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Designing a autotransformer

for  400Vdc, 1000Adc
with 2 parallel connected rectifiers

in 12-pulse operating mode  

General Information

Technical Specification

Input voltage

3 x 400/230V

Autotransformer output voltage for Udc = 400Vdc

3 x 314/182V, +15
3 x 314/182V, -15

Line output current per secondary: (Ia1,Ib1,Ic1,Ia2,Ib2,Ic2)

I1 = 388Arms
I5  = 77.5Arms
I7  = 55.5Arms
I11 = 35Arms
I13 = 30Arms
continuous operating mode

Frequency

50Hz

Ambient temperature

40C

Temperature rise

Max. 120K, insulation class H

--Steel & Core

M6, annealed, strips for alternated stacking (45),
Oval cross section

 

Creating Input

4 input screens are used to set the input parameters for the designing of a transformer:

         Winding parameters per limb

         Core

          Environment

         Other parameters

and 3 screens for selection and set up of material :

         wires

         steels

         cores.

Windings parameters per phase

The following 3 phase autotransformer circuit is often used to drive 2 parallel connected 6 pulse bridge rectifiers in order to compensate the 5. and 7. current harmonics on the input voltage side of the autotransformer. The parallel connection of the rectifiers is normally used if the output current Id is over 500-1000Adc.

 

 

 For equal current distribution between 2 parallel connected rectifiers (without the chokes Ld1 and Ld2) the ratio Ucc_out1-out2/Ucc_in-out has to be bigger or equal 4 and Ucc_in-out > 4%. Normally this condition can be not realized by using the autotransformer and you need to consider using of Ld1&Ld2 chokes and/or 2 3-phase commutating chokes at the AC side between the rectifier and the autotransformer. Using of 2 3-phase commutating chokes is very effective for suppressing all current harmonics.

Note that the short-circuit voltage of a rectifier autotransformer is a complex issue reflecting:

         the rectifier protection in a short circuit operation mode of all secondary winding, a group of windings or of only one winding.

         the commutation operation mode of a group of windings

         the voltage drop of the dc-output voltage

         the current distribution between the parallel connected rectifiers

It has to be prescribed by the user of the transformer

The following 50Hz vector diagram of the autotransformer was created as follows:

         Input voltage per phase is 230V

         The output voltage per phase for 400Vdc rectifier voltage is 182V

         The first current harmonic is 388A (for rectifier DC current 500Adc)

         The zag voltage is 182 x sin(45)/sin(120) = 148.6V

         The zig voltage is 182 x sin(15)/sin(120) = 54.4V

         The voltage on the W1 winding is 230-148.6 = 81.4V

         The input current is 2 x 388 x 148.6/230 = 482A (blue)

         The zig current is 2 x 388 x cos(15) *54.4/230 = 177A (red)

 

 

Windings parameters per leg

The Large Transformers Program supports the input per leg. You have to follow the following rules

  1. You need to declare min. one winding for primary. In this case it is zig winding with the input voltage 81.4V
  2. Wzag1 is the first secondary winding with 54.4V and 388A(first harmonic). Note that the current angle is 135 to the primary zig-winding (view the diagram). Through this winding flow all current harmonics.
  3. Wzag2 is the second secondary winding with 54.4V and 388A(first harmonic). Note that the current angle is 225 to the primary zig-winding (view the diagram). Through this winding flow all current harmonics. Note that the 5. and 7. harmonics flow in opposite direction to the 5. and 7. harmonics of the first zig winding (angle 180)
  4. W1 is the 3. secondary winding with 81.4V and 614A(first harmonic). Note that the 5. and 7. current harmonics do not exist in this winding. 11. and 13. harmonics are 2 x cos(15) bigger than in the zag winding
  5. For check in of your input you need to calculate “ampere-turns” of all winding per leg (without the primary no-load current)
    54.4 x 388 x sin(15) + 54.4 x 388 x sin(345) = 0
    81.4 x (482+177)-148.6 x (267-177) – 2*54.4x388 x (cos(15) = 0

6.      The set current harmonics are calculated for the worst case: Ucc= 0 and Ld = ∞ using the following table:

 

 

In order to get equal short-circuit voltage for both outputs and a good current distribution between 2 parallel connected rectifiers it is recommended to use the following winding configuration per leg 

 

 

 

 

 

uInput & Otput

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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