Uploaded by User33185

209838 MK elda PenyearahTerkendali2017

advertisement
Dosen Pembina MK :
I Nyoman Wahyu Satiawan, ST, M.Sc. Ph.D
NIP. 197009081998021001
Jurusan Teknik Elektro
Fakultas Teknik Universitas Mataram



Normal rectifiers are considered as uncontrolled
rectifiers.
Once the source and load parameters are
established, the dc level of the output and power
transferred to the load are fixed quantities.
A way to control the output is to use SCR instead of
diode. Two condition must be met before SCR can
conduct:
◦ The SCR must be forward biased (VSCR>0)
◦ Current must be applied to the gate of SCR


A phase-controlled thyristor is turned on by
applying a short pulse to its gate, and is turned
off due to natural or line commutation.
They are used extensively in industrial
applications, especially in variable speed drives.
1. dc motor speed control systems, widely used in steel mills,
paper mills, and such.
2. Electrochemical and electro-metallurgical processes.
3. Magnet power supplies.
4. Converters at the input end of dc transmission lines.
5. Portable hand tool drives
3

The phase-control converters are classified
as:
◦ Single-phase converters
◦ Three-phase converters

Each type is divided into:
◦ Semiconverter
◦ Full converter
◦ Dual converter
4



A semiconverter is a one-quadrant converter and
it has one polarity of output voltage and current.
A full converter is a two-quadrant converter and
the polarity of its output voltage can be either
positive or negative.
A dual converter can operate in four quadrants,
and both the output voltage and the output
current can be either positive or negative.
5
Operation Principle

Consider the following circuit, where α is the
delay or firing (ignition) angle.
6
7




If the frequency of the supply is fs, the lowest
frequency of the output ripple voltage will be
fs.
Now:
Vm
1 
Vdc 
Vm sin t d (t ) 
(1  cos  )


2
2
The dc voltage can be varied from Vm/π to 0
by varying from 0 to π.
The average output voltage will be maximum
when α is 0.
8

The rms output voltage is:
1  2 2
Vrms  [  Vm sin t d (t )]1 / 2
2 
Vrms


Vm 1
sin 2 1 / 2

[ (   
)]
2 
2
The rms load current is :
I rms
Vrms

R
The average power absorbed by load
(resistor) :
9

Design a circuit to produce an average voltage
of 40V across 100 load resistor from a
120Vrms 60 Hz ac source. Determine the power
absorbed by the resistor and the power factor.
Briefly describe what happen if the circuit is
replaced by diode to produce the same average
output.

Solution
In such that to achieved 40V average voltage, the delay angle must be
Vs
[1  cos  ]
2
120 2
40 
[1  cos  ]
2
  61.2o  1.07 rad
Vo 
V 2 rms 75.6 2
P

 57.1W
R
100
Vo , rms 
Vm
 sin( 2 )
1 
2

2
120 2
1.07 sin 2(1.07)
1

2

2
 75.6V

pf 
57.1
 0.63
 75.6 
(120)

 100 

Solution
•If an uncontrolled diode is used, the average
voltage would be
Vo 
Vs


2 (120)

 54V
• That means, some reducing average resistor to the design
must be made.
• A series resistor or inductor could be added to an
uncontrolled rectifier, while controlled rectifier has
advantage of not altering the load or introducing the losses
•The analysis of the circuit is very much similar to that of
uncontrolled rectifier.
  t 
 V 


m
  
for   t  
 [sin( wt   )  sin(    )e
i ( wt )   Z 
0
otherwise

and
L
 L 
Z  R 2  (L) 2 ,   tan 1 
,  
R
 R 
rms current ,
I rms
1

2

2
1 2
i
(

t
)
d
(

t
)

i (t ) d (t )


2 
2
and , average current ,

1
Io 
i (t ) d (t )

2 
The average output voltage,
Vm
[cos   cos  ]
1 Vm sin( t )dt 
Vo 
2

2 

The average power absorbed by load ,
2
P  I rms
R ;



The circuit for a single-phase full converter is
shown below.
The load is assumed to be highly inductive,
so the load current is continuous and ripple
free.
The converter operates in the rectification
and inversion modes.
16
17

The average output voltage is:
2
Vdc 
2

 

Vm sin t d (t ) 
2Vm

cos 
The dc voltage can be varied from 2Vm/π to
-2Vm/π by varying α from 0 to π.
18


The average output voltage will be maximum
when α is 0.
The rms output voltage is:
Vrms
2   2 2
 [  Vm sin t d (t )]1/ 2
2 
Vrms
Vm

 Vs
2
19


The operation of the converter can be divided
into two identical modes.
Mode 1 is when T1 and T2 conduct, and mode
2 is when T3 and T4 conduct.
20
Summary of some important
points in analysis
 When
analyzing a thyristor circuit, start from a diode
circuit with the same topology. The behavior of the
diode circuit is exactly the same as the thyristor
circuit when firing angle is 0.
 Take
different principle when dealing with different
load
– For resistive load: current waveform of a resistor is
the same as the voltage waveform
–For inductive load with a large inductor: the inductor
current can be considered constant
I Nyoman Wahyu Satiawan
FT UNRAM






Beroperasi dgn sumber AC 3 –fase.
Dapat menghasilkan tegangan keluaran DC
yang lebih besar.
Menghasilkan daya DC yang lebih besar.
Frekwensi riple yang lebih tinggi pada
tegangan keluaran.
Memerlukan filter yang lebih sederhana untuk
menghaluskan tegangan dan arus beban.
Digunakan secara luas pada high power
variable speed industrial dc drives.
# tegangan fase ke netral di definisikan sbb :
vRN  van  Vm sin  t ;
Vm  Max. Phase Voltage
vYN
2 

 vbn  Vm sin   t 

3


 Vm sin  t  1200 
vBN
2 

 vcn  Vm sin   t 

3 

 Vm sin  t  1200 
 Vm sin  t  2400 

Tegangan keluaran vo merupakan penjumlahan dari
tegangan keluaran dari masing-masing fase


T1 di triger pada t       (30   )
6

 5

T2 di triger pada t  
    (150   )
 6

 7

T3 di triger pada t  
    (270   )
 6

2
Setiap Thyristor konduksi pada 120 atau
radian
3

Jika tegangan fase ref. adalah vRN = van = Vm sin ωt,
maka tegangan keluaran DC atau tegangan rata2 untuk
arus beban kontinyu dihitung dengan :
 56 

3 



Vdc 
V
sin

td

t
m

2   
 6 

 56 

3Vm 



Vdc 
sin

td

t

2   
 6 

5

3Vm
6
( cos t ) 
Vdc 
2
6
Vdc 
3Vm
2

 5




cos



cos







6
6





Dengan menerapkan aturan trigonometri,
 Cos (A+B) = (Cos A. Cos B – Sin A. Sin B) , maka
3Vm
Vdc 
2


 5 
 5 
 
 
 cos 6  cos( )  sin  6  sin(  )  cos 6  cos( )  sin  6  sin(  )
 
 
 
 




3Vm
 cos(150 ) cos( )  sin( 150 ) sin(  )  cos(30 ) cos( )  sin( 30 ) sin(  )
2
3Vm
Vdc 
cos(180  30 ) cos( )  sin( 180  30 ) sin(  )  cos(30 ) cos( )  sin( 30 ) sin(  )
2
Note : cos(180  30 )   cos(30 ) dan sin(  )  cos(180  30 )  sin( 30 )
Vdc 

Vdc 

3Vm
cos(30 ) cos( )  sin( 30 ) sin(  )  cos(30 ) cos( )  sin( 30 ) sin(  )
2



 
3Vm
Vdc 
2 cos 30 cos( )
2

3Vm 
3
Vdc 
cos( )
2 x
2 
2




3Vm
3 3
Vdc 
3 cos( ) 
cos( )
2
2
3V
Vdc  Lm 3 cos( )
2
Dimana, VLm  3Vm  tegangan masukan line to line


#Tegangan keluaran DC atau tegangan rata-rata
maks. pada sudut penyalaan α= 0 adalah
Vdc(max)
3 3Vm
 Vdm
2
#Tegangan keluaran rata-rata yang dinormalisasi
adalah
Vdcn
Vdc
 Vn 
 cos 
Vdm
# Nilai rms dari tegangan keluaran dicari dengan
persamaan :
Vo ( RMS )

 3

2

5

6


2
2
V
sin

t
.
d
(

t
)
m





6
1
2
Shg didapat,
Vo ( RMS )
1

3
 3Vm  
cos 2 
 6 8

1
2

Van
Vbn
Vcn

V0
=300
=30
0
0
30
0
60
90
0
0
120

0
150
0
180
0
210
Van
0
240
0
270
0
300
0
330
0
360
Vbn
0
390
=60
0
30
0
60
90
0
0
120
0
150
0
180
0
210
0
240
t
0
420
Vcn

V0
0
0
0
270
0
300
0
330
0
360
0
390
0
420
0
=600
t

Vbn
Van
Vcn

V0
=900
=90
0
0
0
30
0
60
0
90
0
120
0
0
0
0
0
0
0
0
150 180 210 240 270 300 330 360
0
0
390 420
t



Dikenal sebagai konverter / penyearah 6pulsa.
Digunakan pada aplikasi dengan daya output
tinggi s.d. 120kW.
Dapat melakukan operasi dua kuadran.




Thyristor2 di triger dgn interval  /3.
Frequency ripple tegangan keluaran 6fS.
T1 ditriger pada t = (/6 + ), T6 sudah dlm
keadaan konduksi ketika T1 ON.
Pada interval (/6 + ) sampai (/2 + ),
T1 dan T6 konduksi bersamaan & tegangan
keluaran sama dengan vab = (van – vbn)




T2 di trigger pada t = (/2 + ), dan T6 OFF
secara alamiah ketika dlm kondisi as it is
reverse biased ketika T2 di trigger.
Selama interval (/2 + ) sampe (5/6 + ),
T1 dan T2 konduksi bersama & tegangan pada
beban vO = vac = (van – vcn)
Thyristor2 di berikan nomor sesuai dengan
urutan dimana mereka di triger.
Urutan triger dari thyristor adalah 12, 23, 34,
45, 56, 61, 12, 23, 34, ………

Kita definisikan tegangan 3 fase (tegangan
fase ke netral) sbb;
vRN  van  Vm sin t ;
2
)  Vm sin( t  120)
3
2
vBN  vcn  Vm sin( t 
)  Vm sin( t  120)
3
..........................................  Vm sin( t  240)
vYN  vbn  Vm sin( t 
Vm adalah teg angan fase puncak untuk beban
yang terhubung bintang

Tegangan antar fase dari sumber adalah;

vRY  vab  (van  vbn )  3Vm sin( t  );
6

vYB  vbc  (vbn  vcn )  3Vm sin( t  );
2

vBR  vca  (vcn  van )  3Vm sin( t  );
2

Tegangan keluaran pada beban terdiri dari 6
pulsa tegangan pada setiap periode 2 radians,
sehingga tegangan keluaran rata-rata dihitung
dengan :

VO dc 
6
 Vdc 
2
2



6
vO .d t ;



vO  vab  3Vm sin   t  
6

 2 

3


Vdc    3Vm sin( t  )t 
 
6
 6 

Vdc 
3 3Vm

cos 
dimana VmL  3 Vm adalah teg angan antar fase masukan

Tegangan keluaran DC rata-rata diperoleh pada sudut
penyalaan α= 0
Vdc(max)  Vdc 
3 3Vm


3VmL


Tegangan keluaran DC rata-rata yang
dinormalisasi adalah
Vdcn  Vdc 

3 3Vm


3VmL

Nilai rms tegangan keluaran adalah
Vo ( rms)

 6

2





V
d

t
o





6

2

1
2
VO rms 
VO rms 
VO rms 

 6

2


 3

2




2
v
.
d

t
 
ab




6
2


1
2



2
2
3
V
sin

t

.
d

t
m

  

6




6
2

1 3 3

 3Vm  
cos 2 
 2 4

1
2
1
2
ECE 442
47
1
Vdc 
T
T
 f (t )dt
0

2
Vdc 
2
6
Vdc 

3 3

6
3Vm cos td (t )
0
Vm  1.654Vm
Vrms

 2

2

 6

6

2
2
0 3Vm cos td (t ) 



1
2
Vrms
Vrms
3 9 3
  
 Vm
 2 4 
 1.6554Vm
1
2
3Vm
Im 
 peak
R



6
 4

2
2
Ir  
I m cos td (t ) 

2 0


I r  0.5518 I m
ECE 442
1
2
51
Download
Random flashcards
hardi

0 Cards oauth2_google_0810629b-edb6-401f-b28c-674c45d34d87

Rekening Agen Resmi De Nature Indonesia

9 Cards denaturerumahsehat

Nomor Rekening Asli Agen De Nature Indonesia

2 Cards denaturerumahsehat

Secuplik Kuliner Sepanjang Danau Babakan

2 Cards oauth2_google_2e219703-8a29-4353-9cf2-b8dae956302e

Create flashcards