L2. Single Phase Ac Voltage Controllers

1

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

2

Lecture - 2

3

1 – phase AC Voltage controller

4

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.

6

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.

7

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.

8

Soft start of single phase induction motors

9

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;

10

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)

11

Applications 1- phase ac controller

Woodhog (1.2 kW saving on 16 kW motor)

12

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)

13

Applications single phase ac controller

Drill presses

14


Cuttoff saw

15


CNC flat bed lathes

16

Radial drilling machine

Machine tools

17

Industrial sewing machine

18

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;

19

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;

20

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

21

Speed control of 1- phase ac motors usingtriac based controller


Fan regulator

22


23


Blowers

24


Twist frames

25


Light vacuum system

26


Pumps

27


Crane drives

28


Cutting machines

29


Forming machines

30


Grinders

31


Abrasive planer

32


Plastic extruder

33

Lamp dimmer


34


Thyristor controlled reactor

35


Thyristor switched capacitors

36


Heating control

37


Heating chamber for drying

38

Electric boiler

39

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).

40

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

41

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

42

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

43

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.

44

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.

45

1

1

o

om

On – off control

Circuit waveforms

46

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

47

( ) ( )

( ) ( )

= 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

48

On – off control

Power factor

TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF OUTPUT VOLTAGE, FOR ON-OFF CONTROL METHOD.

49

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( )

( )

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 ⎣ ⎦

50

( ) ( )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

51


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.

52


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.

53


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.

54

55

Schematic drive for rectified, integral-cycle control of fractional horse power series dc motor

56

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

57

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


( ) ( )

( )

( )( ) ( )

( )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

= = =+ +

=+

= = =

= ×

59


( )

( ) ( )( )

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

Documents

L2. Single Phase Ac Voltage Controllers