Post on 20-Jul-2016
description
AC CONTROLLERS
(EEL 744)
Department of Electrical Engineering,
Indian Institute of Technology Delhi,
Hauz Khas, New Delhi-10016, India- 110016
email: bsingh@ee.iitd.ac.in, bhimsingh1956@gmail.com
Ph.:011-2659-1045
Prof. Bhim Singh
LECTURE - 2
2
Single – Phase AC Voltage Controller
Single–phase AC Voltage controller
3
The power flow can be controlled by varying the rms value
of AC voltage applied to the load, a thyristor switch is
connected between the AC source and load.
This type of power circuit is known as ac voltage controller
Single
Phase
Supply
Load
AC Voltage Controller
Converts fixed AC voltage to variable ACvoltage without change in frequency.
Power flow can be controlled by varying therms voltage applied to the load.
High efficiency, flexible Control, lessmaintenance and compact size.
5
AC voltage controllers can be divided in two
broad categories
Single phase;
Three phase.
These controllers can be used for an induction motor as:
Soft starters;
Energy saving controllers;
Solid state speed controllers.
Salient features as soft starter
6
Step less control of motor voltage.
Control flexibility due to low power control
circuitry.
Smooth acceleration and deceleration of the motor.
Easy implementation of current control.
Simple protection against single phasing or
unbalanced operation in case of three phase motors.
Absence of current inrush.
Low maintenance for applications requiring
frequent starting and stopping.
7
Soft start of Single Phase Induction Motors
AC Voltage Controllers
8
Salient features as energy saver:
In such applications, voltage control is used for reduction of
losses not for speed control;
The motor losses primarily depend on three factors:
o Loading on the motor;
o Magnitude of applied voltage; and
o 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 more
energy savings
AC Voltage Controllers
9
Following applications have significant no load operationand so voltage control can serve as energy 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
Applications of Single- phase AC controller
Gang ripsaws
10
11
Wood hogs: Designed to provide a one-step operation reducing byimpact rather than by cutting. No knives are used. Hammers aredesigned to produce maximum reduction of fibrous materials andare built to withstand the impact of such foreign materials as nails,cleats etc.
Applications of Single- phase AC controller
Applications single phase ac controller
Air compressors
12
Applications single phase ac controller
Drill presses
13
Applications single phase ac controller
Cuttoff saw
14
CNC flat bed lathes
15
Radial drilling machine
Machine tools
Applications single phase ac controller
Industrial sewing machine
16
Applications single phase ac controller
AC Voltage Controllers
17
The applied voltage is directly related with core losses;
therefore, optimum voltage shall reduce the losses:
the motor operating near the distribution substation will
have higher voltage than the one at the end of the
distribution line.
Therefore, the voltage reduction will allow energy savings.
A badly designed motor or a rewound motor with uneven air
gap will draw more magnetizing current and will have higher
core losses
Reduced voltage operation of such motor will certainly
improve energy utilization at all loads
18
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
19
Following applications use AC voltage controller
as speed controller:
Grinders
Plastic extruders
Abrasive planer
Lamp dimmer
Thyristor controlled reactor
Thyristor switched capacitor
Heating control
Heating chamber
Electric Boiler
Applications single phase ac controller
20
Speed control of 1- phase ac motors using TRIAC based
controller
Applications single phase AC controller
Fan regulator
21
Applications single phase ac controller
22
Blowers
Applications single phase ac controller
23
Twist frames
Applications single phase ac controller
24
Light vacuum system
Applications single phase ac controller
25
Pumps
Applications single phase ac controller
26
Crane drives
Applications single phase ac controller
27
Cutting machines
Applications single phase ac controller
28
Forming machines
Applications single phase ac controller
29
Grinders
Applications single phase ac controller
30
Abrasive planer
Applications single phase ac controller
31
Plastic extruder
Applications single phase ac controller
32
Lamp dimmer
Applications single phase ac controller
33
Thyristor controlled reactor
Applications single phase ac controller
34
Thyristor switched capacitors
Applications single phase ac controller
35
Heating control
Applications single phase ac controller
36
Heating chamber for drying
37
Electric boiler
Applications single phase ac controller
For power transfer two types of control are used:
On – off control
Phase angle control
On – off control:- Thyristor switches connect load to acsource for few cycles of input voltage and then disconnectfor few cycles. This type of control is also called as burstfiring, zero voltage switching, cycle selection or integral cycleswitching.
The Thyristors thus act as a high speed contactor (or highspeed ac switch).
38
Phase angle control:- Thyristor switches connect the load to theac source for a portion of each cycle of input voltage or if eachthyristor is triggered at some non-zero point on its respectiveanode voltage cycle, the load voltage waveform is described as‘phase angle controlled’
By controlling the phase angle or the trigger angle ‘ ’ (delayangle), the output RMS voltage across the load can be controlled.
The trigger delay angle ‘ ’ is defined as the phase angle (thevalue of t) at which the thyristor turns on and the load currentbegins to flow.
The load voltage and current have identical positive and negativealterations with frequency spectra containing only oddharmonics
39
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 to meetthe 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 orcomponents and the circuits for ac voltage controllers arevery simple.
40
Phase angle control
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
41
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.
42
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.
43
44
T1
T1
vo
+
-v=Vmsinωt
io
On – off control
Circuit waveforms
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
45
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 t
VV
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
46
= 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
f
t n T
t m T
T t t nT mT
t tV V V
T T
here is the RMS input supply voltage = Si RMSV V
On – off control
47
On – off control
Power factor
EXPRESSION FOR THE RMS VALUE OF OUTPUT VOLTAGE, FOR ON-OFF CONTROL METHOD.
48
2 2
0
22
0
2
2
0
2
0 0
1Output RMS voltage . .
.
1 2Substituting for
2
1 2
2
2 .2
ON
ON
ON
ON ON
t
mO RMS
O t
t
m
O RMS
O
t
m
O RMS
O
t t
m
O RMS
O
O RMS
V V Sin t d tT
VV Sin t d t
T
CosSin
V Cos tV d t
T
VV d t Cos t d t
T
V2
0 0
2
22
ON ONt t
m
O
V Sin tt
T
49
2 sin 2 sin 00
2 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
T
t
t T T T T T t
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 ON
O 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 T
VV V
t t nT nk
T t t nT mT n m).
S SO RMS
nV V V k
m n
Applications of on – off control or integral cycle control
Incandescent lighting control: An irritating flicker is noticedeven when only supply cycle is omitted from each controlperiod 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
50
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 was foundbetween 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 this form ofapplication.
51
Applications of on – off control or integral cycle control
Control of ¼ HP universal motor : Control may be effected atno load with fixed control period of 10 supply cycles andvariable ON/OFF, N/T – N.
With N/T – N ≥ 1, the test motor ran smoothly and a smalldegree 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 conduction periods.
52
Applications of on – off control or integral cycle control
Speed control of a FHP, dc series motor: The speed of a dcseries 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 pulses withconstant control period to a 1/8 HP motor
Any integral number of consecutive conduction cycles up tofull conduction could be applied so that this constitutes a formof pulse width modulation.
53
54
Schematic drive for rectified, integral-cycle control of fractional horse power series dc motor
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/OFF ratiohas a minimum value in which the maximum permissible OFFtime is determined by the electrical energy storage capabilityof the motor.
When the energy recovered during OFF period is insufficient,motor current decays to zero.
By the nature of the controller , the voltage and input powerapplied to the motor are pulsating.
Speed pulsating were obtained by means of a tachometer T
55
56
Performance parameters of ac voltage controllers
122
2 2
0
RMS Output (Load) Voltage
sin .2
2
Where = RMS value of input supply voltage.
mO RMS
mSO RMS i RMS
SO RMS i RMS
S i RMS
nV V t d t
n m
V nV V k V k
m n
V V k V k
V V
57
2
Duty Cycle
Where, = duty cycle (d).
RMS Load Current
; for a resistive load .
Output AC (Load) Power
ON ON
O ON OFF
O RMS O RMS
LO RMS
L
O LO RMS
t t nTk
T t t m n T
nk
m n
V VI Z R
Z R
P I R
Performance parameters of ac voltage controllers
58
2
Input Power Factor
output load power
input supply volt amperes
; RMS input supply current.
The input supply current is same as the load current
He
O O
S S
LO RMS
S in RMS
i RMS in RMS
in O L
P PPF
VA 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 k
m n
Performance parameters of ac voltage controllers
59
The Average Current of Thyristor T Avg
I
0
0
sin .2
sin .2
mT Avg
mT Avg
nI I t d t
m n
nI I t d t
m n
60
,
0
cos2
cos cos02
1 12
22
.
duty cycle
.
Where = maximum or peak thyristor cu
m
T Avg
m
T Avg
m
T Avg
mT Avg
m m
T Avg
ON
ON OFF
m m
T Avg
mm
L
nII t
m n
nII
m n
nII
m n
nI I
m n
I n k II
m n
t nk
t t n m
I n k II
m n
VI
Rrrent.
61
,
12
2 2
0
122
2
0
122
0
122
0 0
2
RMS Current of Thyristor
sin .2
sin .2
1 cos 2
2 2
cos 2 .4
T RMS
mT RMS
m
T RMS
m
T RMS
m
T RMS
m
T RMS
I
nI I t d t
n m
nII t d t
n m
tnII d t
n m
nII d t t d t
n m
nII
12
0 0
122
sin 2
24
sin 2 sin 00
4 2
m
T RMS
tt
n m
nII
n m
62
The RMS Current of Thyristor
,
122
1 12 22 2
0 04
4 4
2 2
2
m
T RMS
m m
T RMS
m m
T RMS
m
T RMS
nII
n m
nI nII
n m n m
I InI k
m n
II k
63
Problem: A single phase full wave ac voltage controller working on ON-OFFcontrol technique has supply voltage of 230V, RMS 50Hz, load = 50 . Thecontroller is ON for 30 cycles and off for 40 cycles. Calculate ON & OFF timeintervals, RMS output voltage, Input P.F., Average and RMS thyristorcurrents.
m min RMSSolution: V =230V; V = 2×230V=325.269V; V =325.269V
1 1T= = =0.02sec:T=20ms
f 50Hz
n = number of input cycles during which controller is ON; n = 30.
m = number of input cycles during which controller is OFF
ON
ON
OFF
OFF
; m = 40
t =n×T=30×20ms=600ms=0.6sec
t =n×T=0.6sec= controller ON time.
t =m×T=40×20ms=800ms=0.8sec
t =m×T=0.8sec = controller OFF time.
n 30Dutycycle k= = =0.4285
( ) (40 30)m n
64
2 2
Solution:RMS output voltage
30 3230 230
30 40 7
230 0.42857 230 0.65465 150.570
150.5703.0114
50
3.0114 50 453.426498
Input Power Factor
O RMS i RMS
O RMS
O RMS
O RMS O RMS
O RMS
L
O LO RMS
nV V
m n
V V
V V V
V V VI A
Z R
P I R W
.
300.4285
70
0.654653
P F k
nPF
m n
PF
65
Solution: contd..........
Average Thyristor Current Rating
2 230 325.269where
50 50
6.505382 = Peak (maximum) thyristor current.
6.505382 3
7
RMS Current Rati
m m
T Avg
mm
L
m
T Avg
I k InI
m n
VI
R
I A
I
ng of Thyristor
6.505382 3
2 2 2 7
2.129386
m m
T RMS
T RMS
I InI k
m n
I A
66
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
2
s0
220 nSolution: m + n=48;R= =35.26Ω;V =V ;
1500 n + m
V nP = 500 = ; n =16
(m + n)R
67
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 .
2
-3
2
rms
f
n 2
3 5 7
240Solution: VAR rating =9.167kVAR
2π×50×20×10
V 1 3sin2αi) I = π-2α 0.5+sin α - = 11.23 A
ωL π 2
V 2α sin2αii) I = 1- - =10.71A
π π2ωL
V 4 sinαcosnα-ncosαsinnαiii) I =
ωL π n(n -1)
I =-5.25A iv) I =-1.05A v) I =0.375A
68
T1
D1 io
vo
+
-
v=Vmsinωt
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.
69
70
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 Voltage
0; 0
sin ; 2
Output Loa
s m
mS in RMS
o L
o L m
v V t
VV V
v v for t to
v v V t for t to
d Current
sin; 2
0; 0
o mo L
L L
o L
v V ti i for t to
R R
i i for t to
71
2
2 2
22
22
2 2
1sin .
2
1 cos2.
TO DERIVE AN EXPRESSION FOR R
2 2
MS
1 cos2 .4
OUTPUT VOLTAGE
os2
c .2
mO RMS
m
O RMS
m
O RMS
m
O 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
V2 2
2
sin 2
22
sin 22
22
sin 4 sin 22 ;sin 4 0
2 22
m
m
O RMS
m
O RMS
V tt
V tV
VV
72
sin 22
22
sin 22
22 2
sin 22
22 2
1 sin 22
2 22
1 sin 22
2 2
1 sin 22
2 2
Where, = RMS value of input supply 2
m
O RMS
m
O RMS
m
O RMS
m
O RMS
O RMS i RMS
SO RMS
mSi RMS
VV
VV
VV
VV
V V
V V
VV V voltage
(across the transformer secondary winding).
Note: Output RMS voltage across the load is controlled by
changing as indicated by the expression for O RMS
V
73
PLOT OF VERSUS TRIGGER ANGLE FOR A SINGLE PHASE HALF-WAVE
AC VOLTAGE CONTROLLER
1 sin 22
(UNIDIRECTIONAL CONTR
2 22
1 sin 22
OLLER)
2 2
m
O RM
O RMS
S
SO RMS
V
VV
V V
74
2
2
2
TO CALCULATE THE AVERAGE VALUE (DC VALUE) OF OUTPUT VOLTA
1sin .
2
sin .2
cos2
cos 2 cos :cos 2 12
cos 1 ; 22
2Hence co
GE
s 12
mO dc
m
O dc
m
O dc
m
O dc
mdc m S
Sdc
V V t d t
VV t d t
VV t
VV
VV V V
VV
whe ' 'is varied from 0 to . varies from 0 to mdc
Vn V
75
Problem: A single phase half-wave ac voltage controller has a loadresistance , input ac supply voltage is 230V RMS at 50Hz. The input supplytransformer has a turns ratio of 1:1. If the thyristor is triggered at .Calculate RMS output voltage, Output power. RMS load current andaverage load current, Input power factor and Average and RMS thyristorcurrent.
0
S
Solution:
Given,
230 , primary supply voltage.
Input supply frequency = 50Hz.
50
60 radians.3
V RMS secondary voltage.
11
1
Therefore 230
Where, = Number of turns in the pr
p
L
p p
S S
p S
p
V V RMS
f
R
V N
V N
V V V
N imary winding.
= Number of turns in the secondary winding.SN
76
2
2 2
0
1sin .
2
We have obtained the expression for
RMS Value of
as
1 sin 22
2 2
1 sin120230 2
2 3 2
1230 5.669 230 0.949
Output (Load) Volt
2
age
mO RMS
O RMS
SO
O RMS
RMS
O RMS
O RMS
V V t d t
V
V V
V
V
V
RMS Load Current
86
218.4696 218.47
218.469664.36939
50
O RMS
O RMS
O RM
O R
S
L
MSI
V V V
VI Amps
R
77
22
Output Load Power
Input Power Factor
4.36939 50 954.5799
0.9545799
; RMS secondary supply voltage = 230V.
RMS secondary supply current = RMS load current.
O LO RMS
O
OS
S S
S
S O R
O
P I R Watts
P KW
PPF V
V I
I
I
P
I
2
4.36939
954.5799 W0.9498
230 4.36939 W
1sin .
2
We have obtained the expression for the average / DC output volt
Average Output (Load) Volt
age as,
cos
age
12
MS
mO dc
m
O dc
Amps
PF
V V t d t
VV
78
02 230 325.2691193cos 60 1 0.5 1
2 2
325.26911930.5 25.88409 Volts
2
Average DC Load Current
25.8840940.51768 Amps
50
Average & RMS Thyristor Currents
Referring to the thyristor
O dc
O dc
O dc
O dc
L
V
V
VI
R
current waveform of a single phase half-wave
ac voltage controller circuit, we can calculate the average thyristor current as
Im
iT1
2
(2 + )
3
t
79
1sin .
2
sin .2
cos2
cos cos2
1 cos2
Where, = Peak thyristor current = Peak load current.
2 2306.505382 A
50
mT Avg
m
T Avg
m
T Avg
m
T Avg
m
T Avg
mm
L
m
I I t d t
II t d t
II t
II
II
VI
R
I
0
mps
2 2301 cos 1 cos 60
2 2 50
2 2301 0.5 1.5530 Amps
100
m
T Avg
L
T Avg
VI
R
I
80
2 2
2
2
1sin .
RMS thyristor current can be calculated by using the express
2
1 cos 2.
2 2
cos 2 .4
ion
1 sin 2
24
mT RMS
m
T RMS
m
T RMS
mT RM
RM
S
T S
I I t d t
tII d t
II d t t d t
tI
I
I t
1 sin 2 sin 2
4 2
1 sin 2
4 2
1 sin 2
2 22
mT RMS
mT RMS
m
T RMS
I I
I I
II
81
0sin 1206.50538 1
2 3 22
1 2 0.86602544.6
2 3 2
4.6 0.6342 2.91746
2.91746 Amps
T RMS
T RMS
T RMS
T RMS
I
I
I A
I
Problem: A single phase half wave ac voltage controller hasresistive load 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 rms valueof output voltage Vo , (b) the input PF and (c) the average inputcurrent.
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 A
The load power Po = Io2 R = 10.3922 * 10 = 1079.94 W
Because 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
82
1/2
0 s
1 sin 2V V 2
2 2
The input PF
PF = Po/VA = Vo/Vs =
= √(3/4) = 1079.94/1247.04 = 0.866 (lagging)
The average output voltage
Vdc = -120 *√2/2π = -27V
And the input current
Id = Vdc/R = -27/10 = -2.7V
83
1/2
1 sin 22
2 2
1/2
sdc
2VV cos 1
2
Single phase bidirectional controller with resistive load
84
T1
T2
io
vo
+
v=Vmsinωt -
Triggering circuit
Control signal
circuit waveforms
v- input voltagevg- gate voltagevL – load voltagevT – thyristor voltage
85
Wave forms of 1-phase voltage controller with resistive loads
86
Input supply voltage
sin 2 sin
Output voltage across the load resistor
sin
for to to 2
Output load curren
EQUA
t
sinsin
to
TIONS
to
S m S
L
O L m
O mO m
L L
v V t V t
R
v v V t
t and t
v V ti I t
R R
for t and t 2
87
22 22
2
2 2 2 2
TO DERIVE AN EXPRESSION FOR THE
RMS VALUE OF OUTPUT (LOAD) VOLTAGE
sin ; to to 2
1Hence, sin sin
2
1sin . 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
22
2 22
1 cos 2 1 cos 2
2 2 2
cos 2 . cos 2 .2 2
m
m
V t td t d t
Vd t t d t d t t d t
88
2 22
2
2
2
2
sin 2 sin 2
4 2 2
1 1sin 2 sin 2 sin 4 sin 2
4 2 2
1 12 0 sin 2 0 sin 2
4 2 2
sin 2sin 22
4 2 2
sin 2sin 22
4 2
m
m
m
m
m
V t tt t
V
V
V
V 2
2
89
22
22
2
22
sin 2 12 sin 2 .cos 2 cos 2 .sin 2
4 2 2
sin 2 0 & cos 2 1
sin 2 sin 2Therefore, 2
4 2 2
2 sin 24
2 2 sin 24
Taking the square root, we get
m
L RMS
m
L RMS
m
m
L RMS
L R
VV
VV
V
VV
V 2 2 sin 22
1 sin 22 2 sin 2 2
2 22 2 2
1 sin 2 1 sin 2
2 22
1 sin 2
2
m
MS
m m
L RMS
m
L RMS i RMS
SL RMS
V
V VV
VV V
V V
90
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 R2
mV
0
0
0
MS load voltage decreases.
1 sin 2 00
22
1 0
22
2
The output control characteristic for a single phase full wave ac voltage
controller with resistiv
m
L RMS
m
L RMS
mSL RMS i RMS
VV
VV
VV V V
e load can be obtained by plotting the equation for O RMS
V
91
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 resistive load.
1 sin 2
2
Where RMS value of input supply voltage2
SO RMS
mS
V V
VV
92
20
0
,2
,
2
10
1
1( )
2 2
12 sin( )
2
0
(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
mar
tha
eobtained
t
a v t t d
t
s
t
a
b v2
0
,2
1,
in( )
12 cos( )sin( )
2cos 2 1
2
t d t
In the present case
a V t t d t
V
93
,22
1,
1 1
1
1
2 2
1 1 1
1 11
1
2 2
1 1 1
12 sin ( )
2sin 2 2( )
2
tan
b V t d t
V
Coffiecients a and b arecombined to
givethe peak amplitudec and phase
angle of the fundamental component
of load voltage
c a b
a
b
c a b
94
22
1
1
1
2
10
2
10
2cos2 1 sin 2 2
2
cos2 1tan
sin 2 2
exp
1( )cos( )
1( )sin( )
( )
th
L
L
L
Vc
For n harmoniccomponent the general
fourier ressionsare
a v t n t d t
b v t n t d t
substitutionof v t inthea
1
1
11 1 cos( 1) 1
2 1
121 1 cos( 1) 1
1
n
nn
boveequations
nV n
a
nn
95
1 1
3,5,7.......
2 2
1
2 sin( 1) sin( 1)1 1 1 1
2 1 1
2 2 2cos( 1) 1 cos( 1) 1
2 1 1
2 2 2sin( 1) sin( 1)
2 1 1
tan
0( :
n n
n
n
n
n n n
nn
n
V n nb
n n
For n odd
Va n n
n n
Vb n n
n n
c a b
a
b
For n evenand n i eth ) ,
.
n nedccomponent coefficients a b are zero
The fourier spectrumof theload voltagecontainsonlyodd harmonics
96
1 1
1 1
2 2
1 1 1
1 11
1
2 2
1
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
c
atan
b
2c (cos 2 1) (sin 2 2( ))
2
cos 2
a b
V
1
sin 2 2( )
97
Harmonic content of load voltage/current with resistive load
98
Variation of rms, fundamental and average load voltages with resistive loads
99
Problem: 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 =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).
Solution: Given that, supply rms voltage, Vs = 230 V, Frequency of the supply f=50 Hz, R
= 10 Ω, = 1000.
In a single-phase, phase controlled ac controller, the waveform of the supply current (Is)
has a value of vs/R from angle to π.
AC mains RMS current, Is= Vs [{1/(π)}{( π- )+sin2 /2}]½/R=14.363 A
Fundamental RMS current Is1=Vs/(2π R)[ (cos2 -1)2+{sin2 +2( π- )}2] ½=11.44 A
θ1=tan-1[(cos2 -1)/ {sin2 +2( π- )}]=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- )+sin2 /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
100
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*Vs sin /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- )+sin2 /2}]½/R=14.363 A
Io
πVs
2π
π
2π
T1
T2
Io
vo
+
Vs = 230V
-
α
π+α
Is
101
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.
Io
π
Vs
2π
π
2π
Is
T1
T2
io
vo
+
Vs = 230V
-
α
π+α
Io
Is
Problem: 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.
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.
N. Mohan, T. M. Undeland and W. P. Robbins, “Power Electronics: Converters, Applications and Design”,
John Wiley and Sons Inc, USA, 1995.
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.
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