Waveform Determination of Current Drawn from Mains with Three Crossbars Transformer Feeding a

12 Pulses Rectifier

Codruţ Chivulescu1, Ion Laurentiu-Florian1, Stefan-Serban Cristian1

1SC IPA SA- Bucuresti, codrut@ipa.ro, florian@ipa.ro, cristian@ipa.ro

Abstract - A method for determination the wave shape of the currents in the primary of one transformer with three crossbars is described in this work. An alternating current rectifier with 12 pulses is delivering by this transformer. The currents from the secondary of transformer are continuous functions for one third of period.

INTRODUCTION

The related conditions of the maximum absorbed cur-rent distortion of the curve network for both major and small consumers, requires a careful design of the devices of any kind. Starting from a known waveform for the secondary currents, it is necessary to determine the analytical waveform of current drawn by the transformer from mains, transformer that feeds the rectifier. The paper will look at a transformer with three yokes, a special construction that brings waveform in an amount of steps making up the final sinus.

A. Diagram of the transformer

Fig. 1

The two units together use an intermediate yoke. The rest is as ussual, the construction is using a shared transformer windings.

B. Determination of current waveform absorbed from

the mains of "Unit 1"

If each column is considered a single phase unit and the primary and secondary currents are three-phased and symmetric, then their sum is zero.

I’T1 + i’T2 + i’T3 = 0 I’S1 + i’S2 + i’S3 = 0 (1) I’1 + i’2 + i’3 = 0

Equations are then written for compensation of conduction

current that flows through the open area of primary and sec-ondary windings.

N i’1 = NT i’T1 + NS ( i’T1 – i’T2) = ( NT + NS ) i’T1 - NS i’T2

N i’2 = ( NT + NS ) i’T2 - NS i’T3 (2)

N i’3 = ( NT + NS ) i’T3 - NS i’T1

From this equation system, results intensity currents of star primary wye configuration, currents drawn from the mains.

11NN

1NN

i'i'1NN

i'2NN

NNi'

s

T2

s

T

32s

T1

s

T

SS1

+⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

−⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+−⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+

=

11NN

1NN

i'1NN

i'2NN

i'

NNi'

s

T2

s

T

3s

T2

s

T1

SS2

+⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+−⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛++−

=

(3)

11NN

1NN

i'2NN

i'i'1NN

NNi'

s

T2

s

T

3s

T21

s

T

SS3

+⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛++−⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

=

An area of 120 degrees pulse from the inverter (Fig. 2)

drained from Unit 1.

Fig. 2

The number of turns NS = 114, NT = 196 and N = 61, result: NT / NS = 1.7192982, N / NS = 0.5350877 and then currents: Replacing the numerical values, we obtain:

( )

321

321S

S1

i0,0481458i0,0130922i0,1790889

i0,0899775i0,2446758i0,3346534NNi

⋅−⋅−⋅=

=−−=

( )

321

321S

S2

i0,0130922i0,1790889i0,0481458

i0,2446758i0,3346534i0,0899775NNi

⋅−⋅+⋅−=

=−+−=

( )

321

321S

S3

i0,1790889i0,0481458i0,0130922

i0,3346534i0,0899775i0,2446758NNi

⋅+⋅−⋅−=

=+−−=

(4)

The table below is presented the variation of secondary cur-rents and current i'S1 over a period, over intervals of 30 de-grees.

Table 1.

Angle I1 I2 I3 I’S10-30 -I I 0,1547 I 30-60 I -I 0,579331 I 60-90 I -I 0,579331 I 90-120 I -I 0,424631 I 120-150 I -I 0,424631 I 150-180 I -I -0,1547 I 180-210 I -I -0,1547 I 210-240 -I I -0,579331 I 240-270 -I I -0,579331 I 270-300 -I I -0,424631 I 300-330 -I I -0,424631 I 330-360 -I I 0,1547 I

C. Determination of current waveform absorbed from the mains of "Unit 2"

For Unit 2, the current from first phase of the wye configu-

ration is the difference iT1 - iT3, so the corresponding system equations (1) - (2) are:

I’’T1 + i’’T2 + i’’T3 = 0 I’’S1 + i’’S2 + i’’S3 = 0 (5) I’’1 + i’’2 + i’’3 = 0

N i’’1 = NT i’’T1 + NS ( i’’T1 – i’’T3) = ( NT + NS ) i’’T1 - NS i’’T3 N i’’2 = ( NT + NS ) i’’T2 - NS i’’T1 N i’’3 = ( NT + NS ) i’’T3 - NS i’’T2 (6)

For currents absorbed from the mains by Unit 2, solutions

are obtained:

11NN

1NN

i1NN

ii2NN

NNi

s

T2

s

T

3s

T21

s

T

SS1

+⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+−−⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+

=

11NN

1NN

ii2NN

i1NN

NNi

s

T2

s

T

32s

T1

s

T

SS2

+⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

−⋅⎟⎟⎠

⎞⎜⎜⎝

⎛++⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

=

11NN

1NN

i2NN

i1NN

i

NNi

s

T2

s

T

3s

T2

s

T1

SS3

+⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛++⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+−−

=

(7)

Replacing the numerical values, we obtain

( )

321

321S

S1

i0,0130922i0,0481458i0,1790889

i0,2446758i0,0899775i0,3346534NNi

⋅−⋅−⋅=

=−−=

( )

( )

321

231S

S3

321

321S

S2

i0,1790889i0,0130922i0,0481458

i0,2446758i0,3346534i0,0899775NNi

i0,0481458i0,1790889i0,0130922

i0,0899775i0,3346534i0,2446758NNi

⋅+⋅−⋅−=

=−+−=

⋅−⋅+⋅−=

=−+−=

Currents drained from Unit 2 are out of phase to those of

Unit 1 with 30 degrees forward, as seen in (Fig. 3).

Fig. 3.

The table below presents the variation of secondary currents from Unit 1 and the absorbed current by a phase of Unit 2 dur-ing a period.

Angle I1 I2 I3 IS1 0-30 I -I 0,424631 I 30-60 I -I 0,424631 I 60-90 I -I 0,579331 I 90-120 I -I 0,579331 I 120-150 I -I 0,1547 I 150-180 I -I 0,1547 I 180-210 -I I -0,424631 I 210-240 -I I -0,424631 I 240-270 -I I -0,579331 I 270-300 -I I -0,579331 I 300-330 -I I -0,1547 I 330-360 -I I -0,1547 I

Doing the linking tables 1 and 2, to obtain the total current waveform absorbed from the mains (network) by the two units (Table 3 and Fig. 4)

Tabel 3

Unghi IS1 unit1 IS1 unit2 IS0-30 0,1547 I 0,424631 I 0,579331 I 30-60 0,579331 I 0,424631 I 1,003962 60-90 0,579331 I 0,579331 I 1,158662 90-120 0,424631 I 0,579331 I 1,003962 120-150 0,424631 I 0,1547 I 0,579331 I 150-180 -0,1547 I 0,1547 I 0 180-210 -0,1547 I -0,424631 I -0,579331 I 210-240 -0,579331 I -0,424631 I -1,003962 240-270 -0,579331 I -0,579331 I -1,158662 270-300 -0,424631 I -0,579331 I -1,003962 300-330 -0,424631 I -0,1547 I -0,579331 I 330-360 0,1547 I -0,1547 I 0

Fig.4

RMS current absorbed from the mains:

( ) ( ) ( ) I70,8195147312T2I1,158662

12T4I1,003962

12T4I0,579331

T1I 222

S ⋅=⎥⎦⎤

⎢⎣⎡ ⋅⋅⋅+⋅⋅⋅+⋅⋅⋅=

D. Conclusions Powering the rectifier with 12 pulses through a transformer with three yokes, in a special construction (combination wye and delta) offers the advantage of obtaining a waveform with little distortion (<10% under EU rules) to the current absorbed from the mains. A primary indication of whether sinusoidal current is ab-sorbed gave the ratio of the maximum amplitude of the current absorbed from the network (IS1 max) and the actual amount of that power.

1,413830.81951473

1.158662I

Ik

S1

maxS1 ===

Because of the sine ratio, 2 =1.4142, results that the harmon-ic component waveform is minimal.

Tabel 2