Information On Line-Side Harmonics Generated By Converters In A - Siemens 6RA7086-6KS22 Operating Instructions Manual

05.05
6.1.3

Information on line-side harmonics generated by converters in a

fully-controlled three-phase bridge circuit configuration B6C and (B6)A(B6)C
Converters for the medium power range usually consist of fully-controlled three-phase bridge
circuit configurations. An example of the harmonics generated by a typical system configuration for
two firing angles (α = 20° and α = 60°) is given below.
The values have been taken from an earlier publication entitled "Harmonics in the Line-Side
Current of Six-Pulse Line-Commutated Converters" written by H. Arremann and G. Möltgen,
Siemens Research and Development Dept., Volume 7 (1978) No. 2, © Springer-Verlag 1978.
Formula have been specified with which the short circuit power S K and armature inductance L a of
the motor to which the specified harmonics spectrum applies can be calculated depending on the
applicable operating data [line voltage (no-load voltage U v0 ), line frequency f N and DC current I d ].
A dedicated calculation must be performed if the actual system short circuit power and/or actual
armature reactance deviate from the values determined by this method.
The spectrum of harmonics listed below is obtained if the values for short circuit power S K at the
converter supply connection point and the armature inductance L a of the motor calculated by the
following formula correspond to the actual plant data. If the calculated values differ, the harmonics
must be calculated separately.
a.) α = 20°
Fundamental factor g = 0.962
ν
5
7
11
13
17
19
23
25
The fundamental-frequency current I 1 as a reference quantity is calculated by the following
equation:
where I d
where g
The harmonic currents calculated from the above tables are valid only for
I.) Short-circuit power S K at converter supply connection point
Siemens AG 6RX1700-0AD76
SIMOREG DC Master Operating Instructions
ν
I ν /I 1
0.235 29
0.100 31
0.083 35
0.056 37
0.046 41
0.035 43
0.028 47
0.024 49
= g ×
0.817 I
I
1
DC current of operating point under investigation
Fundamental factor (see above)
2
U
( )
v 0
=
S
K
X
N
b.) α = 60°
Fundamental factor g = 0.953
ν
I ν /I 1
0.018 5
0.016 7
0.011 11
0.010 13
0.006 17
0.006 19
0.003 23
0.003 25
×
d
VA
Connections
ν
I ν /I 1
0.283 29
0.050 31
0.089 35
0.038 37
0.050 41
0.029 43
0.034 47
0.023 49
I ν /I 1
0.026
0.019
0.020
0.016
0.016
0.013
0.013
0.011
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