L3. Single Phase Ac Voltage Controllers

8/12/2019 L3. Single Phase Ac Voltage Controllers

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AC CONTROLLERS

(EEL 744)

Department of Electrical Engineering,Indian Institute of Technology, Delhi,

Hauz Khas, New Delhi-10016, India- 110016email: [email protected], [email protected]

Ph.:011-2659-1045

Prof. Bhim Singh


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Lecture - 3

2

Single phaseAC Voltage controller


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3

Single phase bidirectional controller with series resistive-

inductance load

T1

T2

iL

vL

+

v=V msin t-

Triggering circuit

Control signal

R L

L

vR

vl


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4

Waveforms for thyristor control of series R-L 1 = 90, = 60

The thyristors are triggered at atriggering angle < = (sinusoidal

phase angle) of the load impedance.

The use of single , short triggeringangle < could cause only onethyristor to conduct because of continuation of conduction after theend of the voltages half cyclesensures that gating of the reversethyristor would have no effect. Thethyristor pair then act as a rectifier.


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5

i1

i2

vs vo

ioT2

T1

Single phase bidirectional controller with RL load


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6

To derive an expression for rms output voltage of a single phase full-wave ac voltage controller with RL load.

12

2 2

1

2

When the load current and load voltage waveforms become discontinuous as

shown in the figure above.

1 sin .

Output sin for to , when is ON.

1 cos 2

2

mO RMS

o m

mO RMS

O

V V t d t

v V t t T

t V V d t

12


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7

122

122

12 2

12

cos 2 .2

sin222

sin 2 sin 22 2 2

1 sin 2 sin 22 2 2

1 sin

2

mO RMS

mO RMS

mO RMS

mO RMS

mO RMS

V V d t t d t

V t V t

V V

V V

V V

12

1

2 sin 2

2 2The RMS output voltage across the load can be varied by changing the trigger angle

For a purely resistive load 0, therefore load power factor angle 0

tan 0;Exti

L

L R

0nction angle radians 180


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9

1

Thyristor Conduction Angle

Maximum thyristor conduction angle radians = 1800 for

RMS Output Voltage

1 sin 2 sin 22 22

The Average Thyristor Current

12

mO RMS

T T Avg

V V

I i d t

1sin sin

2

sin . sin2

Maximum value of occur at 0.The thyristors should be rated for maximum

Rt

m LT Avg

Rt

m LT Avg

T Avg

mT Avg

V I t e d t

Z

V I t d t e d t Z

I

I I , where mm

V I

Z


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10

12

RMS Thyristor Current

12

Maximum value of occurs at 0

Thyristors should be rated for maximum2

When a Triac is used in a single phase full wave ac voltage

T RMS

T T RMS

T RMS

mT RMS

I

I i d t

I

I I

controller with RL type of load, then 0

and maximum2

T Avg

mT RMS

I I

I


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11

Waveforms of single phase bidirectional ac voltage controllerwith RL load


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12

When the triggering angle is greater than the load phase angle the current occurs in discontinuous, non sinusoidal pulses .

The waveforms shown in figurefor a load of phase angle = 60 or

power factor 0.5 lagging andtriggering angle = 120If > the onset of the nonsinusoidal load current pulses i Lalways coincides with triggering angle.

Conduction of current is found tocease prior to the end of thesinusoidal current cycle


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13

Theoretical components of load current waveform for series R L circuit1 = 60, = 120


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14

, ,2

0, ,

cot

cot

0

2cot

2sin( )

sin

sin

sin

sin( ) sin(

x x

L

xt

xt

t

Theload current isdescribed bytheequation

V i t

Z

e

e

e

Extinctionangle xisdefined bythetranscendentaleqn x cot ( )

1 2 2 2

) 0

tan ;

xe

L Z R L

R


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15

20

0

2

1 0

cot ( )1

1( ) 0

2 21

( )cos( )

cos(2 ) cos(2 ) sin (2 2 )2

cos( )4sin sin( )2cos( )

i L

i L

xi

The Fourier cofficientsof theload current

waveformareobtained as

ai t d t

a i t t d t

x xV

a x e Z


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16

21 0

cot ( )

1

2 21 1 1

1 11

1

1 ( )sin( )

sin(2 ) sin(2 ) cos (2 2 )2

sin( )4sin sin( )2 sin( )

tan

L

x

i

i i i

ii

i

b i t t d t

x xV

a x e Z

c a b

ab


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17 17

T1

T2

io

vo

+

-

v=V msin t

Performance parameters of a single phase full wave ACvoltage controller with inductive load


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18

, ,2

0, ,

cot

cot

02cot

2sin( )

sin

sin

sin

sin( ) sin(

x x

L

xt

xt

t

Theload current is described by the equation

V i t

Z

e

e

e

Extinction angle x is defined by the transcendental eqn

x cot ( )

1 2 2 2

0

) 0

tan ;

Substititing R = 0 in the equation gives us 90

xe L

Z R L R


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19

2 20

10 0

cot ( )1

1 1( ) 0 ; ( )cos( )2 2

cos(2 ) cos(2 ) sin (2 2 )2 cos( )

4sin sin( )2cos( )

i L i L

xi

The Fourier cofficientsof theload current

waveformareobtained as

a i t d t a i t t d t

x xV a x e Z

0

1

substituting 90 the above equation as R = 02

2( ) sin 2i L

inV

a X


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20

2

1 0

cot ( )1

2 21 1 1 1

1

1 11

1

1( )sin( )

sin(2 ) sin(2 ) cos (2 2 )2

sin( )4sin sin( )2

sin( )

0

is the peak value of fundamental current

tan

L

xi

i i i i

i

ii

i

b i t t d t

x xV

b x e Z

c a b a

ca

b


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Single phase bidirectional controllers fordifferent loads thyristor based

21 21

I M

T1

T2

io

vo

+

-

v=V msin t

T1

T2

io

vo

+

-

v=V msin t

T1

T2

io

vo+

-

v=V msin t


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I M

Tio

vo

+

-

v=V msin ? t

Tio

vo

+

-

v=V msin ? t

T

io

vo

+

-

v=V msin ? t

T

io

vo

+

-

v=V msin ? t

22

Single phase bidirectional controllers fordifferent loads TRIAC based


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I M

23

To have the common cathode for thyristors T 1 and T 2 by addingtwo diodes as shown in figures for different loads.One isolation circuit is required, but at the expense of two power diodes


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Load

24

A single phase full wave controller can also be implemented withone thyristor and four diodes.


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25

Waveforms of single phase full wave controller withone thyristor and four diodes .


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Natural commutation occurs in single phase full wavecontroller with one thyristor and four diodes with resistiveloads .

If large inductance is there in the circuit , thyristor T 1 may

be not turned of in every half cycle of input voltage, and thismay result in a loss of control.

Three power devices conduct at the same time and the

efficiency is also reduced.

26

A single phase full wave controller with onethyristor and four diodes.


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27

Transformer tap changingWith resistive and inductive loads


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I M

28

Transformer tap changingWith RL and motor loads


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30

Waveforms for transformer connection changer


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31

Waveforms without tap changer


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Waveforms with synchronous changer


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Problem: A single phase full wave ac voltage controller hasresistive load of R = 10 and input voltage is V s = 120 V, 60 Hz. Thedelay angle of thyristor T 1 is 1 = 2= = /2. Determine (a) therms value of output voltage V o , (b) the input PF and (c) theaverage current of thyristors I A, and (d) the rms current of thyristors.Solution: R = 10 , Vs = 120 V, = /2 and V m = 2 *120 =169.7 V

(a)

= 120 /(2) 1/2 = 84.85V(b) The rms load current I o = Vo / R = 84.85/10 = 8.485 A

The load power P o = Io2

R = 8.4852

* 10 = 719.95 WBecause the input current is the same as the load current, theinput VA rating is VA = Vs Is = Vs Io = 120 * 8.485 = 1018.2 VA

33

1/2

0 s1 sin 2V V

2


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The input PFPF = Po/VA = Vo/Vs =

= 1/2 = 719.95/1018.2 = 0.707 (lagging)(c) The average thyristor current

IA = 2*120/(2 *10)= 2.7V(d) The rms value of the thyristor current

=120/(2*10) = 6A

34

1/21 sin 2

2

1/2

s

A

2V

I cos 12 R

1/2

sV 1 sin 222R

Solution: contd..........


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35

Problem : A single-phase ac voltage controller has a resistiveload. The input voltage is 230V rms at 50Hz.The delay angle of thyristors is =60 . Determine (a) distortion factor of supply

current (fundamental rms current/net rms current), (b)displacement factor and (c) input power factor.

orms s

orms

2 2

s1

s1

s

-1

230 , 50 , =60

sin 2V V 1- 206.29 V

2

217.48I

(cos 2 ) 1 2( - +sin2V 193.04 + = A

R 2 2 R

193.04I R

DF = = 0.936206.29I

(cos 2 ) 1 = tan

2( - +sin2

orms

s

V V Hz

V R R

I

R

16.54

DPF = cos = 0.959

PF = DPF X DF = 0.959X0.936 = 0.897

Solution:


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36

Problem : A single phase full wave ac voltage controller supplies an RL load. The input supply voltage is 230V, RMS at 50Hz. The load has L = 10mH, R = 10 , the delay angle of thyristors and is equal to 60 0 , where . Determine Conductionangle of the thyristor, RMS output voltage, The input power factor. Comment on the type of operation.

0

1 2

Solution:

Given 230 , 50 , 10 , 10 , 60

radians,3

2 2 230 325.2691193

s

m S

V V f Hz L mH R

V V V


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2 2 22

3

2 2

1

1


Load Impedance 10

2 2 50 10 10 3.14159

10 3.14159 109.8696 10.4818

2 23031.03179

10.4818

Load Impedance Angle tan

tan10

mm

Z R L L

L fL

Z

V I A

Z

L R

1 0

0

tan 0.314159 17.44059

Trigger Angle Hence the type of operation will

be discontinuous load current operation, we get

; 180 60 ; 240

Therefore the range of is from 180 degrees to 240 de0 0

grees.

180 240


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38


Extinction Angle is calculated by using the equation

sin sin

In the exponential term the value of should be

substituted in radians. Hence

sin sin R

R

L

R

L

e

and

e

0

100 0

0 3.183

00

0

0 00

0

;3

60 17.44059 42.5594

sin 17.44 sin 42.5594

sin 17.44 0.676354

180 radians,180

190Assuming 190 ; 3.3161

180 180

ad Rad

Rad

Rad

Rad

e

e


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39

0

3.183 3.316143

0

0 0

0


L.H.S: sin 190 17.44 sin 172.56 0.129487

R.H.S: 0.676354 4.94 10Assuming 183

1833.19395

180 180

3.19395 2.146753

L.H.S: sin si

Rad

e

0

3.183 2.14675 4

0 00

0

n 183 17.44 sin165.56 0.24936

R.H.S: 0.676354 7.2876 10

180Assuming 180 ;

180 1802

3 3

Rad

e


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40

3.18343

0

0 0

0

3.183


L.H.S: sin sin 180 17.44 0.2997

R.H.S: 0.676354 8.6092 10Assuming 196

1963.420845

180 180L.H.S: sin sin 196 17.44 0.02513

R.H.S:0.676354

Rad

e

e3.420845

43

0

0 0

0

3

3.183 3.4382943

3.5394 10

Assuming 197

1973.43829180 180

L.H.S: sin sin 197 17.44 7.69 7.67937 10

R.H.S: 0.676354 4.950386476 10

Rad

e


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41

0

0

0

4

3.183 3.445643

0 0


Assuming 197.42

197.423.4456180 180

L.H.S: sin sin 197.42 17.44 3.4906 10

R.H.S: 0.676354 3.2709 10

Conduction Angle 197.42 60 137

Rad

e

0

0 0

.42RMS Output Voltage

1 sin 2 sin 22 2

sin 2 60 sin 2 197.421230 3.4456

3 2 2

1230 2.39843 0.4330 0.285640

S O RMS

O RMS

O RMS

V V

V

V


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42

22


230 0.9 207.0445 V

Input Power Factor

207.044519.7527 A

10.481819.7527 10 3901.716 W

230 , 19.7527

3901.716230 19.

O RMS

O

S S

O RMS

O RMS

O LO RMS

OS S O RMS

S S

V

P PF

V I

V I

Z P I R

P V V I I PF

V I

0.85887527


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Problem : A single phase TCR has an input 240 V, 50 Hz ACsupply and an inductance of 5mH. Calculate RMS current andfundamental RMS at =30 .

2

L

Solution:The expressions for the RMS current, fundamental current

and THD are

( / ( )) {( 2 )(1 / 2 sin ) 3sin 2 / 2} /

( / ( )){1 2 / sin 2 / } / 2Given that V=240 V, L=5 mH, f=50 Hz.

X =1.570796

Su

RMS

F

I V L

I V L

RMS FUND

RMS

FUND

2

bstituting the values in the expression for I and I ,

I = 44.9374 A

I = 42.2431 A

44.9374THD = 1 *100 36.28%

42.2431THD


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44

Problem : A single phase 25 kVAr TCR has an input of 230 V, 50Hz AC supply. Calculate (a) inductance rating (b) RMS currentunder maximum THD current (c) corresponding delay angle.

2

2

Solution:

The expressions for the RMS current, fundamental current

and THD are

( / ( )) {( 2 )(1 / 2 sin ) 3sin 2 / 2} /

( / ( )){1 2 / sin 2 / } / 2

( / ) 1

Given that V=230 V, Q=25 kVAR, f=50

RMS

F

RMS F

I V L

I V L

THD I I

2L

2 2

Hz.

The inductance rating can be calculated as follows:

X =230 /25k=2.116 , L=6.7354 mH Now, the expression for the harmonic component of the

TCR current is,

4 sin cos( ) cos sin( )( )

( 1)

V n n n I n

L n n


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Solution:contd.....

Differentiating the above equation w.r.t. , we

get the condition as cos cos( ) 0Equating the first term to zero, gives us a trivial

solution when the current in the TCR branch is

n

RMS F

equal to zero. Hence we have to consider the

second term, with the value of n being assignedas the dominant harmonic number i.e. 3

Hence, =30

Substituting the values in the expression for

I and I UND RMS

FUND

, I =31.9690 AI =30.0522 A

Hence THD=36.28%


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46

Problem: A single-phase TCR (thyristor controlled reactor consisting back-to-back connected thyristors with pureinductor) has an input of 240V, 50Hz, AC supply and an

inductance of 20 mH. Calculate maximum VAR rating. Alsocalculate (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

2rms

f

n 2

3 5 7

240Solution: VAR rating =9.167kVAR 2 502010

V 1 3sin2 i) I = -2 0.5+sin - = 11.23 A

L 2

V 2 sin2ii) I = 1- - =10.71A 2 L

V 4 sin cosn-ncossinniii) I =

L n(n -1)

I =-5.25A iv) I =-1.05A v) I =0.375A


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Problem: A single-phase ac switch having anti parallel-connected thyristors is used between a 230 V, 50 Hz ac mainsand a load of 4.0 kW, 0.8 lagging for switching in andinterrupting it. Calculate the (i) peak and rms voltage andcurrent rating of the thyristors and (ii) instant of firing anglesof thyristors for transient free switching.

1

Solution: i) Peak voltage rating for thyristor = 230 2

rms voltage rating= 230 V

4000 2Peak current rating = =30.74A

0.8230thyristor rms current=30.74/2 = 15.37 A

ii) Instant of firing angle = cos 0.8 36.86o


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48

Problem: A single-phase ac switch having anti parallel-connected thyristorsare used between a 240 V, 50 Hz ac mains and three binary weighted TSCs toform 7 kVAR capacitor bank with a minimum of an ac capacitor of 1 kVAR

for switching in and interrupting it. Calculate the (i) peak and rms voltageand current rating of diode and thyristors and (ii) instant of firing angles of thyristors for transient free switching.

Solution: V s= 240 V, Total kVAR = 7, f =50 Hz.

(a) Peak Current (I m) = 2 P o/ V s P.F.= 27000/ 2401 = 41.247 A.

r.m.s. current (I r ) = I m/2 = 20.623 A.

Peak voltage (Vm) = 2 Vs = 339.411 v.

r.m.s. voltage = 240 V.

(b) Firing Angle =90 0 (here, capacitive circuit).


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AC Voltage controllers with PWM control

Naturally commutated thyristor controllersintroduce low order harmonics in both theload and supply side and have low inputpower factor.

The input power factor of AC voltagecontrollers can be improved by PWM controland producing variable output voltage.


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Waveforms of gating signals of AC Voltagecontrollers with PWM control


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Single phase line conditioning unit

IoIi

+

-V i

C1

C2

+

-

+

-

+

-

Vm

+

-

Vx

Vc1

Vc2

Vo

Co+

-

R L

LoL i

D1

D2 S2

S1


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Single phase line conditioning unitDisadvantages:

Input current is seen to have reasonableharmonics.

To control the input current two additionalswitches are required.


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Single phase line conditioning unitLoad regulation for voltage doubler. Vi = 115 V rms . P base = 1KW,C = 1 p.u. (200 F, based on 115 v, 1 KW) and 100 p.u.


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Simulated waveforms for two switch line conditioner.C 1 = C 2 = 1.3 p.u., L i = 0.01 p.u., L o = 0.1 p.u., C O = 0.3 p.u., I o = 1.0 p.u., resistive load, switching frequencyf = 30 kHz. (e)-(g) Boost operation with no phase shift, V o/Vi = 1.2.

Si l h li di i i i


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Experimental waveforms for four switch line conditioner under normaloperation. V i = Vo = 115 V rms , C1 = C 2, = 22.5 F, L i = L o = 3 m H , C o = 100 F,Cs = 40.000 F, R L = 12 , PWM switching frequency = 3 kHz.


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Si l h li diti i it


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Experimental waveforms for two switch line conditioner. V i = 75Vrms , Vo = 100 V rms , C 1 = C 2 = 470 F, C o = 10 F, L o = 1 mH, R L =

50, = 30, switching frequency f = 30 kHz.



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UPS system with sinusoidal input current and 0-2 p.u.

output voltage regulation.

IoIi

+

-Vi

C1

C2

+

-

+

-

+

-

Vm

+

-

Vx

Vc1

Vc2

Vo

Co+

-

R L

Lo

Li

+

-VB

LF

D6S1

S2 S4

S3

S5



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Single phase line conditioning unitCapabilities

All the topologies are transformer less.

They are characterized by common neutral connection

between the output and input.

Buck boost voltage regulation capability.

The input and output filters provide a significant

measure of common mode and normal mode noiserejection.



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Single phase line conditioning unitFeatures

These features are essential to low cost realization.

Isolation is not a significant issue in low power line

conditioners which feed computer and its peripherals

with internally isolated high frequency SMPS.

The four switch topology has significantly superior

operation at the expense of two additional switchesincluding unity power factor operation.


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R f


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References

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 onindustry applications, vol. 30, no. 2, march/april 1994

Documents

L3. Single Phase Ac Voltage Controllers