8/12/2019 L3. Single Phase Ac Voltage Controllers
1/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
2/68
Lecture - 3
2
Single phaseAC Voltage controller
8/12/2019 L3. Single Phase Ac Voltage Controllers
3/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
4/68
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.
8/12/2019 L3. Single Phase Ac Voltage Controllers
5/68
5
i1
i2
vs vo
ioT2
T1
Single phase bidirectional controller with RL load
8/12/2019 L3. Single Phase Ac Voltage Controllers
6/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
7/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
8/68
8/12/2019 L3. Single Phase Ac Voltage Controllers
9/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
10/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
11/68
11
Waveforms of single phase bidirectional ac voltage controllerwith RL load
8/12/2019 L3. Single Phase Ac Voltage Controllers
12/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
13/68
13
Theoretical components of load current waveform for series R L circuit1 = 60, = 120
8/12/2019 L3. Single Phase Ac Voltage Controllers
14/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
15/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
16/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
17/68
17 17
T1
T2
io
vo
+
-
v=V msin t
Performance parameters of a single phase full wave ACvoltage controller with inductive load
8/12/2019 L3. Single Phase Ac Voltage Controllers
18/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
19/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
20/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
21/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
22/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
23/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
24/68
Load
24
A single phase full wave controller can also be implemented withone thyristor and four diodes.
8/12/2019 L3. Single Phase Ac Voltage Controllers
25/68
25
Waveforms of single phase full wave controller withone thyristor and four diodes .
8/12/2019 L3. Single Phase Ac Voltage Controllers
26/68
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.
8/12/2019 L3. Single Phase Ac Voltage Controllers
27/68
27
Transformer tap changingWith resistive and inductive loads
8/12/2019 L3. Single Phase Ac Voltage Controllers
28/68
I M
28
Transformer tap changingWith RL and motor loads
8/12/2019 L3. Single Phase Ac Voltage Controllers
29/68
8/12/2019 L3. Single Phase Ac Voltage Controllers
30/68
30
Waveforms for transformer connection changer
8/12/2019 L3. Single Phase Ac Voltage Controllers
31/68
31
Waveforms without tap changer
8/12/2019 L3. Single Phase Ac Voltage Controllers
32/68
32
Waveforms with synchronous changer
8/12/2019 L3. Single Phase Ac Voltage Controllers
33/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
34/68
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..........
8/12/2019 L3. Single Phase Ac Voltage Controllers
35/68
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:
8/12/2019 L3. Single Phase Ac Voltage Controllers
36/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
37/68
37
2 2 22
3
2 2
1
1
Solution: contd..........
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
38/68
38
Solution: contd..........
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
39/68
39
0
3.183 3.316143
0
0 0
0
Solution: contd..........
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
40/68
40
3.18343
0
0 0
0
3.183
Solution: contd..........
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
41/68
41
0
0
0
4
3.183 3.445643
0 0
Solution: contd..........
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
42/68
42
22
Solution: contd..........
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
43/68
43
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
44/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
45/68
45
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%
8/12/2019 L3. Single Phase Ac Voltage Controllers
46/68
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
47/68
47
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
48/68
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).
8/12/2019 L3. Single Phase Ac Voltage Controllers
49/68
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.
8/12/2019 L3. Single Phase Ac Voltage Controllers
50/68
8/12/2019 L3. Single Phase Ac Voltage Controllers
51/68
Waveforms of gating signals of AC Voltagecontrollers with PWM control
8/12/2019 L3. Single Phase Ac Voltage Controllers
52/68
8/12/2019 L3. Single Phase Ac Voltage Controllers
53/68
Single phase line conditioning unit
IoIi
+
-V i
C1
C2
+
-
+
-
+
-
Vm
+
-
Vx
Vc1
Vc2
Vo
Co+
-
R L
LoL i
D1
D2 S2
S1
8/12/2019 L3. Single Phase Ac Voltage Controllers
54/68
8/12/2019 L3. Single Phase Ac Voltage Controllers
55/68
Single phase line conditioning unitDisadvantages:
Input current is seen to have reasonableharmonics.
To control the input current two additionalswitches are required.
8/12/2019 L3. Single Phase Ac Voltage Controllers
56/68
Single phase line conditioning unit
8/12/2019 L3. Single Phase Ac Voltage Controllers
57/68
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.
8/12/2019 L3. Single Phase Ac Voltage Controllers
58/68
Single phase line conditioning unit
8/12/2019 L3. Single Phase Ac Voltage Controllers
59/68
Single phase line conditioning unit
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
8/12/2019 L3. Single Phase Ac Voltage Controllers
60/68
Single phase line conditioning unit
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.
8/12/2019 L3. Single Phase Ac Voltage Controllers
61/68
8/12/2019 L3. Single Phase Ac Voltage Controllers
62/68
Si l h li diti i it
8/12/2019 L3. Single Phase Ac Voltage Controllers
63/68
Single phase line conditioning unit
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.
Single phase line conditioning unit
8/12/2019 L3. Single Phase Ac Voltage Controllers
64/68
Single phase line conditioning unit
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
Single phase line conditioning unit
8/12/2019 L3. Single Phase Ac Voltage Controllers
65/68
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.
Single phase line conditioning unit
8/12/2019 L3. Single Phase Ac Voltage Controllers
66/68
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.
8/12/2019 L3. Single Phase Ac Voltage Controllers
67/68
R f
8/12/2019 L3. Single Phase Ac Voltage Controllers
68/68
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