hvdc 6 pulse [compatibility mode].pdf
TRANSCRIPT
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DC
C E E
6 E C E E
( 1, 3, 5) ( 2, 4 6). AC , F .
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The wave form for the DC and voltage across valve 1 at = 0 0, 45 0, 90 0, 135 0
and 180 0. are shown in Fig
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DC
The dc voltage waveform contains a ripple whose fundamental frequency is sixtimes the supply frequency.
h=6n The harmonic (rms value) in the dc voltage is given by
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12
12 E C E E 2 transformers 1 0 in star. The currents supplied by the supply are the sums of the currents flowing in the two
primary winding 1 1:1
1:3
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12 pulse converter, 12 intervals in a cycle, duration of 30 0. In each interval, 4 valves (2 each
from two bridges) conduct .
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The current i6p flowing in the primary winding of the star/delta connected.transformer
the rms value of the fundamental component of the supply current in a 12 pulse converter isgiven by
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E A ,
C .
D
,
. F , 3 , 1
3 .
( ) ,
Each interval of the period of supply divided into two subintervals In the first subinterval, three valves are conducting and In the second. subinterval, two valves are conducting. This is based on the
assumption that the overlap angle is less than 60.. As the overlap angle increases to 60 0, there is no instant when only two valves are
conducting. As the overlap angle increases beyond 60 o, there is a finite period of an interval,
when four valves conduct and the rest of the interval three valves -conduct. Thus there are three modes of the converter as follows: 1. 1 ( < 60 0
2. 2 ( = 60 0) 3. 3 ( > 60 0
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A 3 4
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Inverter Equation
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AC A D DC
The reduction factor
Harmonic component with no overlap
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3 4 60 . 3
. ,
. E . 3 , 1,6 2
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A
Voltage and current characteristics are linear. For mode 2, U=60 0 , Characteristics are elliptical
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' F
. 3
30 0
1 2 ( 2). . AC
.
( ) 5 ( ) 6 7 8 7 . A , 5 6 7 6 ( 6 ), ,
.
( AC ).
, ( ).
.