= = × = =
= = = =
( ) ( )
; m = 40 30 20 600 0.6 sec0.6 sec = controller O N tim e.
40 20 800 0.8 sec0.8 sec = controller OFF time.
30 0.428540 30
ON
ON
OFF
OFF
t n T ms m st n Tt m T ms mst m T
nD uty cycle km n
= × = × = =
= × =
= × = × = =
= × =
= = =+ +
65
( ) ( ) ( )
( ) ( )
( )
( )( ) ( )
( )2 2
:
30 3230 23030 40 7
230 0.42857 230 0.65465 150.570
150.570 3.011450
3.0114 50 453.4264
RM S output voltage Solution
Input Power F
98
actor
i RMS
O RMS
O RMS
O RMS O RMSO RMS
L
O L
O RM
O RMS
SnV
m n
V V
V V V
V V VI AZ R
P R W
V
I
= ×+
= × =+
= = × =
= = = =Ω
= × = × =
( )30 0.428570
0.654 53
.
6
P F k
nPFm n
PF
= = =+
=
=
66
( )
( )
2 230 325.269where50
Solutio
50
n: contd.........Average Thyristor Curren
6.50538
t Rating
RMS Curre
2 = Peak (maximu
.
m) thyristor current.6.5
nt
05382 3
i7
Rat
m mT Avg
mm
L
m
T Avg
I k InIm n
VI
RI A
I
×⎛ ⎞= × =⎜ ⎟+⎝ ⎠
×= = =
=
⎛ ⎞= × ⎜ ⎟⎝ ⎠
π π
π
( ) ( )( )
6.505382 32 2 2 7
2.129
ng of Thyristo
38
r
6
m mT RMS
T RMS
I InI km n
I A
= = = ×+
=
67
Problem : A 1.5 kW resistance heating element, fed
from 220 V rms at 50 Hz, is controlled by an ac switch
with integral half cycle control with a base period of 48
half cycles. Determine the number of on half-cycles in a
base period if the output power is to be controlled to a
value of 0.5 kW.
2
o s
2s
0
220 nSolution: m + n=48 ; R= =35.26Ω ; V =V ;1500 n + m
V nP = 500 = ; n =16(m + n)R
68
Problem: A single-phase TCR (thyristor controlled reactor consisting back-to-
back connected thyristors with pure inductor) has an input of 240V, 50Hz, AC
supply and an inductance of 20 mH. Calculate maximum VAR rating. Also
calculate (i) net rms current, (ii) fundamental rms current, (iii) 3rd harmonic
rms current, (iv) 5th harmonic rms current, and (v) 7th harmonic rms current
at delay angle of 30°. [40]
( )( )
2
-3
2rms
f
n 2
3 5 7
240Solution: VAR rating =9.167kVAR2π×50×20×10
V 1 3sin2αi) I = π-2α 0.5+sin α - = 11.23 AωL π 2
V 2α sin2αii) I = 1- - =10.71Aπ π2ωL
V 4 sinαcosnα-ncosαsinnαiii) I =
ωL π n(n -1)I =-5.25A iv) I =-1.05A v) I =
⎡ ⎤⎢ ⎥⎣ ⎦
⎛ ⎞⎜ ⎟⎝ ⎠
0.375A
69
Phase angle control
Circuitwaveforms
The power flow is controlled during the positive half cycle ofinput voltage , this type of controller is also known as aunidirectional controller
This circuit is a single phase half-wave controller and issuitable only for low power resistive loads, such as heatingand lighting.
Half wave controller can vary the output voltage by varyingthe delay angle α, the output contains an desirable dccomponent.
This type of controller is not generally used in practicalapplications.
70
71
Equations of unidirectional or half-wave control of ac controller
( )
Input AC Supply Voltage across the Transformer Secondary Winding.sin
= RMS value of secondary supply voltage.2
Output Load Voltage0; 0
sin ; 2Output Loa
s m
mS in RMS
o L
o L m
v V tVV V
v v for t tov v V t for t to
= ω
= =
= = ω = α
= = ω ω = α π
d Currentsin ; 2
0; 0
o mo L
L L
o L
v V ti i for t toR R
i i for t to
ω= = = ω = α π
= = ω = α
72
( )
( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( )
( )
22 2
22
22
2 2
1 sin .2
1 cos2 .
TO DERIVE AN EXPRESSION FOR R
2 2
MS
1 cos2 .4
OUTPUT VOLTAGE
os2
c .2
mO RMS
mO RMS
mO RMS
mO RMS
O RMS
O RMS
V V t d t
V tV d t
V
VV t d t
VV d t t d t
V
π
α
π
α
π
α
π π
α α
⎡ ⎤= ω ω⎢ ⎥π ⎣ ⎦
⎡ ⎤− ω⎛ ⎞= ω⎢ ⎥⎜ ⎟π ⎝ ⎠⎣ ⎦
⎡ ⎤= − ω ω⎢ ⎥π ⎣ ⎦
⎡ ⎤= ω − ω ω⎢ ⎥
π ⎣ ⎦
∫
∫
∫
∫ ∫
( )
( ) ( )
( ) ( )
2 2
2
sin 222
sin 2222
sin 4 sin 22 ;sin 4 02 22
m
mO RMS
mO RMS
V tt
V tV
VV
π π
α α
π
α
⎡ ⎤ω⎛ ⎞= ω −⎢ ⎥⎜ ⎟π ⎝ ⎠⎣ ⎦
ω⎛ ⎞= π − α − ⎜ ⎟π ⎝ ⎠
π α⎧ ⎫= π − α − − π =⎨ ⎬π ⎩ ⎭
73
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( )
( )
sin 2222sin 22
22 2sin 22
22 2
1 sin 222 22
1 sin 222 2
1 sin 222 2
Where, = RMS value of input supply 2
mO RMS
mO RMS
mO RMS
mO RMS
O RMS i RMS
SO RMS
mSi RMS
VV
VV
VV
VV
V V
V V
VV V
= − +
= − +
= − +
⎡ ⎤= − +⎢ ⎥⎣ ⎦
⎡ ⎤= − +⎢ ⎥⎣ ⎦
⎡ ⎤= − +⎢ ⎥⎣ ⎦
= =
απ απ
απ απ
απ απ
απ απ
απ απ
απ απ
( )
voltage
(across the transformer secondary winding).Note: Output RMS voltage across the load is controlled by changing as indicated by the expression for O RMSV
74
( )
( ) ( )
( ) ( )
PLOT OF VERSUS TRIGGER ANGLE FOR A SINGLE PHASE HALF-WAVE
AC VOLTAGE CONTROLLER
1 sin22
(UNIDIRECTIONAL CONTROLLER)
2 22
1 sin222 2
mO RMS
S
O RMS
O RMS
V
VV
V V
⎡ ⎤= − +⎢ ⎥⎣ ⎦
⎡ ⎤= − +⎢ ⎥⎣ ⎦
απ α
απ απ
α
π
75
( ) ( )
( ) ( )
( )
( ) [ ]
[ ]
( )
2
2
2
1 sin .2
sin .2
cos2
cos 2 cos :cos 2 12
cos 1 ; 22
2Hence cos 1
TO CALCULATE THE AVERAGE VALUE (DC VALUE) OF OUTPUT VOLTA
2
GE
mO dc
mO dc
mO dc
mO dc
mdc m S
Sdc
V V t d t
VV t d t
VV t
VV
VV V V
VV
whe
=
=
⎡ ⎤= −⎢ ⎥
⎣ ⎦
= − + =
= − =
= −
∫
∫
π
α
π
α
π
α
ω ωπ
ω ωπ
ωπ
π α ππ
απ
απ
' ' is varied from 0 to . varies from 0 to mdc
Vn V
−α π
π
76
Problem: A single phase half-wave ac voltage controller has a load resistance , input ac supply voltage is 230V RMS at 50Hz. The input supply transformer has a turns ratio of 1:1. If the thyristor is triggered at . CalculateRMS output voltage, Output power.RMS load current and average load current, Input power factor.Average and RMS thyristor current.
0
S
Given,230 , primary supply voltage.
Input supply frequency = 50Hz.50
60 radians.3
V RMS secondary voltage.
1 11
Therefore 230
Where, = Number of turns in th
Solution:
e pr
p
L
p p
S S
p S
p
V V RMS
fR
V NV N
V V V
N
=
== Ω
= =
=
= = =
= =
πα
imary winding.
= Number of turns in the secondary winding.SN
77
( )
( ) ( )
( )
( ) ( )
( )
( ) [ ]
22 2
0
1 s in .2
W e h a v e o b ta in e d th e e x p r e s s io n f o r
R M S V a lu e o f O u t
a s
1 s in 222 2
1 s in 1 2 02 3 0 22 3 2
12 3 0 5 .6
p u t ( L o a d ) V o lt a
6 9 2 3 0 0 .9 42
e
9
g
mO R M S
O R
O R M
M S
SO R M S
O M
O S
S
R S
R M
V V t d t
V
V
V
V V
V
=
= − +⎡ ⎤⎣ ⎦
⎡ ⎤⎛ ⎞= − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
= = ×
∫π
α
ω ωπ
απ απ
πππ
π
( )
( )
( )( )
R M S L o a d C u r r e n t
8 6
2 1 8 .4 6 9 6 2 1 8 .4 7
2 1 8 .4 6 9 6 6 4 .3 6 9 3 9 5 0
O R M S
O R M SO R M
O R
SL
M SI
V V V
VI A m p s
R
= ≈
= = =
78
( ) ( )22 4.36939 50 954.5799
0.9545799 Input Power Factor
; RMS secondary supply voltage = 230V.
RMS secondary supply current = RMS load
Output L
c
oad Po
ur
w
ren .
er
t
O LO RMS
O
OS
S S
S
S O R
O
P I R Watts
P KW
PPF V
V II
I
P
I
= × = × =
=
= =×
=∴ = ( )
( )
( ) ( )
( ) [ ]
2
4.36939
954.5799 W 0.9498230 4.36939 W
1 sin .2
We have obtained the expression for the average / DC output volt
Average Output (Load) Volt
age as,
cos
age
12
MS
mO dc
mO dc
Amps
PF
V V t d t
VV
=
∴ = =×
⎡ ⎤= ⎢ ⎥
⎣ ⎦
= −
∫π
α
ω ωπ
απ
79
( ) ( ) [ ]
( ) [ ]
( )( )
02 230 325.2691193cos 60 1 0.5 12 2
325.2691193 0.5 25.88409 Volts2
Average DC Load Current
25.884094 0.51768 Amps50
Average & RMS Thyristor CurrentsReferring to the thyristor
O dc
O dc
O dcO dc
L
V
V
VI
R
× ⎡ ⎤= − = −⎣ ⎦
= − = −
−= = = −
π π
π
current waveform of a single phase half-wave ac voltage controller circuit, we can calculate the average thyristor current as
Im
iT1
π 2πα (2 + )π α
3π
α αωt
80
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) [ ]
1 sin .2
sin .2
cos2
cos cos2
1 cos2
Where, = Peak thyristor current = Peak load current.
2 230 6.505382 A50
mT Avg
mT Avg
mT Avg
mT Avg
mT Avg
mm
L
m
I I t d t
II t d t
II t
II
II
VI
R
I
⎡ ⎤= ⎢ ⎥
⎣ ⎦⎡ ⎤
= ⎢ ⎥⎣ ⎦⎡ ⎤
= −⎢ ⎥⎣ ⎦
= − +⎡ ⎤⎣ ⎦
= +
=
×= =
∫
∫
π
α
π
α
π
α
ω ωπ
ω ωπ
ωπ
π απ
απ
( ) [ ] ( )
( ) [ ]
0
mps
2 2301 cos 1 cos 602 2 50
2 230 1 0.5 1.5530 Amps100
mT Avg
L
T Avg
VI
R
I
× ⎡ ⎤= + = +⎣ ⎦×
×= + =
απ π
π
81
( )
( ) ( )
( )( ) ( )
( ) ( ) ( )
( ) ( )
2 2
2
2
RMS thyristor current can be calculated by
1 sin .2
1 cos 2.
2 2
cos 2 .4
1 sin 24 2
using the expression
mT RMS
mT RMS
mT RMS
mT RM
T RM
S
S
I I t d t
tII d t
II d t t d t
tI
I
I t
⎡ ⎤= ⎢ ⎥
⎣ ⎦
−⎡ ⎤= ⎢ ⎥
⎣ ⎦
⎡ ⎤= −⎢ ⎥
⎣ ⎦
⎛ ⎞= − ⎜ ⎟⎝ ⎠
∫
∫
∫ ∫
π
α
π
α
π π
α α
π
α
ω ωπ
ωω
π
ω ω ωπ
ωωπ
( ) ( )
( ) ( )
( ) ( )
1 sin 2 sin 24 2
1 sin 24 2
1 sin 22 22
mT RMS
mT RMS
mT RMS
I I
I I
II
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ − ⎤⎧ ⎫= − − ⎨ ⎬⎢ ⎥⎩ ⎭⎣ ⎦
⎡ ⎤= − +⎢ ⎥⎣ ⎦
⎡ ⎤= − +⎢ ⎥⎣ ⎦
π
α
π απ απ
απ απ
απ απ
82
( )( )
( )
( )
( )
0sin 1206.50538 12 3 22
1 2 0.86602544.62 3 2
4.6 0.6342 2.91746
2.91746 Amps
T RMS
T RMS
T RMS
T RMS
I
I
I A
I
⎡ ⎤⎛ ⎞⎢ ⎥= − +⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
⎡ ⎤⎛ ⎞= +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦= × =
=
πππ
ππ
A single phase half wave ac voltage controller has resistiveload of R = 10 Ω and input voltage is Vs = 120 V, 60 Hz. Thedelay angle of thyristor T1 is α = π/2. Determine (a) the rmsvalue of output voltage Vo , (b) the input PF and (c) theaverage input current.Solution: R = 10 Ω, Vs = 120 V, α = π/2 and Vm = √2*120 =169.7 V(a)
= 120 (3/4)1/2 = 103.92 V(b) The rms load current Io = Vo / R = 103.92/10 = 10.392 AThe load power Po = Io
2 R = 10.3922 * 10 = 1079.94 WBecause the input current is the same as the load current, theinput VA rating is VA = Vs Is = Vs Io = 120 * 10.392 =1247.04 VA
83
1 / 2
0 s1 sin 2V V 2
2 2⎡ α ⎤⎛ ⎞= π −α +⎜ ⎟⎢ ⎥π ⎝ ⎠⎣ ⎦
Problem
Solution
Solution : The input PF
PF = Po/VA = Vo/Vs == √(3/4) = 1079.94/1247.04 = 0.866 (lagging)(c) The average output voltage
Vdc = -120 *√2/2π = -27V
And the input current Id = Vdc/R = -27/10 = -2.7V
84
1 / 21 sin 22
2 2⎡ α ⎤⎛ ⎞π −α +⎜ ⎟⎢ ⎥π ⎝ ⎠⎣ ⎦
( )1/ 2
sdc
2VV cos 1
2⎡ ⎤
= α−⎢ ⎥π⎢ ⎥⎣ ⎦
Single phase bidirectional controller with resistive load
85
1
2
o
o
m
Triggering circuit
circuitwaveforms
86
v- input voltagevg- gate voltagevL – load voltagevT – thyristor voltage
Wave forms of 1-phase voltage controller with resistive loads
87
( )
( )
Input supply voltage
sin 2 sin
sinfor to to 2
sinsin
to
EQUATIONS
Output voltage across the load resistor
Output load cur
o
rent
t
S m S
O L m
O mO m
L
L L
v V t V t
v v V tt and t
v V ti I t
R Rfor t an
R
d t
= =
= =
= = +
= = =
= = +
ω ω
ωω α π ω π α π
ωω
ω α π ω π α 2π
88
( )
( ) ( ) ( ) ( ) ( )
( ) ( )
22 22
22 2 2 2
TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF OUTPUT (LOAD) VOLTAGE
sin ; to to 2
1Hence, sin sin2
1 sin . sin .2
L O m
m mL RMS
m m
v v V t for t and t
V V t d t V t d t
V t d t V t d t
π π
α π α
π π
α π α
ω ω α π ω π α π
ω ω ω ωπ
ω ω ω ωπ
+
+
= = = = +
⎡ ⎤= +⎢ ⎥
⎣ ⎦
= +
∫ ∫
∫
( ) ( )
( ) ( ) ( ) ( )
22
2 22
1 cos 2 1 cos 22 2 2
cos 2 . cos 2 .2 2
m
m
V t td t d t
V d t t d t d t t d t
π π
α π α
π π π π
α α π α π α
ω ωω ωπ
ω ω ω ω ω ωπ
+
+ +
⎡ ⎤⎢ ⎥⎣ ⎦⎡ ⎤− −
= +⎢ ⎥⎣ ⎦⎡ ⎤
= − + −⎢ ⎥× ⎣ ⎦
∫
∫ ∫
∫ ∫ ∫ ∫
89
( ) ( )
( ) ( ) ( ) ( )( )
( ) ( ) ( )( )
( ) ( )
( )
2 22
2
2
2
2
sin 2 sin 24 2 2
1 1sin 2 sin 2 sin 4 sin 24 2 2
1 12 0 sin 2 0 sin 24 2 2
sin 2sin 224 2 2
sin 2sin 224 2
m
m
m
m
m
V t tt t
V
V
V
V
+ +
⎡ ⎤⎡ ⎤ ⎡ ⎤= + − −⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦
⎡ ⎤= − + − − − − − +⎢ ⎥⎣ ⎦
⎡ ⎤= − − − − − +⎢ ⎥⎣ ⎦+⎡ ⎤
= − + +⎢ ⎥⎣ ⎦
= − + +
π π π π
α π α α π α
ω ωω ωπ
π α π α π α π π απ
π α α π απ
π ααπ απ
απ απ
( )22+⎡ ⎤
⎢ ⎥⎣ ⎦
π α
90
( ) ( ) ( )
( ) ( )
( )
( ) ( )
22
22
2
22
sin 2 12 sin 2 .cos 2 cos 2 .sin 24 2 2
sin 2 0 & cos 2 1
sin 2 sin 2Therefore, 24 2 2
2 sin 24
2 2 sin 24
Taking the square root, we get
mL RMS
mL RMS
m
mL RMS
L R
VV
VV
V
VV
V
⎡ ⎤= − + + +⎢ ⎥⎣ ⎦= =
⎡ ⎤= − + +⎢ ⎥⎣ ⎦
= − +⎡ ⎤⎣ ⎦
= − +⎡ ⎤⎣ ⎦
απ α π α π απ
π π
α απ απ
π α απ
π α απ
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
2 2 sin 22
1 sin 22 2 sin 2 22 22 2 2
1 sin 2 1 sin 22 22
1 sin 22
mMS
m mL RMS
mL RMS i RMS
SL RMS
V
V VV
VV V
V V
= − +⎡ ⎤⎣ ⎦
⎡ ⎤⎧ ⎫= − + = − +⎡ ⎤ ⎨ ⎬⎢ ⎥⎣ ⎦ ⎩ ⎭⎣ ⎦
⎡ ⎤ ⎡ ⎤= − + = − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
⎡ ⎤= − +⎢ ⎥⎣ ⎦
π α απ
απ α α π αππ
α απ α π απ π
απ απ
91
Maximum RMS voltage will be applied to the load when , in that case the full sine wave appears across the load. RMS load voltage will be the same as the RMS
supply voltage .When is increased the R2mV
= α
( ) ( )
( ) ( )
( ) ( )
0
0
0
MS load voltage decreases.
1 sin 2 0022
1 022
2
The output control characteristic for a single phase full wave ac voltage controller with resistiv
mL RMS
mL RMS
mSL RMS i RMS
VV
VV
VV V V
=
=
=
×⎡ ⎤= − +⎢ ⎥⎣ ⎦
⎡ ⎤= +⎢ ⎥⎣ ⎦
= = =
α
α
α
ππ
ππ
( )e load can be obtained by plotting the equation for O RMSV
92
CONTROL CHARACTERISTIC OF SINGLE PHASE FULL-WAVE AC VOLTAGE CONTROLLER WITH RESISTIVE LOAD
The control characteristic is the plot of RMS output voltage versus the trigger angle ;which can be obtained by using the expression for the RMS output voltage of a full-wave ac controller with resi
α
( ) ( )
stive load.
1 sin 22
Where RMS value of input supply voltage2
SO RMS
mS
V V
VV
⎡ ⎤= − +⎢ ⎥⎣ ⎦
= =
απ απ
93
20
0
, 2
,
2
1 0
1
1 ( )2 2
1 2 sin( )20
(sup )1 ( )cos( )
1 ( )s
L
L
L
av t d t
V t d t
For the fundament
The Fourier cofficients of theload voltage wavefor
al ply frequency components it is seen
m ar
tha
eobtained
t
a v t t d
t
s
t
a
b v
π
π π
α π+α
π
= ω ωπ
= ω ωπ
=−
= ω ω ωπ
= ωπ
∫
∫
∫
[ ]
2
0
, 2
1 ,
in( )
1 2 cos( )sin( )
2 cos 2 12
t d t
In the present case
a V t t d t
V
π
π π
α π+α
ω ω
= ω ω ωπ
= α−π
∫
∫
94
[ ]
, 2 21 ,
1 1
1
1
2 21 1 1
1 11
1
2 21 1 1
1 2 sin ( )
2 sin 2 2( )2
tan
π π
α π+α
−
= ω ωπ
= α + π − απ
Ψ
= +
Ψ =
= +
∫b V t d t
V
C offiecien ts a and b are com bined tog ive the peak am plitude c and phaseang le o f the fundam en ta l com ponen t
o f load vo ltage
c a bab
c a b
95
[ ] ( )
[ ]( )
221
11
2
1 0
2
1 0
2 cos 2 1 sin 2 22
cos 2 1tan
sin 2 2
exp1 ( ) cos( )
1 ( ) sin ( )
( )
−
π
π
= α − + α + π − α⎡ ⎤⎣ ⎦π⎡ ⎤α −
Ψ = ⎢ ⎥α + π − α⎡ ⎤⎢ ⎥⎣ ⎦⎣ ⎦
= ω ω ωπ
= ω ω ωπ
ω
∫
∫
th
L
L
L
Vc
F or n harm on ic com ponen t the genera lfourier ressions are
a v t n t d t
b v t n t d t
substitu tion o f v t in the a
( )
( )
1
1
1 1 1 cos( 1) 12 1
12 1 1 cos( 1) 11
+
−
⎡ ⎤+ − + α −⎢ ⎥+= ⎢ ⎥π ⎢ ⎥− + − − α −⎢ ⎥−⎣ ⎦
n
nn
bove equa tions
nV na
nn
96
( ) ( )
1 1
3,5,7.......
2 2
1
2 sin( 1) sin( 1)1 1 1 12 1 1
2 2 2cos( 1) 1 cos( 1) 12 1 1
2 2 2sin( 1) sin( 1)2 1 1
tan
0( :
+ −
−
+ α − α⎡ ⎤= + − − + −⎢ ⎥π + −⎣ ⎦=
⎡ ⎤= = + α− − − α−⎢ ⎥π + −⎣ ⎦
⎡ ⎤= + α− − α⎢ ⎥π + −⎣ ⎦
= +
Ψ =
= =
n nn
n
n
n n n
nn
n
V n nbn n
For n odd
Va n nn n
Vb n nn n
c a bab
For n evenand n i eth ) ,.
n nedccomponent coefficients a b are zeroThe fourier spectrumof theload voltagecontainsonly odd harmonics
97
1 1
1 1
2 21 1 1
1 11
1
2 21
1
Coefficients a and b are combined to give the peak amplitude c and phase angle of the fundamental component of load voltage as follows
catanb
2c (cos 2 1) (sin 2 2( ))2
cos 2
a b
V
ψ
ψ
α α π απ
αψ
−
= +
=
= − + + −
−=
1sin 2 2( )α π α+ −
98
Harmonic content of load voltage/current with resistive load
99
Variation of rms, fundamental and average load voltages with resistive loads
100
A single-phase ac voltage controller has a resistive load of 10 ohms. The input voltage is 230V rms at
50Hz.The delay angle of thyristors is α=100°. Calculate (a) rms load voltage, (b) power consumed,
(c) displacement factor (DPF), (d) distortion factor (DF), (e) total harmonic distortion of ac source
current (THDI), (f) power factor (PF), (g) crest factor of ac source current (CF), and (h) ac source rms
current (Is).
Given that, supply rms voltage, Vs = 230 V, Frequency of the supply f=50 Hz, R = 10 Ω, α = 1000.
In a single-phase, phase controlled ac controller, the waveform of the supply current (Is) has a value of
vs/R from angle α to π.
AC mains RMS current, Is= Vs [1/(π)( π-α)+sin2α/2]½/R=14.363 A
Fundamental RMS current Is1=Vs/(2π R)[ (cos2α-1)2+sin2α+2( π-α)2] ½=11.44 A
θ1=tan-1[(cos2α-1)/ sin2α+2( π-α)]=38.3656°
Fundamental active power drawn by the load, P1=VsIs1cos θ1=2062.957 W
Fundamental reactive power drawn by the load, Q1= VsIs1sin θ1=1622.133 VAR
(a) RMS load voltage Vlrms=Vsm[1/(2π)( π-α)+sin2α/2]½=143.63V
(b) Active power drawn by the load, P1=VsIs1cos θ1=2062.957W
(c) Displacement factor, DPF=cos θ1 = 0.784
(d) Distortion Factor, DF= Is1/Is=0.79649
101
Total harmonic distortion of ac source current (THDI)= √(1/DF2)-1=75.9145%
(f) Power factor (PF)=DPF*DF=0.62445
Peak supply current, Ipeak= √ 2*Vssinα/R=32.03A
(g) Crest factor of the supply current, CF=Ipeak/Is=2.23
(h) AC mains RMS current, Is=Vs[1/(π)( π-α)+sin2α/2]½/R=14.363 A
102
A single-phase ac voltage controller is used to control the heating load of a
maximum power of 2 kW fed from single-phase ac mains of 230 V, 50 Hz. Its
power is to be controlled to deliver 1 kW. Calculate (a) load resistance, (b) rms
voltage across the load, (c) supply rms current, (d) supply power factor.
Solution
Given that, Supply rms voltage, Vs = 230 V, Frequency of the supply f=50 Hz,
Pmax=2 kW, P=1kW.
The load resistance, R=Vs2/Pmax=26.45 Ω
The rms voltage across the load, Vls=IsR=162.6346V
The supply rms current, Is=√(P/R)=6.14875A
The supply power factor, PF=P/(VsIs)=0.70711.
References
W.Shepard, “Thyristor control of AC circuits”, Bradford University Press,
1975, ISBN: 0 258 96953 9.
W.Shepard, L. N. Hully and D. T. W. Liang, “Power Electronics and Motor
Control”, Second Edition, Cambridge University Press, 1995, ISBN:
0521472415 0521478138.
M. H. Rashid, “Power Electronics, circuits, Devices and Applications”,
Second Edition, Prentice-Hall, 1995, India, ISBN 81-203-0869-7.
W. Shepherd and L. N. Hully, “Power Electronics and Motor Control”,
Cambridge University Press, 1987, Cambridge, ISBN 0-521-32155-7.
References
P. S. Bhimbra, “Power Electronics”, Third Edition, Khanna
Publishers, 1999, New Delhi, ISBN 81-7409-056-8.
N. G. Hingorani and L. Gyugyi, “Understanding FACTS”,
IEEE Press, Delhi, 2001, ISBN 81-86308-79-2.
Chingchi Chen and Deepakraj M. Divan, “Simple
Topologies for Single Phase AC Line Conditioning” IEEE
transactions on industry applications, vol. 30, no. 2,
march/april 1994