thyristor converters or controlled converters the controlled rectifier circuit is divided into three...
TRANSCRIPT
Thyristor Converters or Controlled Converters
The controlled rectifier circuit is divided into three main circuits-:
Power Circuit Control Circuit Triggering circuit
The commutation of thyristor is
Natural Commutation Forced CommutationHalf Wave Single Phase Controlled Rectifier With Resistive Load
Fig.3.1 Half wave single phase controlled rectifier.
)cos1(2
))cos(cos(2
)sin(2
1
mmmdc
VVtdtVV
/mdm VV )cos1(5.0/ mdcn VVV
2
)2sin(1
2)sin(
2
1 2
mmrms
VtdtVV
Example 1 In the rectifier shown in Fig.3.1 it has a load of R=15 and, Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an average output voltage of 70% of the maximum possible output voltage, calculate:- (a) The firing angle, (b) The efficiency, (c) Ripple factor (d) Peak inverse voltage (PIV) of the thyristor
7.0)cos1(5.0 dm
dcn V
VV
VV
VV mdmdc 02.49*7.0*7.0
A
R
VI dc
dc 268.315
02.49
2
)2sin(1
2
mrms
VV =66.42o, Vrms=95.1217V
, Irms=95.122/15=6.34145A
%56.2634145.6*121.95
268.3*02.49
*
*
rmsrms
dcdc
ac
dc
IV
IV
P
P
94.12202.49
121.95
dc
rms
V
VFF
6624.1194.11 22 FFV
VRF
dc
ac
The PIV is mV
Half Wave Single Phase Controlled Rectifier With RL Load
Fig.3.4 Half wave single phase controlled rectifier with RL load.
Single-Phase Center Tap Controlled Rectifier With Resistive Load
Fig.3.8 Center tap controlled rectifier with resistive load.
a b
)cos1())cos(cos()sin(1
mmmdc
VVtdtVV
)cos1(5.0 dm
dcn V
VV
2
)2sin(
2)sin(
1 2
mmrms
VtdtVV
Example 4 The rectifier shown in Fig.3.8 has load of R=15 and, Vs=220 sin 314 t and unity transformer
ratio. If it is required to obtain an average output voltage of 70 % of the maximum possible output voltage,
calculate:- (a) The delay angle, (b) The efficiency, (c) The ripple factor (d) The peak inverse voltage (PIV) of
the thyristor.
7.0)cos1(5.0 dm
dcn V
VV then, =66.42o
220mV , then, VV
VV mdmdc 04.98
2*7.0*7.0
AR
VI dc
dc 536.615
04.98
2
)2sin(
2
mrms
VV
at =66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A
%04.53976.8*638.134
536.6*04.98
*
*
rmsrms
dcdc
ac
dc
IV
IV
P
P
3733.104.98
638.134
dc
rms
V
VFF
9413.013733.11 22 FFV
VRF
dc
ac
The PIV is 2 Vm
Single-Phase Fully Controlled Rectifier Bridge With Resistive Load
Fig.3.11 Single-phase fully controlled rectifier bridge with resistive load.
)cos1(5.0/ dmdcn VVV
)cos1()cos(cos)sin(1
mmmdc
VVtdtVV
2
)2sin(
2)sin(
1 2
mmrms
VtdtVV
Example 5 The rectifier shown in Fig.3.11 has load of R=15 and, Vs=220 sin 314 t and unity transformer ratio. If it is required
to obtain an average output voltage of 70% of the maximum possible output voltage, calculate:- (a) The delay angle , (b) The efficiency, (c) Ripple factor of output voltage(d) The peak inverse
voltage (PIV) of one thyristor.
7.0)cos1(5.0 dm
dcn V
VV , then, =66.42o
220mV , then, VV
VV mdmdc 04.98
2*7.0*7.0
AR
VI dc
dc 536.615
04.98
2
)2sin(
2
mrms
VV
=66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A
%04.53976.8*638.134
536.6*04.98
*
*
rmsrms
dcdc
ac
dc
IV
IV
P
P
3733.104.98
638.134
dc
rms
V
VFF
9413.013733.11 22 FFV
VRF
dc
ac
The PIV is Vm
Full Wave Fully Controlled Rectifier With RL Load In Continuous Conduction Mode
Fig.3.14 Full wave fully controlled rectifier with RL load.
Full Wave Fully Controlled Rectifier With pure DC Load
)..........9sin9
17sin
7
15sin
5
13sin
3
1(sin*
4)( ttttt
Iti o
cos2
)sin(1 m
mdcV
tdtVV
cos/ dmdcn VVV
2
)2cos(1(2
)sin(1 2 mm
mrmsV
tdtV
tdtVV
Example 6 The rectifier shown in Fig.3.14 has pure DC load current of 50 A and, Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an average output voltage of 70% of the
maximum possible output voltage, calculate:- (a) The delay angle , (b) The efficiency, (c) Ripple factor (d) The peak inverse
voltage (PIV) of the thyristor and (e) Input displacement factor.
7.0cos dm
dcn V
VV then, =45.5731o= 0.7954
220mV , VVVV mdmdc 04.98/2*7.0*7.0 , 2/mrms VV
At =45.5731o Vrms=155.563 V. Then, Irms=50 A
%02.6350*563.155
50*04.98
*
*
rmsrms
dcdc
ac
dc
IV
IV
P
P
587.104.98
563.155
dc
rms
V
VFF
23195.113733.11 22 FFV
VRF
dc
ac
The PIV is Vm
Input displacement factor. 7.0cos
Single Phase Full Wave Fully Controlled Rectifier With Source Inductance
m
s
V
Lu
2coscos 1
osos
rd IfLIL
V 42
4
osm
rdceinducsourcewithoutdcactualdc IfLV
VVV 4cos2
tan
32
2 2 uII o
s
2sin*
2
81
u
u
II o
S
2cos. 1 u
I
Ifp
s
s
Inverter Mode Of Operation
Fig.3.25 Single phase SCR inverter.
Fig.3.27 SCR inverter with a DC voltage source.
dsdodd ILVVE
2cos
Three Phase Half Wave Controlled Rectifier with Resistive Load
Fig.3.31 Three phase half wave controlled rectifier.
30
cos675.0cos2
3
cos827.0cos2
33sin
2
36/5
6/
LLLL
mm
mdc
VV
VV
tdtVV
30
cos*827.0
cos**2
33
R
V
R
VI mm
dc
2cos8
3
6
13sin
2
36/5
6/
2
mmrms VtdtVV
2cos8
3
6
13
R
VI m
rms
2cos8
3
6
1
3
R
VIII mrms
Sr
> 30
6cos14775.0
6cos1
2
3sin
2
3
6/
mm
mdc VV
tdtVV
> 30
6cos1
2
3
R
VI m
dc
)23/sin(8
1
424
53sin
2
3
6/
2
mmrms VtdtVV
)23/sin(8
1
424
53
R
VI m
rms
)23/sin(8
1
424
5
3
R
VIII mrms
Sr
Example 7 Three-phase half-wave controlled rectfier is connected to 380 V three phase supply via delta-way 380/460V transformer. The load of the
rectfier is pure resistance of 5 . The delay angle o25 . Calculate: The rectfication effeciency (b) PIV of thyristors
VVV LLdc 5.28125cos4602
3cos
2
3
AR
VI dc
dc 3.565
5.281
V
VV LLrms
8.29825*2cos8
3
6
1*460*2
2cos8
3
6
1*2
AR
VI rms
rms 76.595
8.298
%75.88100* rmsrms
dcdc
IV
IV
Example 8 Solve the previous example (evample 7) if the firing angle o60
VV
V mdc 33.179
36cos1
2
460*3
23
6cos1
2
3
AR
VI dc
dc 87.355
33.179
From (3.65) we can calculate rmsV as following:
VVV mrms 230)23/sin(8
1
424
53
AR
VI rms
rms 465
230
%79.60100* rmsrms
dcdc
IV
IV
VVPIV LL 54.650460*22
Fig.3.36 Three Phase Half Wave Controlled Rectifier With DC Load Current
Three Phase Half Wave Controlled Rectifier With DC Load Current
t=0
3
2cos1
2)sin(
23/2
0
n
n
ItdtnIb dc
dcn
Then, 2
3*
2
n
Ib dc
n for n=1,2,4,5,7,8,10,…..
And 0nb For n=3,6,9,12
......7sin
7
15sin
5
14sin
4
12sin
2
1sin
3)( ttttt
Iti dc
p
21
21
2
p
pp
I
IITHD
dcp II *
3
2 2
31
dcp
II
%68
2
92
9
3
2
22
22
2
dc
dcdc
I
II
THD
%68....14
1
13
1
11
1
10
1
8
1
7
1
5
1
4
1
2
1222222222
THD
Example 9 Three phase half wave controlled rectfier is connected to 380 V three phase supply via delta-way 380/460V transformer. The load
of the rectfier draws 100 A pure DC current. The delay angle, o30 . Calculate: (a) THD of primary current. (b) Input power factor.
the peak value of primary current is A05.121380
460*100 .
AI
I dcP 74.81
2
05.121*3
2
31
%98.67100*174.81
84.98100*1
22
1
,
P
rmsPI I
ITHD
P
AI rmsP 84.983
2*05.121,
LaggingI
IfP
rmsP
P 414.066
cos*84.98
74.81
6cos*.
,
1
Three Phase Full Wave Fully Controlled Rectifier With Resistive Load
o60
cos33
)6
sin(33
2/
6/
mmdc
VtdtVV
m
dmV
V33
2cos4
33
2
13)
6sin(3
32/
6/
2
mmrms VtdtVV
o60
3/cos133
)6
sin(33
6/5
6/
mmdc
VtdtVV
m
dmV
V33
3/cos1dm
dcn V
VV
6
2cos24
313)
6sin(3
36/5
6/
2
mmrms VtdtVV
VR
VV m
dc 400cos380*3
2*
33cos
33
.
Then o79.38 , AR
VI dc
dc 4010
400
From (3.84) the rms value of the output voltage is:
VVV mrms 412.4122cos4
33
2
13
Then, VVrms 412.412 Then, AR
VI rms
rms 24.4110
412.412
Then, %07.94100*24.41*4.412
40*400100*
*
*
rmsrms
dcdc
IV
IV
The PIV= 3 Vm=537.4V
Example 10 Three-phase full-wave controlled rectifier is connected to 380 V, 50 Hz supply to feed a load of 10 pure resistance. If it is required to get 400 V DC output voltage, calculate the following: (a) The firing angle, (b) The rectfication effeciency (c) PIV of the thyristors.
VV
V mdc 150cos
380*3
2*33
cos33
o73
60
VVdc 1503/cos1380*
3
2*33
o05.75
AR
VI dc
dc 1510
150
3005.75*2cos
180*05.75*2
4
31*380*
3
2*3
62cos2
4
313
mrms VV VVrms 075.198
AR
VI rms
rms 8075.1910
075.198 %35.57100*
81.19*075.198
15*150100*
*
*
rmsrms
dcdc
IV
IV
3
Example 11 Solve the previous example if the required dc voltage is 150V.Solution: From (3.81) the average voltage is :
It is not acceptable result because the above equation valid only for
.Then we have to use the (3.85) to get
From (3.88) the rms value of the output voltage is:
The PIV= Vm=537.4V
Three Phase Full Wave Fully Controlled Rectifier With pure DC Load Current
> 60o
LL
oS
V
ILu
2coscos 1
osLLrdceinducsourcewithoutdcactualdc IfLVVVV 6cos23
tan
63
2 2 uII o
S
2sin
621
u
u
II o
S
2cos
63
2sin*32
2cos
63
2
2sin
62
2cos
21 u
uu
uu
uI
u
u
Iu
I
Ipf
o
o
S
S
Inverter Mode of Operation