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Current equalising transformer for current balancein
parallel-connected 12-pulse converter
F.T. Bennell, CEng, FIEE
Indexing terms: Semiconductor devices and materials,
Transformers, Circuit theory and design
Abstract: There has been a serious problem ofcurrent
out-of-balance in 12-pulse rectifiersbetween two 6-pulse groups.
This has become par-ticularly pronounced with semiconductor
rectifierswhere 12-pulse operation is obtained by
twophase-displaced 3-phase bridge rectifier circuitsoperating in
parallel. A current equalising trans-former (CET) was introduced to
overcome theproblem. A new simpler variation of this has
morerecently been introduced and the paper describesthis new
variation and includes more completeand exact design current and
voltage relationshipsthan have previously been published.
List of symbols
Ea = AC RMS voltage, line-to-lineEpk = AC peak voltage,
line-to-lineIa = AC RMS current/,, = DC mean currentVdo = Ideal
no-load DC voltageXc = Commutating reactance
1 Introduction
12-pulse rectifier operation is desirable for many
applica-tions, to achieve a low value of ripple in the DC outputand
low value current harmonics drawn from the ACsupply. This 12-pulse
operation may be obtained by two3-phase rectifier bridges displaced
in phase by 30, andconnected either in series or in parallel. The
simplest wayof achieving the required phase displacement is to have
arectifier transformer with two secondary windings, oneconnected
star, and the other delta. The choice betweenseries or parallel
connection is preferably decided by rela-tive costs and
efficiencies, and on these criteria parallelbridges are extensively
used. However there are twoserious problems.
The main problem is the current out-of-balance com-monly
experienced between the bridges. This is a particu-lar problem when
to minimise short-circuit current inrelation to DC voltage
regulation over the load range,the bridges are supplied from
closely coupled transformerwindings. There have been cases of this
out-of-balancebeing as high as 3 to 1, and exceptionally 19 to 1,
whichare completely unacceptable if the rectifier is to be
fullyloaded and true 12-pulse operation is necessary.
Paper 58O1B (P6/P4), first received 10th June and in revised
form 3rdSeptember 1987The author is a rectifier consultant working
from 68b Crockford ParkRoad, Addlestone, Weybridge, Surrey KT15
2LU, United Kingdom
These out-of-balances were partly due to the trans-former design
being critical, but largely also due to dis-tortions of the AC
supply voltage waveform, producedparticularly by 6-pulse rectifier
loading on the samesystem. This problem was experienced also in
themercury arc era, using quadruple zigzag connections for12-pulse
operation, where the out-of-balance wasbetween the two 6-pulse
groups each side of a 300 Hz*interphase transformer, the practice
then was to avoidmixing 6-pulse and 12-pulse rectifiers on the same
system.
The second problem is to do with interphase trans-formers
(IPTs), which enable paralleled circuit groups(that do not have at
every instant equal voltages), tocarry current together. Between
parallel bridges, the con-ventional interphase transformer is
required to be con-nected to DC terminals, and is therefore
normally aircooled within the rectifier cubicle. The problem is
noise,due to operation at 300 Hz. Various designs have
beentriedcores built from laminations, wound round cores,cores on
busbars, cores with conventional windings, lowflux densities, noise
barriersbut the problem remains.
A solution to both these problems is the use of acurrent
equalising transformer (CET). Put simply, thisuses the principle
that the primary and secondary wind-ings of a double-wound
transformer maintain equalampere-turns. The CET comprises a 3-limb
core (or 3single-phase cores) carrying two sets of windings, one
inthe AC connections of one rectifier bridge and the otherin the AC
connections of the other bridge.
However, this principle is not directly applicable.There is a 30
phase shift between the currents in the twobridges. In the first
method adopted to look after thisfactor, the windings associated
with one bridge wereinterconnected between phases, to achieve a
resultantphase disposition to correspond with that of the
otherbridge [2, 3]. This method was designed for adding theCET to
an existing rectifier equipment, where the CETwent in the line
connections external to the main trans-former. Due to phase
displacement between series wind-ings, the CET winding VA for one
bridge circuit isincreased by 15.5%.
A new method [4] is to connect one set of windingswithin the
delta secondary of the rectifier transformer asin Fig. 1. These
windings are in phase with the star wind-ings, being together on
common limbs. The advantages ofthis system are that there are only
two windings per limbinstead of three, there are fewer
connectionsnonebetween phases, and the numbers of turns in the
CETwindings are increased by a factor of ^ 3 , facilitating
thechoice of turns with the necessary ^3 relationship. (The
* The frequencies referred to in this paper are all based on a
supplyfrequency of 50 Hz
IEE PROCEEDINGS, Vol. 135, Pt. B, No. 2, MARCH 1988 85
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delta winding turns are J3 times those for the starwinding, and
the current ^ 3 less for ampere-turn equal-ity.)
Fig. 1 12-pulse rectifier circuit comprising parallel bridges
and a CET
2 Current equalising transformer
It is found to be necessary for the CET to span com-pletely the
instantaneous voltage differences between thetwo phase-displaced
bridges, whether there is also a con-ventional IPT or not. The IPT
therefore is omitted andthe situation is that the CET takes the
place of an IPT.
The CET is in the AC connections and can thereforebe
oil-immersed and in the tank of the main transformer,Fig. 2. The
noise of IPT's was not a serious problem
when they were in the transformer tank of a mercury arcrectifier
equipment, although there were then two muchlarger 150 Hz IPT's as
well as the 300 Hz one. Also, andimportantly, the CET has no DC
components of currentto experience out-of-balance and thus cause
fluxsaturationa main factor in the cause of excessive IPTnoise. The
noise is therefore reduced to a level that pre-sents no
problem.
In addition to the main advantages of the CET inrespect of
current balance and noise, further interestingfeatures ensue.
In the conventional parallel bridge system using anIPT, the
transformer primary winding has a basic rating(i.e. excluding
losses and magnetising current) of 1.01times the DC kW. However,
the secondary windingshave the same kVA as for individual bridge
circuits, i.e.1.05 times the DC kW. The reason for this is that
thesecondary windings each carry 5th and 7th harmonics,but of
opposite polarities so that they circulate on thesecondary side and
do not appear in the primary.
With the CET however, each individual bridge circuitoperates in
a 12-pulse mode. The secondary windingkVA's therefore correspond to
a factor of 1.01, the sameas for the primary. Elimination of the
5th and 7th harmo-nics of opposite polarities takes place in the
CET as anatural result of current equalising.
The current equalising effect combines two 120current pulses
with a 30 displacement into a single 150pulse, with a reduced RMS
value.
As a result of the widened current pulses, the CETmust look
after a wider range of AC voltages ( 75 to+ 75 from the peak, see
Fig. 3) with greater voltage dif-ferences than an IPT has to look
after (-60 to +60).Also it has to absorb the voltage differences
individuallyin 3 phases. It is therefore larger in kVA than an IPT
forthe same rectifier rating. However an IPT, having lowvoltage and
a heavy current, is inherently uneconomic inthe use of material for
itself and the busbar connectionsassociated with it. A comparison
of alternative IPT andCET designs for a 12-pulse rectifier for a
rating of 630 V1500 kW, is given in Table 1.
Table 1 : Comparison between an IPT and a CETIPT CET
Equivalent 50 Hz double-wound kVACooling mediumIron, kgCopper or
aluminium, kg
42 150Air Oil375 125
28 (A1) 100 (Cu)
Fig. 3 shows the effective AC voltages, i.e. where theAC
voltages are involved in current-carrying as shown byheavy lines,
for rectifier bridges with and without a CET.In combination with
the relevant currents the AC/DCvoltage relationship can be
determined (see Table 2).
Table 2: AC/DC voltage relationship. Fig. 36
Fig. 2 Transformer assembly for oil immersion including a CETThe
main rectifier transformer is at the bottom, above it lengthwise is
a buck/boost transformer associated with a separate regulating
transformer, and cross-wise the CET
Range
Current, laJ f dtxl.xM/2n=VdE, RMS
EO>L-L =0to15
0.5360.2990.160
1,Z.-/V = 0.577,
15 to 45
0.4640.5170.240
/ ,= 1.Xe = 0
45 to 75
0.2680.2990.080
Sum
0.4800.918 DC0.707 AC0.771
86 IEE PROCEEDINGS, Vol. 135, Pt. B, No. 2, MARCH 1988
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Factors affecting current sharing Current relationships with a
CETThe main transformer for parallel bridge operation withan IPT
must be very carefully designed to avoid seriouscurrent imbalance
even on a perfect supply system. Thefollowing factors are
critical
-75-
XV
x
-75-
30
A
X
Fig. 3 Effective AC voltages at the main transformer fora
Cconventional 3-phase rectifier bridgeb Parallel bridges with a CET
for 12-pulse operation
(a) The y/3 voltage relationship between the two sec-ondaries. A
small discrepancy in voltage has a muchgreater effect on
current.
(b) The individual reactances, primary to each second-ary, must
be closely equal, particularly if the secondariesare very closely
coupled.
(c) The copper losses of the two secondaries must beclosely
equal.
None of these restraints apply with a CET. Currentsharing is
determined solely by the turns ratio of theCET. Transformer
operation ensures that the ampere-turns of the two sets of windings
on the CET must,within flux saturation limits, remain equal. Thus
thetransformer can be to the most economic design and thedesign of
the CET itself is not critical.
The factors determining the current distribution are(a) the 30
phase shift provided by the star and delta
main transformer windings(b) the y/3 turns relationship between
the CET wind-
ings in series with the star secondary windings and theCET
windings connected within the delta
(c) the ampere-turn equivalence between the two setsof CET
windingsThe current values for the first part of a cycle are
derivedin Appendix 14.
0.259
Fig. 4 Transformer secondary voltage vectors for CET voltage
wave-form analysis
4.1 AC currentsApplying the analysis of current changes through
a cycle,to an AC current waveform produces Fig. 5 which
15 30 30 30 30 30 30
0.464
IO.536Id
0.268
Fig. 5 AC current waveform
applies to each line current. The delta winding currentshave the
same waveform but the values of the currentsteps are ^ 3 less. Each
step has a duration of 30, includ-ing the maximum of 0.536 Id. As
there are six of thesesteps per cycle from the star group, and six
from thedelta, the contributions to the total DC current from
IEE PROCEEDINGS, Vol. 135, Pt. B, No. 2, MARCH 1988 87
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each group have a castellated waveform, but the 30phase shift
results in a constant DC current. Each of0.536 Id from one group is
complimented by 0.464 Idfrom the other (see Fig. 6).
Fig. 6 Combination of the current waveforms from the two
bridges
Following the usual rectifier theory practice, thecurrent
waveforms are based and evaluated on a perfectlysmooth DC current,
i.e. as with infinite DC inductance.With no DC inductance, the
current blocks would havecosine crests between 15 and +15, with
mean valuescorresponding to the step values stated in the
precedingnotes. The difference this makes to the RMS value is
only0.0053%, and rectangular current blocks can therefore beassumed
for all cases. The mid-points on the currentsteps lie on a cosine
curve.
The RMS value of the AC current waveform calculatesat 0.379 Id
which compares with 0.408 ld for a simple3-phase bridge circuit.
The mean value is 0.333 Id and theform factor 1.137 compared with
1.225 for the simplebridge.
5 Main transformer voltageThe DC voltage at any point is
obtained by adding theAC voltages after weighting them according to
theircurrent proportions, i.e. by multiplying them by 0.268,0.464
or 0.536 as appropriate (summing to a DC currentof 1).
Referring to Fig. 4 and Table 4,
AC voltage vector position
(a) 'A' L to N at a maximum:(b) 'A' advanced 15:(c) 'A' advanced
a further 15
CalculatedDC voltage
0.928 Epk0.896 0.928 E
The DC voltage is found to comprise a series of cosinecrests
between -15 and -I-15 (0.928 cos 15 = 0.896).
The average DC voltage using 0.928 peak and thefamiliar (p/rc)
sin (rc/p) relationship for 12-pulse operation= 0.928 (12/TT) sin
(TC/12) = 0.918 Epk
Conversely, Epk = VJ0.91S and Ea = 0.771 Vdo. Thiscompares with
0.74 for a bridge circuit with no CET. Thehigher AC voltage
required is due to bringing in tocurrent-carrying service the lower
AC voltages betweencos 60 and cos 75 (see Fig. 3).
5.1 Main transformer secondary VAThe total main transformer
secondary VA = 2^/3x 0.379 x 0.771 Id Vdo = 1.012 Id Vdo, which is
exactly
the same as for the primary of a conventional 12-pulserectifier
transformer. Thus, the nett result of the saving incurrent rating
and the small increase in AC voltagerequired is to reduce the
transformer secondary VA inthe ratio 1.05 to 1.01.
6 CET currentThe CET current ratings are as already determined
forthe corresponding main transformer winding and linecurrents.
6.1 CET voltagesThe voltages to be supported by the CET are the
differ-ences between the individual line-to-line
transformervoltages, and the corresponding mean DC voltages.Tables
3 and 4 give the relevant figures scaled to corre-
Table 3: CET voltages derived from vector diagrams (Fig. 4)The
AC voltages are weighted by their associated currents to obtainthe
resultant DC voltage. An example of when 'A' line-to-neutralvoltage
is at a maximum is given to show how the CET voltages
arederived
MaintransformerwindingabcABC
WindingvoltagesEa+ 1-0.5-0.5+0.577-0.289-0.289
Windingcurrent/+0.309-0.155-0.155+0.536-0.268-0.268
formean Vd+0.309+0.0775+0.0775+0.309+0.0775+0.0775
CETvoltagesAC to Vd-0.072+0.036+0.036+0.042-0.021-0.021
0.928=VdThe CET voltages are the voltages needed to bring the AC
termin-
al voltages to Vd, e.g. for a, 0.928 - 1 = -0.072By CET
transformer action the CET voltages for the star group are
y/3 less than for the delta and of opposite polarity, e.g.
-0.072 for a,therefore +0.042 for A
The total Ea I for the mean Vd is identical for the star and
deltagroups
Table 4: CET voltages for A phase over a cycle. The samesequence
of values is applicable to phases B and C and,with the J3
adjustment, to phases a, b and c, all with therelative dispositions
indicated in Table 3.
Vector position.degrees from A max.
0153045607590
105120135150165180195210225240255270285300315330345360
vd
0.9280.8960.9280.8960.9280.8960.9280.8960.9280.8960.9280.8960.9280.8960.9280.8960.9280.8960.9280.8960.9280.8960.9280.8960.928
StarCET voltage+0.042+0.040-0.036-0.109+0.021+0.149
0-0.149-0.021+0.109+0.036-0.040-0.042-0.040+0.036+0.109-0.021-0.149
0+0.149+0.021-0.109-0.036+0.040+0.042
spond to Epk = 1, and a total DC current Id of 1. The ACvoltages
are weighted by their currents to determine themean DC voltage for
each condition. The complete cycleof the CET voltage for each phase
is shown in Fig. 7.
Analysing Fig. 7, the waveform is built up of 60 sec-tions as
for a 300 Hz IPT. The shapes are for practical
IEE PROCEEDINGS, Vol. 135, Pt. B, No. 2, MARCH 1988
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purposes triangular. They comprise one cycle at a peakvalue of
0.149 Epk, one at a peak value of 0.109 Epk andone virtually
omitted, but with a displacement of (0.04 to
Fig. 7 CET voltage waveform and corresponding main AC voltage
forone cycle
0.042) Epk from zero. The maximum flux required is givenby the
voltage/time integrals or can be derived fromthe average voltages.
The average voltages for the 3parts, over their duration, are
0.0745 Epk, 0.545 Epk and0.041 Epk respectively, and the overall
average, takinginto account the small differences in duration, is
0.05 Epk.Design may be based on the voltage/time integral havingthe
peak value of 0.109 Epk and an average value of0.0545 Epk as this
gives iron losses approximately equalto the total iron losses over
an actual cycle (but note theeffect of transformer reactance, to
follow). The equivalentRMS voltage for CET design is 1.11 x 0.0545
Epk =0.06 Epk.
7 Transformer reactance
The overlap of phases in commutation due to trans-former
reactance adds areas to the basic CET voltages, asshown in the
oscillographs in Fig. 8, taken from actual
Fig. 8 Oscillographs of CET voltages for increasing commutating
reac-tance up to xc = 18% at c
equipment in operation. The volt/time areas addedincrease the
CET voltage rating, and also the DC voltageregulation. The effect
is similar to that of reactanceoverlap on IPT voltage waveforms and
voltage regula-tion.
Reactance produces DC voltage regulation. In conven-tional
12-pulse rectifier circuits supplied by closely
coupled transformer windings, the percentage DC regula-tion is
0.259 times the percentage reactance. With theCET the commutation
current steps for each bridge addup to 0.536 Id instead of 0.5 with
conventional parallelbridge operation. The 0.259 factor therefore
increases to0.277.
The addition to the CET voltage rating due to trans-former
reactance is equal to the DC voltage regulationdue to the
reactance. The same volt/time areas areinvolved in both cases, and
they both relate to averagevalues.
At a reactance of about 18 to 20%, the additional com-mutation
volt/time areas bring the CET basic triangularwaveform to
rectangular form as Fig. 8(c). The voltage/time integral and the
CET voltage rating are thenapproximately doubled.
7.1 CET reactanceThe CET adds a little reactance to that of the
main trans-former. Its own reactance between windings can
beexpressed as a percentage of its own 50 Hz kVA calcu-lated as for
a normal power transformer. To refer thispercentage to the main
circuit kVA, to add it to the maintransformer percentage reactance,
it is decreased ininverse proportion to the kVA. Typical values
would be4%, decreased by a kVA ratio of 150 for the CET to1613 for
the main transformer (to take the examplereferred to in Table 1),
which equals 0.37% reactance,and times 0.277 = 0.1 %
regulation.
8 No-load voltage riseThere is a voltage rise at no-load, as
there is then nomagnetising current available for the CET, and
operationreverts to straightforward 12-pulse, as happens withan
IPT. The no-load voltage is Epk {12/n) sin (TT/12) =0.9886 Epk. the
light-load voltage, i.e. when the CET ismagnetised, is 0.928 Epk
(12/TT) sin (TC/12) = 0.918 Epk, andthe no-load voltage rise is
then 7.7%.
9 Rectifier diode ratingThe 12-pulse operation of the individual
bridges resultsin the rectifier diode current rating being improved
fromthat corresponding to 120 rectangular current pulses, tothe
approximate equivalent for diode rating of 180 halfsine waves.
(Strictly, 150 plus the angle of commutationoverlap.)
10 Example and performanceFig. 2 illustrates a transformer
assembly incorporating aCET for 12-pulse parallel bridge operation.
It includesalso a buck/boost transformer, supplied from a
separatevoltage regulating transformer, to provide a range ofoutput
voltages over which the CET maintains equalbridge currents.
Tests and operation in service have confirmed the per-formance,
the following result being typical.
Current values in the secondary line connections:Without a CET
but with an IPT: Star 35 A Delta
230 AWith a CET but without an IPT: Star 250 A Delta
250 A
11 Conclusions
The CET described in the paper has been shown, both intheory and
in practice, to overcome completely the
IEE PROCEEDINGS, Vol. 135, Pi. B, No. 2, MARCH 1988 89
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current balance problem between parallel phase-displaced 3-phase
rectifier bridges supplied from closelycoupled transformer
windings. This also allows 12-pulserectifiers to be used on a
supply having substantial6-pulse rectifier loading.
The assurance of current balance allows the equipmentto be
loaded up to its full rated value.
The elimination of the IPT gets rid of a source ofobjectionable
noise and provides a saving to offset thecost of the CET.
12 Acknowledgments
Early work leading to UK Patent GB 2001486 wascarried out in
conjunction with M.H.F. Garnham at theWimbledon works of Foster
Transformers Ltd.
Thanks are due to Bonar Brentford Electric Ltd. inproviding
equipment and facilities for the further devel-opment and
performance demonstrations on which thecontent of this paper is
based, and for permitting thepublication of the photograph in Fig.
2.
13 References1 BENNELL, F.T.: 'Current balance in 12-pulse
rectifiers comprising
parallel bridges'. IEE Conf. Publ. 154, 1977, pp. 66-692
BENNELL, F.T.: 'Rectifiers for railway traction substations',
IEE
Proc. B, 1979, pp. 22-263 UK Patent GB 2001486, 3.2.824 UK
Patent GB 2113927, 9.5.85
14 Appendix
Fig. 4 shows the voltage vectors for the transformer starand
delta secondary windings for the first part of a cycle.
(a) A star line-to-neutral voltage at a maximum.A winding
current is at a maximum and it is sharedequally by windings B and
C. For convenient propor-
tioning, values of 2, 1 and 1 respectively are assigned.
Thecurrent in delta winding a on the same core limb aswinding A is,
by the CET ^/3 turns ratio 2/^3, and inwindings b and c a half
this, i.e. 1/^/3. The delta line cur-rents become ^/3 for a and c
and zero for b (b and cwinding currents being equal). Thus, during
these condi-tions which apply while A is between 15 and +15from its
maximum position, the proportions of DCcurrent provided by the star
and delta groups are 2and J3, respectively. For a total DC current
ld of1, 2/(2 + y/3) = 0.536 Id is provided by the star groupthrough
A; by 1/(2 + V3) = 0.268 /,, each from B and C;and 7 3 / ( 2 + V3)
= -4 6 4 l& from the del ta g r o uP 'by 2/^3(2 + 73) = 0.309
ld from a, plus 1/^/3(2 + J3) =0.155 /,, through b and c.
(b) A star line-to-neutral voltage advanced 15.The current in
star winding B and line B (and therefore indelta winding b) ceases.
Star windings A and C by seriesconnection now carry 0.464 Id and by
the 7 3 relationshipdelta windings a and c each carry 0.268 / d ,
addingtogether to 0.536 Id. Thus, the star windings now con-tribute
0.464 Id and the delta windings 0.536 ld.
(c) A now at plus 30. The current proportions remainas for (b)
and continue the same to plus 45.
(d) At 45 star winding B begins to carry current again.The
current in A and B windings is now 0.268 Id, 0.536 Idin C and 0.536
Id total from the star group. In the deltagroup, the current
proportions are 0.155 Id in a and b,0.309 Id in c and 0.464 Id
total.
At 75 current in A ceases.At 90 the voltage reverses.At 105 the
current in A reappears in reverse flow at
0.268 Id.Continuing this procedure for a complete cycle pro-
duces a current waveform for the star connection asshown in Fig.
5, and for the delta windings the samevalues reduced by y/3.
90 IEE PROCEEDINGS, Vol. 135, Pt. B, No. 2, MARCH 1988
