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 (no-load voltage V_V0, line frequency f_N, and DC current I_d), can be used to calculate the short-circuit power S_K and armature inductance L_a 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 power S_K at the point where the unit is connected and the armature inductance L_a 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.

The fundamental component of current I₁ as a reference variable is calculated using the following formula:

I₁ = g ? 0.817 ? I_d

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 point where the converter unit is connected

S_K = V_V0²/X_N (VA)

where

X_N = X_K – X_D = 0.03536 ? U_V0/I_d – 2?  ? f_N ? L_D (?)

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 ? V_V0/(f_N ? I_d) (H)

If the actual values for the short-circuit power S_K and/or armature inductance L_a 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) with I_LN = 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 is required:

L_a = 0.0488 ? 400/(50 ? 150) = 2.0 mH

The harmonic currents I_? (with I₁ = g ? 0.817 ? I_d for firing angles ? = 20° and ? = 60°) that can be taken from the tables, only apply for the values S_K and L_athat 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 values S_K and L_a 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).