L2. Single Phase Ac Voltage Controllers

AC CONTROLLERS

(EEL 744)

Department of Electrical Engineering,

Indian Institute of Technology Delhi,

Hauz Khas, New Delhi-10016, India- 110016

email: [email protected], [email protected]

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


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


Drill presses

13


Cuttoff saw

14

CNC flat bed lathes

15

Radial drilling machine

Machine tools


Industrial sewing machine

16



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


20

Speed control of 1- phase ac motors using TRIAC based

controller

Applications single phase AC controller

Fan regulator

21


22

Blowers


23

Twist frames


24

Light vacuum system


25

Pumps


26

Crane drives


27

Cutting machines


28

Forming machines


29

Grinders


30

Abrasive planer


31

Plastic extruder


32

Lamp dimmer


33

Thyristor controlled reactor


34

Thyristor switched capacitors


35

Heating control


36

Heating chamber for drying

37

Electric boiler


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


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


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


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


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


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

L2. Single Phase Ac Voltage Controllers

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