Line-side harmonics produced by converter units in a fully-controlled three-phase bridge circuit B6C and (B6)A(B6)C
The majority of converter units for medium-power applications have a fully-controlled three-phase bridge circuit. Below is an example of the harmonics that can be found in a typical system configuration for two firing angles ( = 20° and = 60°).
The values have been taken from a previous publication, "Oberschwingungen im netzseitigen Strom sechspulsiger netzgefuhrter Stromrichter (Harmonics in the Line-Side Current of Six-Pulse, Line-Commutated Converters)" by H. Arremann and G. Moltgen, Siemens Research and Development Division, Volume 7 (1978) No. 2, © Springer-Verlag 1978.
In addition, the formulas are specified which, depending on the actual operating data in use, line supply voltage (no-load voltageVV0), line frequencyfN and DC currentId, can be used to calculate the short-circuit powerSK and armature inductanceLa for the motor to which the specified harmonics spectrum applies.
If the actual line short-circuit power and/or actual armature inductance deviate from the values calculated in this way, then they will need to be calculated on a case-by-case basis.
The harmonics spectrum shown below is obtained if the values for the short-circuit powerSK at the point where the unit is connected and the armature inductanceLa of the motor, calculated using the following formulas, match the actual values of the plant or system. If the values do not match, the harmonics will have to be separately calculated.
|
|
I /I1 |
|
|---|---|---|
|
at = 20° |
at = 60° |
|
|
5 |
0.235 |
0.283 |
|
7 |
0.100 |
0.050 |
|
11 |
0.083 |
0.089 |
|
13 |
0.056 |
0.038 |
|
17 |
0.046 |
0.050 |
|
19 |
0.035 |
0.029 |
|
23 |
0.028 |
0.034 |
|
25 |
0.024 |
0.023 |
|
29 |
0.018 |
0.026 |
|
31 |
0.016 |
0.019 |
|
35 |
0.011 |
0.020 |
|
37 |
0.010 |
0.016 |
|
41 |
0.006 |
0.016 |
|
43 |
0.006 |
0.013 |
|
47 |
0.003 |
0.013 |
|
49 |
0.003 |
0.011 |
The fundamental component of currentI1 as a reference variable is calculated using the following formula:
I 1 = g 0.817 Id
I
d DC current of the operating point being investigated
g Basic fundamental content
The harmonics currents calculated according to the table only apply for:
a) Short-circuit power S K at the point where the converter unit is connected
S K = VV02/XN (VA)
where
X N = XK - XD = 0.03536 VV0/Id - 2 fN LD ( )
V V0 No-load voltage at the point where the converter unit is connected in V
I d DC current of the operating point being investigated in A
f N Line frequency in Hz
L D Inductance of the commutating reactor being used in H
b) Armature inductance L a
L a = 0.0488 VV0/(fN Id) (H)
If the actual values for the short-circuit powerSK and/or armature inductanceLa deviate from the values calculated using the formulas above, a separate calculation will need to be made.
Example:
Let us assume a drive with the following data:
V V0 = 400 V
I d = 150 A
f N = 50 Hz
L D = 0.169 mH (4EU2421-7AA10) withILN = 125 A
where
X N = 0.03536 400/150 - 2 0.169 10–3 = 0.0412
The following short-circuit power of the line supply required at the point where the converter is connected:
S K = 4002/0.0412 = 3.88 MVA
and the following armature inductance of the motor required:
L a = 0.0488 400/(50 150) = 2.0 mH
The harmonics currentsI (withI1 = g 0.817 Id for firing angles = 20° and = 60°) that can be taken from the tables, only apply for the valuesSK andLa that have been calculated in this way. If the actual values deviate from these, a separate calculation will have to be made.
For the purpose of dimensioning filters and compensation equipment with reactors, it is only possible to draw on the information provided by the harmonic values calculated in this way if the calculated valuesSK andLa match the actual drive values. In all other cases, a separate calculation will have to be made (this particularly applies when using compensated motors as they have very low armature inductance levels).